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This selection and arrangement of content as a collection is copyrighted by Maxfield Foundation. It is licensed under, the Creative Commons Attribution License 3.0 (http://creativecommons.org/licenses/by/3.0/)., Collection structure revised: July 3, 2012, PDF generated: June 27, 2018, For copyright and attribution information for the modules contained in this collection, see p. 685.
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Table of Contents, Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1, Additional Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5, Author Acknowledgements, , ....................................................... ................... 9, , Student Welcome Letter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11, 1 Sampling and Data, 1.1 Sampling and Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 1.2 Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 1.3 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16, 1.4 Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16, 1.5 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18, 1.6 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24, 1.7 Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31, 1.8 Answers and Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33, 1.9 Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33, 1.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38, 1.11 Practice: Sampling and Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39, 1.12 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42, 1.13 Lab 1: Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50, 1.14 Lab 2: Sampling Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55, 2 Descriptive Statistics, 2.1 Descriptive Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59, 2.2 Displaying Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59, 2.3 Stem and Leaf Graphs (Stemplots), Line Graphs and Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60, 2.4 Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63, 2.5 Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68, 2.6 Measures of the Location of the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71, 2.7 Measures of the Center of the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76, 2.8 Skewness and the Mean, Median, and Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79, 2.9 Measures of the Spread of the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81, 2.10 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89, 2.11 Practice 1: Center of the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90, 2.12 Practice 2: Spread of the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93, 2.13 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95, 2.14 Lab: Descriptive Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114, 3 Probability Topics, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, , Probability Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121, Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122, Independent and Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124, Two Basic Rules of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 127, Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131, Venn Diagrams (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134, Tree Diagrams (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135, Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140, Practice 1: Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
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iv, 3.10 Practice 2: Calculating Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143, 3.11 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144, 3.12 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155, 3.13 Lab: Probability Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160, 4 Discrete Random Variables, 4.1 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167, 4.2 Probability Distribution Function (PDF) for a Discrete Random Variable . . . . . . . . . . . . . . . . . . . . 168, 4.3 Mean or Expected Value and Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169, 4.4 Common Discrete Probability Distribution Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172, 4.5 Binomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172, 4.6 Geometric (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175, 4.7 Hypergeometric (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178, 4.8 Poisson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180, 4.9 Summary of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183, 4.10 Practice 1: Discrete Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185, 4.11 Practice 2: Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186, 4.12 Practice 3: Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188, 4.13 Practice 4: Geometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189, 4.14 Practice 5: Hypergeometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191, 4.15 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192, 4.16 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202, 4.17 Lab 1: Discrete Distribution (Playing Card Experiment) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205, 4.18 Lab 2: Discrete Distribution (Lucky Dice Experiment) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213, 5 Continuous Random Variables, 5.1 Continuous Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221, 5.2 Continuous Probability Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223, 5.3 The Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226, 5.4 The Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233, 5.5 Summary of the Uniform and Exponential Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . 239, 5.6 Practice 1: Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240, 5.7 Practice 2: Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243, 5.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245, 5.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251, 5.10 Lab: Continuous Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257, 6 The Normal Distribution, 6.1 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261, 6.2 The Standard Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262, 6.3 Z-scores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263, 6.4 Areas to the Left and Right of x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 265, 6.5 Calculations of Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265, 6.6 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269, 6.7 Practice: The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270, 6.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272, 6.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278, 6.10 Lab 1: Normal Distribution (Lap Times) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280, 6.11 Lab 2: Normal Distribution (Pinkie Length) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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v, 7 The Central Limit Theorem, 7.1 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 289, 7.2 The Central Limit Theorem for Sample Means (Averages) . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 290, 7.3 The Central Limit Theorem for Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293, 7.4 Using the Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294, 7.5 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301, 7.6 Practice: The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302, 7.7 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305, 7.8 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312, 7.9 Lab 1: Central Limit Theorem (Pocket Change) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314, 7.10 Lab 2: Central Limit Theorem (Cookie Recipes) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323, 8 Confidence Intervals, 8.1 Confidence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327, 8.2 Confidence Interval, Single Population Mean, Population Standard Deviation, Known, Normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329, 8.3 Confidence Interval, Single Population Mean, Standard Deviation Unknown,, Student’s-t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336, 8.4 Confidence Interval for a Population Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339, 8.5 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344, 8.6 Practice 1: Confidence Intervals for Means, Known Population Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 345, 8.7 Practice 2: Confidence Intervals for Means, Unknown Population Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347, 8.8 Practice 3: Confidence Intervals for Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349, 8.9 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351, 8.10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361, 8.11 Lab 1: Confidence Interval (Home Costs) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364, 8.12 Lab 2: Confidence Interval (Place of Birth) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367, 8.13 Lab 3: Confidence Interval (Womens’ Heights) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371, 9 Hypothesis Testing: Single Mean and Single Proportion, 9.1 Hypothesis Testing: Single Mean and Single Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377, 9.2 Null and Alternate Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 378, 9.3 Outcomes and the Type I and Type II Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379, 9.4 Distribution Needed for Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380, 9.5 Assumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 381, 9.6 Rare Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381, 9.7 Using the Sample to Test the Null Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382, 9.8 Decision and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383, 9.9 Additional Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383, 9.10 Summary of the Hypothesis Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385, 9.11 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385, 9.12 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396, 9.13 Practice 1: Single Mean, Known Population Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . 397, 9.14 Practice 2: Single Mean, Unknown Population Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . 399, 9.15 Practice 3: Single Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401, 9.16 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403, 9.17 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416, 9.18 Lab: Hypothesis Testing of a Single Mean and Single Proportion . . . . . . . . . . . . . . .. . . . . . . . . . . . 419, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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vi, 10 Hypothesis Testing: Two Means, Paired Data, Two Proportions, 10.1 Hypothesis Testing: Two Population Means and Two Population Proportions . .. . . . . . . . . . . . 429, 10.2 Comparing Two Independent Population Means with Unknown Population, Standard Deviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430, 10.3 Comparing Two Independent Population Means with Known Population Standard Deviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433, 10.4 Comparing Two Independent Population Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435, 10.5 Matched or Paired Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437, 10.6 Summary of Types of Hypothesis Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442, 10.7 Practice 1: Hypothesis Testing for Two Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443, 10.8 Practice 2: Hypothesis Testing for Two Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445, 10.9 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447, 10.10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459, 10.11 Lab: Hypothesis Testing for Two Means and Two Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466, 11 The Chi-Square Distribution, The Chi-Square Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471, Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472, Facts About the Chi-Square Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472, Goodness-of-Fit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474, Test of Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 481, The Chi-Square Distribution: Test for Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485, The Chi-Square Distribution: Comparison Summary of the Chi-Square Tests, Goodness-of-Fit, Independence and Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488, 11.8 Test of a Single Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488, 11.9 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491, 11.10 Practice 1: Goodness-of-Fit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492, 11.11 Practice 2: Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494, 11.12 Practice 3: Test of a Single Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 496, 11.13 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498, 11.14 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507, 11.15 Lab 1: Chi-Square Goodness-of-Fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511, 11.16 Lab 2: Chi-Square Test for Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518, 11.1, 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, , 12 Linear Regression and Correlation, 12.1 Linear Regression and Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523, 12.2 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523, 12.3 Slope and Y-Intercept of a Linear Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525, 12.4 Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 526, 12.5 The Regression Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528, 12.6 Correlation Coefficient and Coefficient of Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534, 12.7 Testing the Significance of the Correlation Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 536, 12.8 Prediction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 541, 12.9 Outliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541, 12.10 95% Critical Values of the Sample Correlation Coefficient Table . . . . . . . . . . . . . . . . . . . . . . . . . . . 548, 12.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551, 12.12 Practice: Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 552, 12.13 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555, 12.14 Lab 1: Regression (Distance from School) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 571, 12.15 Lab 2: Regression (Textbook Cost) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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vii, 12.16 Lab 3: Regression (Fuel Efficiency) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579, 13 F Distribution and One-Way ANOVA, 13.1 F Distribution and One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583, 13.2 One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584, 13.3 The F Distribution and the F Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 585, 13.4 Facts About the F Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589, 13.5 Test of Two Variances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593, 13.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596, 13.7 Practice: One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597, 13.8 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599, 13.9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604, 13.10 Lab: One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610, 14 Appendix, 14.1 Practice Final Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613, 14.2 Practice Final Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 622, 14.3 Data Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 632, 14.4 Group Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635, 14.5 Solution Sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646, 14.6 English Phrases Written Mathematically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 650, 14.7 Symbols and their Meanings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651, 14.8 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656, 14.9 Notes for the TI-83, 83+, 84 Calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 658, Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668, 15 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671, Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .672, Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 680, Attributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Preface, , 1, , Welcome to Collaborative Statistics, presented by Connexions. The initial section below introduces you to, Connexions. If you are familiar with Connexions, please skip to About "Collaborative Statistics." (Section :, About Connexions), , About Connexions, Connexions Modular Content, Connexions (cnx.org2 ) is an online, open access educational resource dedicated to providing high quality, learning materials free online, free in printable PDF format, and at low cost in bound volumes through, print-on-demand publishing. The Collaborative Statistics textbook is one of many collections available, to Connexions users. Each collection is composed of a number of re-usable learning modules written in, the Connexions XML markup language. Each module may also be re-used (or ’re-purposed’) as part of, other collections and may be used outside of Connexions. Including Collaborative Statistics, Connexions, currently offers over 6500 modules and more than 350 collections., The modules of Collaborative Statistics are derived from the original paper version of the textbook under, the same title, Collaborative Statistics. Each module represents a self-contained concept from the original, work. Together, the modules comprise the original textbook., Re-use and Customization, The Creative Commons (CC) Attribution license3 applies to all Connexions modules. Under this license,, any module in Connexions may be used or modified for any purpose as long as proper attribution to the, original author(s) is maintained. Connexions’ authoring tools make re-use (or re-purposing) easy. Therefore, instructors anywhere are permitted to create customized versions of the Collaborative Statistics textbook by editing modules, deleting unneeded modules, and adding their own supplementary modules., Connexions’ authoring tools keep track of these changes and maintain the CC license’s required attribution, to the original authors. This process creates a new collection that can be viewed online, downloaded as a, single PDF file, or ordered in any quantity by instructors and students as a low-cost printed textbook. To, start building custom collections, please visit the help page, “Create a Collection with Existing Modules”4 ., For a guide to authoring modules, please look at the help page, “Create a Module in Minutes”5 ., Read the book online, print the PDF, or buy a copy of the book., To browse the Collaborative Statistics textbook online, visit the collection home page at, cnx.org/content/col10522/latest6 . You will then have three options., 1 This, , content is available online at <http://cnx.org/content/m16026/1.17/>., , 2 http://cnx.org/, 3 http://creativecommons.org/licenses/by/2.0/, 4 http://cnx.org/help/CreateCollection, 5 http://cnx.org/help/ModuleInMinutes, 6 Collaborative, , Statistics <http://cnx.org/content/col10522/latest/>, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 1
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2, 1. You may obtain a PDF of the entire textbook to print or view offline by clicking on the “Download, PDF” link in the “Content Actions” box., 2. You may order a bound copy of the collection by clicking on the “Order Printed Copy” button., 3. You may view the collection modules online by clicking on the “Start ” link, which takes you to the, first module in the collection. You can then navigate through the subsequent modules by using their, “Next ” and “Previous ” links to move forward and backward in the collection. You can jump to, any module in the collection by clicking on that module’s title in the “Collection Contents” box on the, left side of the window. If these contents are hidden, make them visible by clicking on “[show table, of contents]”., Accessibility and Section 508 Compliance, , • For, information, on, general, Connexions, accessibility, features,, please, visit, http://cnx.org/content/m17212/latest/7 ., • For information on accessibility features specific to the Collaborative Statistics textbook, please visit, http://cnx.org/content/m17211/latest/8 ., Version Change History and Errata, , • For, a, list, of, modifications,, http://cnx.org/content/m17360/latest/9 ., , updates,, , and, , corrections,, , please, , visit, , Adoption and Usage, , • The Collaborative Statistics collection has been adopted and customized by a number of professors and educators for use in their classes. For a list of known versions and adopters, please visit, http://cnx.org/content/m18261/latest/10 ., , About “Collaborative Statistics”, Collaborative Statistics was written by Barbara Illowsky and Susan Dean, faculty members at De Anza College in Cupertino, California. The textbook was developed over several years and has been used in regular, and honors-level classroom settings and in distance learning classes. Courses using this textbook have been, articulated by the University of California for transfer of credit. The textbook contains full materials for, course offerings, including expository text, examples, labs, homework, and projects. A Teacher’s Guide is, currently available in print form and on the Connexions site at http://cnx.org/content/col10547/latest/11 ,, and supplemental course materials including additional problem sets and video lectures are available at, http://cnx.org/content/col10586/latest/12 . The on-line text for each of these collections collections will, meet the Section 508 standards for accessibility., An on-line course based on the textbook was also developed by Illowsky and Dean. It has won an award, as the best on-line California community college course. The on-line course will be available at a later date, as a collection in Connexions, and each lesson in the on-line course will be linked to the on-line textbook, chapter. The on-line course will include, in addition to expository text and examples, videos of course, lectures in captioned and non-captioned format., The original preface to the book as written by professors Illowsky and Dean, now follows:, 7 "Accessibility, , Features of Connexions" <http://cnx.org/content/m17212/latest/>, Statistics: Accessibility" <http://cnx.org/content/m17211/latest/>, 9 "Collaborative Statistics: Change History" <http://cnx.org/content/m17360/latest/>, 10 "Collaborative Statistics: Adoption and Usage" <http://cnx.org/content/m18261/latest/>, 11 Collaborative Statistics Teacher’s Guide <http://cnx.org/content/col10547/latest/>, 12 Collaborative Statistics: Supplemental Course Materials <http://cnx.org/content/col10586/latest/>, 8 "Collaborative, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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3, This book is intended for introductory statistics courses being taken by students at two– and four–year, colleges who are majoring in fields other than math or engineering. Intermediate algebra is the only prerequisite. The book focuses on applications of statistical knowledge rather than the theory behind it. The, text is named Collaborative Statistics because students learn best by doing. In fact, they learn best by, working in small groups. The old saying “two heads are better than one” truly applies here., Our emphasis in this text is on four main concepts:, , •, •, •, •, , thinking statistically, incorporating technology, working collaboratively, writing thoughtfully, , These concepts are integral to our course. Students learn the best by actively participating, not by just, watching and listening. Teaching should be highly interactive. Students need to be thoroughly engaged, in the learning process in order to make sense of statistical concepts. Collaborative Statistics provides, techniques for students to write across the curriculum, to collaborate with their peers, to think statistically,, and to incorporate technology., This book takes students step by step. The text is interactive. Therefore, students can immediately apply, what they read. Once students have completed the process of problem solving, they can tackle interesting, and challenging problems relevant to today’s world. The problems require the students to apply their, newly found skills. In addition, technology (TI-83 graphing calculators are highlighted) is incorporated, throughout the text and the problems, as well as in the special group activities and projects. The book also, contains labs that use real data and practices that lead students step by step through the problem solving, process., At De Anza, along with hundreds of other colleges across the country, the college audience involves a, large number of ESL students as well as students from many disciplines. The ESL students, as well as, the non-ESL students, have been especially appreciative of this text. They find it extremely readable and, understandable. Collaborative Statistics has been used in classes that range from 20 to 120 students, and in, regular, honor, and distance learning classes., Susan Dean, Barbara Illowsky, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Additional Resources, , 13, , Additional Resources Currently Available, , •, •, •, •, •, •, •, •, •, , Glossary (Glossary, p. 5), View or Download This Textbook Online (View or Download This Textbook Online, p. 5), Collaborative Statistics Teacher’s Guide (Collaborative Statistics Teacher’s Guide, p. 6), Supplemental Materials (Supplemental Materials, p. 6), Video Lectures (Video Lectures, p. 6), Version History (Version History, p. 6), Textbook Adoption and Usage (Textbook Adoption and Usage, p. 6), Additional Technologies and Notes (Additional Technologies, p. 7), Accessibility and Section 508 Compliance (Accessibility and Section 508 Compliance, p. 7), , The following section describes some additional resources for learners and educators. These modules and, collections are all available on the Connexions website (http://cnx.org/14 ) and can be viewed online,, downloaded, printed, or ordered as appropriate., Glossary, This module contains the entire glossary for the Collaborative Statistics textbook collection (col10522), since its initial release on 15 July 2008. The glossary is located at http://cnx.org/content/m16129/latest/15 ., Below are links to additional resources:, Link to the Statistics Glossary by Dr. Philip Stark, UC Berkeley, http:// statistics.berkeley.edu/∼stark/SticiGui/Text/gloss.htm 16, Link to Wikipedia, http:// http://www.wikipedia.org/ 17, (Search on "Glossary of probability and statistics."), View or Download This Textbook Online, The complete contents of this book are available at no cost on the Connexions website at, http://cnx.org/content/col10522/latest/18 . Anybody can view this content free of charge either as an, online e-book or a downloadable PDF file. A low-cost printed version of this textbook is also available, here19 ., 13 This, , content is available online at <http://cnx.org/content/m18746/1.7/>., , 14 http://cnx.org/, 15 "Collaborative, , Statistics: Glossary" <http://cnx.org/content/m16129/latest/>, , 16 http://statistics.berkeley.edu/∼stark/SticiGui/Text/gloss.htm, 17 http://www.wikipedia.org/, 18 Collaborative, , Statistics <http://cnx.org/content/col10522/latest/>, , 19 http://my.qoop.com/store/7064943342106149/7781159220340, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 5
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6, Collaborative Statistics Teacher’s Guide, A complementary Teacher’s Guide for Collaborative statistics is available through Connexions at, http://cnx.org/content/col10547/latest/20 . The Teacher’s Guide includes suggestions for presenting concepts found throughout the book as well as recommended homework assignments. A low-cost printed, version of this textbook is also available here21 ., Supplemental Materials, This companion to Collaborative Statistics provides a number of additional resources for use by students, and instructors based on the award winning Elementary Statistics Sofia online course22 , also by textbook, authors Barbara Illowsky and Susan Dean. This content is designed to complement the textbook by providing video tutorials, course management materials, and sample problem sets. The Supplemental Materials, collection can be found at http://cnx.org/content/col10586/latest/23 ., Video Lectures, , •, •, •, •, •, •, •, •, •, •, •, •, , Video Lecture 1: Sampling and Data24, Video Lecture 2: Descriptive Statistics25, Video Lecture 3: Probability Topics26, Video Lecture 4: Discrete Distributions27, Video Lecture 5: Continuous Random Variables28, Video Lecture 6: The Normal Distribution29, Video Lecture 7: The Central Limit Theorem30, Video Lecture 8: Confidence Intervals31, Video Lecture 9: Hypothesis Testing with a Single Mean32, Video Lecture 10: Hypothesis Testing with Two Means33, Video Lecture 11: The Chi-Square Distribution34, Video Lecture 12: Linear Regression and Correlation35, , Version History, This module contains a listing of changes, updates, and corrections made to the Collaborative Statistics, textbook collection (col10522) since its initial release on 15 July 2008. The Version History is located at, http://cnx.org/content/m17360/latest/36 ., Textbook Adoption and Usage, This module is designed to track the various derivations of the Collaborative Statistics textbook and its, various companion resources, as well as keep track of educators who have adopted various versions for, their courses. New adopters are encouraged to provide their contact information and describe how they, will use this book for their courses. The goal is to provide a list that will allow educators using this book, 20 Collaborative, , Statistics Teacher’s Guide <http://cnx.org/content/col10547/latest/>, , 21 http://my.qoop.com/store/7064943342106149/8791310589747, 22 http://sofia.fhda.edu/gallery/statistics/index.html, 23 Collaborative, , Statistics: Supplemental Course Materials <http://cnx.org/content/col10586/latest/>, Statistics: Video Lecture - Sampling and Data" <http://cnx.org/content/m17561/latest/>, 25 "Elementary Statistics: Video Lecture - Descriptive Statistics" <http://cnx.org/content/m17562/latest/>, 26 "Elementary Statistics: Video Lecture - Probability Topics" <http://cnx.org/content/m17563/latest/>, 27 "Elementary Statistics: Video Lecture - Discrete Distributions" <http://cnx.org/content/m17565/latest/>, 28 "Elementary Statistics: Video Lecture - Continuous Random Variables" <http://cnx.org/content/m17566/latest/>, 29 "Elementary Statistics: Video Lecture - The Normal Distribution" <http://cnx.org/content/m17567/latest/>, 30 "Elementary Statistics: Video Lecture - The Central Limit Theorem" <http://cnx.org/content/m17568/latest/>, 31 "Elementary Statistics: Video Lecture - Confidence Intervals" <http://cnx.org/content/m17569/latest/>, 32 "Elementary Statistics: Video Lecture - Hypothesis Testing with a Single Mean" <http://cnx.org/content/m17570/latest/>, 33 "Elementary Statistics: Video Lecture - Hypothesis Testing with Two Means" <http://cnx.org/content/m17577/latest/>, 34 "Elementary Statistics: Video Lecture - The Chi-Square Distribution" <http://cnx.org/content/m17571/latest/>, 35 "Elementary Statistics: Video Lecture - Linear Regression and Correlation" <http://cnx.org/content/m17572/latest/>, 36 "Collaborative Statistics: Change History" <http://cnx.org/content/m17360/latest/>, 24 "Elementary, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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7, to collaborate, share ideas, and make suggestions for future development of this text. The Adoption and, Usage module is located at http://cnx.org/content/m18261/latest/37 ., Additional Technologies, In order to provide the most flexible learning resources possible, we invite collaboration from all instructors, wishing to create customized versions of this content for use with other technologies. For instance, you may, be interested in creating a set of instructions similar to this collection’s calculator notes. If you would like to, contribute to this collection, please use the contact the authors with any ideas or materials you have created., Accessibility and Section 508 Compliance, , • For, information, on, general, Connexions, accessibility, features,, please, visit, http://cnx.org/content/m17212/latest/38 ., • For information on accessibility features specific to the Collaborative Statistics textbook, please visit, http://cnx.org/content/m17211/latest/39 ., , 37 "Collaborative, , Statistics: Adoption and Usage" <http://cnx.org/content/m18261/latest/>, Features of Connexions" <http://cnx.org/content/m17212/latest/>, 39 "Collaborative Statistics: Accessibility" <http://cnx.org/content/m17211/latest/>, 38 "Accessibility, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Author Acknowledgements, , 40, , For this second edition, we appreciate the tremendous feedback from De Anza College colleagues and, students, as well as from the dozens of faculty around the world who taught out of the first and preliminary, editions. We have updated Collaborative Statistics with contributions from many faculty and students. We, especially thank Roberta Bloom, who wrote new problems and additional text., So many students and colleagues have contributed to the text, both the hard copy and open version. We, thank the following people for their contributions to the first and/or second editions., At De Anza College:, Dr. Inna Grushko (deceased), who wrote the glossary; Diane Mathios, who checked every homework, problem in the first edition; Kathy Plum, Lenore Desilets, Charles Klein, Janice Hector, Frank Snow, Dr., Lisa Markus, Dr. Vladimir Logvinenko (deceased), Mo Geraghty, Rupinder Sekhon, Javier Rueda, Carol, Olmstead; Also, Dr. Jim Lucas and Valerie Hauber of De Anza’s Office of Institutional Research, Mary Jo, Kane of Health Services; and the thousands of students who have used this text. Many of the students gave, us permission to include their outstanding word problems as homework., Additional thanks:, Dr. Larry Green of Lake Tahoe Community College, Terrie Teegarden of San Diego Mesa College, Ann, Flanigan of Kapiolani Community College, Birgit Aquilonius of West Valley College., The conversion from a for-profit hard copy text to a free open textbook is the result of many individuals and, organizations. We particularly thank Dr. Martha Kanter, Hal Plotkin, Dr. Judy Baker, Dr. Robert Maxfield, of Maxfield Foundation, Hewlett Foundation, and Connexions., Finally, we owe much to Frank, Jeffrey, and Jessica Dean and to Dan, Rachel, Matthew, and Rebecca Illowsky, who encouraged us to continue with our work and who had to hear more than their share of “I’m, sorry, I can’t” and “Just a minute, I’m working.”, , 40 This, , content is available online at <http://cnx.org/content/m16308/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 9
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Student Welcome Letter, , 41, , Dear Student:, Have you heard others say, “You’re taking statistics? That’s the hardest course I ever took!” They say that,, because they probably spent the entire course confused and struggling. They were probably lectured to, and never had the chance to experience the subject. You will not have that problem. Let’s find out why., There is a Chinese Proverb that describes our feelings about the field of statistics:, I HEAR, AND I FORGET, I SEE, AND I REMEMBER, I DO, AND I UNDERSTAND, Statistics is a “do” field. In order to learn it, you must “do” it. We have structured this book so that you will, have hands-on experiences. They will enable you to truly understand the concepts instead of merely going, through the requirements for the course., What makes this book different from other texts? First, we have eliminated the drudgery of tedious calculations. You might be using computers or graphing calculators so that you do not need to struggle with, algebraic manipulations. Second, this course is taught as a collaborative activity. With others in your class,, you will work toward the common goal of learning this material., Here are some hints for success in your class:, , •, •, •, •, •, •, •, •, •, , Work hard and work every night., Form a study group and learn together., Don’t get discouraged - you can do it!, As you solve problems, ask yourself, “Does this answer make sense?”, Many statistics words have the same meaning as in everyday English., Go to your teacher for help as soon as you need it., Don’t get behind., Read the newspaper and ask yourself, “Does this article make sense?”, Draw pictures - they truly help!, , Good luck and don’t give up!, Sincerely,, Susan Dean and Barbara Illowsky, 41 This, , content is available online at <http://cnx.org/content/m16305/1.5/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 11
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Chapter 1, , Sampling and Data, 1.1 Sampling and Data1, 1.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Recognize and differentiate between key terms., • Apply various types of sampling methods to data collection., • Create and interpret frequency tables., , 1.1.2 Introduction, You are probably asking yourself the question, "When and where will I use statistics?". If you read any, newspaper or watch television, or use the Internet, you will see statistical information. There are statistics, about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or, watch a news program on television, you are given sample information. With this information, you may, make a decision about the correctness of a statement, claim, or "fact." Statistical methods can help you make, the "best educated guess.", Since you will undoubtedly be given statistical information at some point in your life, you need to know, some techniques to analyze the information thoughtfully. Think about buying a house or managing a, budget. Think about your chosen profession. The fields of economics, business, psychology, education,, biology, law, computer science, police science, and early childhood development require at least one course, in statistics., Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what, "good" data are., , 1.2 Statistics2, The science of statistics deals with the collection, analysis, interpretation, and presentation of data. We see, and use data in our everyday lives., 1 This, 2 This, , content is available online at <http://cnx.org/content/m16008/1.9/>., content is available online at <http://cnx.org/content/m16020/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 13
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14, , CHAPTER 1. SAMPLING AND DATA, , 1.2.1 Optional Collaborative Classroom Exercise, In your classroom, try this exercise. Have class members write down the average time (in hours, to the, nearest half-hour) they sleep per night. Your instructor will record the data. Then create a simple graph, (called a dot plot) of the data. A dot plot consists of a number line and dots (or points) positioned above, the number line. For example, consider the following data:, 5; 5.5; 6; 6; 6; 6.5; 6.5; 6.5; 6.5; 7; 7; 8; 8; 9, The dot plot for this data would be as follows:, , Frequency of Average Time (in Hours) Spent Sleeping per Night, , Figure 1.1, , Does your dot plot look the same as or different from the example? Why? If you did the same example in, an English class with the same number of students, do you think the results would be the same? Why or, why not?, Where do your data appear to cluster? How could you interpret the clustering?, The questions above ask you to analyze and interpret your data. With this example, you have begun your, study of statistics., In this course, you will learn how to organize and summarize data. Organizing and summarizing data is, called descriptive statistics. Two ways to summarize data are by graphing and by numbers (for example,, finding an average). After you have studied probability and probability distributions, you will use formal, methods for drawing conclusions from "good" data. The formal methods are called inferential statistics., Statistical inference uses probability to determine how confident we can be that the conclusions are correct., Effective interpretation of data (inference) is based on good procedures for producing data and thoughtful, examination of the data. You will encounter what will seem to be too many mathematical formulas for, interpreting data. The goal of statistics is not to perform numerous calculations using the formulas, but to, gain an understanding of your data. The calculations can be done using a calculator or a computer. The, understanding must come from you. If you can thoroughly grasp the basics of statistics, you can be more, confident in the decisions you make in life., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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15, , 1.2.2 Levels of Measurement and Statistical Operations, The way a set of data is measured is called its level of measurement. Correct statistical procedures depend, on a researcher being familiar with levels of measurement. Not every statistical operation can be used with, every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest, level):, , •, •, •, •, , Nominal scale level, Ordinal scale level, Interval scale level, Ratio scale level, , Data that is measured using a nominal scale is qualitative. Categories, colors, names, labels and favorite, foods along with yes or no responses are examples of nominal level data. Nominal scale data are not, ordered. For example, trying to classify people according to their favorite food does not make any sense., Putting pizza first and sushi second is not meaningful., Smartphone companies are another example of nominal scale data. Some examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list and there is no agreed upon order. Some people may, favor Apple but that is a matter of opinion. Nominal scale data cannot be used in calculations., Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five, national parks in the United States. The top five national parks in the United States can be ranked from one, to five but we cannot measure differences between the data., Another example using the ordinal scale is a cruise survey where the responses to questions about, the cruise are “excellent,” “good,” “satisfactory” and “unsatisfactory.” These responses are ordered from, the most desired response by the cruise lines to the least desired. But the differences between two pieces of, data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations., Data that is measured using the interval scale is similar to ordinal level data because it has a definite ordering but there is a difference between data. The differences between interval scale data can be, measured though the data does not have a starting point., Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In, both temperature measurements, 40 degrees is equal to 100 degrees minus 60 degrees. Differences, make sense. But 0 degrees does not because, in both scales, 0 is not the absolute lowest temperature., Temperatures like -10o F and -15o C exist and are colder than 0., Interval level data can be used in calculations but one type of comparison cannot be done. Eighty, degrees C is not 4 times as hot as 20o C (nor is 80o F 4 times as hot as 20o F). There is no meaning to the, ratio of 80 to 20 (or 4 to 1)., Data that is measured using the ratio scale takes care of the ratio problem and gives you the most, information. Ratio scale data is like interval scale data but, in addition, it has a 0 point and ratios can be, calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a, possible 100 points). The exams were machine-graded., The data can be put in order from lowest to highest: 20, 68, 80, 92., The differences between the data have meaning. The score 92 is more than the score 68 by 24 points., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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16, , CHAPTER 1. SAMPLING AND DATA, , Ratios can be calculated. The smallest score for ratio data is 0. So 80 is 4 times 20. The score of 80, is 4 times better than the score of 20., Exercises, What type of measure scale is being used? Nominal, Ordinal, Interval or Ratio., 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., , High school men soccer players classified by their athletic ability: Superior, Average, Above average., Baking temperatures for various main dishes: 350, 400, 325, 250, 300, The colors of crayons in a 24-crayon box., Social security numbers., Incomes measured in dollars, A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not, satisfied., Political outlook: extreme left, left-of-center, right-of-center, extreme right., Time of day on an analog watch., The distance in miles to the closest grocery store., The dates 1066, 1492, 1644, 1947, 1944., The heights of 21 – 65 year-old women., Common letter grades A, B, C, D, F., , Answers 1. ordinal, 2. interval, 3. nominal, 4. nominal, 5. ratio, 6. ordinal, 7. nominal, 8. interval, 9. ratio,, 10. interval, 11. ratio, 12. ordinal, , 1.3 Probability3, Probability is a mathematical tool used to study randomness. It deals with the chance (the likelihood) of, an event occurring. For example, if you toss a fair coin 4 times, the outcomes may not be 2 heads and 2, tails. However, if you toss the same coin 4,000 times, the outcomes will be close to half heads and half tails., The expected theoretical probability of heads in any one toss is 12 or 0.5. Even though the outcomes of a, few repetitions are uncertain, there is a regular pattern of outcomes when there are many repetitions. After, reading about the English statistician Karl Pearson who tossed a coin 24,000 times with a result of 12,012, 996, heads, one of the authors tossed a coin 2,000 times. The results were 996 heads. The fraction 2000, is equal, to 0.498 which is very close to 0.5, the expected probability., The theory of probability began with the study of games of chance such as poker. Predictions take the form, of probabilities. To predict the likelihood of an earthquake, of rain, or whether you will get an A in this, course, we use probabilities. Doctors use probability to determine the chance of a vaccination causing the, disease the vaccination is supposed to prevent. A stockbroker uses probability to determine the rate of, return on a client’s investments. You might use probability to decide to buy a lottery ticket or not. In your, study of statistics, you will use the power of mathematics through probability calculations to analyze and, interpret your data., , 1.4 Key Terms4, In statistics, we generally want to study a population. You can think of a population as an entire collection, of persons, things, or objects under study. To study the larger population, we select a sample. The idea of, sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to, gain information about the population. Data are the result of sampling from a population., 3 This, 4 This, , content is available online at <http://cnx.org/content/m16015/1.11/>., content is available online at <http://cnx.org/content/m16007/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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17, Because it takes a lot of time and money to examine an entire population, sampling is a very practical, technique. If you wished to compute the overall grade point average at your school, it would make sense, to select a sample of students who attend the school. The data collected from the sample would be the, students’ grade point averages. In presidential elections, opinion poll samples of 1,000 to 2,000 people are, taken. The opinion poll is supposed to represent the views of the people in the entire country. Manufacturers of canned carbonated drinks take samples to determine if a 16 ounce can contains 16 ounces of, carbonated drink., From the sample data, we can calculate a statistic. A statistic is a number that is a property of the sample., For example, if we consider one math class to be a sample of the population of all math classes, then the, average number of points earned by students in that one math class at the end of the term is an example of, a statistic. The statistic is an estimate of a population parameter. A parameter is a number that is a property, of the population. Since we considered all math classes to be the population, then the average number of, points earned per student over all the math classes is an example of a parameter., One of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The, accuracy really depends on how well the sample represents the population. The sample must contain the, characteristics of the population in order to be a representative sample. We are interested in both the, sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the, sample statistic to test the validity of the established population parameter., A variable, notated by capital letters like X and Y, is a characteristic of interest for each person or thing in, a population. Variables may be numerical or categorical. Numerical variables take on values with equal, units such as weight in pounds and time in hours. Categorical variables place the person or thing into a, category. If we let X equal the number of points earned by one math student at the end of a term, then X, is a numerical variable. If we let Y be a person’s party affiliation, then examples of Y include Republican,, Democrat, and Independent. Y is a categorical variable. We could do some math with values of X (calculate, the average number of points earned, for example), but it makes no sense to do math with values of Y, (calculating an average party affiliation makes no sense)., Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single, value., Two words that come up often in statistics are mean and proportion. If you were to take three exams in, your math classes and obtained scores of 86, 75, and 92, you calculate your mean score by adding the three, exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math, 22, class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is 40, 18, and the proportion of women students is 40, . Mean and proportion are discussed in more detail in later, chapters., NOTE : The words "mean" and "average" are often used interchangeably. The substitution of one, word for the other is common practice. The technical term is "arithmetic mean" and "average" is, technically a center location. However, in practice among non-statisticians, "average" is commonly, accepted for "arithmetic mean.", , Example 1.1, Define the key terms from the following study: We want to know the average (mean) amount, of money first year college students spend at ABC College on school supplies that do not include, books. We randomly survey 100 first year students at the college. Three of those students spent, $150, $200, and $225, respectively., Solution, The population is all first year students attending ABC College this term., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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18, , CHAPTER 1. SAMPLING AND DATA, The sample could be all students enrolled in one section of a beginning statistics course at ABC, College (although this sample may not represent the entire population)., The parameter is the average (mean) amount of money spent (excluding books) by first year college students at ABC College this term., The statistic is the average (mean) amount of money spent (excluding books) by first year college, students in the sample., The variable could be the amount of money spent (excluding books) by one first year student., Let X = the amount of money spent (excluding books) by one first year student attending ABC, College., The data are the dollar amounts spent by the first year students. Examples of the data are $150,, $200, and $225., , 1.4.1 Optional Collaborative Classroom Exercise, Do the following exercise collaboratively with up to four people per group. Find a population, a sample,, the parameter, the statistic, a variable, and data for the following study: You want to determine the average, (mean) number of glasses of milk college students drink per day. Suppose yesterday, in your English class,, you asked five students how many glasses of milk they drank the day before. The answers were 1, 0, 1, 3,, and 4 glasses of milk., , 1.5 Data5, Data may come from a population or from a sample. Small letters like x or y generally are used to represent, data values. Most data can be put into the following categories:, , • Qualitative, • Quantitative, Qualitative data are the result of categorizing or describing attributes of a population. Hair color, blood, type, ethnic group, the car a person drives, and the street a person lives on are examples of qualitative data., Qualitative data are generally described by words or letters. For instance, hair color might be black, dark, brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Researchers often prefer to, use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For, example, it does not make sense to find an average hair color or blood type., Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes, of a population. Amount of money, pulse rate, weight, number of people living in your town, and the, number of students who take statistics are examples of quantitative data. Quantitative data may be either, discrete or continuous., All data that are the result of counting are called quantitative discrete data. These data take on only certain, numerical values. If you count the number of phone calls you receive for each day of the week, you might, get 0, 1, 2, 3, etc., 5 This, , content is available online at <http://cnx.org/content/m16005/1.18/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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19, All data that are the result of measuring are quantitative continuous data assuming that we can measure, accurately. Measuring angles in radians might result in the numbers π6 , π3 , π2 , π , 3π, 4 , etc. If you and your, friends carry backpacks with books in them to school, the numbers of books in the backpacks are discrete, data and the weights of the backpacks are continuous data., NOTE : In this course, the data used is mainly quantitative. It is easy to calculate statistics (like the, mean or proportion) from numbers. In the chapter Descriptive Statistics, you will be introduced, to stem plots, histograms and box plots all of which display quantitative data. Qualitative data is, discussed at the end of this section through graphs., , Example 1.2: Data Sample of Quantitative Discrete Data, The data are the number of books students carry in their backpacks. You sample five students., Two students carry 3 books, one student carries 4 books, one student carries 2 books, and one, student carries 1 book. The numbers of books (3, 4, 2, and 1) are the quantitative discrete data., Example 1.3: Data Sample of Quantitative Continuous Data, The data are the weights of the backpacks with the books in it. You sample the same five students., The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying, three books can have different weights. Weights are quantitative continuous data because weights, are measured., Example 1.4: Data Sample of Qualitative Data, The data are the colors of backpacks. Again, you sample the same five students. One student has, a red backpack, two students have black backpacks, one student has a green backpack, and one, student has a gray backpack. The colors red, black, black, green, and gray are qualitative data., , NOTE : You may collect data as numbers and report it categorically. For example, the quiz scores, for each student are recorded throughout the term. At the end of the term, the quiz scores are, reported as A, B, C, D, or F., , Example 1.5, Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate, whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with, the words "the number of.", 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., 14., , The number of pairs of shoes you own., The type of car you drive., Where you go on vacation., The distance it is from your home to the nearest grocery store., The number of classes you take per school year., The tuition for your classes, The type of calculator you use., Movie ratings., Political party preferences., Weight of sumo wrestlers., Amount of money won playing poker., Number of correct answers on a quiz., Peoples’ attitudes toward the government., IQ scores. (This may cause some discussion.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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20, , CHAPTER 1. SAMPLING AND DATA, , Qualitative Data Discussion, Below are tables of part-time vs full-time students at De Anza College in Cupertino, CA and Foothill College in Los Altos, CA for the Spring 2010 quarter. The tables display counts (frequencies) and percentages, or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important, when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is, compared to De Anza College., De Anza College, Number, , Percent, , Full-time, , 9,200, , 40.9%, , Part-time, , 13,296, , 59.1%, , Total, , 22,496, , 100%, , Table 1.1, , Foothill College, Number, , Percent, , Full-time, , 4,059, , 28.6%, , Part-time, , 10,124, , 71.4%, , Total, , 14,183, , 100%, , Table 1.2, , Tables are a good way of organizing and displaying data. But graphs can be even more helpful in, understanding the data. There are no strict rules concerning what graphs to use. Below are pie charts and, bar graphs, two graphs that are used to display qualitative data., In a pie chart, categories of data are represented by wedges in the circle and are proportional in size, to the percent of individuals in each category., In a bar graph, the length of the bar for each category is proportional to the number or percent of, individuals in each category. Bars may be vertical or horizontal., A Pareto chart consists of bars that are sorted into order by category size (largest to smallest)., Look at the graphs and determine which graph (pie or bar) you think displays the comparisons better. This is a matter of preference., It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the, data. We might make different choices of what we think is the "best" graph depending on the data and the, context. Our choice also depends on what we are using the data for., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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21, , Table 1.3, , Table 1.4, , Percentages That Add to More (or Less) Than 100%, Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages, add to more than 100% because students can be in more than one category. A bar graph is appropriate, to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the, percentages added to less than 100%., De Anza College Spring 2010, Characteristic/Category, , Percent, , Full-time Students, , 40.9%, , Students who intend to transfer to a 4-year educational institution, , 48.6%, , Students under age 25, , 61.0%, , TOTAL, , 150.5%, Table 1.5, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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22, , CHAPTER 1. SAMPLING AND DATA, , Table 1.6, , Omitting Categories/Missing Data, The table displays Ethnicity of Students but is missing the "Other/Unknown" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that, the frequencies do not add up to the total number of students. Create a bar graph and not a pie chart., Missing Data: Ethnicity of Students De Anza College Fall Term 2007 (Census Day), Frequency, , Percent, , Asian, , 8,794, , 36.1%, , Black, , 1,412, , 5.8%, , Filipino, , 1,298, , 5.3%, , Hispanic, , 4,180, , 17.1%, , Native American, , 146, , 0.6%, , Pacific Islander, , 236, , 1.0%, , White, , 5,978, , 24.5%, , TOTAL, , 22,044 out of 24,382, , 90.4% out of 100%, , Table 1.7, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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23, Bar graph Without Other/Unknown Category, , Table 1.8, , The following graph is the same as the previous graph but the "Other/Unknown" percent (9.6%) has been, added back in. The "Other/Unknown" category is large compared to some of the other categories (Native, American, 0.6%, Pacific Islander 1.0% particularly). This is important to know when we think about what, the data are telling us., This particular bar graph can be hard to understand visually. The graph below it is a Pareto chart., The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret., Bar Graph With Other/Unknown Category, , Table 1.9, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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24, , CHAPTER 1. SAMPLING AND DATA, Pareto Chart With Bars Sorted By Size, , Table 1.10, , Pie Charts: No Missing Data, The following pie charts have the "Other/Unknown" category added back in (since the percentages must, add to 100%). The chart on the right is organized having the wedges by size and makes for a more visually, informative graph than the unsorted, alphabetical graph on the left., , Table 1.11, , 1.6 Sampling6, Gathering information about an entire population often costs too much or is virtually impossible. Instead,, we use a sample of the population. A sample should have the same characteristics as the population it, is representing. Most statisticians use various methods of random sampling in an attempt to achieve this, goal. This section will describe a few of the most common methods., There are several different methods of random sampling. In each form of random sampling, each member, of a population initially has an equal chance of being selected for the sample. Each method has pros and, cons. The easiest method to describe is called a simple random sample. Any group of n individuals is, 6 This, , content is available online at <http://cnx.org/content/m16014/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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25, equally likely to be chosen by any other group of n individuals if the simple random sampling technique is, used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus, class, which has 31 members not including Lisa. To choose a simple random sample of size 3 from the other, members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out 3, names. A more technological way is for Lisa to first list the last names of the members of her class together, with a two-digit number as shown below., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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26, , CHAPTER 1. SAMPLING AND DATA, Class Roster, ID, , Name, , 00, , Anselmo, , 01, , Bautista, , 02, , Bayani, , 03, , Cheng, , 04, , Cuarismo, , 05, , Cuningham, , 06, , Fontecha, , 07, , Hong, , 08, , Hoobler, , 09, , Jiao, , 10, , Khan, , 11, , King, , 12, , Legeny, , 13, , Lundquist, , 14, , Macierz, , 15, , Motogawa, , 16, , Okimoto, , 17, , Patel, , 18, , Price, , 19, , Quizon, , 20, , Reyes, , 21, , Roquero, , 22, , Roth, , 23, , Rowell, , 24, , Salangsang, , 25, , Slade, , 26, , Stracher, , 27, , Tallai, , 28, , Tran, , 29, , Wai, , 30, , Wood, Table 1.12, , Lisa can either use a table of random numbers (found in many statistics books as well as mathematical, handbooks) or a calculator or computer to generate random numbers. For this example, suppose Lisa, chooses to generate random numbers from a calculator. The numbers generated are:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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27, .94360; .99832; .14669; .51470; .40581; .73381; .04399, Lisa reads two-digit groups until she has chosen three class members (that is, she reads .94360 as the groups, 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could, have generated more random numbers., The random numbers .94360 and .99832 do not contain appropriate two digit numbers. However the third, random number, .14669, contains 14 (the fourth random number also contains 14), the fifth random number, contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz,, 05 corresponds to Cunningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist, of Marcierz, and Cunningham, and Cuarismo., Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster, sample, and the systematic sample., To choose a stratified sample, divide the population into groups called strata and then take a proportionate, number from each stratum. For example, you could stratify (group) your college population by department, and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of, the first department, number each member of the second department and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department, and do the same for each of the remaining departments. Those numbers picked from the first department,, picked from the second department and so on represent the members who make up the stratified sample., To choose a cluster sample, divide the population into clusters (groups) and then randomly select some of, the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly, sample four departments from your college population, the four departments make up the cluster sample., For example, divide your college faculty by department. The departments are the clusters. Number each, department and then choose four different numbers using simple random sampling. All members of the, four departments with those numbers are the cluster sample., To choose a systematic sample, randomly select a starting point and take every nth piece of data from a, listing of the population. For example, suppose you have to do a phone survey. Your phone book contains, 20,000 residence listings. You must choose 400 names for the sample. Number the population 1 - 20,000, and then use a simple random sample to pick a number that represents the first name of the sample. Then, choose every 50th name thereafter until you have a total of 400 names (you might have to go back to the of, your phone list). Systematic sampling is frequently chosen because it is a simple method., A type of sampling that is nonrandom is convenience sampling. Convenience sampling involves using, results that are readily available. For example, a computer software store conducts a marketing study by, interviewing potential customers who happen to be in the store browsing through the available software., The results of convenience sampling may be very good in some cases and highly biased (favors certain, outcomes) in others., Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (for example, they may favor a certain, group). It is better for the person conducting the survey to select the sample respondents., True random sampling is done with replacement. That is, once a member is picked that member goes, back into the population and thus may be chosen more than once. However for practical reasons, in most, populations, simple random sampling is done without replacement. Surveys are typically done without, replacement. That is, a member of the population may be chosen only once. Most samples are taken from, large populations and the sample tends to be small in comparison to the population. Since this is the case,, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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28, , CHAPTER 1. SAMPLING AND DATA, , sampling without replacement is approximately the same as sampling with replacement because the chance, of picking the same individual more than once using with replacement is very low., For example, in a college population of 10,000 people, suppose you want to randomly pick a sample of 1000, for a survey. For any particular sample of 1000, if you are sampling with replacement,, , • the chance of picking the first person is 1000 out of 10,000 (0.1000);, • the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999);, • the chance of picking the same person again is 1 out of 10,000 (very low)., If you are sampling without replacement,, , • the chance of picking the first person for any particular sample is 1000 out of 10,000 (0.1000);, • the chance of picking a different second person is 999 out of 9,999 (0.0999);, • you do not replace the first person before picking the next person., Compare the fractions 999/10,000 and 999/9,999. For accuracy, carry the decimal answers to 4 place decimals. To 4 decimal places, these numbers are equivalent (0.0999)., Sampling without replacement instead of sampling with replacement only becomes a mathematics issue, when the population is small which is not that common. For example, if the population is 25 people, the, sample is 10 and you are sampling with replacement for any particular sample,, , • the chance of picking the first person is 10 out of 25 and a different second person is 9 out of 25 (you, replace the first person)., If you sample without replacement,, , • the chance of picking the first person is 10 out of 25 and then the second person (which is different) is, 9 out of 24 (you do not replace the first person)., Compare the fractions 9/25 and 9/24. To 4 decimal places, 9/25 = 0.3600 and 9/24 = 0.3750. To 4 decimal, places, these numbers are not equivalent., When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual, process of sampling causes sampling errors. For example, the sample may not be large enough. Factors, not related to the sampling process cause nonsampling errors. A defective counting device can cause a, nonsampling error., In reality, a sample will never be exactly representative of the population so there will always be, some sampling error. As a rule, the larger the sample, the smaller the sampling error., In statistics, a sampling bias is created when a sample is collected from a population and some, members of the population are not as likely to be chosen as others (remember, each member of the, population should have an equally likely chance of being chosen). When a sampling bias happens, there, can be incorrect conclusions drawn about the population that is being studied., Example 1.6, Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience)., 1. A soccer coach selects 6 players from a group of boys aged 8 to 10, 7 players from a group of, boys aged 11 to 12, and 3 players from a group of boys aged 13 to 14 to form a recreational, soccer team., 2. A pollster interviews all human resource personnel in five different high tech companies., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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29, 3. A high school educational researcher interviews 50 high school female teachers and 50 high, school male teachers., 4. A medical researcher interviews every third cancer patient from a list of cancer patients at a, local hospital., 5. A high school counselor uses a computer to generate 50 random numbers and then picks, students whose names correspond to the numbers., 6. A student interviews classmates in his algebra class to determine how many pairs of jeans a, student owns, on the average., Solution, 1., 2., 3., 4., 5., 6., , stratified, cluster, stratified, systematic, simple random, convenience, , If we were to examine two samples representing the same population, even if we used random sampling, methods for the samples, they would not be exactly the same. Just as there is variation in data, there is, variation in samples. As you become accustomed to sampling, the variability will seem natural., Example 1.7, Suppose ABC College has 10,000 part-time students (the population). We are interested in the, average amount of money a part-time student spends on books in the fall term. Asking all 10,000, students is an almost impossible task., Suppose we take two different samples., First, we use convenience sampling and survey 10 students from a first term organic chemistry, class. Many of these students are taking first term calculus in addition to the organic chemistry, class . The amount of money they spend is as follows:, $128; $87; $173; $116; $130; $204; $147; $189; $93; $153, The second sample is taken by using a list from the P.E. department of senior citizens who take, P.E. classes and taking every 5th senior citizen on the list, for a total of 10 senior citizens. They, spend:, $50; $40; $36; $15; $50; $100; $40; $53; $22; $22, Problem 1, Do you think that either of these samples is representative of (or is characteristic of) the entire, 10,000 part-time student population?, Solution, No. The first sample probably consists of science-oriented students. Besides the chemistry course,, some of them are taking first-term calculus. Books for these classes tend to be expensive. Most, of these students are, more than likely, paying more than the average part-time student for their, books. The second sample is a group of senior citizens who are, more than likely, taking courses, for health and interest. The amount of money they spend on books is probably much less than the, average part-time student. Both samples are biased. Also, in both cases, not all students have a, chance to be in either sample., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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30, , CHAPTER 1. SAMPLING AND DATA, , Problem 2, Since these samples are not representative of the entire population, is it wise to use the results to, describe the entire population?, Solution, No. For these samples, each member of the population did not have an equally likely chance of, being chosen., Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education,, art, and early childhood development. (We assume that these are the only disciplines in which, part-time students at ABC College are enrolled and that an equal number of part-time students, are enrolled in each of the disciplines.) Each student is chosen using simple random sampling., Using a calculator, random numbers are generated and a student from a particular discipline is, selected if he/she has a corresponding number. The students spend:, $180; $50; $150; $85; $260; $75; $180; $200; $200; $150, Problem 3, Is the sample biased?, Solution, The sample is unbiased, but a larger sample would be recommended to increase the likelihood, that the sample will be close to representative of the population. However, for a biased sampling, technique, even a large sample runs the risk of not being representative of the population., Students often ask if it is "good enough" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes., , 1.6.1 Optional Collaborative Classroom Exercise, Exercise 1.6.1, As a class, determine whether or not the following samples are representative. If they are not,, discuss the reasons., 1. To find the average GPA of all students in a university, use all honor students at the university as the sample., 2. To find out the most popular cereal among young people under the age of 10, stand outside, a large supermarket for three hours and speak to every 20th child under age 10 who enters, the supermarket., 3. To find the average annual income of all adults in the United States, sample U.S. congressmen. Create a cluster sample by considering each state as a stratum (group). By using simple, random sampling, select states to be part of the cluster. Then survey every U.S. congressman, in the cluster., 4. To determine the proportion of people taking public transportation to work, survey 20 people in New York City. Conduct the survey by sitting in Central Park on a bench and interviewing every person who sits next to you., 5. To determine the average cost of a two day stay in a hospital in Massachusetts, survey 100, hospitals across the state using simple random sampling., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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31, , 1.7 Variation7, 1.7.1 Variation in Data, Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less, than 16 ounces of liquid. In one study, eight 16 ounce cans were measured and produced the following, amount (in ounces) of beverage:, 15.8; 16.1; 15.2; 14.8; 15.8; 15.9; 16.0; 15.5, Measurements of the amount of beverage in a 16-ounce can may vary because different people make the, measurements or because the exact amount, 16 ounces of liquid, was not put into the cans. Manufacturers, regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range., Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the, same purpose. This is completely natural. However, if two or more of you are taking the same data and, get very different results, it is time for you and the others to reevaluate your data-taking methods and your, accuracy., , 1.7.2 Variation in Samples, It was mentioned previously that two or more samples from the same population, taken randomly, and, having close to the same characteristics of the population are different from each other. Suppose Doreen and, Jung both decide to study the average amount of time students at their college sleep each night. Doreen and, Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling., Doreen’s sample will be different from Jung’s sample. Even if Doreen and Jung used the same sampling, method, in all likelihood their samples would be different. Neither would be wrong, however., Think about what contributes to making Doreen’s and Jung’s samples different., If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results, (the average amount of time a student sleeps) might be closer to the actual population average. But still,, their samples would be, in all likelihood, different from each other. This variability in samples cannot be, stressed enough., 1.7.2.1 Size of a Sample, The size of a sample (often called the number of observations) is important. The examples you have seen, in this book so far have been small. Samples of only a few hundred observations, or even smaller, are, sufficient for many purposes. In polling, samples that are from 1200 to 1500 observations are considered, large enough and good enough if the survey is random and is well done. You will learn why when you, study confidence intervals., Be aware that many large samples are biased., because people choose to respond or not., 7 This, , For example, call-in surveys are invariable biased, , content is available online at <http://cnx.org/content/m16021/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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32, , CHAPTER 1. SAMPLING AND DATA, , 1.7.2.2 Optional Collaborative Classroom Exercise, Exercise 1.7.1, Divide into groups of two, three, or four. Your instructor will give each group one 6-sided die., Try this experiment twice. Roll one fair die (6-sided) 20 times. Record the number of ones, twos,, threes, fours, fives, and sixes you get below ("frequency" is the number of times a particular face, of the die occurs):, First Experiment (20 rolls), Face on Die, , Frequency, , 1, 2, 3, 4, 5, 6, Table 1.13, , Second Experiment (20 rolls), Face on Die, , Frequency, , 1, 2, 3, 4, 5, 6, Table 1.14, , Did the two experiments have the same results? Probably not. If you did the experiment a third, time, do you expect the results to be identical to the first or second experiment? (Answer yes or, no.) Why or why not?, Which experiment had the correct results? They both did. The job of the statistician is to see, through the variability and draw appropriate conclusions., , 1.7.3 Critical Evaluation, We need to critically evaluate the statistical studies we read about and analyze before accepting the results, of the study. Common problems to be aware of include, , • Problems with Samples: A sample should be representative of the population. A sample that is not, representative of the population is biased. Biased samples that are not representative of the population give results that are inaccurate and not valid., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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33, , • Self-Selected Samples: Responses only by people who choose to respond, such as call-in surveys are, often unreliable., • Sample Size Issues: Samples that are too small may be unreliable. Larger samples are better if possible., In some situations, small samples are unavoidable and can still be used to draw conclusions, even, though larger samples are better. Examples: Crash testing cars, medical testing for rare conditions., • Undue influence: Collecting data or asking questions in a way that influences the response., • Non-response or refusal of subject to participate: The collected responses may no longer be representative of the population. Often, people with strong positive or negative opinions may answer surveys,, which can affect the results., • Causality: A relationship between two variables does not mean that one causes the other to occur., They may both be related (correlated) because of their relationship through a different variable., • Self-Funded or Self-Interest Studies: A study performed by a person or organization in order to support their claim. Is the study impartial? Read the study carefully to evaluate the work. Do not, automatically assume that the study is good but do not automatically assume the study is bad either., Evaluate it on its merits and the work done., • Misleading Use of Data: Improperly displayed graphs, incomplete data, lack of context., • Confounding: When the effects of multiple factors on a response cannot be separated. Confounding, makes it difficult or impossible to draw valid conclusions about the effect of each factor., , 1.8 Answers and Rounding Off8, A simple way to round off answers is to carry your final answer one more decimal place than was present, in the original data. Round only the final answer. Do not round any intermediate results, if possible. If it, becomes necessary to round intermediate results, carry them to at least twice as many decimal places as the, final answer. For example, the average of the three quiz scores 4, 6, 9 is 6.3, rounded to the nearest tenth,, because the data are whole numbers. Most answers will be rounded in this manner., It is not necessary to reduce most fractions in this course. Especially in Probability Topics (Section 3.1), the, chapter on probability, it is more helpful to leave an answer as an unreduced fraction., , 1.9 Frequency9, Twenty students were asked how many hours they worked per day. Their responses, in hours, are listed, below:, 5; 6; 3; 3; 2; 4; 7; 5; 2; 3; 5; 6; 5; 4; 4; 3; 5; 2; 5; 3, Below is a frequency table listing the different data values in ascending order and their frequencies., 8 This, 9 This, , content is available online at <http://cnx.org/content/m16006/1.8/>., content is available online at <http://cnx.org/content/m16012/1.20/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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34, , CHAPTER 1. SAMPLING AND DATA, Frequency Table of Student Work Hours, DATA VALUE, , FREQUENCY, , 2, , 3, , 3, , 5, , 4, , 3, , 5, , 6, , 6, , 2, , 7, , 1, Table 1.15, , A frequency is the number of times a given datum occurs in a data set. According to the table above,, there are three students who work 2 hours, five students who work 3 hours, etc. The total of the frequency, column, 20, represents the total number of students included in the sample., A relative frequency is the fraction or proportion of times an answer occurs. To find the relative frequencies, divide each frequency by the total number of students in the sample - in this case, 20. Relative, frequencies can be written as fractions, percents, or decimals., Frequency Table of Student Work Hours w/ Relative Frequency, DATA VALUE, , FREQUENCY, , RELATIVE FREQUENCY, , 2, , 3, , 3, , 5, , 4, , 3, , 5, , 6, , 6, , 2, , 7, , 1, , 3, 20, 5, 20, 3, 20, 6, 20, 2, 20, 1, 20, , or 0.15, or 0.25, or 0.15, or 0.30, or 0.10, or 0.05, , Table 1.16, , The sum of the relative frequency column is, , 20, 20 ,, , or 1., , Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current, row., Frequency Table of Student Work Hours w/ Relative and Cumulative Relative Frequency, DATA VALUE, , FREQUENCY, , RELATIVE, FREQUENCY, , CUMULATIVE RELATIVE, FREQUENCY, continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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35, 2, , 3, , 3, , 5, , 4, , 3, , 5, , 6, , 6, , 2, , 7, , 1, , 3, 20, 5, 20, 3, 20, 6, 20, 2, 20, 1, 20, , or 0.15, , 0.15, , or 0.25, , 0.15 + 0.25 = 0.40, , or 0.15, , 0.40 + 0.15 = 0.55, , or 0.30, , 0.55 + 0.30 = 0.85, , or 0.10, , 0.85 + 0.10 = 0.95, , or 0.05, , 0.95 + 0.05 = 1.00, , Table 1.17, , The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of, the data has been accumulated., NOTE : Because of rounding, the relative frequency column may not always sum to one and the last, entry in the cumulative relative frequency column may not be one. However, they each should be, close to one., , The following table represents the heights, in inches, of a sample of 100 male semiprofessional soccer players., Frequency Table of Soccer Player Height, HEIGHTS, (INCHES), , FREQUENCY, , RELATIVE, FREQUENCY, , CUMULATIVE, RELATIVE, FREQUENCY, , 59.95 - 61.95, , 5, , = 0.05, , 0.05, , 61.95 - 63.95, , 3, , = 0.03, , 0.05 + 0.03 = 0.08, , 63.95 - 65.95, , 15, , = 0.15, , 0.08 + 0.15 = 0.23, , 65.95 - 67.95, , 40, , = 0.40, , 0.23 + 0.40 = 0.63, , 67.95 - 69.95, , 17, , = 0.17, , 0.63 + 0.17 = 0.80, , 69.95 - 71.95, , 12, , = 0.12, , 0.80 + 0.12 = 0.92, , 71.95 - 73.95, , 7, , = 0.07, , 0.92 + 0.07 = 0.99, , 73.95 - 75.95, , 1, , 5, 100, 3, 100, 15, 100, 40, 100, 17, 100, 12, 100, 7, 100, 1, 100, , = 0.01, , 0.99 + 0.01 = 1.00, , Total = 100, , Total = 1.00, Table 1.18, , The data in this table has been grouped into the following intervals:, , •, •, •, •, •, •, •, •, , 59.95 - 61.95 inches, 61.95 - 63.95 inches, 63.95 - 65.95 inches, 65.95 - 67.95 inches, 67.95 - 69.95 inches, 69.95 - 71.95 inches, 71.95 - 73.95 inches, 73.95 - 75.95 inches, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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36, , CHAPTER 1. SAMPLING AND DATA, NOTE : This example is used again in the Descriptive Statistics (Section 2.1) chapter, where the, method used to compute the intervals will be explained., , In this sample, there are 5 players whose heights are between 59.95 - 61.95 inches, 3 players whose heights, fall within the interval 61.95 - 63.95 inches, 15 players whose heights fall within the interval 63.95 - 65.95, inches, 40 players whose heights fall within the interval 65.95 - 67.95 inches, 17 players whose heights, fall within the interval 67.95 - 69.95 inches, 12 players whose heights fall within the interval 69.95 - 71.95,, 7 players whose height falls within the interval 71.95 - 73.95, and 1 player whose height falls within the, interval 73.95 - 75.95. All heights fall between the endpoints of an interval and not at the endpoints., Example 1.8, From the table, find the percentage of heights that are less than 65.95 inches., Solution, If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are, 5 + 3 + 15 = 23 males whose heights are less than 65.95 inches. The percentage of heights less than, 23, or 23%. This percentage is the cumulative relative frequency entry in the, 65.95 inches is then 100, third row., , Example 1.9, From the table, find the percentage of heights that fall between 61.95 and 65.95 inches., Solution, Add the relative frequencies in the second and third rows: 0.03 + 0.15 = 0.18 or 18%., , Example 1.10, Use the table of heights of the 100 male semiprofessional soccer players. Fill in the blanks and, check your answers., 1., 2., 3., 4., 5., 6., , The percentage of heights that are from 67.95 to 71.95 inches is:, The percentage of heights that are from 67.95 to 73.95 inches is:, The percentage of heights that are more than 65.95 inches is:, The number of players in the sample who are between 61.95 and 71.95 inches tall is:, What kind of data are the heights?, Describe how you could gather this data (the heights) so that the data are characteristic of all, male semiprofessional soccer players., , Remember, you count frequencies. To find the relative frequency, divide the frequency by the, total number of data values. To find the cumulative relative frequency, add all of the previous, relative frequencies to the relative frequency for the current row., , 1.9.1 Optional Collaborative Classroom Exercise, Exercise 1.9.1, In your class, have someone conduct a survey of the number of siblings (brothers and sisters) each, student has. Create a frequency table. Add to it a relative frequency column and a cumulative, relative frequency column. Answer the following questions:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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37, 1. What percentage of the students in your class has 0 siblings?, 2. What percentage of the students has from 1 to 3 siblings?, 3. What percentage of the students has fewer than 3 siblings?, Example 1.11, Nineteen people were asked how many miles, to the nearest mile they commute to work each, day. The data are as follows:, 2; 5; 7; 3; 2; 10; 18; 15; 20; 7; 10; 18; 5; 12; 13; 12; 4; 5; 10, The following table was produced:, Frequency of Commuting Distances, DATA, , FREQUENCY, , RELATIVEFREQUENCY, , CUMULATIVERELATIVEFREQUENCY, , 3, , 3, , 0.1579, , 4, , 1, , 5, , 3, , 7, , 2, , 10, , 3, , 12, , 2, , 13, , 1, , 15, , 1, , 18, , 1, , 20, , 1, , 3, 19, 1, 19, 3, 19, 2, 19, 4, 19, 2, 19, 1, 19, 1, 19, 1, 19, 1, 19, , 0.2105, 0.1579, 0.2632, 0.4737, 0.7895, 0.8421, 0.8948, 0.9474, 1.0000, Table 1.19, , Problem, , (Solution on p. 55.), , 1. Is the table correct? If it is not correct, what is wrong?, 2. True or False: Three percent of the people surveyed commute 3 miles. If the statement is not, correct, what should it be? If the table is incorrect, make the corrections., 3. What fraction of the people surveyed commute 5 or 7 miles?, 4. What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between 5 and 13 miles (does not include 5 and 13 miles)?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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38, , CHAPTER 1. SAMPLING AND DATA, , 1.10 Summary10, Statistics, , • Deals with the collection, analysis, interpretation, and presentation of data, Probability, , • Mathematical tool used to study randomness, Key Terms, , •, •, •, •, •, •, , Population, Parameter, Sample, Statistic, Variable, Data, , Types of Data, , • Quantitative Data (a number), · Discrete (You count it.), · Continuous (You measure it.), • Qualitative Data (a category, words), Sampling, , • With Replacement: A member of the population may be chosen more than once, • Without Replacement: A member of the population may be chosen only once, Random Sampling, , • Each member of the population has an equal chance of being selected, Sampling Methods, , • Random, · Simple random sample, · Stratified sample, · Cluster sample, · Systematic sample, • Not Random, · Convenience sample, Frequency (freq. or f), , • The number of times an answer occurs, Relative Frequency (rel. freq. or RF), , • The proportion of times an answer occurs, • Can be interpreted as a fraction, decimal, or percent, Cumulative Relative Frequencies (cum. rel. freq. or cum RF), , • An accumulation of the previous relative frequencies, 10 This, , content is available online at <http://cnx.org/content/m16023/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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39, , 1.11 Practice: Sampling and Data11, 1.11.1 Student Learning Outcomes, • The student will construct frequency tables., • The student will differentiate between key terms., • The student will compare sampling techniques., , 1.11.2 Given, Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program., Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS, symptoms have revealed themselves. Of interest is the average(mean) length of time in months patients, live once starting the treatment. Two researchers each follow a different set of 40 AIDS patients from the, start of treatment until their deaths. The following data (in months) are collected., Researcher A 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32;, 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34, Researcher B 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22;, 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29, , 1.11.3 Organize the Data, Complete the tables below using the data provided., Researcher A, Survival Frequency, Length, (in, months), , Relative Frequency, , Cumulative Relative Frequency, , 0.5 - 6.5, 6.5, 12.5, , -, , 12.5, 18.5, , -, , 18.5, 24.5, , -, , 24.5, 30.5, , continued on next page, , 11 This, , content is available online at <http://cnx.org/content/m16016/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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40, , CHAPTER 1. SAMPLING AND DATA, , 30.5, 36.5, , -, , 36.5, 42.5, , -, , 42.5, 48.5, , -, , Table 1.20, , Researcher B, Survival, months), , Length, , (in, , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , 0.5 - 6.5, 6.5 - 12.5, 12.5 - 18.5, 18.5 - 24.5, 24.5 - 30.5, 30.5 - 36.5, 36.5 - 42.5, 42.5 - 48.5, Table 1.21, , 1.11.4 Key Terms, Define the key terms based upon the above example for Researcher A., Exercise 1.11.1, Population, Exercise 1.11.2, Sample, Exercise 1.11.3, Parameter, Exercise 1.11.4, Statistic, Exercise 1.11.5, Variable, Exercise 1.11.6, Data, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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41, , 1.11.5 Discussion Questions, Discuss the following questions and then answer in complete sentences., Exercise 1.11.7, List two reasons why the data may differ., Exercise 1.11.8, Can you tell if one researcher is correct and the other one is incorrect? Why?, Exercise 1.11.9, Would you expect the data to be identical? Why or why not?, Exercise 1.11.10, How could the researchers gather random data?, Exercise 1.11.11, Suppose that the first researcher conducted his survey by randomly choosing one state in the, nation and then randomly picking 40 patients from that state. What sampling method would that, researcher have used?, Exercise 1.11.12, Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What, sampling method would that researcher have used? What concerns would you have about this, data set, based upon the data collection method?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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42, , CHAPTER 1. SAMPLING AND DATA, , 1.12 Homework12, Exercise 1.12.1, For each item below:, , (Solution on p. 55.), , i. Identify the type of data (quantitative - discrete, quantitative - continuous, or qualitative) that, would be used to describe a response., ii. Give an example of the data., a. Number of tickets sold to a concert, b. Amount of body fat, c. Favorite baseball team, d. Time in line to buy groceries, e. Number of students enrolled at Evergreen Valley College, f. Most–watched television show, g. Brand of toothpaste, h. Distance to the closest movie theatre, i. Age of executives in Fortune 500 companies, j. Number of competing computer spreadsheet software packages, Exercise 1.12.2, Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below:, Part-time Student Course Loads, # of Courses, , Frequency, , Relative Frequency, , 1, , 30, , 0.6, , 2, , 15, , Cumulative, Frequency, , Relative, , 3, Table 1.22, , a. Fill in the blanks in the table above., b. What percent of students take exactly two courses?, c. What percent of students take one or two courses?, Exercise 1.12.3, (Solution on p. 55.), Sixty adults with gum disease were asked the number of times per week they used to floss before, their diagnoses. The (incomplete) results are shown below:, Flossing Frequency for Adults with Gum Disease, # Flossing per Week, , Frequency, , Relative Frequency, , 0, , 27, , 0.4500, , 1, , 18, , 3, , 12 This, , Cumulative Relative Freq., , 0.9333, , 6, , 3, , 0.0500, , 7, , 1, , 0.0167, , content is available online at <http://cnx.org/content/m16010/1.19/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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43, Table 1.23, , a. Fill in the blanks in the table above., b. What percent of adults flossed six times per week?, c. What percent flossed at most three times per week?, Exercise 1.12.4, A fitness center is interested in the mean amount of time a client exercises in the center each week., Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.5, (Solution on p. 55.), Ski resorts are interested in the mean age that children take their first ski and snowboard lessons., They need this information to optimally plan their ski classes. Define the following in terms of the, study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.6, A cardiologist is interested in the mean recovery period for her patients who have had heart, attacks. Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.7, (Solution on p. 56.), Insurance companies are interested in the mean health costs each year for their clients, so that, they can determine the costs of health insurance. Define the following in terms of the study. Give, examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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44, , CHAPTER 1. SAMPLING AND DATA, Exercise 1.12.8, A politician is interested in the proportion of voters in his district that think he is doing a good, job. Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.9, (Solution on p. 56.), A marriage counselor is interested in the proportion the clients she counsels that stay married., Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.10, Political pollsters may be interested in the proportion of people that will vote for a particular, cause. Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.11, (Solution on p. 56.), A marketing company is interested in the proportion of people that will buy a particular product., Define the following in terms of the study. Give examples where appropriate., a. Population, b. Sample, c. Parameter, d. Statistic, e. Variable, f. Data, Exercise 1.12.12, Airline companies are interested in the consistency of the number of babies on each flight, so that, they have adequate safety equipment. Suppose an airline conducts a survey. Over Thanksgiving, weekend, it surveys 6 flights from Boston to Salt Lake City to determine the number of babies on, the flights. It determines the amount of safety equipment needed by the result of that study., a. Using complete sentences, list three things wrong with the way the survey was conducted., b. Using complete sentences, list three ways that you would improve the survey if it were to be, repeated., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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45, Exercise 1.12.13, Suppose you want to determine the mean number of students per statistics class in your state., Describe a possible sampling method in 3 – 5 complete sentences. Make the description detailed., Exercise 1.12.14, Suppose you want to determine the mean number of cans of soda drunk each month by persons, in their twenties. Describe a possible sampling method in 3 - 5 complete sentences. Make the, description detailed., Exercise 1.12.15, (Solution on p. 56.), 771 distance learning students at Long Beach City College responded to surveys in the 201011 academic year. Highlights of the summary report are listed in the table below. (Source:, http://de.lbcc.edu/reports/2010-11/future/highlights.html#focus)., LBCC Distance Learning Survey Results, Have computer at home, , 96%, , Unable to come to campus for classes, , 65%, , Age 41 or over, , 24%, , Would like LBCC to offer more DL courses, , 95%, , Took DL classes due to a disability, , 17%, , Live at least 16 miles from campus, , 13%, , Took DL courses to fulfill transfer requirements, , 71%, , Table 1.24, , a. What percent of the students surveyed do not have a computer at home?, b. About how many students in the survey live at least 16 miles from campus?, c. If the same survey was done at Great Basin College in Elko, Nevada, do you think the percentages would be the same? Why?, Exercise 1.12.16, Nineteen immigrants to the U.S were asked how many years, to the nearest year, they have lived, in the U.S. The data are as follows:, 2; 5; 7; 2; 2; 10; 20; 15; 0; 7; 0; 20; 5; 12; 15; 12; 4; 5; 10, The following table was produced:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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46, , CHAPTER 1. SAMPLING AND DATA, Frequency of Immigrant Survey Responses, Data, , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , 0, , 2, , 0.1053, , 2, , 3, , 4, , 1, , 5, , 3, , 7, , 2, , 10, , 2, , 12, , 2, , 15, , 1, , 20, , 1, , 2, 19, 3, 19, 1, 19, 3, 19, 2, 19, 2, 19, 2, 19, 1, 19, 1, 19, , 0.2632, 0.3158, 0.1579, 0.5789, 0.6842, 0.7895, 0.8421, 1.0000, Table 1.25, , a. Fix the errors on the table. Also, explain how someone might have arrived at the incorrect, number(s)., b. Explain what is wrong with this statement: “47 percent of the people surveyed have lived in, the U.S. for 5 years.”, c. Fix the statement above to make it correct., d. What fraction of the people surveyed have lived in the U.S. 5 or 7 years?, e. What fraction of the people surveyed have lived in the U.S. at most 12 years?, f. What fraction of the people surveyed have lived in the U.S. fewer than 12 years?, g. What fraction of the people surveyed have lived in the U.S. from 5 to 20 years, inclusive?, Exercise 1.12.17, A “random survey” was conducted of 3274 people of the “microprocessor generation” (people, born since 1971, the year the microprocessor was invented). It was reported that 48% of those, individuals surveyed stated that if they had $2000 to spend, they would use it for computer, equipment. Also, 66% of those surveyed considered themselves relatively savvy computer users., (Source: San Jose Mercury News), a. Do you consider the sample size large enough for a study of this type? Why or why not?, b. Based on your “gut feeling,” do you believe the percents accurately reflect the U.S. population, for those individuals born since 1971? If not, do you think the percents of the population are, actually higher or lower than the sample statistics? Why?, Additional information: The survey was reported by Intel Corporation of individuals who visited, the Los Angeles Convention Center to see the Smithsonian Institure’s road show called “America’s, Smithsonian.”, c. With this additional information, do you feel that all demographic and ethnic groups were, equally represented at the event? Why or why not?, d. With the additional information, comment on how accurately you think the sample statistics, reflect the population parameters., Exercise 1.12.18, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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47, a. List some practical difficulties involved in getting accurate results from a telephone survey., b. List some practical difficulties involved in getting accurate results from a mailed survey., c. With your classmates, brainstorm some ways to overcome these problems if you needed to, conduct a phone or mail survey., , 1.12.1 Try these multiple choice questions, The next four questions refer to the following: A Lake Tahoe Community College instructor is interested, in the mean number of days Lake Tahoe Community College math students are absent from class during a, quarter., Exercise 1.12.19, What is the population she is interested in?, A., B., C., D., , (Solution on p. 56.), , All Lake Tahoe Community College students, All Lake Tahoe Community College English students, All Lake Tahoe Community College students in her classes, All Lake Tahoe Community College math students, , Exercise 1.12.20, Consider the following:, , (Solution on p. 56.), , X = number of days a Lake Tahoe Community College math student is absent, In this case, X is an example of a:, A., B., C., D., , Variable, Population, Statistic, Data, , Exercise 1.12.21, (Solution on p. 56.), The instructor takes her sample by gathering data on 5 randomly selected students from each, Lake Tahoe Community College math class. The type of sampling she used is, A., B., C., D., , Cluster sampling, Stratified sampling, Simple random sampling, Convenience sampling, , Exercise 1.12.22, (Solution on p. 56.), The instructor’s sample produces an mean number of days absent of 3.5 days. This value is an, example of a, A., B., C., D., , Parameter, Data, Statistic, Variable, , The next two questions refer to the following relative frequency table on hurricanes that have made direct, hits on the U.S between 1851 and 2004. Hurricanes are given a strength category rating based on the, minimum wind speed generated by the storm. (http://www.nhc.noaa.gov/gifs/table5.gif 13 ), 13 http://www.nhc.noaa.gov/gifs/table5.gif, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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48, , CHAPTER 1. SAMPLING AND DATA, Frequency of Hurricane Direct Hits, Category, , Number of Direct Hits, , Relative Frequency, , Cumulative Frequency, , 1, , 109, , 0.3993, , 0.3993, , 2, , 72, , 0.2637, , 0.6630, , 3, , 71, , 0.2601, , 4, , 18, , 5, , 3, , 0.9890, 0.0110, , 1.0000, , Total = 273, Table 1.26, , Exercise 1.12.23, What is the relative frequency of direct hits that were category 4 hurricanes?, A., B., C., D., , (Solution on p. 56.), , 0.0768, 0.0659, 0.2601, Not enough information to calculate, , Exercise 1.12.24, (Solution on p. 56.), What is the relative frequency of direct hits that were AT MOST a category 3 storm?, A., B., C., D., , 0.3480, 0.9231, 0.2601, 0.3370, , The next three questions refer to the following: A study was done to determine the age, number of times, per week and the duration (amount of time) of resident use of a local park in San Jose. The first house in, the neighborhood around the park was selected randomly and then every 8th house in the neighborhood, around the park was interviewed., Exercise 1.12.25, “‘Number of times per week”’ is what type of data?, , (Solution on p. 56.), , A. qualitative, B. quantitative - discrete, C. quantitative - continuous, Exercise 1.12.26, The sampling method was:, A., B., C., D., , (Solution on p. 56.), , simple random, systematic, stratified, cluster, , Exercise 1.12.27, “‘Duration (amount of time)”’ is what type of data?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (Solution on p. 56.)
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49, A. qualitative, B. quantitative - discrete, C. quantitative - continuous, Exercises 28 and 29 are not multiple choice exercises., Exercise 1.12.28, Name the sampling method used in each of the following situations:, , (Solution on p. 56.), , A. A woman in the airport is handing out questionnaires to travelers asking them to evaluate the, airport’s service. She does not ask travelers who are hurrying through the airport with their, hands full of luggage, but instead asks all travelers sitting near gates and who are not taking, naps while they wait., B. A teacher wants to know if her students are doing homework so she randomly selects rows 2, and 5, and then calls on all students in row 2 and all students in row 5 to present the solution, to homework problems to the class., C. The marketing manager for an electronics chain store wants information about the ages of its, customers. Over the next two weeks, at each store location, 100 randomly selected customers, are given questionnaires to fill out which asks for information about age, as well as about, other variables of interest., D. The librarian at a public library wants to determine what proportion of the library users are, children. The librarian has a tally sheet on which she marks whether the books are checked, out by an adult or a child. She records this data for every 4th patron who checks out books., E. A political party wants to know the reaction of voters to a debate between the candidates. The, day after the debate, the party’s polling staff calls 1200 randomly selected phone numbers., If a registered voter answers the phone or is available to come to the phone, that registered, voter is asked who he/she intends to vote for and whether the debate changed his/her, opinion of the candidates., ** Contributed by Roberta Bloom, Exercise 1.12.29, (Solution on p. 57.), Several online textbook retailers advertise that they have lower prices than on-campus bookstores. However, an important factor is whether the internet retailers actually have the textbooks, that students need in stock. Students need to be able to get textbooks promptly at the beginning of, the college term. If the book is not available, then a student would not be able to get the textbook, at all, or might get a delayed delivery if the book is back ordered., A college newspaper reporter is investigating textbook availability at online retailers. He, decides to investigate one textbook for each of the following 7 subjects: calculus, biology,, chemistry, physics, statistics, geology, and general engineering. He consults textbook industry, sales data and selects the most popular nationally used textbook in each of these subjects. He, visits websites for a random sample of major online textbook sellers and looks up each of these 7, textbooks to see if they are available in stock for quick delivery through these retailers. Based on, his investigation, he writes an article in which he draws conclusions about the overall availability, of all college textbooks through online textbook retailers., Write an analysis of his study that addresses the following issues: Is his sample representative of the population of all college textbooks? Explain why or why not. Describe some possible, sources of bias in this study, and how it might affect the results of the study. Give some suggestions about what could be done to improve the study., ** Contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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51, Frequency of Number of Movies Viewed, Number of Movies, , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , 0, 1, 2, 3, 4, 5, 6, 7+, Table 1.28, , Frequency of Number of Movies Viewed, Number of Movies, , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , 0-1, 2-3, 4-5, 6-7+, Table 1.29, , 1., 2., 3., 4., , Using the tables, find the percent of data that is at most 2. Which table did you use and why?, Using the tables, find the percent of data that is at most 3. Which table did you use and why?, Using the tables, find the percent of data that is more than 2. Which table did you use and why?, Using the tables, find the percent of data that is more than 3. Which table did you use and why?, , 1.13.4 Discussion Questions, 1. Is one of the tables above "more correct" than the other? Why or why not?, 2. In general, why would someone group the data in different ways? Are there any advantages to either, way of grouping the data?, 3. Why did you switch between tables, if you did, when answering the question above?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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52, , CHAPTER 1. SAMPLING AND DATA, , 1.14 Lab 2: Sampling Experiment15, Class Time:, Names:, , 1.14.1 Student Learning Outcomes, • The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques., • The student will explain each of the details of each procedure used., In this lab, you will be asked to pick several random samples. In each case, describe your procedure briefly,, including how you might have used the random number generator, and then list the restaurants in the, sample you obtained, NOTE : The following section contains restaurants stratified by city into columns and grouped, horizontally by entree cost (clusters)., , 1.14.2 A Simple Random Sample, Pick a simple random sample of 15 restaurants., 1. Describe the procedure:, 2., 1. __________ 6. __________, , 11. __________, , 2. __________, , 7. __________, , 12. __________, , 3. __________, , 8. __________, , 13. __________, , 4. __________, , 9. __________, , 14. __________, , 5. __________, , 10. __________, , 15. __________, Table 1.30, , 1.14.3 A Systematic Sample, Pick a systematic sample of 15 restaurants., 1. Describe the procedure:, 2., 1. __________ 6. __________, , 11. __________, , 2. __________, , 7. __________, , 12. __________, , 3. __________, , 8. __________, , 13. __________, , 4. __________, , 9. __________, , 14. __________, , 5. __________, , 10. __________, , 15. __________, Table 1.31, , 15 This, , content is available online at <http://cnx.org/content/m16013/1.15/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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55, , Solutions to Exercises in Chapter 1, Solution to Example 1.5, Problem (p. 19), Items 1, 5, 11, and 12 are quantitative discrete; items 4, 6, 10, and 14 are quantitative continuous; and items, 2, 3, 7, 8, 9, and 13 are qualitative., Solution to Example 1.10, Problem (p. 36), 1., 2., 3., 4., 5., 6., , 29%, 36%, 77%, 87, quantitative continuous, get rosters from each team and choose a simple random sample from each, , Solution to Example 1.11, Problem (p. 37), 1. No. Frequency column sums to 18, not 19. Not all cumulative relative frequencies are correct., 2. False. Frequency for 3 miles should be 1; for 2 miles (left out), 2. Cumulative relative frequency, column should read: 0.1052, 0.1579, 0.2105, 0.3684, 0.4737, 0.6316, 0.7368, 0.7895, 0.8421, 0.9474, 1., 5, 3. 19, 7 12 7, 4. 19, , 19 , 19, , Solutions to Homework, Solution to Exercise 1.12.1 (p. 42), a. quantitative - discrete, b. quantitative - continuous, c. qualitative, d. quantitative - continuous, e. quantitative - discrete, f. qualitative, g. qualitative, h. quantitative - continuous, i. quantitative - continuous, j. quantitative - discrete, Solution to Exercise 1.12.3 (p. 42), a. Cum. Rel. Freq. for 0 is 0.4500, Rel. Freq. for 1 is 0.3000 and Cum. Rel. Freq. for 1 or less is 0.7500, Freq. for 3 is 11 and Rel. Freq. is 0.1833, Cum. Rel. Freq. for 6 or less is 0.9833, Cum. Rel. Freq. for 7 or less is 1, b. 5.00%, c. 93.33%, Solution to Exercise 1.12.5 (p. 43), a., b., c., d., e., , Children who take ski or snowboard lessons, A group of these children, The population mean, The sample mean, X = the age of one child who takes the first ski or snowboard lesson, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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56, , CHAPTER 1. SAMPLING AND DATA, , f. Values for X, such as 3, 7, etc., Solution to Exercise 1.12.7 (p. 43), a. The clients of the insurance companies, b. A group of the clients, c. The mean health costs of the clients, d. The mean health costs of the sample, e. X = the health costs of one client, f. Values for X, such as 34, 9, 82, etc., Solution to Exercise 1.12.9 (p. 44), a. All the clients of the counselor, b. A group of the clients, c. The proportion of all her clients who stay married, d. The proportion of the sample who stay married, e. X = the number of couples who stay married, f. yes, no, Solution to Exercise 1.12.11 (p. 44), a. All people (maybe in a certain geographic area, such as the United States), b. A group of the people, c. The proportion of all people who will buy the product, d. The proportion of the sample who will buy the product, e. X = the number of people who will buy it, f. buy, not buy, Solution to Exercise 1.12.15 (p. 45), a: 4%, b: 100, Solution to Exercise 1.12.19 (p., D, Solution to Exercise 1.12.20 (p., A, Solution to Exercise 1.12.21 (p., B, Solution to Exercise 1.12.22 (p., C, Solution to Exercise 1.12.23 (p., B, Solution to Exercise 1.12.24 (p., B, Solution to Exercise 1.12.25 (p., B, Solution to Exercise 1.12.26 (p., B, Solution to Exercise 1.12.27 (p., C, Solution to Exercise 1.12.28 (p., , 47), 47), 47), 47), 48), 48), 48), 48), 48), 49), , A. Convenience, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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57, B. Cluster, C. Stratified, D. Systematic, E. Simple Random, Solution to Exercise 1.12.29 (p. 49), The answer below contains some of the issues that students might discuss for this problem. Individual, student’s answers may also identify other issues that pertain to this problem that are not included in the, answer below., The sample is not representative of the population of all college textbooks. Two reasons why it is, not representative are that he only sampled 7 subjects and he only investigated one textbook in each, subject. There are several possible sources of bias in the study. The 7 subjects that he investigated are, all in mathematics and the sciences; there are many subjects in the humanities, social sciences, and many, other subject areas, (for example: literature, art, history, psychology, sociology, business) that he did not, investigate at all. It may be that different subject areas exhibit different patterns of textbook availability,, but his sample would not detect such results., He also only looked at the most popular textbook in each of the subjects he investigated. The availability of the most popular textbooks may differ from the availability of other textbooks in one of two, ways:, , • the most popular textbooks may be more readily available online, because more new copies are, printed and more students nationwide selling back their used copies OR, • the most popular textbooks may be harder to find available online, because more student demand, exhausts the supply more quickly., In reality, many college students do not use the most popular textbook in their subject, and this study gives, no useful information about the situation for those less popular textbooks., He could improve this study by, , • expanding the selection of subjects he investigates so that it is more representative of all subjects, studied by college students and, • expanding the selection of textbooks he investigates within each subject to include a mixed representation of both the popular and less popular textbooks., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 2, , Descriptive Statistics, 2.1 Descriptive Statistics1, 2.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, , Display data graphically and interpret graphs: stemplots, histograms and boxplots., Recognize, describe, and calculate the measures of location of data: quartiles and percentiles., Recognize, describe, and calculate the measures of the center of data: mean, median, and mode., Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation,, and range., , 2.1.2 Introduction, Once you have collected data, what will you do with it? Data can be described and presented in many, different formats. For example, suppose you are interested in buying a house in a particular area. You may, have no clue about the house prices, so you might ask your real estate agent to give you a sample data set, of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look, at the median price and the variation of prices. The median and variation are just two ways that you will, learn to describe data. Your agent might also provide you with a graph of the data., In this chapter, you will study numerical and graphical ways to describe and display your data. This area, of statistics is called "Descriptive Statistics". You will learn to calculate, and even more importantly, to, interpret these measurements and graphs., , 2.2 Displaying Data2, A statistical graph is a tool that helps you learn about the shape or distribution of a sample. The graph can, be a more effective way of presenting data than a mass of numbers because we can see where data clusters, and where there are only a few data values. Newspapers and the Internet use graphs to show trends and, to enable readers to compare facts and figures quickly., Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied., 1 This, 2 This, , content is available online at <http://cnx.org/content/m16300/1.9/>., content is available online at <http://cnx.org/content/m16297/1.9/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 59
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60, , CHAPTER 2. DESCRIPTIVE STATISTICS, , Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar chart,, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), pie charts, and, the boxplot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs and bar graphs. Our, emphasis will be on histograms and boxplots., , 2.3 Stem and Leaf Graphs (Stemplots), Line Graphs and Bar Graphs3, One simple graph, the stem-and-leaf graph or stem plot, comes from the field of exploratory data analysis.It is a good choice when the data sets are small. To create the plot, divide each observation of data into, a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem 2 and leaf 3. Four, hundred thirty-two (432) has stem 43 and leaf 2. Five thousand four hundred thirty-two (5,432) has stem, 543 and leaf 2. The decimal 9.3 has stem 9 and leaf 3. Write the stems in a vertical line from smallest the, largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their, corresponding stem., Example 2.1, For Susan Dean’s spring pre-calculus class, scores for the first exam were as follows (smallest to, largest):, 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94;, 96; 100, Stem-and-Leaf Diagram, Stem, , Leaf, , 3, , 3, , 4, , 299, , 5, , 355, , 6, , 1378899, , 7, , 2348, , 8, , 03888, , 9, , 0244446, , 10, , 0, Table 2.1, , The stem plot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or, approximately 26% of the scores were in the 90’s or 100, a fairly high number of As., The stem plot is a quick way to graph and gives an exact picture of the data. You want to look for an overall, pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is, sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the, graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may, indicate that something unusual is happening. It takes some background information to explain outliers., In the example above, there were no outliers., Example 2.2, Create a stem plot using the data:, 3 This, , content is available online at <http://cnx.org/content/m16849/1.17/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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61, 1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3, The data are the distance (in kilometers) from a home to the nearest supermarket., Problem, , (Solution on p. 114.), , 1. Are there any values that might possibly be outliers?, 2. Do the data seem to have any concentration of values?, H INT: The leaves are to the right of the decimal., , Another type of graph that is useful for specific data values is a line graph. In the particular line graph, shown in the example, the x-axis consists of data values and the y-axis consists of frequency points. The, frequency points are connected., Example 2.3, In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do, his/her chores. The results are shown in the table and the line graph., Number of times teenager is reminded, , Frequency, , 0, , 2, , 1, , 5, , 2, , 8, , 3, , 14, , 4, , 7, , 5, , 4, Table 2.2, , Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be, rectangular boxes and they can be vertical or horizontal., The bar graph shown in Example 4 has age groups represented on the x-axis and proportions on the y-axis., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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62, , CHAPTER 2. DESCRIPTIVE STATISTICS, Example 2.4, By the end of 2011, in the United States, Facebook had over 146 million users. The table, shows three age groups, the number of users in each age group and the proportion (%) of, users in each age group. Source: http://www.kenburbary.com/2011/03/facebook-demographicsrevisited-2011-statistics-2/, Age groups, , Number of Facebook users, , Proportion (%) of Facebook users, , 13 - 25, , 65,082,280, , 45%, , 26 - 44, , 53,300,200, , 36%, , 45 - 64, , 27,885,100, , 19%, Table 2.3, , Example 2.5, The columns in the table below contain the race/ethnicity of U.S. Public Schools: High School, Class of 2011, percentages for the Advanced Placement Examinee Population for that class, and percentages for the Overall Student Population. The 3-dimensional graph shows the, Race/Ethnicity of U.S. Public Schools (qualitative data) on the x-axis and Advanced Placement, Examinee Population percentages on the y-axis. (Source: http://www.collegeboard.com and, Source: http://apreport.collegeboard.org/goals-and-findings/promoting-equity), , Race/Ethnicity, , AP Examinee Population, , Overall Student Population, , 1 = Asian, Asian American or Pacific Islander, , 10.3%, , 5.7%, continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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63, 2 = Black or African American, , 9.0%, , 14.7%, , 3 = Hispanic or Latino, , 17.0%, , 17.6%, , 4 = American Indian or Alaska, Native, , 0.6%, , 1.1%, , 5 = White, , 57.1%, , 59.2%, , 6 = Not reported/other, , 6.0%, , 1.7%, Table 2.4, , Go to Outcomes of Education Figure 224 for an example of a bar graph that shows unemployment rates of, persons 25 years and older for 2009., NOTE : This book contains instructions for constructing a histogram and a box plot for the TI-83+, and TI-84 calculators. You can find additional instructions for using these calculators on the Texas, Instruments (TI) website5 ., , 2.4 Histograms6, For most of the work you do in this book, you will use a histogram to display the data. One advantage of a, histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data, set consists of 100 values or more., A histogram consists of contiguous boxes. It has both a horizontal axis and a vertical axis. The horizontal, axis is labeled with what the data represents (for instance, distance from your home to school). The vertical, axis is labeled either Frequency or relative frequency. The graph will have the same shape with either, label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the, data. (The next section tells you how to calculate the center and the spread.), 4 http://nces.ed.gov/pubs2011/2011015_5.pdf, 5 http://education.ti.com/educationportal/sites/US/sectionHome/support.html, 6 This, , content is available online at <http://cnx.org/content/m16298/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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64, , CHAPTER 2. DESCRIPTIVE STATISTICS, , The relative frequency is equal to the frequency for an observed value of the data divided by the total, number of data values in the sample. (In the chapter on Sampling and Data (Section 1.1), we defined, frequency as the number of times an answer occurs.) If:, , • f = frequency, • n = total number of data values (or the sum of the individual frequencies), and, • RF = relative frequency,, then:, f, (2.0), n, For example, if 3 students in Mr. Ahab’s English class of 40 students received from 90% to 100%, then,, RF =, , f = 3 , n = 40 , and RF =, , f, n, , =, , 3, 40, , = 0.075, , Seven and a half percent of the students received 90% to 100%. Ninety percent to 100 % are quantitative, measures., To construct a histogram, first decide how many bars or intervals, also called classes, represent the data., Many histograms consist of from 5 to 15 bars or classes for clarity. Choose a starting point for the first, interval to be less than the smallest data value. A convenient starting point is a lower value carried out, to one more decimal place than the value with the most decimal places. For example, if the value with the, most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 - 0.05 = 6.05)., We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value, is 1.5, a convenient starting point is 1.495 (1.5 - 0.005 = 1.495). If the value with the most decimal places is, 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 - .0005 = 0.9995). If all the data, happen to be integers and the smallest value is 2, then a convenient starting point is 1.5 (2 - 0.5 = 1.5). Also,, when the starting point and other boundaries are carried to one additional decimal place, no data value, will fall on a boundary., Example 2.6, The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional, soccer players. The heights are continuous data since height is measured., 60; 60.5; 61; 61; 61.5, 63.5; 63.5; 63.5, 64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5, 66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67;, 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5, 68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5, 70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71, 72; 72; 72; 72.5; 72.5; 73; 73.5, 74, The smallest data value is 60. Since the data with the most decimal places has one decimal (for, instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5,, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for, the convenient starting point., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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65, 60 - 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is,, then, 59.95., The largest value is 74. 74+ 0.05 = 74.05 is the ending value., Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting, point from the ending value and divide by the number of bars (you must choose the number of, bars you desire). Suppose you choose 8 bars., 74.05 − 59.95, = 1.76, 8, NOTE : We will round up to 2 and make each bar or class interval 2 units wide. Rounding up to 2 is, one way to prevent a value from falling on a boundary. Rounding to the next number is necessary, even if it goes against the standard rules of rounding. For this example, using 1.76 as the width, would also work., , The boundaries are:, , •, •, •, •, •, •, •, •, •, , 59.95, 59.95 + 2 = 61.95, 61.95 + 2 = 63.95, 63.95 + 2 = 65.95, 65.95 + 2 = 67.95, 67.95 + 2 = 69.95, 69.95 + 2 = 71.95, 71.95 + 2 = 73.95, 73.95 + 2 = 75.95, , The heights 60 through 61.5 inches are in the interval 59.95 - 61.95. The heights that are 63.5 are, in the interval 61.95 - 63.95. The heights that are 64 through 64.5 are in the interval 63.95 - 65.95., The heights 66 through 67.5 are in the interval 65.95 - 67.95. The heights 68 through 69.5 are in the, interval 67.95 - 69.95. The heights 70 through 71 are in the interval 69.95 - 71.95. The heights 72, through 73.5 are in the interval 71.95 - 73.95. The height 74 is in the interval 73.95 - 75.95., The following histogram displays the heights on the x-axis and relative frequency on the y-axis., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (2.0)
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66, , CHAPTER 2. DESCRIPTIVE STATISTICS, , Example 2.7, The following data are the number of books bought by 50 part-time college students at ABC, College. The number of books is discrete data since books are counted., 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1, 2; 2; 2; 2; 2; 2; 2; 2; 2; 2, 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3, 4; 4; 4; 4; 4; 4, 5; 5; 5; 5; 5, 6; 6, Eleven students buy 1 book. Ten students buy 2 books. Sixteen students buy 3 books. Six students, buy 4 books. Five students buy 5 books. Two students buy 6 books., Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the, largest data value. Then the starting point is 0.5 and the ending value is 6.5., Problem, (Solution on p. 114.), Next, calculate the width of each bar or class interval. If the data are discrete and there are not too, many different values, a width that places the data values in the middle of the bar or class interval, is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6 and the starting point is, 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of, the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of, the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______,, and the _______ in the middle of the interval from _______ to _______ ., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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67, Calculate the number of bars as follows:, 6.5 − 0.5, =1, bars, where 1 is the width of a bar. Therefore, bars = 6., , (2.0), , The following histogram displays the number of books on the x-axis and the frequency on the, y-axis., , Using the TI-83, 83+, 84, 84+ Calculator Instructions, Go to the Appendix (14:Appendix) in the menu on the left. There are calculator instructions for entering, data and for creating a customized histogram. Create the histogram for Example 2., , • Press Y=. Press CLEAR to clear out any equations., • Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR and arrow down. If, necessary, do the same for L2., • Into L1, enter 1, 2, 3, 4, 5, 6, • Into L2, enter 11, 10, 16, 6, 5, 2, • Press WINDOW. Make Xmin = .5, Xmax = 6.5, Xscl = (6.5 - .5)/6, Ymin = -1, Ymax = 20, Yscl = 1, Xres, =1, • Press 2nd Y=. Start by pressing 4:Plotsoff ENTER., • Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3rd picture (histogram). Press ENTER., • Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2 (2nd 2)., • Press GRAPH, • Use the TRACE key and the arrow keys to examine the histogram., , 2.4.1 Optional Collaborative Exercise, Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a, class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You, may want to experiment with the number of intervals. Discuss, also, the shape of the histogram., Record the data, in dollars (for example, 1.25 dollars)., Construct a histogram., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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68, , CHAPTER 2. DESCRIPTIVE STATISTICS, , 2.5 Box Plots7, Box plots or box-whisker plots give a good graphical image of the concentration of the data. They also, show how far from most of the data the extreme values are. The box plot is constructed from five values:, the smallest value, the first quartile, the median, the third quartile, and the largest value. The median, the, first quartile, and the third quartile will be discussed here, and then again in the section on measuring data, in this chapter. We use these values to compare how close other data values are to them., The median, a number, is a way of measuring the "center" of the data. You can think of the median as the, "middle value," although it does not actually have to be one of the observed values. It is a number that, separates ordered data into halves. Half the values are the same number or smaller than the median and, half the values are the same number or larger. For example, consider the following data:, 1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1, Ordered from smallest to largest:, 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5, The median is between the 7th value, 6.8, and the 8th value 7.2. To find the median, add the two values, together and divide by 2., 6.8 + 7.2, =7, 2, The median is 7. Half of the values are smaller than 7 and half of the values are larger than 7., , (2.0), , Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data., To find the quartiles, first find the median or second quartile. The first quartile is the middle value of the, lower half of the data and the third quartile is the middle value of the upper half of the data. To get the, idea, consider the same data set shown above:, 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5, The median or second quartile is 7. The lower half of the data is 1, 1, 2, 2, 4, 6, 6.8. The middle value of the, lower half is 2., 1; 1; 2; 2; 4; 6; 6.8, The number 2, which is part of the data, is the first quartile. One-fourth of the values are the same or less, than 2 and three-fourths of the values are more than 2., The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is 9., 7.2; 8; 8.3; 9; 10; 10; 11.5, The number 9, which is part of the data, is the third quartile. Three-fourths of the values are less than 9, and one-fourth of the values are more than 9., To construct a box plot, use a horizontal number line and a rectangular box. The smallest and largest data, values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile, marks the other end of the box. The middle fifty percent of the data fall inside the box. The "whiskers", extend from the ends of the box to the smallest and largest data values. The box plot gives a good quick, picture of the data., 7 This, , content is available online at <http://cnx.org/content/m16296/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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69, NOTE : You may encounter box and whisker plots that have dots marking outlier values. In those, cases, the whiskers are not extending to the minimum and maximum values., , Consider the following data:, 1; 1; 2; 2; 4; 6; 6.8 ; 7.2; 8; 8.3; 9; 10; 10; 11.5, The first quartile is 2, the median is 7, and the third quartile is 9. The smallest value is 1 and the largest, value is 11.5. The box plot is constructed as follows (see calculator instructions in the back of this book or, on the TI web site8 ):, , The two whiskers extend from the first quartile to the smallest value and from the third quartile to the, largest value. The median is shown with a dashed line., Example 2.8, The following data are the heights of 40 students in a statistics class., 59; 60; 61; 62; 62; 63; 63; 64; 64; 64; 65; 65; 65; 65; 65; 65; 65; 65; 65; 66; 66; 67; 67; 68; 68; 69; 70; 70; 70;, 70; 70; 71; 71; 72; 72; 73; 74; 74; 75; 77, Construct a box plot:, Using the TI-83, 83+, 84, 84+ Calculator, , • Enter data into the list editor (Press STAT 1:EDIT). If you need to clear the list, arrow up to, the name L1, press CLEAR, arrow down., • Put the data values in list L1., • Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1., • Press ENTER, • Use the down and up arrow keys to scroll., •, •, •, •, •, , Smallest value = 59, Largest value = 77, Q1: First quartile = 64.5, Q2: Second quartile or median= 66, Q3: Third quartile = 70, , Using the TI-83, 83+, 84, 84+ to Construct the Box Plot, Go to 14:Appendix for Notes for the TI-83, 83+, 84, 84+ Calculator. To create the box plot:, , • Press Y=. If there are any equations, press CLEAR to clear them., • Press 2nd Y=., • Press 4:Plotsoff. Press ENTER, 8 http://education.ti.com/educationportal/sites/US/sectionHome/support.html, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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70, , CHAPTER 2. DESCRIPTIVE STATISTICS, , • Press 2nd Y=, • Press 1:Plot1. Press ENTER., • Arrow down and then use the right arrow key to go to the 5th picture which is the box plot., Press ENTER., • Arrow down to Xlist: Press 2nd 1 for L1, • Arrow down to Freq: Press ALPHA. Press 1., • Press ZOOM. Press 9:ZoomStat., • Press TRACE and use the arrow keys to examine the box plot., , a. Each quarter has 25% of the data., b. The spreads of the four quarters are 64.5 - 59 = 5.5 (first quarter), 66 - 64.5 = 1.5 (second quarter),, 70 - 66 = 4 (3rd quarter), and 77 - 70 = 7 (fourth quarter). So, the second quarter has the, smallest spread and the fourth quarter has the largest spread., c. Interquartile Range: IQR = Q3 − Q1 = 70 − 64.5 = 5.5., d. The interval 59 through 65 has more than 25% of the data so it has more data in it than the, interval 66 through 70 which has 25% of the data., e. The middle 50% (middle half) of the data has a range of 5.5 inches., For some sets of data, some of the largest value, smallest value, first quartile, median, and third, quartile may be the same. For instance, you might have a data set in which the median and the, third quartile are the same. In this case, the diagram would not have a dotted line inside the box, displaying the median. The right side of the box would display both the third quartile and the, median. For example, if the smallest value and the first quartile were both 1, the median and the, third quartile were both 5, and the largest value was 7, the box plot would look as follows:, , Example 2.9, Test scores for a college statistics class held during the day are:, 99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90, Test scores for a college statistics class held during the evening are:, 98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5, Problem, , (Solution on p. 114.), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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71, , •, •, •, •, , What are the smallest and largest data values for each data set?, What is the median, the first quartile, and the third quartile for each data set?, Create a boxplot for each set of data., Which boxplot has the widest spread for the middle 50% of the data (the data between the, first and third quartiles)? What does this mean for that set of data in comparison to the other, set of data?, • For each data set, what percent of the data is between the smallest value and the first quartile? (Answer: 25%) the first quartile and the median? (Answer: 25%) the median and the, third quartile? the third quartile and the largest value? What percent of the data is between, the first quartile and the largest value? (Answer: 75%), The first data set (the top box plot) has the widest spread for the middle 50% of the data. IQR =, Q3 − Q1 is 82.5 − 56 = 26.5 for the first data set and 89 − 78 = 11 for the second data set., So, the first set of data has its middle 50% of scores more spread out., 25% of the data is between M and Q3 and 25% is between Q3 and Xmax., , 2.6 Measures of the Location of the Data9, The common measures of location are quartiles and percentiles (%iles). Quartiles are special percentiles., The first quartile, Q1 is the same as the 25th percentile (25th %ile) and the third quartile, Q3 , is the same as, the 75th percentile (75th %ile). The median, M, is called both the second quartile and the 50th percentile, (50th %ile)., NOTE :, , Quartiles are given special attention in the Box Plots module in this chapter., , To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Recall that, quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in, the 90th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that, 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater, than your test score., Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles, extensively., Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of, the test scores are less (and not the same or less) than your score, it would be acceptable because removing, one particular data value is not significant., The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the, data. It is the difference between the third quartile (Q3 ) and the first quartile (Q1 )., IQR = Q3 − Q1, , (2.0), , The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is, less than (1.5) ( IQR) below the first quartile or more than (1.5) ( IQR) above the third quartile. Potential, outliers always need further investigation., 9 This, , content is available online at <http://cnx.org/content/m16314/1.18/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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72, , CHAPTER 2. DESCRIPTIVE STATISTICS, Example 2.10, For the following 13 real estate prices, calculate the IQR and determine if any prices are outliers., Prices are in dollars. (Source: San Jose Mercury News), 389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000;, 488,800; 1,095,000, Solution, Order the data from smallest to largest., 114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000;, 1,095,000; 5,500,000, M = 488, 800, Q1 =, , 230500+387000, 2, , = 308750, , Q3 =, , 639000+659000, 2, , = 649000, , IQR = 649000 − 308750 = 340250, , (1.5) ( IQR) = (1.5) (340250) = 510375, Q1 − (1.5) ( IQR) = 308750 − 510375 = −201625, Q3 + (1.5) ( IQR) = 649000 + 510375 = 1159375, No house price is less than -201625. However, 5,500,000 is more than 1,159,375. Therefore,, 5,500,000 is a potential outlier., , Example 2.11, For the two data sets in the test scores example (p. 70), find the following:, a. The interquartile range. Compare the two interquartile ranges., b. Any outliers in either set., c. The 30th percentile and the 80th percentile for each set. How much data falls below the 30th, percentile? Above the 80th percentile?, , Example 2.12: Finding Quartiles and Percentiles Using a Table, Fifty statistics students were asked how much sleep they get per school night (rounded to the, nearest hour). The results were (student data):, , AMOUNT OF SLEEP, PER SCHOOL NIGHT, (HOURS), , FREQUENCY, , RELATIVE, QUENCY, , FRE-, , CUMULATIVE RELATIVE FREQUENCY, , continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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73, 4, , 2, , 0.04, , 0.04, , 5, , 5, , 0.10, , 0.14, , 6, , 7, , 0.14, , 0.28, , 7, , 12, , 0.24, , 0.52, , 8, , 14, , 0.28, , 0.80, , 9, , 7, , 0.14, , 0.94, , 10, , 3, , 0.06, , 1.00, , Table 2.5, , Find the 28th percentile: Notice the 0.28 in the "cumulative relative frequency" column. 28% of 50, data values = 14. There are 14 values less than the 28th %ile. They include the two 4s, the five 5s,, and the seven 6s. The 28th %ile is between the last 6 and the first 7. The 28th %ile is 6.5., Find the median: Look again at the "cumulative relative frequency " column and find 0.52. The, median is the 50th %ile or the second quartile. 50% of 50 = 25. There are 25 values less than the, median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or, 50th %ile is between the 25th (7) and 26th (7) values. The median is 7., Find the third quartile: The third quartile is the same as the 75th percentile. You can "eyeball" this, answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When, you have all the 4s, 5s, 6s and 7s, you have 52% of the data. When you include all the 8s, you have, 80% of the data. The 75th %ile, then, must be an 8 . Another way to look at the problem is to find, 75% of 50 (= 37.5) and round up to 38. The third quartile, Q3 , is the 38th value which is an 8. You, can check this answer by counting the values. (There are 37 values below the third quartile and 12, values above.), Example 2.13, Using the table:, 1., 2., 3., 4., , Find the 80th percentile., Find the 90th percentile., Find the first quartile., What is another name for the first quartile?, , Collaborative Classroom Exercise: Your instructor or a member of the class will ask everyone in class how, many sweaters they own. Answer the following questions., 1., 2., 3., 4., 5., 6., 7., , How many students were surveyed?, What kind of sampling did you do?, Construct a table of the data., Construct 2 different histograms. For each, starting value = _____ ending value = ____., Use the table to find the median, first quartile, and third quartile., Construct a box plot., Use the table to find the following:, , • The 10th percentile, • The 70th percentile, • The percent of students who own less than 4 sweaters, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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74, , CHAPTER 2. DESCRIPTIVE STATISTICS, , Interpreting Percentiles, Quartiles, and Median, A percentile indicates the relative standing of a data value when data are sorted into numerical order, from, smallest to largest. p% of data values are less than or equal to the pth percentile. For example, 15% of data, values are less than or equal to the 15th percentile., , • Low percentiles always correspond to lower data values., • High percentiles always correspond to higher data values., A percentile may or may not correspond to a value judgment about whether it is "good" or "bad". The, interpretation of whether a certain percentile is good or bad depends on the context of the situation to, which the data applies. In some situations, a low percentile would be considered "good’; in other contexts, a high percentile might be considered "good". In many situations, there is no value judgment that applies., Understanding how to properly interpret percentiles is important not only when describing data,, but is also important in later chapters of this textbook when calculating probabilities., Guideline:, When writing the interpretation of a percentile in the context of the given data, the sentence should, contain the following information:, , •, •, •, •, , information about the context of the situation being considered,, the data value (value of the variable) that represents the percentile,, the percent of individuals or items with data values below the percentile., Additionally, you may also choose to state the percent of individuals or items with data values above, the percentile., Example 2.14, On a timed math test, the first quartile for times for finishing the exam was 35 minutes. Interpret, the first quartile in the context of this situation., , • 25% of students finished the exam in 35 minutes or less., • 75% of students finished the exam in 35 minutes or more., • A low percentile could be considered good, as finishing more quickly on a timed exam is, desirable. (If you take too long, you might not be able to finish.), Example 2.15, On a 20 question math test, the 70th percentile for number of correct answers was 16. Interpret, the 70th percentile in the context of this situation., , • 70% of students answered 16 or fewer questions correctly., • 30% of students answered 16 or more questions correctly., • Note: A high percentile could be considered good, as answering more questions correctly is, desirable., Example 2.16, At a certain community college, it was found that the 30th percentile of credit units that students, are enrolled for is 7 units. Interpret the 30th percentile in the context of this situation., , • 30% of students are enrolled in 7 or fewer credit units, • 70% of students are enrolled in 7 or more credit units, • In this example, there is no "good" or "bad" value judgment associated with a higher or, lower percentile. Students attend community college for varied reasons and needs, and their, course load varies according to their needs., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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75, , Do the following Practice Problems for Interpreting Percentiles, Exercise 2.6.1, , (Solution on p. 115.), , a. For runners in a race, a low time means a faster run. The winners in a race have the shortest, running times. Is it more desirable to have a finish time with a high or a low percentile when, running a race?, b. The 20th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting, the 20th percentile in the context of the situation., c. A bicyclist in the 90th percentile of a bicycle race between two towns completed the race in 1, hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence, interpreting the 90th percentile in the context of the situation., Exercise 2.6.2, , (Solution on p. 116.), , a. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed, with a high or a low percentile when running a race?, b. The 40th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40th percentile in the context of the situation., Exercise 2.6.3, (Solution on p. 116.), On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain., Exercise 2.6.4, (Solution on p. 116.), Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes, is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th, percentile in the context of this situation., Exercise 2.6.5, (Solution on p. 116.), In a survey collecting data about the salaries earned by recent college graduates, Li found that her, salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain., Exercise 2.6.6, (Solution on p. 116.), In a study collecting data about the repair costs of damage to automobiles in a certain type of, crash tests, a certain model of car had $1700 in damage and was in the 90th percentile. Should the, manufacturer and/or a consumer be pleased or upset by this result? Explain. Write a sentence, that interprets the 90th percentile in the context of this problem., Exercise 2.6.7, , (Solution on p. 116.), , The University of California has two criteria used to set admission standards for freshman to be, admitted to a college in the UC system:, a. Students’ GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula, that calculates an "admissions index" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students, in the state. In this context, what percentile does the top 12% represent?, b. Students whose GPAs are at or above the 96th percentile of all students at their high school, are eligible (called eligible in the local context), even if they are not in the top 12% of all, students in the state. What percent of students from each high school are "eligible in the, local context"?, Exercise 2.6.8, (Solution on p. 116.), Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34th percentile. The 34th percentile of housing prices is $240,000, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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76, , CHAPTER 2. DESCRIPTIVE STATISTICS, in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the, houses?, , **With contributions from Roberta Bloom, , 2.7 Measures of the Center of the Data10, The "center" of a data set is also a way of describing location. The two most widely used measures of the, "center" of the data are the mean (average) and the median. To calculate the mean weight of 50 people,, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data, and find the number that splits the data into two equal parts (previously discussed under box plots in this, chapter). The median is generally a better measure of the center when there are extreme values or outliers, because it is not affected by the precise numerical values of the outliers. The mean is the most common, measure of the center., NOTE : The words "mean" and "average" are often used interchangeably. The substitution of one, word for the other is common practice. The technical term is "arithmetic mean" and "average" is, technically a center location. However, in practice among non-statisticians, "average" is commonly, accepted for "arithmetic mean.", , The mean can also be calculated by multiplying each distinct value by its frequency and then dividing the, sum by the total number of data values. The letter used to represent the sample mean is an x with a bar, over it (pronounced "x bar"): x., The Greek letter µ (pronounced "mew") represents the population mean. One of the requirements for the, sample mean to be a good estimate of the population mean is for the sample taken to be truly random., To see that both ways of calculating the mean are the same, consider the sample:, 1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4, x=, , 1+1+1+2+2+3+4+4+4+4+4, = 2.7, 11, , 3×1+2×2+1×3+5×4, = 2.7, 11, In the second calculation for the sample mean, the frequencies are 3, 2, 1, and 5., x=, , You can quickly find the location of the median by using the expression, , (2.0), (2.0), , n +1, 2 ., , The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle, value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the, two middle values added together and divided by 2 after the data has been ordered. For example, if the, 1, 97+1, total number of data values is 97, then n+, 2 = 2 = 49. The median is the 49th value in the ordered data., 1 100+1, If the total number of data values is 100, then n+, 2 =, 2 = 50.5. The median occurs midway between the, 50th and 51st values. The location of the median and the value of the median are not the same. The upper, case letter M is often used to represent the median. The next example illustrates the location of the median, and the value of the median., Example 2.17, AIDS data indicating the number of months an AIDS patient lives after taking a new antibody, drug are as follows (smallest to largest):, 10 This, , content is available online at <http://cnx.org/content/m17102/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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77, 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32;, 33; 33; 34; 34; 35; 37; 40; 44; 44; 47, Calculate the mean and the median., Solution, The calculation for the mean is:, x=, , [3+4+(8)(2)+10+11+12+13+14+(15)(2)+(16)(2)+...+35+37+40+(44)(2)+47], 40, , = 23.6, , To find the median, M, first use the formula for the location. The location is:, n +1, 2, , =, , 40+1, 2, , = 20.5, , Starting at the smallest value, the median is located between the 20th and 21st values (the two, 24s):, 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32;, 33; 33; 34; 34; 35; 37; 40; 44; 44; 47, M=, , 24+24, 2, , = 24, , The median is 24., Using the TI-83,83+,84, 84+ Calculators, Calculator Instructions are located in the menu item 14:Appendix (Notes for the TI-83, 83+, 84,, 84+ Calculators)., , •, •, •, •, , Enter data into the list editor. Press STAT 1:EDIT, Put the data values in list L1., Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and ENTER., Press the down and up arrow keys to scroll., , x = 23.6, M = 24, Example 2.18, Suppose that, in a small town of 50 people, one person earns $5,000,000 per year and the other 49, each earn $30,000. Which is the better measure of the "center," the mean or the median?, Solution, x = 5000000+5049×30000 = 129400, M = 30000, (There are 49 people who earn $30,000 and one person who earns $5,000,000.), The median is a better measure of the "center" than the mean because 49 of the values are 30,000, and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of, the data., , Another measure of the center is the mode. The mode is the most frequent value. If a data set has two, values that occur the same number of times, then the set is bimodal., Example 2.19: Statistics exam scores for 20 students are as follows, Statistics exam scores for 20 students are as follows:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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78, , CHAPTER 2. DESCRIPTIVE STATISTICS, 50 ; 53 ; 59 ; 59 ; 63 ; 63 ; 72 ; 72 ; 72 ; 72 ; 72 ; 76 ; 78 ; 81 ; 83 ; 84 ; 84 ; 84 ; 90 ; 93, Problem, Find the mode., Solution, The most frequent score is 72, which occurs five times. Mode = 72., , Example 2.20, Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores, 430 and 480 each occur twice., When is the mode the best measure of the "center"? Consider a weight loss program that advertises, a mean weight loss of six pounds the first week of the program. The mode might indicate that most, people lose two pounds the first week, making the program less appealing., NOTE :, , The mode can be calculated for qualitative data as well as for quantitative data., , Statistical software will easily calculate the mean, the median, and the mode. Some graphing, calculators can also make these calculations. In the real world, people make these calculations, using software., , 2.7.1 The Law of Large Numbers and the Mean, The Law of Large Numbers says that if you take samples of larger and larger size from any population,, then the mean x of the sample is very likely to get closer and closer to µ. This is discussed in more detail in, The Central Limit Theorem., NOTE : The formula for the mean is located in the Summary of Formulas (Section 2.10) section, course., , 2.7.2 Sampling Distributions and Statistic of a Sampling Distribution, You can think of a sampling distribution as a relative frequency distribution with a great many samples., (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students, were asked the number of movies they watched the previous week. The results are in the relative frequency, table shown below., # of movies, , Relative Frequency, , 0, , 5/30, , 1, , 15/30, , 2, , 6/30, , 3, , 4/30, , 4, , 1/30, Table 2.6, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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79, If you let the number of samples get very large (say, 300 million or more), the relative frequency table, becomes a relative frequency distribution., A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the, mode as well as others. The sample mean x is an example of a statistic which estimates the population, mean µ., , 2.8 Skewness and the Mean, Median, and Mode11, Consider the following data set:, 4 ; 5 ; 6 ; 6 ; 6 ; 7 ; 7 ; 7 ; 7 ; 7 ; 7 ; 8 ; 8 ; 8 ; 9 ; 10, This data set produces the histogram shown below. Each interval has width one and each value is located, in the middle of an interval., , The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line, can be drawn at some point in the histogram such that the shape to the left and the right of the vertical, line are mirror images of each other. The mean, the median, and the mode are each 7 for these data. In a, perfectly symmetrical distribution, the mean and the median are the same. This example has one mode, (unimodal) and the mode is the same as the mean and median. In a symmetrical distribution that has two, modes (bimodal), the two modes would be different from the mean and median., The histogram for the data:, 4;5;6;6;6;7;7;7;7;8, is not symmetrical. The right-hand side seems "chopped off" compared to the left side. The shape distribution is called skewed to the left because it is pulled out to the left., 11 This, , content is available online at <http://cnx.org/content/m17104/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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80, , CHAPTER 2. DESCRIPTIVE STATISTICS, , The mean is 6.3, the median is 6.5, and the mode is 7. Notice that the mean is less than the median and, they are both less than the mode. The mean and the median both reflect the skewing but the mean more, so., The histogram for the data:, 6 ; 7 ; 7 ; 7 ; 7 ; 8 ; 8 ; 8 ; 9 ; 10, is also not symmetrical. It is skewed to the right., , The mean is 7.7, the median is 7.5, and the mode is 7. Of the three statistics, the mean is the largest, while, the mode is the smallest. Again, the mean reflects the skewing the most., To summarize, generally if the distribution of data is skewed to the left, the mean is less than the median,, which is often less than the mode. If the distribution of data is skewed to the right, the mode is often less, than the median, which is less than the mean., Skewness and symmetry become important when we discuss probability distributions in later chapters., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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81, , 2.9 Measures of the Spread of the Data12, An important characteristic of any set of data is the variation in the data. In some data sets, the data values, are concentrated closely near the mean; in other data sets, the data values are more widely spread out from, the mean. The most common measure of variation, or spread, is the standard deviation., The standard deviation is a number that measures how far data values are from their mean., The standard deviation, , • provides a numerical measure of the overall amount of variation in a data set, • can be used to determine whether a particular data value is close to or far from the mean, The standard deviation provides a measure of the overall variation in a data set, The standard deviation is always positive or 0. The standard deviation is small when the data are all, concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when, the data values are more spread out from the mean, exhibiting more variation., Suppose that we are studying waiting times at the checkout line for customers at supermarket A and, supermarket B; the average wait time at both markets is 5 minutes. At market A, the standard deviation, for the waiting time is 2 minutes; at market B the standard deviation for the waiting time is 4 minutes., Because market B has a higher standard deviation, we know that there is more variation in the waiting times at market B. Overall, wait times at market B are more spread out from the average; wait times at, market A are more concentrated near the average., The standard deviation can be used to determine whether a data value is close to or far from the mean., Suppose that Rosa and Binh both shop at Market A. Rosa waits for 7 minutes and Binh waits for 1 minute, at the checkout counter. At market A, the mean wait time is 5 minutes and the standard deviation is 2, minutes. The standard deviation can be used to determine whether a data value is close to or far from the, mean., Rosa waits for 7 minutes:, , • 7 is 2 minutes longer than the average of 5; 2 minutes is equal to one standard deviation., • Rosa’s wait time of 7 minutes is 2 minutes longer than the average of 5 minutes., • Rosa’s wait time of 7 minutes is one standard deviation above the average of 5 minutes., Binh waits for 1 minute., , •, •, •, •, , 1 is 4 minutes less than the average of 5; 4 minutes is equal to two standard deviations., Binh’s wait time of 1 minute is 4 minutes less than the average of 5 minutes., Binh’s wait time of 1 minute is two standard deviations below the average of 5 minutes., A data value that is two standard deviations from the average is just on the borderline for what many, statisticians would consider to be far from the average. Considering data to be far from the mean if it, is more than 2 standard deviations away is more of an approximate "rule of thumb" than a rigid rule., In general, the shape of the distribution of the data affects how much of the data is further away than, 2 standard deviations. (We will learn more about this in later chapters.), , The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line,, 7 is to the right of 5. We say, then, that 7 is one standard deviation to the right of 5 because, 5 + (1) (2) = 7., 12 This, , content is available online at <http://cnx.org/content/m17103/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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82, , CHAPTER 2. DESCRIPTIVE STATISTICS, , If 1 were also part of the data set, then 1 is two standard deviations to the left of 5 because, 5 + (−2) (2) = 1., , •, •, •, •, , In general, a value = mean + (#ofSTDEV)(standard deviation), where #ofSTDEVs = the number of standard deviations, 7 is one standard deviation more than the mean of 5 because: 7=5+(1)(2), 1 is two standard deviations less than the mean of 5 because: 1=5+(−2)(2), , The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a, population:, , • sample: x = x + (#o f STDEV ) (s), • Population: x = µ + (#o f STDEV ) (σ), The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case), represents the population standard deviation., The symbol x is the sample mean and the Greek symbol µ is the population mean., Calculating the Standard Deviation, If x is a number, then the difference "x - mean" is called its deviation. In a data set, there are as many, deviations as there are items in the data set. The deviations are used to calculate the standard deviation., If the numbers belong to a population, in symbols a deviation is x − µ . For sample data, in symbols a, deviation is x − x ., The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used, to represent the standard deviation depends on whether it is calculated from a population or a sample., The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case), represents the population standard deviation. If the sample has the same characteristics as the population,, then s should be a good estimate of σ., To calculate the standard deviation, we need to calculate the variance first. The variance is an average of, the squares of the deviations (the x − x values for a sample, or the x − µ values for a population). The, symbol σ2 represents the population variance; the population standard deviation σ is the square root of, the population variance. The symbol s2 represents the sample variance; the sample standard deviation s is, the square root of the sample variance. You can think of the standard deviation as a special average of the, deviations., If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population., If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n-1, one less than the number of items in the sample. You can see that in the formulas, below., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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83, Formulas for the Sample Standard Deviation, q, q, Σ f ·( x − x )2, − x )2, • s = Σ(nx−, or, s, =, 1, n −1, • For the sample standard deviation, the denominator is n-1, that is the sample size MINUS 1., Formulas for the Population Standard Deviation, q, q, − µ )2, Σ f ·( x −µ)2, or σ =, • σ = Σ(xN, N, • For the population standard deviation, the denominator is N, the number of items in the population., In these formulas, f represents the frequency with which a value appears. For example, if a value appears, once, f is 1. If a value appears three times in the data set or population, f is 3., Sampling Variability of a Statistic, The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of, the Data. How much the statistic varies from one sample to another is known as the sampling variability of, a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard, error of the mean is an example of a standard error. It is a special standard deviation and is known as the, standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean, in The Central Limit Theorem (not now). The notation for the standard error of the mean is √σn where σ is, the standard deviation of the population and n is the size of the sample., NOTE : In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE, THE STANDARD DEVIATION. If you are using a TI-83,83+,84+ calculator, you need to select, the appropriate standard deviation σx or s x from the summary statistics. We will concentrate on, using and interpreting the information that the standard deviation gives us. However you should, study the following step-by-step example to help you understand how the standard deviation, measures variation from the mean., , Example 2.21, In a fifth grade class, the teacher was interested in the average age and the sample standard, deviation of the ages of her students. The following data are the ages for a SAMPLE of n = 20 fifth, grade students. The ages are rounded to the nearest half year:, 9 ; 9.5 ; 9.5 ; 10 ; 10 ; 10 ; 10 ; 10.5 ; 10.5 ; 10.5 ; 10.5 ; 11 ; 11 ; 11 ; 11 ; 11 ; 11 ; 11.5 ; 11.5 ; 11.5, 9 + 9.5 × 2 + 10 × 4 + 10.5 × 4 + 11 × 6 + 11.5 × 3, = 10.525, 20, The average age is 10.53 years, rounded to 2 places., x=, , The variance may be calculated by using a table. Then the standard deviation is calculated by, taking the square root of the variance. We will explain the parts of the table after calculating s., Data, , Freq., , Deviations, , Deviations2, , (Freq.)(Deviations2 ), , x, , f, , (x − x), , ( x − x )2, , ( f ) ( x − x )2, , 9, , 1, , 9 − 10.525 = −1.525, , (−1.525)2 = 2.325625, , 9.5, 10, 10.5, 11, 11.5, , 2, 4, 4, 6, 3, , 9.5 − 10.525 = −1.025, 10 − 10.525 = −0.525, 10.5 − 10.525 = −0.025, 11 − 10.525 = 0.475, 11.5 − 10.525 = 0.975, , 1 × 2.325625 = 2.325625, , 2, , 2 × 1.050625 = 2.101250, , 2, , 4 × .275625 = 1.1025, , 2, , 4 × .000625 = .0025, , (−1.025) = 1.050625, (−0.525) = 0.275625, (−0.025) = 0.000625, 2, , 6 × .225625 = 1.35375, , 2, , 3 × .950625 = 2.851875, , (0.475) = 0.225625, (0.975) = 0.950625, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (2.0)
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84, , CHAPTER 2. DESCRIPTIVE STATISTICS, Table 2.7, , The sample variance, s2 , is equal to the sum of the last column (9.7375) divided by the total number, of data values minus one (20 - 1):, s2 =, , 9.7375, 20−1, , = 0.5125, , The sample standard deviation s is equal to the square root of the sample variance:, √, s = 0.5125 = .0715891 Rounded to two decimal places, s = 0.72, Typically, you do the calculation for the standard deviation on your calculator or computer. The, intermediate results are not rounded. This is done for accuracy., Problem 1, Verify the mean and standard deviation calculated above on your calculator or computer., Solution, Using the TI-83,83+,84+ Calculators, • Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up, into the name. Press CLEAR and arrow down., • Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3), into list L2. Use the arrow keys to move around., • Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not, forget the comma. Press ENTER., • x=10.525, • Use Sx because this is sample data (not a population): Sx=0.715891, , •, •, •, •, , For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation), For a sample: x = x + (#ofSTDEVs)(s), For a population: x = µ + (#ofSTDEVs)( σ), For this example, use x = x + (#ofSTDEVs)(s) because the data is from a sample, , Problem 2, Find the value that is 1 standard deviation above the mean. Find ( x + 1s)., Solution, ( x + 1s) = 10.53 + (1) (0.72) = 11.25, , Problem 3, Find the value that is two standard deviations below the mean. Find ( x − 2s)., Solution, ( x − 2s) = 10.53 − (2) (0.72) = 9.09, , Problem 4, Find the values that are 1.5 standard deviations from (below and above) the mean., Solution, , • ( x − 1.5s) = 10.53 − (1.5) (0.72) = 9.45, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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85, , • ( x + 1.5s) = 10.53 + (1.5) (0.72) = 11.61, , Explanation of the standard deviation calculation shown in the table, The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the, mean than is the data value 11. The deviations 0.97 and 0.47 indicate that. A positive deviation occurs when, the data value is greater than the mean. A negative deviation occurs when the data value is less than the, mean; the deviation is -1.525 for the data value 9. If you add the deviations, the sum is always zero. (For, this example, there are n=20 deviations.) So you cannot simply add the deviations to get the spread of the, data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The, variance, then, is the average squared deviation., The variance is a squared measure and does not have the same units as the data. Taking the square root, solves the problem. The standard deviation measures the spread in the same units as the data., Notice that instead of dividing by n=20, the calculation divided by n-1=20-1=19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n − 1). Why not divide by n? The, answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n − 1) gives a, better estimate of the population variance., NOTE : Your concentration should be on what the standard deviation tells us about the data. The, standard deviation is a number which measures how far the data are spread from the mean. Let a, calculator or computer do the arithmetic., , The standard deviation, s or σ, is either zero or larger than zero. When the standard deviation is 0, there is, no spread; that is, the all the data values are equal to each other. The standard deviation is small when the, data are all concentrated close to the mean, and is larger when the data values show more variation from, the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about, the mean; outliers can make s or σ very large., The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a, better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions,, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be, much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed, distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and, the largest value. Because numbers can be confusing, always graph your data., NOTE :, , The formula for the standard deviation is at the end of the chapter., , Example 2.22, Use the following data (first exam scores) from Susan Dean’s spring pre-calculus class:, 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94;, 96; 100, a. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative, frequencies to three decimal places., b. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:, i. The sample mean, ii. The sample standard deviation, iii. The median, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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86, , CHAPTER 2. DESCRIPTIVE STATISTICS, iv. The first quartile, v. The third quartile, vi. IQR, c. Construct a box plot and a histogram on the same set of axes. Make comments about the box, plot, the histogram, and the chart., Solution, a., Data, , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , 33, , 1, , 0.032, , 0.032, , 42, , 1, , 0.032, , 0.064, , 49, , 2, , 0.065, , 0.129, , 53, , 1, , 0.032, , 0.161, , 55, , 2, , 0.065, , 0.226, , 61, , 1, , 0.032, , 0.258, , 63, , 1, , 0.032, , 0.29, , 67, , 1, , 0.032, , 0.322, , 68, , 2, , 0.065, , 0.387, , 69, , 2, , 0.065, , 0.452, , 72, , 1, , 0.032, , 0.484, , 73, , 1, , 0.032, , 0.516, , 74, , 1, , 0.032, , 0.548, , 78, , 1, , 0.032, , 0.580, , 80, , 1, , 0.032, , 0.612, , 83, , 1, , 0.032, , 0.644, , 88, , 3, , 0.097, , 0.741, , 90, , 1, , 0.032, , 0.773, , 92, , 1, , 0.032, , 0.805, , 94, , 4, , 0.129, , 0.934, , 96, , 1, , 0.032, , 0.966, , 100, , 1, , 0.032, , 0.998 (Why isn’t this value 1?), Table 2.8, , b. i. The sample mean = 73.5, ii. The sample standard deviation = 17.9, iii. The median = 73, iv. The first quartile = 61, v. The third quartile = 90, vi. IQR = 90 - 61 = 29, c. The x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram; number of, intervals is 5 for the histogram so the width of an interval is (100.5 - 32.5) divided by 5 which, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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87, is equal to 13.6. Endpoints of the intervals: starting point is 32.5, 32.5+13.6 = 46.1, 46.1+13.6 =, 59.7, 59.7+13.6 = 73.3, 73.3+13.6 = 86.9, 86.9+13.6 = 100.5 = the ending value; No data values, fall on an interval boundary., , Figure 2.1, , The long left whisker in the box plot is reflected in the left side of the histogram. The spread of, the exam scores in the lower 50% is greater (73 - 33 = 40) than the spread in the upper 50% (100 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A, and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the, middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the, lower 25% of the exam scores are Ds and Fs., Comparing Values from Different Data Sets, The standard deviation is useful when comparing data values that come from different data sets. If the data, sets have different means and standard deviations, it can be misleading to compare the data values directly., , •, •, •, •, , For each data value, calculate how many standard deviations the value is away from its mean., Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs., value−mean, #o f STDEVs = standard, deviation, Compare the results of this calculation., , #ofSTDEVs is often called a "z-score"; we can use the symbol z. In symbols, the formulas become:, Sample, , x=x+zs, , z=, , Population, , x=µ+zσ, , z=, , x−x, s, x −µ, σ, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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88, , CHAPTER 2. DESCRIPTIVE STATISTICS, Table 2.9, , Example 2.23, Two students, John and Ali, from different high schools, wanted to find out who had the highest, G.P.A. when compared to his school. Which student had the highest G.P.A. when compared to his, school?, Student, , GPA, , School Mean GPA, , School Standard Deviation, , John, , 2.85, , 3.0, , 0.7, , Ali, , 77, , 80, , 10, Table 2.10, , Solution, For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from, the average, for his school. Pay careful attention to signs when comparing and interpreting the, answer., #o f STDEVs =, , value−mean, standard deviation, , For John, z = #o f STDEVs =, For Ali, z = #ofSTDEVs =, , ;z=, , x −µ, σ, , 2.85−3.0, 0.7, , 77−80, 10, , = −0.21, , = −0.3, , John has the better G.P.A. when compared to his school because his G.P.A. is 0.21 standard, deviations below his school’s mean while Ali’s G.P.A. is 0.3 standard deviations below his, school’s mean., John’s z-score of −0.21 is higher than Ali’s z-score of −0.3 . For GPA, higher values are, better, so we conclude that John has the better GPA when compared to his school., , The following lists give a few facts that provide a little more insight into what the standard deviation tells, us about the distribution of the data., For ANY data set, no matter what the distribution of the data is:, , •, •, •, •, , At least 75% of the data is within 2 standard deviations of the mean., At least 89% of the data is within 3 standard deviations of the mean., At least 95% of the data is within 4 1/2 standard deviations of the mean., This is known as Chebyshev’s Rule., , For data having a distribution that is MOUND-SHAPED and SYMMETRIC:, , •, •, •, •, •, , Approximately 68% of the data is within 1 standard deviation of the mean., Approximately 95% of the data is within 2 standard deviations of the mean., More than 99% of the data is within 3 standard deviations of the mean., This is known as the Empirical Rule., It is important to note that this rule only applies when the shape of the distribution of the data is, mound-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters., , **With contributions from Roberta Bloom, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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89, , 2.10 Summary of Formulas13, Commonly Used Symbols, , •, •, •, •, •, •, •, •, •, , The symbol Σ means to add or to find the sum., n = the number of data values in a sample, N = the number of people, things, etc. in the population, x = the sample mean, s = the sample standard deviation, µ = the population mean, σ = the population standard deviation, f = frequency, x = numerical value, , Commonly Used Expressions, , •, •, •, •, •, •, , x ∗ f = A value multiplied by its respective frequency, ∑ x = The sum of the values, ∑ x ∗ f = The sum of values multiplied by their respective frequencies, ( x − x ) or ( x − µ) = Deviations from the mean (how far a value is from the mean), ( x − x )2 or ( x − µ)2 = Deviations squared, f ( x − x )2 or f ( x − µ)2 = The deviations squared and multiplied by their frequencies, , Mean Formulas:, , • x = ∑nx or x = ∑ nf · x, • µ = ∑Nx or µ= ∑Nf · x, Standard Deviation Formulas:, q, q, Σ f ·( x − x )2, − x )2, • s = Σ(nx−, or, s, =, 1, q n −1 2, q, Σ ( x − µ )2, Σ f ·( x −µ), or σ =, • σ=, N, N, Formulas Relating a Value, the Mean, and the Standard Deviation:, , • value = mean + (#ofSTDEVs)(standard deviation), where #ofSTDEVs = the number of standard deviations, • x = x+ (#ofSTDEVs)(s), • x = µ + (#ofSTDEVs)(σ), , 13 This, , content is available online at <http://cnx.org/content/m16310/1.9/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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90, , CHAPTER 2. DESCRIPTIVE STATISTICS, , 2.11 Practice 1: Center of the Data14, 2.11.1 Student Learning Outcomes, • The student will calculate and interpret the center, spread, and location of the data., • The student will construct and interpret histograms an box plots., , 2.11.2 Given, Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one, week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve, generally sell five cars; nine generally sell six cars; eleven generally sell seven cars., , 2.11.3 Complete the Table, Data Value (# cars), , Frequency, , Relative Frequency, , Cumulative Relative Frequency, , Table 2.11, , 2.11.4 Discussion Questions, Exercise 2.11.1, What does the frequency column sum to? Why?, , (Solution on p. 116.), , Exercise 2.11.2, (Solution on p. 116.), What does the relative frequency column sum to? Why?, Exercise 2.11.3, What is the difference between relative frequency and frequency for each data value?, Exercise 2.11.4, What is the difference between cumulative relative frequency and relative frequency for each data, value?, , 2.11.5 Enter the Data, Enter your data into your calculator or computer., 14 This, , content is available online at <http://cnx.org/content/m16312/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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91, , 2.11.6 Construct a Histogram, Determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram, below. Label the horizontal and vertical axes with words. Include numerical scaling., , 2.11.7 Data Statistics, Calculate the following values:, Exercise 2.11.5, Sample mean = x =, Exercise 2.11.6, Sample standard deviation = s x =, , (Solution on p. 116.), , Exercise 2.11.7, Sample size = n =, , (Solution on p. 116.), , (Solution on p. 116.), , 2.11.8 Calculations, Use the table in section 2.11.3 to calculate the following values:, Exercise 2.11.8, Median =, , (Solution on p. 116.), , Exercise 2.11.9, Mode =, , (Solution on p. 116.), , Exercise 2.11.10, First quartile =, , (Solution on p. 116.), , Exercise 2.11.11, Second quartile = median = 50th percentile =, , (Solution on p. 117.), , Exercise 2.11.12, Third quartile =, , (Solution on p. 117.), , Exercise 2.11.13, Interquartile range (IQR) = _____ - _____ = _____, , (Solution on p. 117.), , Exercise 2.11.14, 10th percentile =, , (Solution on p. 117.), , Exercise 2.11.15, 70th percentile =, , (Solution on p. 117.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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92, , CHAPTER 2. DESCRIPTIVE STATISTICS, Exercise 2.11.16, Find the value that is 3 standard deviations:, , (Solution on p. 117.), , a. Above the mean, b. Below the mean, , 2.11.9 Box Plot, Construct a box plot below. Use a ruler to measure and scale accurately., , 2.11.10 Interpretation, Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or, concentrated in some areas, but not in others? How can you tell?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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93, , 2.12 Practice 2: Spread of the Data15, 2.12.1 Student Learning Outcomes, • The student will calculate measures of the center of the data., • The student will calculate the spread of the data., , 2.12.2 Given, The population parameters below describe the full-time equivalent number of students (FTES) each year, at Lake Tahoe Community College from 1976-77 through 2004-2005. (Source: Graphically Speaking by Bill, King, LTCC Institutional Research, December 2005 )., Use these values to answer the following questions:, , •, •, •, •, •, •, , µ = 1000 FTES, Median = 1014 FTES, σ = 474 FTES, First quartile = 528.5 FTES, Third quartile = 1447.5 FTES, n = 29 years, , 2.12.3 Calculate the Values, Exercise 2.12.1, (Solution on p. 117.), A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above?, Explain how you determined your answer., Exercise 2.12.2, 75% of all years have a FTES:, , (Solution on p. 117.), , a. At or below:, b. At or above:, Exercise 2.12.3, The population standard deviation =, , (Solution on p. 117.), , Exercise 2.12.4, What percent of the FTES were from 528.5 to 1447.5? How do you know?, , (Solution on p. 117.), , Exercise 2.12.5, What is the IQR? What does the IQR represent?, , (Solution on p. 117.), , Exercise 2.12.6, How many standard deviations away from the mean is the median?, , (Solution on p. 117.), , Additional Information: The population FTES for 2005-2006 through 2010-2011 was given in an updated, report. (Source: http://www.ltcc.edu/data/ResourcePDF/LTCC_FactBook_2010-11.pdf). The data are reported here., , 15 This, , Year, , 2005-06, , 2006-07, , 2007-08, , 2008-09, , 2009-10, , 2010-11, , Total FTES, , 1585, , 1690, , 1735, , 1935, , 2021, , 1890, , content is available online at <http://cnx.org/content/m17105/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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94, , CHAPTER 2. DESCRIPTIVE STATISTICS, Table 2.12, , Exercise 2.12.7, (Solution on p. 117.), Calculate the mean, median, standard deviation, first quartile, the third quartile and the IQR., Round to one decimal place., Exercise 2.12.8, Construct a boxplot for the FTES for 2005-2006 through 2010-2011 and a boxplot for the FTES for, 1976-1977 through 2004-2005., Exercise 2.12.9, (Solution on p. 117.), Compare the IQR for the FTES for 1976-77 through 2004-2005 with the IQR for the FTES for 20052006 through 2010-2011. Why do you suppose the IQRs are so different?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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95, , 2.13 Homework16, Exercise 2.13.1, (Solution on p. 117.), Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:, # of movies, , Frequency, , 0, , 5, , 1, , 9, , 2, , 6, , 3, , 4, , 4, , 1, , Relative Frequency, , Cumulative Relative Frequency, , Table 2.13, , a. Find the sample mean x, b. Find the sample standard deviation, s, c. Construct a histogram of the data., d. Complete the columns of the chart., e. Find the first quartile., f. Find the median., g. Find the third quartile., h. Construct a box plot of the data., i. What percent of the students saw fewer than three movies?, j. Find the 40th percentile., k. Find the 90th percentile., l. Construct a line graph of the data., m. Construct a stem plot of the data., Exercise 2.13.2, The median age for U.S. blacks currently is 30.9 years; for U.S. whites it is 42.3, years. ((Source: http://www.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1) ), a. Based upon this information, give two reasons why the black median age could be lower than, the white median age., b. Does the lower median age for blacks necessarily mean that blacks die younger than whites?, Why or why not?, c. How might it be possible for blacks and whites to die at approximately the same age, but for, the median age for whites to be higher?, Exercise 2.13.3, (Solution on p. 118.), Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X, = the number of pairs of sneakers owned. The results are as follows:, 16 This, , content is available online at <http://cnx.org/content/m16801/1.25/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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96, , CHAPTER 2. DESCRIPTIVE STATISTICS, , X, , Frequency, , 1, , 2, , 2, , 5, , 3, , 8, , 4, , 12, , 5, , 12, , 7, , 1, , Relative Frequency, , Cumulative Relative Frequency, , Table 2.14, , a. Find the sample mean x, b. Find the sample standard deviation, s, c. Construct a histogram of the data., d. Complete the columns of the chart., e. Find the first quartile., f. Find the median., g. Find the third quartile., h. Construct a box plot of the data., i. What percent of the students owned at least five pairs?, j. Find the 40th percentile., k. Find the 90th percentile., l. Construct a line graph of the data, m. Construct a stem plot of the data, Exercise 2.13.4, 600 adult Americans were asked by telephone poll, What do you think constitutes a middle-class, income? The results are below. Also, include left endpoint, but not the right endpoint. (Source:, Time magazine; survey by Yankelovich Partners, Inc.), NOTE :, , "Not sure" answers were omitted from the results., Salary ($), , Relative Frequency, , < 20,000, , 0.02, , 20,000 - 25,000, , 0.09, , 25,000 - 30,000, , 0.19, , 30,000 - 40,000, , 0.26, , 40,000 - 50,000, , 0.18, , 50,000 - 75,000, , 0.17, , 75,000 - 99,999, , 0.02, , 100,000+, , 0.01, Table 2.15, , a. What percent of the survey answered "not sure" ?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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97, b. What percent think that middle-class is from $25,000 - $50,000 ?, c. Construct a histogram of the data, a. Should all bars have the same width, based on the data? Why or why not?, b. How should the <20,000 and the 100,000+ intervals be handled? Why?, d. Find the 40th and 80th percentiles, e. Construct a bar graph of the data, Exercise 2.13.5, (Solution on p. 118.), Following are the published weights (in pounds) of all of the team members of the San Francisco, 49ers from a previous year (Source: San Jose Mercury News), 177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241;, 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230;, 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265, a. Organize the data from smallest to largest value., b. Find the median., c. Find the first quartile., d. Find the third quartile., e. Construct a box plot of the data., f. The middle 50% of the weights are from _______ to _______., g. If our population were all professional football players, would the above data be a sample of, weights or the population of weights? Why?, h. If our population were the San Francisco 49ers, would the above data be a sample of weights, or the population of weights? Why?, i. Assume the population was the San Francisco 49ers. Find:, i. the population mean, µ., ii. the population standard deviation, σ., iii. the weight that is 2 standard deviations below the mean., iv. When Steve Young, quarterback, played football, he weighed 205 pounds. How many, standard deviations above or below the mean was he?, j. That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard, deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team,, who was lighter, Smith or Young? How did you determine your answer?, Exercise 2.13.6, An elementary school class ran 1 mile with a mean of 11 minutes and a standard deviation of 3, minutes. Rachel, a student in the class, ran 1 mile in 8 minutes. A junior high school class ran 1, mile with a mean of 9 minutes and a standard deviation of 2 minutes. Kenji, a student in the class,, ran 1 mile in 8.5 minutes. A high school class ran 1 mile with a mean of 7 minutes and a standard, deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes., a. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?, b. Who is the fastest runner with respect to his or her class? Explain why., Exercise 2.13.7, In a survey of 20 year olds in China, Germany and America, people were asked the number of, foreign countries they had visited in their lifetime. The following box plots display the results., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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98, , CHAPTER 2. DESCRIPTIVE STATISTICS, , a. In complete sentences, describe what the shape of each box plot implies about the distribution, of the data collected., b. Explain how it is possible that more Americans than Germans surveyed have been to over eight, foreign countries., c. Compare the three box plots. What do they imply about the foreign travel of twenty year old, residents of the three countries when compared to each other?, Exercise 2.13.8, One hundred teachers attended a seminar on mathematical problem solving. The attitudes of, a representative sample of 12 of the teachers were measured before and after the seminar. A, positive number for change in attitude indicates that a teacher’s attitude toward math became, more positive. The twelve change scores are as follows:, 3; 8; -1; 2; 0; 5; -3; 1; -1; 6; 5; -2, a., b., c., d., , What is the mean change score?, What is the standard deviation for this population?, What is the median change score?, Find the change score that is 2.2 standard deviations below the mean., , Exercise 2.13.9, (Solution on p. 118.), Three students were applying to the same graduate school. They came from schools with different, grading systems. Which student had the best G.P.A. when compared to his school? Explain how, you determined your answer., Student, , G.P.A., , School Ave. G.P.A., , School Standard Deviation, , Thuy, , 2.7, , 3.2, , 0.8, , Vichet, , 87, , 75, , 20, , Kamala, , 8.6, , 8, , 0.4, Table 2.16, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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99, Exercise 2.13.10, Given the following box plot:, , a., b., c., d., e., , Which quarter has the smallest spread of data? What is that spread?, Which quarter has the largest spread of data? What is that spread?, Find the Inter Quartile Range (IQR)., Are there more data in the interval 5 - 10 or in the interval 10 - 13? How do you know this?, Which interval has the fewest data in it? How do you know this?, I. 0-2, II. 2-4, III. 10-12, IV. 12-13, , Exercise 2.13.11, Given the following box plot:, , a. Think of an example (in words) where the data might fit into the above box plot. In 2-5 sentences, write down the example., b. What does it mean to have the first and second quartiles so close together, while the second to, fourth quartiles are far apart?, Exercise 2.13.12, Santa Clara County, CA, has approximately 27,873 Japanese-Americans. Their ages are as follows., (Source: West magazine), Age Group, , Percent of Community, , 0-17, , 18.9, , 18-24, , 8.0, , 25-34, , 22.8, , 35-44, , 15.0, , 45-54, , 13.1, , 55-64, , 11.9, , 65+, , 10.3, Table 2.17, , a. Construct a histogram of the Japanese-American community in Santa Clara County, CA. The, bars will not be the same width for this example. Why not?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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100, , CHAPTER 2. DESCRIPTIVE STATISTICS, b. What percent of the community is under age 35?, c. Which box plot most resembles the information above?, , Exercise 2.13.13, Suppose that three book publishers were interested in the number of fiction paperbacks adult, consumers purchase per month. Each publisher conducted a survey. In the survey, each asked, adult consumers the number of fiction paperbacks they had purchased the previous month. The, results are below., Publisher A, # of books, , Freq., , 0, , 10, , 1, , 12, , 2, , 16, , 3, , 12, , 4, , 8, , 5, , 6, , 6, , 2, , 8, , 2, , Rel. Freq., , Table 2.18, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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101, Publisher B, # of books, , Freq., , 0, , 18, , 1, , 24, , 2, , 24, , 3, , 22, , 4, , 15, , 5, , 10, , 7, , 5, , 9, , 1, , Rel. Freq., , Table 2.19, , Publisher C, # of books, , Freq., , 0-1, , 20, , 2-3, , 35, , 4-5, , 12, , 6-7, , 2, , 8-9, , 1, , Rel. Freq., , Table 2.20, , a. Find the relative frequencies for each survey. Write them in the charts., b. Using either a graphing calculator, computer, or by hand, use the frequency column to construct, a histogram for each publisher’s survey. For Publishers A and B, make bar widths of 1. For, Publisher C, make bar widths of 2., c. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical., d. Would you have expected the graph for Publisher C to look like the other two graphs? Why or, why not?, e. Make new histograms for Publisher A and Publisher B. This time, make bar widths of 2., f. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the, graphs more similar or more different? Explain your answer., Exercise 2.13.14, Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all on-board transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day, cruise from Los Angeles to the Mexican Riviera. Below is a summary of the bills for each group., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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102, , CHAPTER 2. DESCRIPTIVE STATISTICS, Singles, Amount($), , Frequency, , 51-100, , 5, , 101-150, , 10, , 151-200, , 15, , 201-250, , 15, , 251-300, , 10, , 301-350, , 5, , Rel. Frequency, , Table 2.21, , Couples, Amount($), , Frequency, , 100-150, , 5, , 201-250, , 5, , 251-300, , 5, , 301-350, , 5, , 351-400, , 10, , 401-450, , 10, , 451-500, , 10, , 501-550, , 10, , 551-600, , 5, , 601-650, , 5, , Rel. Frequency, , Table 2.22, , a. Fill in the relative frequency for each group., b. Construct a histogram for the Singles group. Scale the x-axis by $50. widths. Use relative, frequency on the y-axis., c. Construct a histogram for the Couples group. Scale the x-axis by $50. Use relative frequency on, the y-axis., d. Compare the two graphs:, i. List two similarities between the graphs., ii. List two differences between the graphs., iii. Overall, are the graphs more similar or different?, e. Construct a new graph for the Couples by hand. Since each couple is paying for two individuals, instead of scaling the x-axis by $50, scale it by $100. Use relative frequency on the, y-axis., f. Compare the graph for the Singles with the new graph for the Couples:, i. List two similarities between the graphs., ii. Overall, are the graphs more similar or different?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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103, i. By scaling the Couples graph differently, how did it change the way you compared it to the, Singles?, j. Based on the graphs, do you think that individuals spend the same amount, more or less, as, singles as they do person by person in a couple? Explain why in one or two complete sentences., Exercise 2.13.15, (Solution on p. 118.), Refer to the following histograms and box plot. Determine which of the following are true and, which are false. Explain your solution to each part in complete sentences., , a., b., c., d., , The medians for all three graphs are the same., We cannot determine if any of the means for the three graphs is different., The standard deviation for (b) is larger than the standard deviation for (a)., We cannot determine if any of the third quartiles for the three graphs is different., , Exercise 2.13.16, Refer to the following box plots., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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104, , CHAPTER 2. DESCRIPTIVE STATISTICS, , a. In complete sentences, explain why each statement is false., i. Data 1 has more data values above 2 than Data 2 has above 2., ii. The data sets cannot have the same mode., iii. For Data 1, there are more data values below 4 than there are above 4., b. For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why, in complete sentences, Exercise 2.13.17, (Solution on p. 119.), In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted, five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine, days. Let X = the length (in days) of an engineering conference., a. Organize the data in a chart., b. Find the median, the first quartile, and the third quartile., c. Find the 65th percentile., d. Find the 10th percentile., e. Construct a box plot of the data., f. The middle 50% of the conferences last from _______ days to _______ days., g. Calculate the sample mean of days of engineering conferences., h. Calculate the sample standard deviation of days of engineering conferences., i. Find the mode., j. If you were planning an engineering conference, which would you choose as the length of the, conference: mean; median; or mode? Explain why you made that choice., k. Give two reasons why you think that 3 - 5 days seem to be popular lengths of engineering, conferences., Exercise 2.13.18, A survey of enrollment at 35 community colleges across the United States yielded the following, figures (source: Microsoft Bookshelf ):, 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357;, 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285;, 28165; 5080; 11622, a. Organize the data into a chart with five intervals of equal width. Label the two columns "Enrollment" and "Frequency.", b. Construct a histogram of the data., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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105, c. If you were to build a new community college, which piece of information would be more, valuable: the mode or the mean?, d. Calculate the sample mean., e. Calculate the sample standard deviation., f. A school with an enrollment of 8000 would be how many standard deviations away from the, mean?, Exercise 2.13.19, (Solution on p. 119.), The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1, years. (Source: Bureau of the Census), a. What does it mean for the median age to rise?, b. Give two reasons why the median age could rise., c. For the median age to rise, is the actual number of children less in 1991 than it was in 1980?, Why or why not?, Exercise 2.13.20, A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new, BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age, they were when they purchased their car. The following box plots display the results., , a. In complete sentences, describe what the shape of each box plot implies about the distribution, of the data collected for that car series., b. Which group is most likely to have an outlier? Explain how you determined that., c. Compare the three box plots. What do they imply about the age of purchasing a BMW from the, series when compared to each other?, d. Look at the BMW 5 series. Which quarter has the smallest spread of data? What is that spread?, e. Look at the BMW 5 series. Which quarter has the largest spread of data? What is that spread?, f. Look at the BMW 5 series. Estimate the Inter Quartile Range (IQR)., g. Look at the BMW 5 series. Are there more data in the interval 31-38 or in the interval 45-55?, How do you know this?, h. Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this?, i. 31-35, ii. 38-41, iii. 41-64, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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106, , CHAPTER 2. DESCRIPTIVE STATISTICS, Exercise 2.13.21, (Solution on p. 119.), The following box plot shows the U.S. population for 1990, the latest available year. (Source:, Bureau of the Census, 1990 Census), , a. Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)?, How do you know?, b. 12.6% are age 65 and over. Approximately what percent of the population are of working age, adults (above age 17 to age 65)?, Exercise 2.13.22, Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the, mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The, samples yielded the following information:, Javier, , Ercilla, , x, , 6.0 miles, , 6.0 miles, , s, , 4.0 miles, , 7.0 miles, , Table 2.23, , a. How can you determine which survey was correct ?, b. Explain what the difference in the results of the surveys implies about the data., c. If the two histograms depict the distribution of values for each supervisor, which one depicts, Ercilia’s sample? How do you know?, , Figure 2.2, , d. If the two box plots depict the distribution of values for each supervisor, which one depicts, Ercilia’s sample? How do you know?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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107, , Figure 2.3, , Exercise 2.13.23, Student grades on a chemistry exam were:, , (Solution on p. 119.), , 77, 78, 76, 81, 86, 51, 79, 82, 84, 99, a. Construct a stem-and-leaf plot of the data., b. Are there any potential outliers? If so, which scores are they? Why do you consider them, outliers?, , 2.13.1 Try these multiple choice questions (Exercises 24 - 30)., The next three questions refer to the following information. We are interested in the number of years, students in a particular elementary statistics class have lived in California. The information in the following, table is from the entire section., Number of years, , Frequency, , 7, , 1, , 14, , 3, , 15, , 1, , 18, , 1, , 19, , 4, , 20, , 3, , 22, , 1, , 23, , 1, , 26, , 1, , 40, , 2, , 42, , 2, Total = 20, Table 2.24, , Exercise 2.13.24, What is the IQR?, , (Solution on p. 119.), , A. 8, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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108, , CHAPTER 2. DESCRIPTIVE STATISTICS, B. 11, C. 15, D. 35, Exercise 2.13.25, What is the mode?, A., B., C., D., , (Solution on p. 119.), , 19, 19.5, 14 and 20, 22.65, , Exercise 2.13.26, Is this a sample or the entire population?, , (Solution on p. 119.), , A. sample, B. entire population, C. neither, The next two questions refer to the following table. X = the number of days per week that 100 clients use, a particular exercise facility., x, , Frequency, , 0, , 3, , 1, , 12, , 2, , 33, , 3, , 28, , 4, , 11, , 5, , 9, , 6, , 4, Table 2.25, , Exercise 2.13.27, The 80th percentile is:, A., B., C., D., , (Solution on p. 119.), , 5, 80, 3, 4, , Exercise 2.13.28, (Solution on p. 119.), The number that is 1.5 standard deviations BELOW the mean is approximately:, A., B., C., D., , 0.7, 4.8, -2.8, Cannot be determined, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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109, The next two questions refer to the following histogram. Suppose one hundred eleven people who, shopped in a special T-shirt store were asked the number of T-shirts they own costing more than $19 each., , Exercise 2.13.29, (Solution on p. 119.), The percent of people that own at most three (3) T-shirts costing more than $19 each is approximately:, A., B., C., D., , 21, 59, 41, Cannot be determined, , Exercise 2.13.30, (Solution on p. 119.), If the data were collected by asking the first 111 people who entered the store, then the type of, sampling is:, A., B., C., D., , cluster, simple random, stratified, convenience, , Exercise 2.13.31, (Solution on p. 119.), Below are the 2010 obesity rates by U.S. states and Washington, DC.(Source:, http://www.cdc.gov/obesity/data/adult.html) ), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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110, , CHAPTER 2. DESCRIPTIVE STATISTICS, , State, , Percent (%), , State, , Percent (%), , Alabama, , 32.2, , Montana, , 23.0, , Alaska, , 24.5, , Nebraska, , 26.9, , Arizona, , 24.3, , Nevada, , 22.4, , Arkansas, , 30.1, , New Hampshire, , 25.0, , California, , 24.0, , New Jersey, , 23.8, , Colorado, , 21.0, , New Mexico, , 25.1, , Connecticut, , 22.5, , New York, , 23.9, , Delaware, , 28.0, , North Carolina, , 27.8, , Washington, DC, , 22.2, , North Dakota, , 27.2, , Florida, , 26.6, , Ohio, , 29.2, , Georgia, , 29.6, , Oklahoma, , 30.4, , Hawaii, , 22.7, , Oregon, , 26.8, , Idaho, , 26.5, , Pennsylvania, , 28.6, , Illinois, , 28.2, , Rhode Island, , 25.5, , Indiana, , 29.6, , South Carolina, , 31.5, , Iowa, , 28.4, , South Dakota, , 27.3, , Kansas, , 29.4, , Tennessee, , 30.8, , Kentucky, , 31.3, , Texas, , 31.0, , Louisiana, , 31.0, , Utah, , 22.5, , Maine, , 26.8, , Vermont, , 23.2, , Maryland, , 27.1, , Virginia, , 26.0, , Massachusetts, , 23.0, , Washington, , 25.5, , Michigan, , 30.9, , West Virginia, , 32.5, , Minnesota, , 24.8, , Wisconsin, , 26.3, , Mississippi, , 34.0, , Wyoming, , 25.1, , Missouri, , 30.5, Table 2.26, , a.. Construct a bar graph of obesity rates of your state and the four states closest to your state., Hint: Label the x-axis with the states., b.. Use a random number generator to randomly pick 8 states. Construct a bar graph of the obesity, rates of those 8 states., c.. Construct a bar graph for all the states beginning with the letter "A.", d.. Construct a bar graph for all the states beginning with the letter "M.", Exercise 2.13.32, (Solution on p. 120.), A music school has budgeted to purchase 3 musical instruments. They plan to purchase a piano, costing $3000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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111, $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard, deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost, is the lowest, when compared to other instruments of the same type? Which cost is the highest, when compared to other instruments of the same type. Justify your answer numerically., Exercise 2.13.33, (Solution on p. 120.), Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the table, below. (Note that this is the data presented for publisher B in homework exercise 13)., Publisher B, # of books, , Freq., , 0, , 18, , 1, , 24, , 2, , 24, , 3, , 22, , 4, , 15, , 5, , 10, , 7, , 5, , 9, , 1, , Rel. Freq., , Table 2.27, , a. Are there any outliers in the data? Use an appropriate numerical test involving the IQR to, identify outliers, if any, and clearly state your conclusion., b. If a data value is identified as an outlier, what should be done about it?, c. Are any data values further than 2 standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared, to the other data values. (Note that this criteria is most appropriate to use for data that is, mound-shaped and symmetric, rather than for skewed data.), d. Do parts (a) and (c) of this problem give the same answer?, e. Examine the shape of the data. Which part, (a) or (c), of this question gives a more appropriate result for this data?, f. Based on the shape of the data which is the most appropriate measure of center for this data:, mean, median or mode?, **Exercises 32 and 33 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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113, , Figure 2.4, , 3. Calculate the following:, , • x=, • s=, 4. Are the data discrete or continuous? How do you know?, 5. Describe the shape of the histogram. Use complete sentences., 6. Are there any potential outliers? Which value(s) is (are) it (they)? Use a formula to check the end, values to determine if they are potential outliers., , 2.14.3 Analyze the Data, 1. Determine the following:, , •, •, •, •, •, •, , Minimum value =, Median =, Maximum value =, First quartile =, Third quartile =, IQR =, , 2., 3., 4., 5., , Construct a box plot of data, What does the shape of the box plot imply about the concentration of data? Use complete sentences., Using the box plot, how can you determine if there are potential outliers?, How does the standard deviation help you to determine concentration of the data and whether or not, there are potential outliers?, 6. What does the IQR represent in this problem?, 7. Show your work to find the value that is 1.5 standard deviations:, a. Above the mean:, b. Below the mean:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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114, , CHAPTER 2. DESCRIPTIVE STATISTICS, , Solutions to Exercises in Chapter 2, Solution to Example 2.2, Problem (p. 61), The value 12.3 may be an outlier. Values appear to concentrate at 3 and 4 kilometers., Stem, , Leaf, , 1, , 15, , 2, , 357, , 3, , 23358, , 4, , 025578, , 5, , 56, , 6, , 57, , 7, 8, 9, 10, 11, 12, , 3, Table 2.29, , Solution to Example 2.7, Problem (p. 66), , •, •, •, •, , 3.5 to 4.5, 4.5 to 5.5, 6, 5.5 to 6.5, , Solution to Example 2.9, Problem (p. 70), First Data Set, , •, •, •, •, •, , Xmin = 32, Q1 = 56, M = 74.5, Q3 = 82.5, Xmax = 99, , Second Data Set, , •, •, •, •, •, , Xmin = 25.5, Q1 = 78, M = 81, Q3 = 89, Xmax = 98, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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115, , Solution to Example 2.11, Problem (p. 72), For the IQRs, see the answer to the test scores example (Solution to Example 2.9: p. 114). The first data set, has the larger IQR, so the scores between Q3 and Q1 (middle 50%) for the first data set are more spread out, and not clustered about the median., First Data Set, , , • 23 · ( IQR) = 23 · (26.5) = 39.75, • Xmax − Q3 = 99 − 82.5 = 16.5, • Q1 − Xmin = 56 − 32 = 24, , 3, 2 · ( IQR ) = 39.75 is larger than 16.5 and larger than 24, so the first set has no outliers., Second Data Set, , , • 32 · ( IQR) = 23 · (11) = 16.5, • Xmax − Q3 = 98 − 89 = 9, • Q1 − Xmin = 78 − 25.5 = 52.5, , 3, 2 · ( IQR ) = 16.5 is larger than 9 but smaller than 52.5, so for the second set 45 and 25.5 are outliers., To find the percentiles, create a frequency, relative frequency, and cumulative relative frequency chart (see, "Frequency" from the Sampling and Data Chapter (Section 1.9)). Get the percentiles from that chart., First Data Set, , • 30th %ile (between the 6th and 7th values) = (56 +2 59) = 57.5, • 80th %ile (between the 16th and 17th values) = (84 +2 84.5) = 84.25, Second Data Set, , • 30th %ile (7th value) = 78, • 80th %ile (18th value) = 90, 30% of the data falls below the 30th %ile, and 20% falls above the 80th %ile., Solution to Example 2.13, Problem (p. 73), (8 + 9), 2, , = 8.5, Look where cum. rel. freq. = 0.80. 80% of the data is 8 or less. 80th %ile is between the last 8 and first, 9., 2. 9, 3. 6, 4. First Quartile = 25th %ile, 1., , Solution to Exercise 2.6.1 (p. 75), a. For runners in a race it is more desirable to have a low percentile for finish time. A low percentile means, a short time, which is faster., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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116, , CHAPTER 2. DESCRIPTIVE STATISTICS, , b. INTERPRETATION: 20% of runners finished the race in 5.2 minutes or less. 80% of runners finished the, race in 5.2 minutes or longer., c. He is among the slowest cyclists (90% of cyclists were faster than him.) INTERPRETATION: 90% of, cyclists had a finish time of 1 hour, 12 minutes or less.Only 10% of cyclists had a finish time of 1 hour,, 12 minutes or longer, Solution to Exercise 2.6.2 (p. 75), a. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a, higher speed, which is faster., b. INTERPRETATION: 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners, ran at speeds of 7.5 miles per hour or more (faster)., Solution to Exercise 2.6.3 (p. 75), On an exam you would prefer a high percentile; higher percentiles correspond to higher grades on the, exam., Solution to Exercise 2.6.4 (p. 75), When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other, people waiting. 85% of people had shorter wait times than you did. In this context, you would prefer a, wait time corresponding to a lower percentile. INTERPRETATION: 85% of people at the DMV waited 32, minutes or less. 15% of people at the DMV waited 32 minutes or longer., Solution to Exercise 2.6.5 (p. 75), Li should be pleased. Her salary is relatively high compared to other recent college grads. 78% of recent, college graduates earn less than Li does. 22% of recent college graduates earn more than Li does., Solution to Exercise 2.6.6 (p. 75), The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared, to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs, of $1700 or less; only 10% had damage repair costs of $1700 or more., Solution to Exercise 2.6.7 (p. 75), a. The top 12% of students are those who are at or above the 88th percentile of admissions index scores., b. The top 4% of students’ GPAs are at or above the 96th percentile, making the top 4% of students "eligible, in the local context"., Solution to Exercise 2.6.8 (p. 75), You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION:, 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more., , Solutions to Practice 1: Center of the Data, Solution to Exercise 2.11.1 (p., 65, Solution to Exercise 2.11.2 (p., 1, Solution to Exercise 2.11.5 (p., 4.75, Solution to Exercise 2.11.6 (p., 1.39, Solution to Exercise 2.11.7 (p., 65, Solution to Exercise 2.11.8 (p., 4, Solution to Exercise 2.11.9 (p., 4, , 90), 90), 91), 91), 91), 91), 91), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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117, Solution to Exercise 2.11.10 (p., 4, Solution to Exercise 2.11.11 (p., 4, Solution to Exercise 2.11.12 (p., 6, Solution to Exercise 2.11.13 (p., 6 − 4 = 2, Solution to Exercise 2.11.14 (p., 3, Solution to Exercise 2.11.15 (p., 6, Solution to Exercise 2.11.16 (p., , 91), 91), 91), 91), 91), 91), 92), , a. 8.93, b. 0.58, , Solutions to Practice 2: Spread of the Data, Solution to Exercise 2.12.1 (p. 93), 6, Solution to Exercise 2.12.2 (p. 93), a. 1447.5, b. 528.5, Solution to Exercise 2.12.3 (p. 93), 474 FTES, Solution to Exercise 2.12.4 (p. 93), 50%, Solution to Exercise 2.12.5 (p. 93), 919, Solution to Exercise 2.12.6 (p. 93), 0.03, Solution to Exercise 2.12.7 (p. 94), mean = 1809.3, median = 1812.5, standard deviation = 151.2, First quartile = 1690, Third quartile = 1935, IQR = 245, Solution to Exercise 2.12.9 (p. 94), Hint: Think about the number of years covered by each time period and what happened to higher education during those periods., , Solutions to Homework, Solution to Exercise 2.13.1 (p. 95), a. 1.48, b. 1.12, e. 1, f. 1, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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118, , CHAPTER 2. DESCRIPTIVE STATISTICS, , g. 2, , h., i. 80%, j. 1, k. 3, Solution to Exercise 2.13.3 (p. 95), a. 3.78, b. 1.29, e. 3, f. 4, g. 5, , h., i. 32.5%, j. 4, k. 5, Solution to Exercise 2.13.5 (p. 97), b. 241, c. 205.5, d. 272.5, , e., f. 205.5, 272.5, g. sample, h. population, i. i. 236.34, ii. 37.50, iii. 161.34, iv. 0.84 std. dev. below the mean, j. Young, Solution to Exercise 2.13.9 (p. 98), Kamala, Solution to Exercise 2.13.15 (p. 103), a. True, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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119, b. True, c. True, d. False, Solution to Exercise 2.13.17 (p. 104), b. 4,3,5, c. 4, d. 3, , e., f. 3,5, g. 3.94, h. 1.28, i. 3, j. mode, Solution to Exercise 2.13.19 (p. 105), c. Maybe, Solution to Exercise 2.13.21 (p. 106), a. more children, b. 62.4%, Solution to Exercise 2.13.23 (p. 107), b. 51,99, Solution to Exercise 2.13.24 (p. 107), A, Solution to Exercise 2.13.25 (p. 108), A, Solution to Exercise 2.13.26 (p. 108), B, Solution to Exercise 2.13.27 (p. 108), D, Solution to Exercise 2.13.28 (p. 108), A, Solution to Exercise 2.13.29 (p. 109), C, Solution to Exercise 2.13.30 (p. 109), D, Solution to Exercise 2.13.31 (p. 109), Example solution for b using the random number generator for the Ti-84 Plus to generate a simple random, sample of 8 states. Instructions are below., Number the entries in the table 1 - 51 (Includes Washington, DC; Numbered vertically), Press MATH, Arrow over to PRB, Press 5:randInt(, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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120, , CHAPTER 2. DESCRIPTIVE STATISTICS, , Enter 51,1,8), Eight numbers are generated (use the right arrow key to scroll through the numbers)., The, numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}., If, any numbers are repeated, generate a different number by using 5:randInt(51,1))., Here, the, states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}., Corresponding percents are {28.7 21.8 24.5 26 28.9 32.8 25 24.6}., , Solution to Exercise 2.13.32 (p. 110), For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the, guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard, deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other, instruments of the same type. The guitar cost the most in comparison to the cost of other instruments of the, same type., Solution to Exercise 2.13.33 (p. 111), , • IQR = 4 – 1 = 3 ; Q1 – 1.5*IQR = 1 – 1.5(3) = -3.5 ; Q3 + 1.5*IQR = 4 + 1.5(3) = 8.5 ;The data value of 9 is, larger than 8.5. The purchase of 9 books in one month is an outlier., • The outlier should be investigated to see if there is an error or some other problem in the data; then a, decision whether to include or exclude it should be made based on the particular situation. If it was, a correct value then the data value should remain in the data set. If there is a problem with this data, value, then it should be corrected or removed from the data. For example: If the data was recorded, incorrectly (perhaps a 9 was miscoded and the correct value was 6) then the data should be corrected., If it was an error but the correct value is not known it should be removed from the data set., • xbar – 2s = 2.45 – 2*1.88 = -1.31 ; xbar + 2s = 2.45 + 2*1.88 = 6.21 ; Using this method, the five data values, of 7 books purchased and the one data value of 9 books purchased would be considered unusual., • No: part (a) identifies only the value of 9 to be an outlier but part (c) identifies both 7 and 9., • The data is skewed (to the right). It would be more appropriate to use the method involving the IQR, in part (a), identifying only the one value of 9 books purchased as an outlier. Note that part (c) remarks, that identifying unusual data values by using the criteria of being further than 2 standard deviations, away from the mean is most appropriate when the data are mound-shaped and symmetric., • The data are skewed to the right. For skewed data it is more appropriate to use the median as a, measure of center., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 3, , Probability Topics, 3.1 Probability Topics1, 3.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, •, •, , Understand and use the terminology of probability., Determine whether two events are mutually exclusive and whether two events are independent., Calculate probabilities using the Addition Rules and Multiplication Rules., Construct and interpret Contingency Tables., Construct and interpret Venn Diagrams (optional)., Construct and interpret Tree Diagrams (optional)., , 3.1.2 Introduction, It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study, polls to guess their likelihood of winning an election. Teachers choose a particular course of study based, on what they think students can comprehend. Doctors choose the treatments needed for various diseases, based on their assessment of likely results. You may have visited a casino where people play games chosen, because of the belief that the likelihood of winning is good. You may have chosen your course of study, based on the probable availability of jobs., You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability., Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not, to do your homework or to study for an exam, you are using probability. In this chapter, you will learn to, solve probability problems using a systematic approach., , 3.1.3 Optional Collaborative Classroom Exercise, Your instructor will survey your class. Count the number of students in the class today., , • Raise your hand if you have any change in your pocket or purse. Record the number of raised hands., • Raise your hand if you rode a bus within the past month. Record the number of raised hands., • Raise your hand if you answered "yes" to BOTH of the first two questions. Record the number of, raised hands., 1 This, , content is available online at <http://cnx.org/content/m16838/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 121
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122, , CHAPTER 3. PROBABILITY TOPICS, , Use the class data as estimates of the following probabilities. P(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that, a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers., , • Find P(change)., • Find P(bus)., • Find P(change and bus) Find the probability that a randomly chosen student in your class has change, in his/her pocket or purse and rode a bus within the last month., • Find P(change| bus) Find the probability that a randomly chosen student has change given that, he/she rode a bus within the last month. Count all the students that rode a bus. From the group, of students who rode a bus, count those who have change. The probability is equal to those who have, change and rode a bus divided by those who rode a bus., , 3.2 Terminology2, Probability is a measure that is associated with how certain we are of outcomes of a particular experiment, or activity. An experiment is a planned operation carried out under controlled conditions. If the result is, not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is, an example of an experiment., The result of an experiment is called an outcome. A sample space is a set of all possible outcomes. Three, ways to represent a sample space are to list the possible outcomes, to create a tree diagram, or to create a, Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair, coin, S = { H, T } where H = heads and T = tails are the outcomes., An event is any combination of outcomes. Upper case letters like A and B represent events. For example,, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an, event A is written P ( A)., The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are, between 0 and 1, inclusive (includes 0 and 1 and all numbers between these values). P ( A) = 0 means, the event A can never happen. P ( A) = 1 means the event A always happens. P ( A) = 0.5 means the, event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20, to 2,000 to 20,000 times) the relative fequency of heads approaches 0.5 (the probability of heads)., Equally likely means that each outcome of an experiment occurs with equal probability. For example, if, you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you, toss a fair coin, a Head(H) and a Tail(T) are equally likely to occur. If you randomly guess the answer to a, true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer., To calculate the probability of an event A when all outcomes in the sample space are equally likely,, count the number of outcomes for event A and divide by the total number of outcomes in the sample space., For example, if you toss a fair dime and a fair nickel, the sample space is { HH, TH, HT, TT } where T =, tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes, { HT, TH }. P ( A) = 42 ., Suppose you roll one fair six-sided die, with the numbers {1,2,3,4,5,6} on its faces. Let event E = rolling a, number that is at least 5. There are two outcomes {5, 6}. P ( E) = 26 . If you were to roll the die only a few, times, you would not be surprised if your observed results did not match the probability. If you were to, roll the die a very large number of times, you would expect that, overall, 2/6 of the rolls would result in an, 2 This, , content is available online at <http://cnx.org/content/m16845/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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123, outcome of "at least 5". You would not expect exactly 2/6. The long-term relative frequency of obtaining, this result would approach the theoretical probability of 2/6 as the number of repetitions grows larger and, larger., This important characteristic of probability experiments is the known as the Law of Large Numbers: as, the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment, tends to become closer and closer to the theoretical probability. Even though the outcomes don’t happen, according to any set pattern or order, overall, the long-term observed relative frequency will approach the, theoretical probability. (The word empirical is often used instead of the word observed.) The Law of Large, Numbers will be discussed again in Chapter 7., It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may, be unfair, or biased . Two math professors in Europe had their statistics students test the Belgian 1 Euro, coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44%, of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to, draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you, have at home; the spots on each face are usually small holes carved out and then painted to make the spots, visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight, weight differences due to the different numbers of holes in the faces. Gambling casinos have a lot of money, depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice, have flat faces; the holes are completely filled with paint having the same density as the material that the, dice are made out of so that each face is equally likely to occur. Later in this chapter we will learn techniques, to use to work with probabilities for events that are not equally likely., "OR" Event:, An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let, A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are, NOT listed twice., "AND" Event:, An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let, A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}., The complement of event A is denoted A’ (read "A prime"). A’ consists of all outcomes that are NOT in A., Notice that P ( A) + P ( A’) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then,, A’ = {5, 6}. P ( A) = 64 , P ( A’) = 26 , and P ( A) + P ( A’) = 64 + 26 = 1, The conditional probability of A given B is written P ( A| B). P ( A| B) is the probability that event A will, occur given that the event B has already occurred. A conditional reduces the sample space. We calculate, the probability of A from the reduced sample space B. The formula to calculate P ( A| B) is, P ( A| B) =, , P( AANDB), P( B), , where P ( B) is greater than 0., For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A =, face is 2 or 3 and B = face is even (2, 4, 6). To calculate P ( A| B), we count the number of outcomes 2 or 3 in, the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes in B (and not S)., We get the same result by using the formula. Remember that S has 6 outcomes., P ( A| B) =, , P(A and B), P(B), , =, , (the number of outcomes that are 2 or 3 and even in S) / 6, (the number of outcomes that are even in S) / 6, , =, , 1/6, 3/6, , =, , 1, 3, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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124, , CHAPTER 3. PROBABILITY TOPICS, , Understanding Terminology and Symbols, It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem, several times if necessary. Clearly identify the event of interest. Determine whether there is a condition, stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any., Exercise 3.2.1, (Solution on p. 160.), In a particular college class, there are male and female students. Some students have long hair and, some students have short hair. Write the symbols for the probabilities of the events for parts (a), through (j) below. (Note that you can’t find numerical answers here. You were not given enough, information to find any probability values yet; concentrate on understanding the symbols.), , •, •, •, •, , Let F be the event that a student is female., Let M be the event that a student is male., Let S be the event that a student has short hair., Let L be the event that a student has long hair., , a. The probability that a student does not have long hair., b. The probability that a student is male or has short hair., c. The probability that a student is a female and has long hair., d. The probability that a student is male, given that the student has long hair., e. The probability that a student has long hair, given that the student is male., f. Of all the female students, the probability that a student has short hair., g. Of all students with long hair, the probability that a student is female., h. The probability that a student is female or has long hair., i. The probability that a randomly selected student is a male student with short hair., j. The probability that a student is female., **With contributions from Roberta Bloom, , 3.3 Independent and Mutually Exclusive Events3, Independent and mutually exclusive do not mean the same thing., , 3.3.1 Independent Events, Two events are independent if the following are true:, , • P ( A| B) = P ( A), • P ( B| A) = P ( B), • P ( A AND B) = P ( A) · P ( B), Two events A and B are independent if the knowledge that one occurred does not affect the chance the, other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome, of the first roll does not change the probability for the outcome of the second roll. To show two events are, independent, you must show only one of the above conditions. If two events are NOT independent, then, we say that they are dependent., Sampling may be done with replacement or without replacement., 3 This, , content is available online at <http://cnx.org/content/m16837/1.14/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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125, , • With replacement: If each member of a population is replaced after it is picked, then that member has, the possibility of being chosen more than once. When sampling is done with replacement, then events, are considered to be independent, meaning the result of the first pick will not change the probabilities, for the second pick., • Without replacement:: When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the, result of the first pick. The events are considered to be dependent or not independent., If it is not known whether A and B are independent or dependent, assume they are dependent until you, can show otherwise., , 3.3.2 Mutually Exclusive Events, A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do, not share any outcomes and P(A AND B) = 0., For, example,, suppose, the, sample, space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}., Let, A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B, = {4, 5}. P(A AND B) =, 2, 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any, numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive., If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise., The following examples illustrate these definitions and terms., Example 3.1, Flip two fair coins. (This is an experiment.), The sample space is { HH, HT, TH, TT } where T = tails and H = heads. The outcomes are HH,, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin, showed heads and the second coin showed tails. The TH means that the first coin showed tails, and the second coin showed heads., , • Let A = the event of getting at most one tail. (At most one tail means 0 or 1 tail.) Then A can, be written as { HH, HT, TH }. The outcome HH shows 0 tails. HT and TH each show 1 tail., • Let B = the event of getting all tails. B can be written as { TT }. B is the complement of A. So,, B = A’. Also, P ( A) + P ( B) = P ( A) + P ( A’) = 1., • The probabilities for A and for B are P ( A) = 43 and P ( B) = 14 ., • Let C = the event of getting all heads. C = { HH }. Since B = { TT }, P ( B AND C ) = 0., B and C are mutually exclusive. (B and C have no members in common because you cannot, have all tails and all heads at the same time.), • Let D = event of getting more than one tail. D = { TT }. P ( D ) = 41 ., • Let E = event of getting a head on the first roll. (This implies you can get either a head or tail, on the second roll.) E = { HT, HH }. P ( E) = 42 ., • Find the probability of getting at least one (1 or 2) tail in two flips. Let F = event of getting, at least one tail in two flips. F = { HT, TH, TT }. P(F) = 34, Example 3.2, Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then, A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}., , • Find the complement of A, A’. The complement of A, A’, is B because A and B together, make up the sample space. P(A) + P(B) = P(A) + P(A’) = 1. Also, P(A) = 63 and P(B) = 36, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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126, , CHAPTER 3. PROBABILITY TOPICS, , • Let event C = odd faces larger than 2. Then C = {3, 5}. Let event D = all even faces smaller, than 5. Then D = {2, 4}. P(C and D) = 0 because you cannot have an odd and even face at, the same time. Therefore, C and D are mutually exclusive events., • Let event E = all faces less than 5. E = {1, 2, 3, 4}., Problem, (Solution on p. 160.), Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?, • Find P(C|A). This is a conditional. Recall that the event C is {3, 5} and event A is {1, 3, 5}., To find P(C|A), find the probability of C using the sample space A. You have reduced the, sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P(C|A) = 23, Example 3.3, Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking, a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are, G and H independent?, If G and H are independent, then you must show ONE of the following:, , • P(G|H) = P(G), • P(H|G) = P(H), • P(G AND H) = P(G) · P(H), NOTE : The choice you make depends on the information you have. You could choose any of the, methods here because you have the necessary information., , Problem 1, Show that P(G|H) = P(G)., Solution, P(G|H) =, , P(G AND H), P(H), , =, , 0.3, 0.5, , = 0.6 = P(G), , Problem 2, Show P(G AND H) = P(G) · P(H)., Solution, P ( G ) · P ( H ) = 0.6 · 0.5 = 0.3 = P(G AND H), Since G and H are independent, then, knowing that a person is taking a science class does not, change the chance that he/she is taking math. If the two events had not been independent (that, is, they are dependent) then knowing that a person is taking a science class would change the, chance he/she is taking math. For practice, show that P(H|G) = P(H) to show that G and H are, independent events., Example 3.4, In a box there are 3 red cards and 5 blue cards. The red cards are marked with the numbers 1, 2,, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled., You reach into the box (you cannot see into it) and draw one card., Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn., The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has 8 outcomes., , • P(R) = 83 . P(B) = 85 . P(R AND B) = 0. (You cannot draw one card that is both red and blue.), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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127, , • P(E) = 38 . (There are 3 even-numbered cards, R2, B2, and B4.), • P(E|B) = 25 . (There are 5 blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are, 2 even cards: B2 and B4.), • P(B|E) = 23 . (There are 3 even-numbered cards: R2, B2, and B4. Out of the even-numbered, cards, 2 are blue: B2 and B4.), • The events R and B are mutually exclusive because P(R AND B) = 0., • Let G = card with a number greater than 3. G = { B4, B5}. P(G) = 82 . Let H = blue card, numbered between 1 and 4, inclusive. H = { B1, B2, B3, B4}. P(G|H) = 41 . (The only card in, H that has a number greater than 3 is B4.) Since 82 = 14 , P(G) = P(G|H) which means that, G and H are independent., Example 3.5, In a particular college class, 60% of the students are female. 50 % of all students in the class have, long hair. 45% of the students are female and have long hair. Of the female students, 75% have, long hair. Let F be the event that the student is female. Let L be the event that the student has, long hair. One student is picked randomly. Are the events of being female and having long hair, independent?, , •, •, •, •, , The following probabilities are given in this example:, P(F ) = 0.60 ; P(L ) = 0.50, P(F AND L) = 0.45, P(L|F) = 0.75, , NOTE : The choice you make depends on the information you have. You could use the first or, last condition on the list for this example. You do not know P(F|L) yet, so you can not use the, second condition., , Solution 1, Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) =, (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because, P(F and L) does not equal P(F)P(L)., Solution 2, check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not, equal. The events of being female and having long hair are not independent., Interpretation of Results, The events of being female and having long hair are not independent; knowing that a student is, female changes the probability that a student has long hair., **Example 5 contributed by Roberta Bloom, , 3.4 Two Basic Rules of Probability4, 3.4.1 The Multiplication Rule, If A and B are two events defined on a sample space, then: P(A AND B) = P(B) · P(A|B)., This rule may also be written as : P ( A| B) =, 4 This, , P( A AND B), P( B), , content is available online at <http://cnx.org/content/m16847/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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128, , CHAPTER 3. PROBABILITY TOPICS, , (The probability of A given B equals the probability of A and B divided by the probability of B.), If A and B are independent, then P(A|B), P(A AND B) = P(A) P(B)., , = P(A). Then P(A AND B), , = P(A|B) P(B) becomes, , 3.4.2 The Addition Rule, If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) − P(A AND B)., If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B), becomes P(A OR B) = P(A) + P(B)., Example 3.6, Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B, = Alaska, , • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the, probability that he chooses B is P(B) = 0.35., • P(A and B) = 0 because Klaus can only afford to take one vacation, • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) =, P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go, anywhere on vacation must be 0.05., Example 3.7, Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt, two goals in a row in the next game., A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is, successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability, that he makes the second goal GIVEN that he made the first goal is 0.90., Problem 1, What is the probability that he makes both goals?, Solution, The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90:, P(B AND A) = P(B|A) P(A) = 0.90 ∗ 0.65 = 0.585, , (3.0), , Carlos makes the first and second goals with probability 0.585., , Problem 2, What is the probability that Carlos makes either the first goal or the second goal?, Solution, The problem is asking you to find P(A OR B)., P(A OR B) = P(A) + P(B) − P(A AND B) = 0.65 + 0.65 − 0.585 = 0.715, Carlos makes either the first goal or the second goal with probability 0.715., , Problem 3, Are A and B independent?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (3.0)
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129, Solution, No, they are not, because P(B AND A) = 0.585., P(B) · P(A) = (0.65) · (0.65) = 0.423, , (3.0), , 0.423 6= 0.585 = P(B AND A), , (3.0), , So, P(B AND A) is not equal to P(B) · P(A)., , Problem 4, Are A and B mutually exclusive?, Solution, No, they are not because P(A and B) = 0.585., To be mutually exclusive, P(A AND B) must equal 0., , Example 3.8, A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice 4 times a week. Thirty of the intermediate swimmers practice 4 times a week. Ten of the novice swimmers practice 4 times a week. Suppose one, member of the swim team is randomly chosen. Answer the questions (Verify the answers):, Problem 1, What is the probability that the member is a novice swimmer?, Solution, 28, 150, , Problem 2, What is the probability that the member practices 4 times a week?, Solution, 80, 150, , Problem 3, What is the probability that the member is an advanced swimmer and practices 4 times a week?, Solution, 40, 150, , Problem 4, What is the probability that a member is an advanced swimmer and an intermediate swimmer?, Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why, not?, Solution, P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot, be an advanced swimmer and an intermediate swimmer at the same time., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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130, , CHAPTER 3. PROBABILITY TOPICS, , Problem 5, Are being a novice swimmer and practicing 4 times a week independent events? Why or why, not?, Solution, No, these are not independent events., P(novice AND practices 4 times per week) = 0.0667, , (3.0), , P(novice) · P(practices 4 times per week) = 0.0996, , (3.0), , 0.0667 6= 0.0996, , (3.0), , Example 3.9, Studies show that, if she lives to be 90, about 1 woman in 7 (approximately 14.3%) will develop, breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the, time. Also suppose that in the general population of women, the test for breast cancer is negative, about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose, one woman is selected at random., Problem 1, What is the probability that the woman develops breast cancer? What is the probability that, woman tests negative?, Solution, P(B) = 0.143 ; P(N) = 0.85, , Problem 2, Given that the woman has breast cancer, what is the probability that she tests negative?, Solution, P(N|B) = 0.02, , Problem 3, What is the probability that the woman has breast cancer AND tests negative?, Solution, P(B AND N) = P(B) · P(N|B) = (0.143) · (0.02) = 0.0029, , Problem 4, What is the probability that the woman has breast cancer or tests negative?, Solution, P(B OR N) = P(B) + P(N) − P(B AND N) = 0.143 + 0.85 − 0.0029 = 0.9901, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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131, Problem 5, Are having breast cancer and testing negative independent events?, Solution, No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N), , Problem 6, Are having breast cancer and testing negative mutually exclusive?, Solution, No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be 0., , 3.5 Contingency Tables5, A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table, helps in determining conditional probabilities quite easily. The table displays sample values in relation, to two different variables that may be dependent or contingent on one another. Later on, we will use, contingency tables again, but in another manner. Contingincy tables provide a way of portraying data that, can facilitate calculating probabilities., Example 3.10, Suppose a study of speeding violations and drivers who use car phones produced the following, fictional data:, , Speeding violation in, the last year, , No speeding violation, in the last year, , Total, , Car phone user, , 25, , 280, , 305, , Not a car phone user, , 45, , 405, , 450, , Total, , 70, , 685, , 755, , Table 3.1, , The total number of people in the sample is 755. The row totals are 305 and 450. The column totals, are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755., Calculate the following probabilities using the table, Problem 1, P(person is a car phone user) =, Solution, , number of car phone users, total number in study, , =, , 305, 755, , Problem 2, P(person had no violation in the last year) =, 5 This, , content is available online at <http://cnx.org/content/m16835/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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132, , CHAPTER 3. PROBABILITY TOPICS, Solution, , number that had no violation, total number in study, , =, , 685, 755, , Problem 3, P(person had no violation in the last year AND was a car phone user) =, Solution, 280, 755, , Problem 4, P(person is a car phone user OR person had no violation in the last year) =, Solution , 305, 685, 755 + 755 −, , 280, 755, , =, , 710, 755, , Problem 5, P(person is a car phone user GIVEN person had a violation in the last year) =, Solution, 25, 70 (The sample space is reduced to the number of persons who had a violation.), Problem 6, P(person had no violation last year GIVEN person was not a car phone user) =, Solution, 405, 450 (The sample space is reduced to the number of persons who were not car phone users.), , Example 3.11, The following table shows a random sample of 100 hikers and the areas of hiking preferred:, Hiking Area Preference, Sex, , The Coastline, , Near Lakes and Streams, , On Mountain Peaks, , Total, , Female, , 18, , 16, , ___, , 45, , Male, , ___, , ___, , 14, , 55, , Total, , ___, , 41, , ___, , ___, , Table 3.2, , Problem 1, Complete the table., , (Solution on p. 160.), , Problem 2, (Solution on p. 160.), Are the events "being female" and "preferring the coastline" independent events?, Let F = being female and let C = preferring the coastline., a. P(F AND C) =, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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133, b. P ( F ) · P (C ) =, Are these two numbers the same? If they are, then F and C are independent. If they are not, then, F and C are not independent., Problem 3, (Solution on p. 160.), Find the probability that a person is male given that the person prefers hiking near lakes and, streams. Let M = being male and let L = prefers hiking near lakes and streams., a. What word tells you this is a conditional?, b. Fill in the blanks and calculate the probability: P(___|___) = ___., c. Is the sample space for this problem all 100 hikers? If not, what is it?, Problem 4, (Solution on p. 160.), Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being, female and let P = prefers mountain peaks., a., b., c., d., , P(F) =, P(P) =, P(F AND P) =, Therefore, P(F OR P) =, , Example 3.12, Muddy Mouse lives in a cage with 3 doors. If Muddy goes out the first door, the probability that, he gets caught by Alissa the cat is 15 and the probability he is not caught is 45 . If he goes out the, second door, the probability he gets caught by Alissa is 41 and the probability he is not caught is 34 ., The probability that Alissa catches Muddy coming out of the third door is 21 and the probability, she does not catch Muddy is 12 . It is equally likely that Muddy will choose any of the three doors, so the probability of choosing each door is 31 ., Door Choice, Caught or Not, , Door One, , Door Two, , Door Three, , Total, , Caught, , 1, 12, 3, 12, , 1, 6, 1, 6, , ____, , Not Caught, , 1, 15, 4, 15, , Total, , ____, , ____, , ____, , 1, , ____, , Table 3.3, , , 1, = 15, is P(Door One AND Caught)., 3, 4, 1, = 5, 3 is P(Door One AND Not Caught)., , • The first entry, • The entry, , 4, 15, , 1, 15, , Verify the remaining entries., Problem 1, (Solution on p. 160.), Complete the probability contingency table. Calculate the entries for the totals. Verify that the, lower-right corner entry is 1., Problem 2, What is the probability that Alissa does not catch Muddy?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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134, , CHAPTER 3. PROBABILITY TOPICS, Solution, 41, 60, , Problem 3, What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught, by Alissa?, Solution, 9, 19, , NOTE : You could also do this problem by using a probability tree. See the Tree Diagrams (Optional) (Section 3.7) section of this chapter for examples., , 3.6 Venn Diagrams (optional)6, A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box, that represents the sample space S together with circles or ovals. The circles or ovals represent events., Example 3.13, Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance, of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6}, and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:, , Example 3.14, Flip 2 fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = { TT, TH }, and B = { TT, HT }. Therefore, A AND B = { TT }. A OR B = { TH, TT, HT }., The sample space when you flip two fair coins is S = { HH, HT, TH, TT }. The outcome HH is in, neither A nor B. The Venn diagram is as follows:, 6 This, , content is available online at <http://cnx.org/content/m16848/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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135, , Example 3.15, Forty percent of the students at a local college belong to a club and 50% work part time. Five, percent of the students work part time and belong to a club. Draw a Venn diagram showing the, relationships. Let C = student belongs to a club and PT = student works part time., , If a student is selected at random find, , • The probability that the student belongs to a club. P(C) = 0.40., • The probability that the student works part time. P(PT) = 0.50., • The probability that the student belongs to a club AND works part time. P(C AND PT) =, 0.05., • The probability that the student belongs to a club given that the student works part time., P(C|PT) =, , P(C AND PT), 0.05, =, = 0.1, P(PT), 0.50, , (3.0), , • The probability that the student belongs to a club OR works part time., P(C OR PT) = P(C) + P(PT) − P(C AND PT) = 0.40 + 0.50 − 0.05 = 0.85, , (3.0), , 3.7 Tree Diagrams (optional)7, A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of, "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram., 7 This, , content is available online at <http://cnx.org/content/m16846/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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136, , CHAPTER 3. PROBABILITY TOPICS, Example 3.16, In an urn, there are 11 balls. Three balls are red (R) and 8 balls are blue (B). Draw two balls, one, at a time, with replacement. "With replacement" means that you put the first ball back in the urn, before you select the second ball. The tree diagram using frequencies that show all the possible, outcomes follows., , Figure 3.1: Total = 64 + 24 + 24 + 9 = 121, , The first set of branches represents the first draw. The second set of branches represents the second, draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each, blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the 9 RR outcomes can be written as:, R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3, The other outcomes are similar., There are a total of 11 balls in the urn. Draw two balls, one at a time, and with replacement. There, are 11 · 11 = 121 outcomes, the size of the sample space., Problem 1, List the 24 BR outcomes: B1R1, B1R2, B1R3, ..., , (Solution on p. 161.), , Problem 2, Using the tree diagram, calculate P(RR)., Solution, P(RR) =, , 3, 11, , ·, , 3, 11, , =, , 9, 121, , Problem 3, Using the tree diagram, calculate P(RB OR BR)., Solution, P(RB OR BR) =, , 3, 11, , ·, , 8, 11, , +, , 8, 11, , ·, , 3, 11, , =, , 48, 121, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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137, Problem 4, Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw)., Solution, P(R on 1st draw AND B on 2nd draw) = P(RB) =, , 3, 11, , ·, , 8, 11, , =, , 24, 121, , Problem 5, Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw)., Solution, P(R on 2nd draw given B on 1st draw) = P(R on 2nd | B on 1st) =, , 24, 88, , =, , 3, 11, , This problem is a conditional. The sample space has been reduced to those outcomes that already, have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB)., 3, Twenty-four of the 88 possible outcomes are BR. 24, 88 = 11 ., Problem 6, Using the tree diagram, calculate P(BB)., , (Solution on p. 161.), , Problem 7, (Solution on p. 161.), Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw)., Example 3.17, An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time, this time, without replacement from the urn. "Without replacement" means that you do not put the first, ball back before you select the second ball. Below is a tree diagram. The branches are labeled with, probabilities instead of frequencies. The numbers at the ends of the branches are calculated by, 3, 2, 6, multiplying the numbers on the two corresponding branches, for example, 11, · 10, = 110, ., , Figure 3.2: Total =, , 56 + 24 + 24 + 6, 110, , =, , 110, 110, , = 1, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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138, , CHAPTER 3. PROBABILITY TOPICS, NOTE : If you draw a red on the first draw from the 3 red possibilities, there are 2 red left to draw, on the second draw. You do not put back or replace the first ball after you have drawn it. You, draw without replacement, so that on the second draw there are 10 marbles left in the urn., , Calculate the following probabilities using the tree diagram., Problem 1, P(RR) =, Solution, P(RR) =, , 3, 11, , ·, , 2, 10, , 6, 110, , =, , Problem 2, Fill in the blanks:, P(RB OR BR) =, , 3, 11, , (Solution on p. 161.), , ·, , 8, 10, , + (___)(___) =, , 48, 110, , Problem 3, P(R on 2d | B on 1st) =, , (Solution on p. 161.), , Problem 4, Fill in the blanks:, , (Solution on p. 161.), , P(R on 1st and B on 2nd) = P(RB) = (___)(___) =, Problem 5, P(BB) =, , 24, 110, , (Solution on p. 161.), , Problem 6, P(B on 2nd | R on 1st) =, Solution, There are 6 + 24 outcomes that have R on the first draw (6 RR and 24 RB). The 6 and the 24, 24, 6, are frequencies. They are also the numerators of the fractions 110, and 110, . The sample space is no, longer 110 but 6 + 24 = 30. Twenty-four of the 30 outcomes have B on the second draw. The, 24, probability is then 30, . Did you get this answer?, If we are using probabilities, we can label the tree in the following general way., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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139, , •, •, •, •, , P(R|R) here means P(R on 2nd | R on 1st), P(B|R) here means P(B on 2nd | R on 1st), P(R|B) here means P(R on 2nd | B on 1st), P(B|B) here means P(B on 2nd | B on 1st), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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140, , CHAPTER 3. PROBABILITY TOPICS, , 3.8 Summary of Formulas8, Formula 3.1: Complement, If A and A’ are complements then P ( A) + P(A’ ) = 1, Formula 3.2: Addition Rule, P(A OR B) = P(A) + P(B) − P(A AND B), Formula 3.3: Mutually Exclusive, If A and B are mutually exclusive then P(A AND B) = 0 ; so P(A OR B) = P(A) + P(B)., Formula 3.4: Multiplication Rule, , • P(A AND B) = P(B)P(A|B), • P(A AND B) = P(A)P(B|A), Formula 3.5: Independence, If A and B are independent then:, , • P(A|B) = P(A), • P(B|A) = P(B), • P(A AND B) = P(A)P(B), , 8 This, , content is available online at <http://cnx.org/content/m16843/1.5/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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141, , 3.9 Practice 1: Contingency Tables9, 3.9.1 Student Learning Outcomes, • The student will construct and interpret contingency tables., , 3.9.2 Given, An article in the New England Journal of Medicine , reported about a study of smokers in California and, Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the, people smoking at most 10 cigarettes per day, there were 9886 African Americans, 2745 Native Hawaiians,, 12,831 Latinos, 8378 Japanese Americans, and 7650 Whites. Of the people smoking 11-20 cigarettes per, day, there were 6514 African Americans, 3062 Native Hawaiians, 4932 Latinos, 10,680 Japanese Americans,, and 9877 Whites. Of the people smoking 21-30 cigarettes per day, there were 1671 African Americans, 1419, Native Hawaiians, 1406 Latinos, 4715 Japanese Americans, and 6062 Whites. Of the people smoking at least, 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2305 Japanese, Americans, and 3970 Whites. ((Source: http://www.nejm.org/doi/full/10.1056/NEJMoa033250) ), , 3.9.3 Complete the Table, Complete the table below using the data provided., Smoking Levels by Ethnicity, Smoking, Level, , African, American, , Native, Hawaiian, , Latino, , Japanese, Americans, , White, , TOTALS, , 1-10, 11-20, 21-30, 31+, TOTALS, Table 3.4, , 3.9.4 Analyze the Data, Suppose that one person from the study is randomly selected., Exercise 3.9.1, Find the probability that person smoked 11-20 cigarettes per day., , (Solution on p. 161.), , Exercise 3.9.2, Find the probability that person was Latino., , (Solution on p. 161.), , 9 This, , content is available online at <http://cnx.org/content/m16839/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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142, , CHAPTER 3. PROBABILITY TOPICS, , 3.9.5 Discussion Questions, Exercise 3.9.3, (Solution on p. 161.), In words, explain what it means to pick one person from the study and that person is “Japanese, American AND smokes 21-30 cigarettes per day.” Also, find the probability., Exercise 3.9.4, (Solution on p. 161.), In words, explain what it means to pick one person from the study and that person is “Japanese, American OR smokes 21-30 cigarettes per day.” Also, find the probability., Exercise 3.9.5, (Solution on p. 161.), In words, explain what it means to pick one person from the study and that person is “Japanese, American GIVEN that person smokes 21-30 cigarettes per day.” Also, find the probability., Exercise 3.9.6, Prove that smoking level/day and ethnicity are dependent events., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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143, , 3.10 Practice 2: Calculating Probabilities10, 3.10.1 Student Learning Outcomes, • Students will define basic probability terms., • Students will calculate probabilities., • Students will determine whether two events are mutually exclusive or whether two events are independent., NOTE :, , Use probability rules to solve the problems below. Show your work., , 3.10.2 Given, 48% of all Californians registered voters prefer life in prison without parole over the death penalty for, a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life, in prison without parole over the death penalty for a person convicted of first degree murder. (Source:, http://field.com/fieldpollonline/subscribers/Rls2393.pdf )., 37.6% of all Californians are Latino (Source: U.S. Census Bureau)., In this problem, let:, , • C = Cali f ornians (registered voters) pre f erring li f e in prison without parole over the death penalty f or a person con, • L = Latino Cali f ornians, Suppose that one Californian is randomly selected., , 3.10.3 Analyze the Data, Exercise 3.10.1, P (C ) =, , (Solution on p. 161.), , Exercise 3.10.2, P ( L) =, , (Solution on p. 161.), , Exercise 3.10.3, P (C | L) =, Exercise 3.10.4, In words, what is " C | L"?, , (Solution on p. 161.), , Exercise 3.10.5, P ( L AND C ) =, , (Solution on p. 161.), , Exercise 3.10.6, In words, what is “L and C”?, Exercise 3.10.7, Are L and C independent events? Show why or why not., , (Solution on p. 162.), , Exercise 3.10.8, P ( L OR C ) =, , (Solution on p. 162.), , Exercise 3.10.9, In words, what is “L or C”?, Exercise 3.10.10, Are L and C mutually exclusive events? Show why or why not., 10 This, , (Solution on p. 162.), , content is available online at <http://cnx.org/content/m16840/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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144, , CHAPTER 3. PROBABILITY TOPICS, , 3.11 Homework11, Exercise 3.11.1, (Solution on p. 162.), Suppose that you have 8 cards. 5 are green and 3 are yellow. The 5 green cards are numbered, 1, 2, 3, 4, and 5. The 3 yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You, randomly draw one card., , • G = card drawn is green, • E = card drawn is even-numbered, a. List the sample space., b. P ( G ) =, c. P ( G | E) =, d. P ( G AND E) =, e. P ( G OR E) =, f. Are G and E mutually exclusive? Justify your answer numerically., Exercise 3.11.2, Refer to the previous problem. Suppose that this time you randomly draw two cards, one at a, time, and with replacement., , • G1 = first card is green, • G2 = second card is green, a., b., c., d., e., , Draw a tree diagram of the situation., P ( G1 AND G2 ) =, P (at least one green) =, P ( G2 | G1 ) =, Are G2 and G1 independent events? Explain why or why not., , Exercise 3.11.3, (Solution on p. 162.), Refer to the previous problems. Suppose that this time you randomly draw two cards, one at a, time, and without replacement., , • G1 = first card is green, • G2 = second card is green, a., b., c., d., e., , Draw a tree diagram of the situation., P ( G1 AND G2 ) =, P(at least one green) =, P ( G2 | G1 ) =, Are G2 and G1 independent events? Explain why or why not., , Exercise 3.11.4, Roll two fair dice. Each die has 6 faces., a. List the sample space., b. Let A be the event that either a 3 or 4 is rolled first, followed by an even number. Find P ( A)., c. Let B be the event that the sum of the two rolls is at most 7. Find P ( B)., 11 This, , content is available online at <http://cnx.org/content/m16836/1.21/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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145, d. In words, explain what “P ( A| B)” represents. Find P ( A| B)., e. Are A and B mutually exclusive events? Explain your answer in 1 - 3 complete sentences,, including numerical justification., f. Are A and B independent events? Explain your answer in 1 - 3 complete sentences, including, numerical justification., Exercise 3.11.5, (Solution on p. 162.), A special deck of cards has 10 cards. Four are green, three are blue, and three are red. When a, card is picked, the color of it is recorded. An experiment consists of first picking a card and then, tossing a coin., a. List the sample space., b. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss., Find P(A)., c. Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are, the events A and B mutually exclusive? Explain your answer in 1 - 3 complete sentences,, including numerical justification., d. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are, the events A and C mutually exclusive? Explain your answer in 1 - 3 complete sentences,, including numerical justification., Exercise 3.11.6, An experiment consists of first rolling a die and then tossing a coin:, a. List the sample space., b. Let A be the event that either a 3 or 4 is rolled first, followed by landing a head on the coin toss., Find P(A)., c. Let B be the event that a number less than 2 is rolled, followed by landing a head on the coin, toss. Are the events A and B mutually exclusive? Explain your answer in 1 - 3 complete, sentences, including numerical justification., Exercise 3.11.7, (Solution on p. 162.), An experiment consists of tossing a nickel, a dime and a quarter. Of interest is the side the coin, lands on., a. List the sample space., b. Let A be the event that there are at least two tails. Find P(A)., c. Let B be the event that the first and second tosses land on heads. Are the events A and B, mutually exclusive? Explain your answer in 1 - 3 complete sentences, including justification., Exercise 3.11.8, Consider the following scenario:, , • Let P(C) = 0.4, • Let P(D) = 0.5, • Let P(C|D) = 0.6, a., b., c., d., e., , Find P(C AND D) ., Are C and D mutually exclusive? Why or why not?, Are C and D independent events? Why or why not?, Find P(C OR D) ., Find P(D|C)., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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146, , CHAPTER 3. PROBABILITY TOPICS, Exercise 3.11.9, E and F mutually exclusive events. P ( E) = 0.4; P ( F ) = 0.5. Find P ( E | F )., , (Solution on p. 162.), , Exercise 3.11.10, J and K are independent events. P(J | K) = 0.3. Find P ( J ) ., Exercise 3.11.11, U and V are mutually exclusive events. P (U ) = 0.26; P (V ) = 0.37. Find:, , (Solution on p. 162.), , a. P(U AND V) =, b. P(U | V) =, c. P(U OR V) =, Exercise 3.11.12, Q and R are independent events. P ( Q) = 0.4 ; P ( Q AND R) = 0.1 . Find P ( R)., Exercise 3.11.13, Y and Z are independent events., , (Solution on p. 162.), , a. Rewrite the basic Addition Rule P(Y OR Z) = P (Y ) + P ( Z ) − P (Y AND Z ) using the, information that Y and Z are independent events., b. Use the rewritten rule to find P ( Z ) if P (Y OR Z ) = 0.71 and P (Y ) = 0.42 ., Exercise 3.11.14, G and H are mutually exclusive events. P ( G ) = 0.5; P ( H ) = 0.3, a. Explain why the following statement MUST be false: P ( H | G ) = 0.4 ., b. Find: P(H OR G)., c. Are G and H independent or dependent events? Explain in a complete sentence., Exercise 3.11.15, (Solution on p. 162.), The following are real data from Santa Clara County, CA. As of a certain time, there had been, a total of 3059 documented cases of AIDS in the county. They were grouped into the following, categories (Source: Santa Clara County Public H.D.):, Homosexual/Bisexual, , IV Drug User*, , Heterosexual Contact, , Other, , Totals, , Female, , 0, , 70, , 136, , 49, , ____, , Male, , 2146, , 463, , 60, , 135, , ____, , Totals, , ____, , ____, , ____, , ____, , ____, , Table 3.5: * includes homosexual/bisexual IV drug users, , Suppose one of the persons with AIDS in Santa Clara County is randomly selected. Compute the, following:, a. P(person is female) =, b. P(person has a risk factor Heterosexual Contact) =, c. P(person is female OR has a risk factor of IV Drug User) =, d. P(person is female AND has a risk factor of Homosexual/Bisexual) =, e. P(person is male AND has a risk factor of IV Drug User) =, f. P(female GIVEN person got the disease from heterosexual contact) =, g. Construct a Venn Diagram. Make one group females and the other group heterosexual contact., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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147, Exercise 3.11.16, Solve these questions using probability rules. Do NOT use the contingency table above. 3059, cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases, will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact, and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual, contact., a., b., c., d., , P(person is female) =, P(person obtained the disease through heterosexual contact) =, P(female GIVEN person got the disease from heterosexual contact) =, Construct a Venn Diagram. Make one group females and the other group heterosexual contact., Fill in all values as probabilities., , Exercise 3.11.17, (Solution on p. 163.), The following table identifies a group of children by one of four hair colors, and by type of hair., Hair Type, , Brown, , Wavy, , 20, , Straight, , 80, , Totals, , Blond, , Black, , Red, , Totals, , 15, , 3, , 43, , 15, , 12, , 20, , 215, , Table 3.6, , Complete the table above., What is the probability that a randomly selected child will have wavy hair?, What is the probability that a randomly selected child will have either brown or blond hair?, What is the probability that a randomly selected child will have wavy brown hair?, What is the probability that a randomly selected child will have red hair, given that he has, straight hair?, f. If B is the event of a child having brown hair, find the probability of the complement of B., g. In words, what does the complement of B represent?, , a., b., c., d., e., , Exercise 3.11.18, A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys, were published in the San Jose Mercury News. The factual data are compiled into the following, table., Shirt#, , ≤ 210, , 211-250, , 251-290, , 290≤, , 1-33, , 21, , 5, , 0, , 0, , 34-66, , 6, , 18, , 7, , 4, , 66-99, , 6, , 12, , 22, , 5, , Table 3.7, , For the following, suppose that you randomly select one player from the 49ers or Cowboys., a. Find the probability that his shirt number is from 1 to 33., b. Find the probability that he weighs at most 210 pounds., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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148, , CHAPTER 3. PROBABILITY TOPICS, c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds., d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds., e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210, pounds., f. If having a shirt number from 1 to 33 and weighing at most 210 pounds were independent, events, then what should be true about P(Shirt# 1-33 | ≤ 210 pounds)?, Exercise 3.11.19, (Solution on p. 163.), Approximately 281,000,000 people over age 5 live in the United States., Of, these people, 55,000,000 speak a language other than English at home., Of, those who speak another language at home, 62.3% speak Spanish., (Source:, http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf ), Let: E = speak English at home; E’ = speak another language at home; S = speak Spanish;, Finish each probability statement by matching the correct answer., Probability Statements, , Answers, , a. P(E’) =, , i. 0.8043, , b. P(E) =, , ii. 0.623, , c. P(S and E’) =, , iii. 0.1957, , d. P(S|E’) =, , iv. 0.1219, Table 3.8, , Exercise 3.11.20, The probability that a male develops some form of cancer in his lifetime is 0.4567 (Source: American Cancer Society). The probability that a male has at least one false positive test result (meaning, the test comes back for cancer when the man does not have it) is 0.51 (Source: USA Today). Some of, the questions below do not have enough information for you to answer them. Write “not enough, information” for those answers., Let: C = a man develops cancer in his lifetime; P = man has at least one false positive, a., b., c., d., e., , Construct a tree diagram of the situation., P (C ) =, P( P|C ) =, P( P|C’ ) =, If a test comes up positive, based upon numerical values, can you assume that man has cancer?, Justify numerically and explain why or why not., , Exercise 3.11.21, (Solution on p. 163.), In 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens, to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million, people who entered this lottery. Let G = won Green Card., a. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement., b. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen., Once the finalists were chosen, assuming that each finalist had an equal chance to win, what, was Renate’s chance of winning a Green Card? Let F = was a finalist. Write your answer as, a conditional probability statement., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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149, c. Are G and F independent or dependent events? Justify your answer numerically and also, explain why., d. Are G and F mutually exclusive events? Justify your answer numerically and also explain why., NOTE : P.S. Amazingly, on 2/1/95, Renate learned that she would receive her Green Card – true, story!, , Exercise 3.11.22, Three professors at George Washington University did an experiment to determine if economists, are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash, in different classrooms on the George Washington campus. 44% were returned overall. From the, economics classes 56% of the envelopes were returned. From the business, psychology, and history, classes 31% were returned. (Source: Wall Street Journal ), Let: R = money returned; E = economics classes; O = other classes, Write a probability statement for the overall percent of money returned., Write a probability statement for the percent of money returned out of the economics classes., Write a probability statement for the percent of money returned out of the other classes., Is money being returned independent of the class? Justify your answer numerically and explain, it., e. Based upon this study, do you think that economists are more selfish than other people? Explain, why or why not. Include numbers to justify your answer., a., b., c., d., , Exercise 3.11.23, (Solution on p. 163.), The chart below gives the number of suicides estimated in the U.S. for a recent year by age, race, (black and white), and sex. We are interested in possible relationships between age, race, and sex., We will let suicide victims be our population. (Source: The National Center for Health Statistics,, U.S. Dept. of Health and Human Services), Race and Sex, , 1 - 14, , 15 - 24, , 25 - 64, , over 64, , TOTALS, , white, male, , 210, , 3360, , 13,610, , 22,050, , white, female, , 80, , 580, , 3380, , 4930, , black, male, , 10, , 460, , 1060, , 1670, , black, female, , 0, , 40, , 270, , 330, , 310, , 4650, , 18,780, , 29,760, , all others, TOTALS, , Table 3.9, NOTE :, , a., b., c., d., e., , Do not include "all others" for parts (f), (g), and (i)., , Fill in the column for the suicides for individuals over age 64., Fill in the row for all other races., Find the probability that a randomly selected individual was a white male., Find the probability that a randomly selected individual was a black female., Find the probability that a randomly selected individual was black, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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150, , CHAPTER 3. PROBABILITY TOPICS, f. Find the probability that a randomly selected individual was male., g. Out of the individuals over age 64, find the probability that a randomly selected individual was, a black or white male., h. Comparing “Race and Sex” to “Age,” which two groups are mutually exclusive? How do you, know?, i. Are being male and committing suicide over age 64 independent events? How do you know?, , The next two questions refer to the following: The percent of licensed U.S. drivers (from a recent year), that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20 - 64; 13.61% are age, 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20 - 64; 13.53% are, age 65 or over. (Source: Federal Highway Administration, U.S. Dept. of Transportation), Exercise 3.11.24, Complete the following:, a. Construct a table or a tree diagram of the situation., b. P(driver is female) =, c. P(driver is age 65 or over | driver is female) =, d. P(driver is age 65 or over AND female) =, e. In words, explain the difference between the probabilities in part (c) and part (d)., f. P(driver is age 65 or over) =, g. Are being age 65 or over and being female mutually exclusive events? How do you know, Exercise 3.11.25, Suppose that 10,000 U.S. licensed drivers are randomly selected., , (Solution on p. 163.), , a. How many would you expect to be male?, b. Using the table or tree diagram from the previous exercise, construct a contingency table of, gender versus age group., c. Using the contingency table, find the probability that out of the age 20 - 64 group, a randomly, selected driver is female., Exercise 3.11.26, Approximately 86.5% of Americans commute to work by car, truck or van. Out of that group,, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. (Source: Bureau of the Census, U.S. Dept. of Commerce., Disregard rounding approximations.), a. Construct a table or a tree diagram of the situation. Include a branch for all other modes of, transportation to work., b. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?, c. Suppose that 1000 workers are randomly selected. How many would you expect to travel alone, to work?, d. Suppose that 1000 workers are randomly selected. How many would you expect to drive in a, carpool?, Exercise 3.11.27, Explain what is wrong with the following statements. Use complete sentences., a. If there’s a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there’s a, 130% chance of rain over the weekend., b. The probability that a baseball player hits a home run is greater than the probability that he, gets a successful hit., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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151, , 3.11.1 Try these multiple choice questions., The next two questions refer to the following probability tree diagram which shows tossing an unfair, coin FOLLOWED BY drawing one bead from a cup containing 3 red (R), 4 yellow (Y) and 5 blue (B) beads., For the coin, P ( H ) = 23 and P ( T ) = 13 where H = "heads" and T = "tails”., , Figure 3.3, , Exercise 3.11.28, Find P(tossing a Head on the coin AND a Red bead), A., B., C., D., , 2, 3, 5, 15, 6, 36, 5, 36, , Exercise 3.11.29, Find P(Blue bead)., A., B., C., D., , (Solution on p. 164.), , (Solution on p. 164.), , 15, 36, 10, 36, 10, 12, 6, 36, , The next three questions refer to the following table of data obtained from www.baseball-almanac.com12, showing hit information for 4 well known baseball players. Suppose that one hit from the table is randomly, selected., 12 http://cnx.org/content/m16836/latest/www.baseball-almanac.com, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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152, , CHAPTER 3. PROBABILITY TOPICS, , NAME, , Single, , Double, , Triple, , Home Run, , TOTAL HITS, , Babe Ruth, , 1517, , 506, , 136, , 714, , 2873, , Jackie Robinson, , 1054, , 273, , 54, , 137, , 1518, , Ty Cobb, , 3603, , 174, , 295, , 114, , 4189, , Hank Aaron, , 2294, , 624, , 98, , 755, , 3771, , TOTAL, , 8471, , 1577, , 583, , 1720, , 12351, , Table 3.10, , Exercise 3.11.30, Find P(hit was made by Babe Ruth)., A., B., C., D., , 1518, 2873, 2873, 12351, 583, 12351, 4189, 12351, , Exercise 3.11.31, Find P(hit was made by Ty Cobb | The hit was a Home Run), A., B., C., D., , (Solution on p. 164.), , (Solution on p. 164.), , 4189, 12351, 114, 1720, 1720, 4189, 114, 12351, , Exercise 3.11.32, (Solution on p. 164.), Are the hit being made by Hank Aaron and the hit being a double independent events?, A., B., C., D., , Yes, because P(hit by Hank Aaron | hit is a double) = P(hit by Hank Aaron), No, because P(hit by Hank Aaron | hit is a double) 6= P(hit is a double), No, because P(hit is by Hank Aaron | hit is a double) 6= P(hit by Hank Aaron), Yes, because P(hit is by Hank Aaron | hit is a double) = P(hit is a double), , Exercise 3.11.33, Given events G and H: P(G) = 0.43 ; P(H) = 0.26 ; P(H and G) = 0.14, , (Solution on p. 164.), , A. Find P(H or G), B. Find the probability of the complement of event (H and G), C. Find the probability of the complement of event (H or G), Exercise 3.11.34, Given events J and K: P(J) = 0.18 ; P(K) = 0.37 ; P(J or K) = 0.45, , (Solution on p. 164.), , A. Find P(J and K), B. Find the probability of the complement of event (J and K), C. Find the probability of the complement of event (J or K), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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153, Exercise 3.11.35, (Solution on p. 164.), United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, http://www.unitedbloodservices.org/humanbloodtypes.html, a person with, type O blood and a negative Rh factor (Rh−) can donate blood to any person with any bloodtype., Their data show that 43% of people have type O blood and 15% of people have Rh− factor; 52%, of people have type O or Rh− factor., A. Find the probability that a person has both type O blood and the Rh− factor, B. Find the probability that a person does NOT have both type O blood and the Rh− factor., Exercise 3.11.36, (Solution on p. 164.), At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose, that 32% of courses have a research paper and a final exam. Let F be the event that a course has a, final exam. Let R be the event that a course requires a research paper., A. Find the probability that a course has a final exam or a research project., B. Find the probability that a course has NEITHER of these two requirements., Exercise 3.11.37, (Solution on p. 164.), In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain, both chocolate and nuts. Sean is allergic to both chocolate and nuts., A. Find the probability that a cookie contains chocolate or nuts (he can’t eat it)., B. Find the probability that a cookie does not contain chocolate or nuts (he can eat it)., Exercise 3.11.38, (Solution on p. 164.), A college finds that 10% of students have taken a distance learning class and that 40% of students, are part time students. Of the part time students, 20% have taken a distance learning class. Let D, = event that a student takes a distance learning class and E = event that a student is a part time, student, A. Find P(D and E), B. Find P(E | D), C. Find P(D or E), D. Using an appropriate test, show whether D and E are independent., E. Using an appropriate test, show whether D and E are mutually exclusive., Exercise 3.11.39, (Solution on p. 164.), When the Euro coin was introduced in 2002, two math professors had their statistics students test, whether the Belgian 1 Euro coin was a fair coin. They spun the coin rather than tossing it, and it, was found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T)., Therefore, they claim that this is not a fair coin., A. Based on the data above, find P(H) and P(T)., B. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the, coin twice., C. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin., D. Use the tree to find the probability of obtaining at least one head., Exercise 3.11.40, A box of cookies contains 3 chocolate and 7 butter cookies. Miguel randomly selects a cookie and, eats it. Then he randomly selects another cookie and eats it also. (How many cookies did he take?), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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154, , CHAPTER 3. PROBABILITY TOPICS, A. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities, along each branch of the tree., B. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of, his first selection? Explain., C. For each complete path through the tree, write the event it represents and find the probabilities., D. Let S be the event that both cookies selected were the same flavor. Find P(S)., E. Let T be the event that both cookies selected were different flavors. Find P(T) by two different, methods: by using the complement rule and by using the branches of the tree. Your answers, should be the same with both methods., F. Let U be the event that the second cookie selected is a butter cookie. Find P(U)., , **Exercises 33 - 40 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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155, , 3.12 Review13, The first six exercises refer to the following study: In a survey of 100 stocks on NASDAQ, the average, percent increase for the past year was 9% for NASDAQ stocks. Answer the following:, Exercise 3.12.1, The “average increase” for all NASDAQ stocks is the:, , (Solution on p. 165.), , A. Population, B. Statistic, C. Parameter, D. Sample, E. Variable, Exercise 3.12.2, All of the NASDAQ stocks are the:, , (Solution on p. 165.), , A. Population, B. Statistic, C. Parameter, D. Sample, E. Variable, Exercise 3.12.3, 9% is the:, , (Solution on p. 165.), , A. Population, B. Statistic, C. Parameter, D. Sample, E. Variable, Exercise 3.12.4, The 100 NASDAQ stocks in the survey are the:, , (Solution on p. 165.), , A. Population, B. Statistic, C. Parameter, D. Sample, E. Variable, Exercise 3.12.5, The percent increase for one stock in the survey is the:, , (Solution on p. 165.), , A. Population, B. Statistic, C. Parameter, D. Sample, E. Variable, 13 This, , content is available online at <http://cnx.org/content/m16842/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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156, , CHAPTER 3. PROBABILITY TOPICS, , Exercise 3.12.6, (Solution on p. 165.), Would the data collected be qualitative, quantitative – discrete, or quantitative – continuous?, The next two questions refer to the following study: Thirty people spent two weeks around Mardi Gras, in New Orleans. Their two-week weight gain is below. (Note: a loss is shown by a negative weight gain.), Weight Gain, , Frequency, , -2, , 3, , -1, , 5, , 0, , 2, , 1, , 4, , 4, , 13, , 6, , 2, , 11, , 1, Table 3.11, , Exercise 3.12.7, Calculate the following values:, , (Solution on p. 165.), , a. The average weight gain for the two weeks, b. The standard deviation, c. The first, second, and third quartiles, Exercise 3.12.8, Construct a histogram and a boxplot of the data., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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157, , 3.13 Lab: Probability Topics14, Class time:, Names:, , 3.13.1 Student Learning Outcomes:, • The student will use theoretical and empirical methods to estimate probabilities., • The student will appraise the differences between the two estimates., • The student will demonstrate an understanding of long-term relative frequencies., , 3.13.2 Do the Experiment:, Count out 40 mixed-color M&M’s® which is approximately 1 small bag’s worth (distance learning classes, using the virtual lab would want to count out 25 M&M’s®). Record the number of each color in the "Population" table. Use the information from this table to complete the theoretical probability questions. Next,, put the M&M’s in a cup. The experiment is to pick 2 M&M’s, one at a time. Do not look at them as you, pick them. The first time through, replace the first M&M before picking the second one. Record the results, in the “With Replacement” column of the empirical table. Do this 24 times. The second time through, after, picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record, the results in the “Without Replacement” column section of the "Empirical Results" table. After you record, the pick, put both M&M’s back. Do this a total of 24 times, also. Use the data from the "Empirical Results", table to calculate the empirical probability questions. Leave your answers in unreduced fractional form., Do not multiply out any fractions., Population, Color, , Quantity, , Yellow (Y), Green (G), Blue (BL), Brown (B), Orange (O), Red (R), Table 3.12, 14 This, , content is available online at <http://cnx.org/content/m16841/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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159, Table 3.15: Note:, , 3.13.3 Discussion Questions, 1. Why are the “With Replacement” and “Without Replacement” probabilities different?, 2. Convert P(no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical, “With Replacement”. Round to 4 decimal places., a. Theoretical “With Replacement”: P(no yellows) =, b. Empirical “With Replacement”: P(no yellows) =, c. Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why?, 3. If you increased the number of times you picked 2 M&M’s to 240 times, why would empirical probability values change?, 4. Would this change (see (3) above) cause the empirical probabilities and theoretical probabilities to be, closer together or farther apart? How do you know?, 5. Explain the differences in what P ( G1 AND R2 ) and P ( R1 | G2 ) represent. Hint: Think about the, sample space for each probability., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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160, , CHAPTER 3. PROBABILITY TOPICS, , Solutions to Exercises in Chapter 3, Solution to Exercise 3.2.1 (p. 124), a. P(L’)=P(S), b. P(M or S), c. P(F and L), d. P(M|L), e. P(L|M), f. P(S|F), g. P(F|L), h. P(F or L), i. P(M and S), j. P(F), Solution to Example 3.2, Problem (p. 126), No. C = {3, 5} and E = {1, 2, 3, 4}. P (C AND E) = 61 . To be mutually exclusive, P (C AND E) must be, 0., Solution to Example 3.11, Problem 1 (p. 132), Hiking Area Preference, Sex, , The Coastline, , Near Lakes and Streams, , On Mountain Peaks, , Total, , Female, , 18, , 16, , 11, , 45, , Male, , 16, , 25, , 14, , 55, , Total, , 34, , 41, , 25, , 100, , Table 3.16, , Solution to Example 3.11, Problem 2 (p. 132), 18, a. P(F AND C) = 100, = 0.18, 45, 34, b. P ( F ) · P (C ) = 100 · 100, = 0.45 · 0.34 = 0.153, , P(F AND C) 6= P ( F ) · P (C ), so the events F and C are not independent., Solution to Example 3.11, Problem 3 (p. 133), a. The word ’given’ tells you that this is a conditional., 25, b. P(M|L) = 41, c. No, the sample space for this problem is 41., Solution to Example 3.11, Problem 4 (p. 133), a., b., c., d., , 45, P(F) = 100, 25, P(P) = 100, 11, P(F AND P) = 100, 45, P(F OR P) = 100, +, , 25, 100, , −, , 11, 100, , =, , 59, 100, , Solution to Example 3.12, Problem 1 (p. 133), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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161, Door Choice, Caught or Not, , Door One, , Door Two, , Door Three, , Total, , Caught, , 1, 15, 4, 15, 5, 15, , 1, 12, 3, 12, 4, 12, , 1, 6, 1, 6, 2, 6, , 19, 60, 41, 60, , Not Caught, Total, , 1, , Table 3.17, , Solution to Example 3.16, Problem 1 (p. 136), B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1;, B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3, Solution to Example 3.16, Problem 6 (p. 137), 64, P(BB) = 121, Solution to Example 3.16, Problem 7 (p. 137), 8, P(B on 2nd draw | R on 1st draw) = 11, There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then, 9 + 24 = 33. Twenty-four of the 33 outcomes have B on the second draw. The probability is then 24, 33 ., Solution to Example 3.17, Problem, 2, (p., 138), 3, 3, 8, 8, 48, P(RB or BR) = 11, · 10, + 11, 10 = 110, Solution to Example 3.17, Problem 3 (p. 138), 3, P(R on 2d | B on 1st) = 10, Solution to Example 3.17, Problem 4 (p. 138), , 3, 8, 24, P(R on 1st and B on 2nd) = P(RB) =, = 110, 11, 10, Solution to Example 3.17, Problem 5 (p. 138), 8, 7, P(BB) = 11, · 10, , Solutions to Practice 1: Contingency Tables, Solution to Exercise 3.9.1 (p. 141), 35,065, 100,450, , Solution to Exercise 3.9.2 (p. 141), 19,969, 100,450, , Solution to Exercise 3.9.3 (p. 142), 4,715, 100,450, , Solution to Exercise 3.9.4 (p. 142), 36,636, 100,450, , Solution to Exercise 3.9.5 (p. 142), 4715, 15,273, , Solutions to Practice 2: Calculating Probabilities, Solution to Exercise 3.10.1 (p. 143), 0.48, Solution to Exercise 3.10.2 (p. 143), 0.376, Solution to Exercise 3.10.3 (p. 143), 0.55, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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162, , CHAPTER 3. PROBABILITY TOPICS, , Solution to Exercise 3.10.5 (p. 143), 0.2068, Solution to Exercise 3.10.7 (p. 143), No, Solution to Exercise 3.10.8 (p. 143), 0.6492, Solution to Exercise 3.10.10 (p. 143), No, , Solutions to Homework, Solution to Exercise 3.11.1 (p. 144), a. { G1, G2, G3, G4, G5, Y1, Y2, Y3}, b. 58, c. 23, d. 82, e. 68, f. No, Solution to Exercise 3.11.3 (p. 144), 4, b. 85, 7, 3, , 4, 5, c. 8 7 + 38 57 + 58, 7, d. 74, e. No, Solution to Exercise 3.11.5 (p. 145), a. {GH, GT, BH, BT, RH, RT}, 3, b. 20, c. Yes, d. No, Solution to Exercise 3.11.7 (p. 145), a. {(HHH) , (HHT) , (HTH) , (HTT) , (THH) , (THT) , (TTH) , (TTT)}, b. 84, c. Yes, Solution to Exercise 3.11.9 (p. 146), 0, Solution to Exercise 3.11.11 (p. 146), a. 0, b. 0, c. 0.63, Solution to Exercise 3.11.13 (p. 146), b. 0.5, Solution to Exercise 3.11.15 (p. 146), The completed contingency table is as follows:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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163, , Homosexual/Bisexual, , IV Drug User*, , Heterosexual Contact, , Other, , Totals, , Female, , 0, , 70, , 136, , 49, , 255, , Male, , 2146, , 463, , 60, , 135, , 2804, , Totals, , 2146, , 533, , 196, , 184, , 3059, , Table 3.18: * includes homosexual/bisexual IV drug users, 255, a. 3059, 196, b. 3059, 718, c. 3059, d. 0, 463, e. 3059, 136, f. 196, , Solution to Exercise 3.11.17 (p. 147), b., c., d., e., f., , 43, 215, 120, 215, 20, 215, 12, 172, 115, 215, , Solution to Exercise 3.11.19 (p. 148), a., b., c., d., , iii, i, iv, ii, , Solution to Exercise 3.11.21 (p. 148), a., b., c., d., , P ( G ) = 0.008, 0.5, dependent, No, , Solution to Exercise 3.11.23 (p. 149), c. 22050, 29760, 330, d. 29760, 2000, e. 29760, 23720, f. 29760, g. 5010, 6020, h. Black females and ages 1-14, i. No, Solution to Exercise 3.11.25 (p. 150), a. 5140, c. 0.49, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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164, , CHAPTER 3. PROBABILITY TOPICS, , Solution to Exercise 3.11.28 (p., C, Solution to Exercise 3.11.29 (p., A, Solution to Exercise 3.11.30 (p., B, Solution to Exercise 3.11.31 (p., B, Solution to Exercise 3.11.32 (p., C, Solution to Exercise 3.11.33 (p., , 151), 151), 152), 152), 152), 152), , A. P(H or G) = P(H) + P(G) − P(H and G) = 0.26 + 0.43 − 0.14 = 0.55, B. P( NOT (H and G) ) = 1 − P(H and G) = 1 − 0.14 = 0.86, C. P( NOT (H or G) ) = 1 − P(H or G) = 1 − 0.55 = 0.45, Solution to Exercise 3.11.34 (p. 152), A. P(J or K) = P(J) + P(K) − P(J and K); 0.45 = 0.18 + 0.37 − P(J and K) ; solve to find P(J and K) = 0.10, B. P( NOT (J and K) ) = 1 − P(J and K) = 1 − 0.10 = 0.90, C. P( NOT (J or K) ) = 1 − P(J or K) = 1 − 0.45 = 0.55, Solution to Exercise 3.11.35 (p. 153), A. P(Type O or Rh−) = P(Type O) + P(Rh−) − P(Type O and Rh−), 0.52 = 0.43 + 0.15 − P(Type O and Rh−); solve to find P(Type O and Rh−) = 0.06, 6% of people have type O Rh− blood, B. P( NOT (Type O and Rh−) ) = 1 − P(Type O and Rh−) = 1 − 0.06 = 0.94, 94% of people do not have type O Rh− blood, Solution to Exercise 3.11.36 (p. 153), A. P(R or F) = P(R) + P(F) − P(R and F) = 0.72 + 0.46 − 0.32 = 0.86, B. P( Neither R nor F ) = 1 − P(R or F) = 1 − 0.86 = 0.14, Solution to Exercise 3.11.37 (p. 153), Let C be the event that the cookie contains chocolate. Let N be the event that the cookie contains nuts., A. P(C or N) = P(C) + P(N) − P(C and N) = 0.36 + 0.12 − 0.08 = 0.40, B. P( neither chocolate nor nuts) = 1 − P(C or N) = 1 − 0.40 = 0.60, Solution to Exercise 3.11.38 (p. 153), A. P(D and E) = P(D|E)P(E) = (0.20)(0.40) = 0.08, B. P(E|D) = P(D and E) / P(D) = 0.08/0.10 = 0.80, C. P(D or E) = P(D) + P(E) − P(D and E) = 0.10 + 0.40 − 0.08 = 0.42, D. Not Independent: P(D|E) = 0.20 which does not equal P(D) = .10, E. Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we would need to have, P(D and E) = 0, which is not true here., Solution to Exercise 3.11.39 (p. 153), A. P(H) = 140/250; P(T) = 110/250, C. 308/625, D. 504/625, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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165, , Solutions to Review, Solution to Exercise 3.12.1 (p. 155), C. Parameter, Solution to Exercise 3.12.2 (p. 155), A. Population, Solution to Exercise 3.12.3 (p. 155), B. Statistic, Solution to Exercise 3.12.4 (p. 155), D. Sample, Solution to Exercise 3.12.5 (p. 155), E. Variable, Solution to Exercise 3.12.6 (p. 156), quantitative - continuous, Solution to Exercise 3.12.7 (p. 156), a. 2.27, b. 3.04, c. -1, 4, 4, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 4, , Discrete Random Variables, 4.1 Discrete Random Variables1, 4.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, •, •, •, , Recognize and understand discrete probability distribution functions, in general., Calculate and interpret expected values., Recognize the binomial probability distribution and apply it appropriately., Recognize the Poisson probability distribution and apply it appropriately (optional)., Recognize the geometric probability distribution and apply it appropriately (optional)., Recognize the hypergeometric probability distribution and apply it appropriately (optional)., Classify discrete word problems by their distributions., , 4.1.2 Introduction, A student takes a 10 question true-false quiz. Because the student had such a busy schedule, he or she, could not study and randomly guesses at each answer. What is the probability of the student passing the, test with at least a 70%?, Small companies might be interested in the number of long distance phone calls their employees make, during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees, make more than 20 long distance phone calls during the peak time?, These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes, of a statistical experiment in words. The values of a random variable can vary with each repetition of an, experiment., In this chapter, you will study probability problems involving discrete random distributions. You will also, study long-term averages associated with them., , 4.1.3 Random Variable Notation, Upper case letters like X or Y denote a random variable. Lower case letters like x or y denote the value of a, random variable. If X is a random variable, then X is written in words. and x is given as a number., 1 This, , content is available online at <http://cnx.org/content/m16825/1.14/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 167
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168, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , For example, let X = the number of heads you get when you toss three fair coins. The sample space for the, toss of three fair coins is TTT; THH; HTH; HHT; HTT; THT; TTH; HHH. Then, x = 0, 1, 2, 3. X is in, words and x is a number. Notice that for this example, the x values are countable outcomes. Because you, can count the possible values that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is, a discrete random variable., , 4.1.4 Optional Collaborative Classroom Activity, Toss a coin 10 times and record the number of heads. After all members of the class have completed the, experiment (tossed a coin 10 times and counted the number of heads), fill in the chart using a heading like, the one below. Let X = the number of heads in 10 tosses of the coin., x, , Frequency of x, , Relative Frequency of x, , Table 4.1, , • Which value(s) of x occurred most frequently?, • If you tossed the coin 1,000 times, what values could x take on? Which value(s) of x do you think, would occur most frequently?, • What does the relative frequency column sum to?, , 4.2 Probability Distribution Function (PDF) for a Discrete Random, Variable2, A discrete probability distribution function has two characteristics:, , • Each probability is between 0 and 1, inclusive., • The sum of the probabilities is 1., Example 4.1, A child psychologist is interested in the number of times a newborn baby’s crying wakes its mother, after midnight. For a random sample of 50 mothers, the following information was obtained. Let, X = the number of times a newborn wakes its mother after midnight. For this example, x = 0, 1, 2,, 3, 4, 5., P(x) = probability that X takes on a value x., 2 This, , content is available online at <http://cnx.org/content/m16831/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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169, , x, , P(x), , 0, , P(x=0) =, , 1, , P(x=1) =, , 2, , P(x=2) =, , 3, , P(x=3) =, , 4, , P(x=4) =, , 5, , P(x=5) =, , 2, 50, 11, 50, 23, 50, 9, 50, 4, 50, 1, 50, , Table 4.2, , X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because, 1. Each P(x) is between 0 and 1, inclusive., 2. The sum of the probabilities is 1, that is,, 2, 11 23, 9, 4, 1, +, +, +, +, +, =1, 50 50 50 50 50 50, , (4.0), , Example 4.2, Suppose Nancy has classes 3 days a week. She attends classes 3 days a week 80% of the time, 2, days 15% of the time, 1 day 4% of the time, and no days 1% of the time. Suppose one week is, randomly selected., Problem 1, Let X = the number of days Nancy ____________________ ., , (Solution on p. 213.), , Problem 2, X takes on what values?, , (Solution on p. 213.), , Problem 3, (Solution on p. 213.), Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF, table) like the one in the previous example. The table should have two columns labeled x and P(x)., What does the P(x) column sum to?, , 4.3 Mean or Expected Value and Standard Deviation3, The expected value is often referred to as the "long-term"average or mean . This means that over the long, term of doing an experiment over and over, you would expect this average., The mean of a random variable X is µ. If we do an experiment many times (for instance, flip a fair coin, as, Karl Pearson did, 24,000 times and let X = the number of heads) and record the value of X each time, the, average is likely to get closer and closer to µ as we keep repeating the experiment. This is known as the, Law of Large Numbers., NOTE : To find the expected value or long term average, µ, simply multiply each value of the, random variable by its probability and add the products., 3 This, , content is available online at <http://cnx.org/content/m16828/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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170, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , A Step-by-Step Example, A men’s soccer team plays soccer 0, 1, or 2 days a week. The probability that they play 0 days is 0.2, the, probability that they play 1 day is 0.5, and the probability that they play 2 days is 0.3. Find the long-term, average, µ, or expected value of the days per week the men’s soccer team plays soccer., To do the problem, first let the random variable X = the number of days the men’s soccer team plays soccer, per week. X takes on the values 0, 1, 2. Construct a PDF table, adding a column xP ( x ). In this column,, you will multiply each x value by its probability., Expected Value Table, x, , P(x), , xP(x), , 0, , 0.2, , (0)(0.2) = 0, , 1, , 0.5, , (1)(0.5) = 0.5, , 2, , 0.3, , (2)(0.3) = 0.6, , Table 4.4: This table is called an expected value table. The table helps you calculate the expected value or, , long-term average., Add, the, last, column, to, find, the, long, (0) (0.2) + (1) (0.5) + (2) (0.3) = 0 + 0.5 + 0.6 = 1.1., , term, , average, , or, , expected, , value:, , The expected value is 1.1. The men’s soccer team would, on the average, expect to play soccer 1.1 days, per week. The number 1.1 is the long term average or expected value if the men’s soccer team plays soccer, week after week after week. We say µ = 1.1, Example 4.3, Find the expected value for the example about the number of times a newborn baby’s crying, wakes its mother after midnight. The expected value is the expected number of times a newborn, wakes its mother after midnight., x, , xP(X), , P(X), , 0, , P(x=0), , 1, , P(x=1), , 2, , P(x=2), , 3, , P(x=3), , 4, , P(x=4), , 5, , P(x=5), , 2, = 50, 11, = 50, 23, = 50, 9, = 50, 4, = 50, 1, = 50, , 2, =0, 50 , 11, (1) 11, 50 = 50, , 46, (2) 23, 50 = 50, , 9, = 27, (3) 50, 50, 4, (4) 50, = 16, 50, 5, 1, (5) 50, = 50, , (0), , , , Table 4.5: You expect a newborn to wake its mother after midnight 2.1 times, on the average., , Add the last column to find the expected value. µ = Expected Value =, , 105, 50, , = 2.1, , Problem, Go back and calculate the expected value for the number of days Nancy attends classes a week., Construct the third column to do so., Solution, 2.74 days a week., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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171, , Example 4.4, Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8,, 9. A computer randomly selects five numbers from 0 to 9 with replacement. You pay $2 to play, and could profit $100,000 if you match all 5 numbers in order (you get your $2 back plus $100,000)., Over the long term, what is your expected profit of playing the game?, To do this problem, set up an expected value table for the amount of money you can profit., Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you, are interested in your profit (or loss), the values of x are 100,000 dollars and -2 dollars., To win, you must get all 5 numbers correct, in order. The probability of choosing one correct, 1, because there are 10 numbers. You may choose a number more than once. The, number is 10, probability of choosing all 5 numbers correctly and in order is:, 1, 1, 1, 1, 1, ∗, ∗, ∗, ∗ ∗ = 1 ∗ 10−5 = 0.00001, 10 10 10 10 10, Therefore, the probability of winning is 0.00001 and the probability of losing is, 1 − 0.00001 = 0.99999, , (4.0), , The expected value table is as follows., x, , P(x), , xP(x), , Loss, , -2, , 0.99999, , (-2)(0.99999)=-1.99998, , Profit, , 100,000, , 0.00001, , (100000)(0.00001)=1, , Table 4.6: Add the last column. -1.99998 + 1 = -0.99998, , Since −0.99998 is about −1, you would, on the average, expect to lose approximately one dollar, for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1, is the average or expected LOSS per game after playing this game over and over., Example 4.5, Suppose you play a game with a biased coin. You play each game by tossing the coin once., P(heads) = 23 and P(tails) = 13 . If you toss a head, you pay $6. If you toss a tail, you win $10., If you play this game many times, will you come out ahead?, Problem 1, Define a random variable X., , (Solution on p. 213.), , Problem 2, Complete the following expected value table., , (Solution on p. 213.), , x, , ____, , ____, , WIN, , 10, , 1, 3, , ____, , LOSE, , ____, , ____, , −12, 3, , (4.0), , Table 4.7, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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172, , CHAPTER 4. DISCRETE RANDOM VARIABLES, Problem 3, What is the expected value, µ? Do you come out ahead?, , (Solution on p. 213.), , Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a, probability distribution, find each deviation from its expected value, square it, multiply it by its probability,, add the products, and take the square root . To understand how to do the calculation, look at the table for, the number of days per week a men’s soccer team plays soccer. To find the standard deviation, add the, entries in the column labeled ( x − µ)2 · P ( x ) and take the square root., x, , P(x), , xP(x), , (x -µ)2 P(x), , 0, , 0.2, , (0)(0.2) = 0, , (0 − 1.1)2 (.2) = 0.242, , 1, , 0.5, , (1)(0.5) = 0.5, , (1 − 1.1)2 (.5) = 0.005, , 2, , 0.3, , (2)(0.3) = 0.6, , (2 − 1.1)2 (.3) = 0.243, , Table 4.8, , Add the last√, column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root, of 0.49. σ = 0.49 = 0.7, Generally for probability distributions, we use a calculator or a computer to calculate µ and σ to reduce, roundoff error. For some probability distributions, there are short-cut formulas that calculate µ and σ., , 4.4 Common Discrete Probability Distribution Functions4, Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and, Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions., A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics, enables you to distinguish among the different distributions., , 4.5 Binomial5, The characteristics of a binomial experiment are:, 1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes, the number of trials., 2. There are only 2 possible outcomes, called "success" and, "failure" for each trial. The letter p denotes, the probability of a success on one trial and q denotes the probability of a failure on one trial. p + q = 1., 3. The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another, way of saying this is that for each individual trial, the probability, p, of a success and probability, q,, of a failure remain the same. For example, randomly guessing at a true - false statistics question has, only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose, 4 This, 5 This, , content is available online at <http://cnx.org/content/m16821/1.6/>., content is available online at <http://cnx.org/content/m16820/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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173, Joe always guesses correctly on any statistics true - false question with probability p = 0.6. Then,, q = 0.4 .This means that for every true - false statistics question Joe answers, his probability of success, (p = 0.6) and his probability of failure (q = 0.4) remain the same., The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X =, the number of successes obtained in the n independent trials., The mean, µ, and variance, σ2 , for the binomial probability distribution is µ = np and σ2 = npq. The, √, standard deviation, σ, is then σ = npq., Any experiment that has characteristics 2 and 3 and where n = 1 is called a Bernoulli Trial (named after, Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when, the number of successes is counted in one or more Bernoulli Trials., Example 4.6, At ABC College, the withdrawal rate from an elementary physics course is 30% for any given, term. This implies that, for any given term, 70% of the students stay in the class for the entire, term. A "success" could be defined as an individual who withdrew. The random variable is X =, the number of students who withdraw from the randomly selected elementary physics class., Example 4.7, Suppose you play a game that you can only either win or lose. The probability that you win any, game is 55% and the probability that you lose is 45%. Each game you play is independent. If you, play the game 20 times, what is the probability that you win 15 of the 20 games? Here, if you, define X = the number of wins, then X takes on the values 0, 1, 2, 3, ..., 20. The probability of a, success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The, probability question can be stated mathematically as P ( x = 15)., Example 4.8, A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more, than 10 heads? Let X = the number of heads in 15 flips of the fair coin. X takes on the values 0, 1,, 2, 3, ..., 15. Since the coin is fair, p = 0.5 and q = 0.5. The number of trials is n = 15. The probability, question can be stated mathematically as P ( x > 10)., Example 4.9, Approximately 70% of statistics students do their homework in time for it to be collected and, graded. Each student does homework independently. In a statistics class of 50 students, what is, the probability that at least 40 will do their homework on time? Students are selected randomly., Problem 1, (Solution on p. 213.), This is a binomial problem because there is only a success or a __________, there are a definite, number of trials, and the probability of a success is 0.70 for each trial., Problem 2, (Solution on p. 213.), If we are interested in the number of students who do their homework, then how do we define, X?, Problem 3, What values does x take on?, , (Solution on p. 213.), , Problem 4, What is a "failure", in words?, , (Solution on p. 213.), , The probability of a success is p = 0.70. The number of trial is n = 50., Problem 5, If p + q = 1, then what is q?, , (Solution on p. 213.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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174, , CHAPTER 4. DISCRETE RANDOM VARIABLES, Problem 6, (Solution on p. 213.), The words "at least" translate as what kind of inequality for the probability question P ( x____40)., The probability question is P ( x ≥ 40)., , 4.5.1 Notation for the Binomial: B = Binomial Probability Distribution Function, X ∼ B (n, p), Read this as "X is a random variable with a binomial distribution." The parameters are n and p. n = number, of trials p = probability of a success on each trial, Example 4.10, It has been stated that about 41% of adult workers have a high school diploma but do not pursue, any further education. If 20 adult workers are randomly selected, find the probability that at most, 12 of them have a high school diploma but do not pursue any further education. How many adult, workers do you expect to have a high school diploma but do not pursue any further education?, Let X = the number of workers who have a high school diploma but do not pursue any further, education., X takes on the values 0, 1, 2, ..., 20 where n = 20 and p = 0.41. q = 1 - 0.41 = 0.59. X ∼ B (20, 0.41), Find P ( x ≤ 12) . P ( x ≤ 12) = 0.9738. (calculator or computer), Using the TI-83+ or the TI-84 calculators, the calculations are as follows. Go into 2nd DISTR. The, syntax for the instructions are, To calculate (x = value): binompdf(n, p, number) If "number" is left out, the result is the binomial, probability table., To calculate P ( x ≤ value): binomcdf(n, p, number) If "number" is left out, the result is the cumulative binomial probability table., For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter, 20,.41,12). The result is P ( x ≤ 12) = 0.9738., NOTE : If you want to find P ( x = 12), use the pdf (binompdf). If you want to find P(x>12), use 1 binomcdf(20,.41,12)., , The probability at most 12 workers have a high school diploma but do not pursue any further, education is 0.9738, The graph of x ∼ B (20, 0.41) is:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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175, , The y-axis contains the probability of x, where X = the number of workers who have only a high, school diploma., The number of adult workers that you expect to have a high school diploma but not pursue any, further education is the mean, µ = np = (20) (0.41) = 8.2., √, The formula for the variance is σ2 = npq. The standard deviation is σ =, npq. σ =, p, (20) (0.41) (0.59) = 2.20., Example 4.11, The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of 10 staff members and, 6 students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? All names of the committee are put, into a box and two names are drawn without replacement. The first name drawn determines the, chairperson and the second name the recorder. There are two trials. However, the trials are not, independent because the outcome of the first trial affects the outcome of the second trial. The, 6, probability of a student on the first draw is 16, . The probability of a student on the second draw, 5, 6, is 15 , when the first draw produces a student. The probability is 15, when the first draw produces, a staff member. The probability of drawing a student’s name changes for each of the trials and,, therefore, violates the condition of independence., , 4.6 Geometric (optional)6, The characteristics of a geometric experiment are:, 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other, words, you keep repeating what you are doing until the first success. Then you stop. For example,, you throw a dart at a bull’s eye until you hit the bull’s eye. The first time you hit the bull’s eye is a, "success" so you stop throwing the dart. It might take you 6 tries until you hit the bull’s eye. You can, think of the trials as failure, failure, failure, failure, failure, success. STOP., 2. In theory, the number of trials could go on forever. There must be at least one trial., 3. The probability,p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1, and q = 1 − p. For example, the probability of rolling a 3 when you throw one fair die is 61 . This is, 6 This, , content is available online at <http://cnx.org/content/m16822/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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176, , CHAPTER 4. DISCRETE RANDOM VARIABLES, true no matter how many times you roll the die. Suppose you want to know the probability of getting, the first 3 on the fifth roll. On rolls 1, 2, 3, and 4, you do not get a face with a 3. The probability for, each of rolls 1, 2, 3, and 4 is q = 56 , the probability of a failure. The probability of getting a 3 on the, fifth roll is 56 · 65 · 56 · 56 · 61 = 0.0804, , X = the number of independent trials until the first success. The mean and variance are in the summary in, this chapter., Example 4.12, You play a game of chance that you can either win or lose (there are no other possibilities) until, you lose. Your probability of losing is p = 0.57. What is the probability that it takes 5 games until, you lose? Let X = the number of games you play until you lose (includes the losing game). Then, X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P ( x = 5)., Example 4.13, A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of, employees to follow instructions. She decides to look at the accident reports (selected randomly, and replaced in the pile after reading) until she finds one that shows an accident caused by failure, of employees to follow instructions. On the average, how many reports would the safety engineer, expect to look at until she finds a report showing an accident caused by employee failure to follow, instructions? What is the probability that the safety engineer will have to examine at least 3 reports, until she finds a report showing an accident caused by employee failure to follow instructions?, Let X = the number of accidents the safety engineer must examine until she finds a report showing, an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The, first question asks you to find the expected value or the mean. The second question asks you to, find P ( x ≥ 3). ("At least" translates as a "greater than or equal to" symbol)., Example 4.14, Suppose that you are looking for a student at your college who lives within five miles of you. You, know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he/she lives within five miles of you. What is the probability, that you need to contact four people?, This is a geometric problem because you may have a number of failures before you have the one, success you desire. Also, the probability of a success stays the same each time you ask a student if, he/she lives within five miles of you. There is no definite number of trials (number of times you, ask a student)., Problem 1, Let X = the number of ____________ you must ask ____________ one says yes., Solution, Let X = the number of students you must ask until one says yes., , Problem 2, What values does X take on?, , (Solution on p. 213.), , Problem 3, What are p and q?, , (Solution on p. 213.), , Problem 4, The probability question is P(_______)., , (Solution on p. 214.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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177, , 4.6.1 Notation for the Geometric: G = Geometric Probability Distribution Function, X ∼ G ( p), Read this as "X is a random variable with a geometric distribution." The parameter is p. p = the probability, of a success for each trial., Example 4.15, Assume that the probability of a defective computer component is 0.02. Components are randomly, selected. Find the probability that the first defect is caused by the 7th component tested. How, many components do you expect to test until one is found to be defective?, Let X = the number of computer components tested until the first defect is found., X takes on the values 1, 2, 3, ... where p = 0.02. X ∼ G(0.02), Find P ( x = 7). P ( x = 7) = 0.0177. (calculator or computer), TI-83+ and TI-84: For a general discussion, see this example (binomial). The syntax is similar., The geometric parameter list is (p, number) If "number" is left out, the result is the geometric, probability table. For this problem: After you are in 2nd DISTR, arrow down to D:geometpdf., Press ENTER. Enter .02,7). The result is P ( x = 7) = 0.0177., The probability that the 7th component is the first defect is 0.0177., The graph of X ∼ G(0.02) is:, , The y-axis contains the probability of x, where X = the number of computer components tested., The number of components that you would expect to test until you find the first defective one is, the mean, µ = 50., The formula for the mean is µ =, , 1, p, , The formula for the variance is σ2 =, r, The standard deviation is σ =, , 1, p, , 1, 0.02, , =, , ·, , 1, p, , , , 1, p, , ·, , , , = 50, 1, p, , , −1 =, , 1, 0.02, , ·, , , , 1, 0.02, , , − 1 = 2450, , r, , , 1, 1, − 1 = 0.02, · 0.02, − 1 = 49.5, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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178, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.7 Hypergeometric (optional)7, The characteristics of a hypergeometric experiment are:, 1. You take samples from 2 groups., 2. You are concerned with a group of interest, called the first group., 3. You sample without replacement from the combined groups. For example, you want to choose a, softball team from a combined group of 11 men and 13 women. The team consists of 10 players., 4. Each pick is not independent, since sampling is without replacement. In the softball example, the, 11, probability of picking a women first is 13, 24 . The probability of picking a man second is 23 if a woman, 10, was picked first. It is 23 if a man was picked first. The probability of the second pick depends on what, happened in the first pick., 5. You are not dealing with Bernoulli Trials., The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random, variable X = the number of items from the group of interest. The mean and variance are given in the, summary., Example 4.16, A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What, is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops., Since the probability question asks for the probability of picking gumdrops, the group of interest, (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the, second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let X = the number, of gumdrops in the sample of 50. X takes on the values x = 0, 1, 2, ..., 50. The probability question, is P ( x = 35)., Example 4.17, Suppose a shipment of 100 VCRs is known to have 10 defective VCRs. An inspector randomly, chooses 12 for inspection. He is interested in determining the probability that, among the 12, at, most 2 are defective. The two groups are the 90 non-defective VCRs and the 10 defective VCRs., The group of interest (first group) is the defective group because the probability question asks for, the probability of at most 2 defective VCRs. The size of the sample is 12 VCRs. (They may be, non-defective or defective.) Let X = the number of defective VCRs in the sample of 12. X takes on, the values 0, 1, 2, ..., 10. X may not take on the values 11 or 12. The sample size is 12, but there, are only 10 defective VCRs. The inspector wants to know P ( x ≤ 2) ("At most" means "less than or, equal to")., Example 4.18, You are president of an on-campus special events organization. You need a committee of 7 to plan, a special birthday party for the president of the college. Your organization consists of 18 women, and 15 men. You are interested in the number of men on your committee. If the members of the, committee are randomly selected, what is the probability that your committee has more than 4, men?, This is a hypergeometric problem because you are choosing your committee from two groups, (men and women)., Problem 1, Are you choosing with or without replacement?, , (Solution on p. 214.), , Problem 2, What is the group of interest?, , (Solution on p. 214.), , 7 This, , content is available online at <http://cnx.org/content/m16824/1.16/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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179, Problem 3, How many are in the group of interest?, , (Solution on p. 214.), , Problem 4, How many are in the other group?, , (Solution on p. 214.), , Problem 5, Let X = _________ on the committee. What values does X take on?, , (Solution on p. 214.), , Problem 6, The probability question is P(_______)., , (Solution on p. 214.), , 4.7.1 Notation for the Hypergeometric: H = Hypergeometric Probability Distribution, Function, X ∼ H (r, b, n), Read this as "X is a random variable with a hypergeometric distribution." The parameters are r, b, and n. r, = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen, sample, Example 4.19, A school site committee is to be chosen randomly from 6 men and 5 women. If the committee, consists of 4 members chosen randomly, what is the probability that 2 of them are men? How, many men do you expect to be on the committee?, Let X = the number of men on the committee of 4. The men are the group of interest (first group)., X takes on the values 0, 1, 2, 3, 4, where r = 6, b = 5 , and n = 4. X ∼ H (6, 5, 4), Find P ( x = 2). P ( x = 2) = 0.4545 (calculator or computer), NOTE : Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There, are a number of computer packages, including Microsoft Excel, that do., , The probability that there are 2 men on the committee is about 0.45., The graph of X ∼ H (6, 5, 4) is:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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180, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , The y-axis contains the probability of X, where X = the number of men on the committee., You would expect m = 2.18 (about 2) men on the committee., The formula for the mean is µ =, , n ·r, r +b, , =, , 4·6, 6+5, , = 2.18, , The formula for the variance is fairly complex. You will find it in the Summary of the Discrete, Probability Functions Chapter (Section 4.9)., , 4.8 Poisson8, Characteristics of a Poisson experiment:, 1. The Poisson gives the probability of a number of events occurring in a fixed interval of time or space if, these events happen with a known average rate and independently of the time since the last event. For, example, a book editor might be interested in the number of words spelled incorrectly in a particular, book. It might be that, on the average, there are 5 words spelled incorrectly in 100 pages. The interval, is the 100 pages., 2. The Poisson may be used to approximate the binomial if the probability of success is "small" (such, as 0.01) and the number of trials is "large" (such as 1000). You will verify the relationship in the, homework exercises. n is the number of trials and p is the probability of a "success.", Poisson probability distribution. The random variable X = the number of occurrences in the interval of, interest. The mean and variance are given in the summary., Example 4.20, The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of, interest is the number of loaves of bread put on the shelf in 5 minutes. The time interval of interest, is 5 minutes. What is the probability that the number of loaves, selected randomly, put on the shelf, in 5 minutes is 3?, Let X = the number of loaves of bread put on the shelf in 5 minutes. If the average number of, loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on, the shelf in 5 minutes is, 8 This, , content is available online at <http://cnx.org/content/m16829/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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181, 5, 30, , , , · 12 = 2 loaves of bread, , The probability question asks you to find P ( x = 3)., Example 4.21, A certain bank expects to receive 6 bad checks per day, on average. What is the probability of the, bank getting fewer than 5 bad checks on any given day? Of interest is the number of checks the, bank receives in 1 day, so the time interval of interest is 1 day. Let X = the number of bad checks, the bank receives in one day. If the bank expects to receive 6 bad checks per day then the average, is 6 checks per day. The probability question asks for P ( x < 5)., Example 4.22, You notice that a news reporter says "uh", on average, 2 times per broadcast. What is the probability that the news reporter says "uh" more than 2 times per broadcast., This is a Poisson problem because you are interested in knowing the number of times the news, reporter says "uh" during a broadcast., Problem 1, What is the interval of interest?, , (Solution on p. 214.), , Problem 2, (Solution on p. 214.), What is the average number of times the news reporter says "uh" during one broadcast?, Problem 3, Let X = ____________. What values does X take on?, , (Solution on p. 214.), , Problem 4, The probability question is P(______)., , (Solution on p. 214.), , 4.8.1 Notation for the Poisson: P = Poisson Probability Distribution Function, X ∼ P(µ), Read this as "X is a random variable with a Poisson distribution." The parameter is µ (or λ). µ (or λ) = the, mean for the interval of interest., Example 4.23, Leah’s answering machine receives about 6 telephone calls between 8 a.m. and 10 a.m. What is, the probability that Leah receives more than 1 call in the next 15 minutes?, Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or, 1, 4 hour.), x = 0, 1, 2, 3, ..., If Leah receives, on the average, 6 telephone calls in 2 hours, and there are eight 15 minutes intervals in 2 hours, then Leah receives, 1, 8, , · 6 = 0.75, , calls in 15 minutes, on the average. So, µ = 0.75 for this problem., X ∼ P(0.75), Find P ( x > 1). P ( x > 1) = 0.1734 (calculator or computer), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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182, , CHAPTER 4. DISCRETE RANDOM VARIABLES, TI-83+ and TI-84: For a general discussion, see this example (Binomial). The syntax is similar., The Poisson parameter list is (µ for the interval of interest, number). For this problem:, Press 1- and then press 2nd DISTR. Arrow down to C:poissoncdf. Press ENTER. Enter .75,1)., The result is P ( x > 1) = 0.1734. NOTE: The TI calculators use λ (lambda) for the mean., The probability that Leah receives more than 1 telephone call in the next fifteen minutes is about, 0.1734., The graph of X ∼ P(0.75) is:, , The y-axis contains the probability of x where X = the number of calls in 15 minutes., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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183, , 4.9 Summary of Functions9, Formula 4.1: Binomial, X ∼ B (n, p), X = the number of successes in n independent trials, n = the number of independent trials, X takes on the values x = 0,1, 2, 3, ...,n, p = the probability of a success for any trial, q = the probability of a failure for any trial, p+q = 1, , q = 1− p, , The mean is µ = np. The standard deviation is σ =, , √, , npq., , Formula 4.2: Geometric, X ∼ G ( p), X = the number of independent trials until the first success (count the failures and the first success), X takes on the values x= 1, 2, 3, ..., p = the probability of a success for any trial, q = the probability of a failure for any trial, p+q = 1, q = 1− p, The mean is µ =, , 1, p, , The standard deviation is σ =, , r , 1, p, , 1, p, , −1, , , , Formula 4.3: Hypergeometric, X ∼ H (r, b, n), X = the number of items from the group of interest that are in the chosen sample., X may take on the values x= 0, 1, ..., up to the size of the group of interest. (The minimum value, for X may be larger than 0 in some instances.), r = the size of the group of interest (first group), b= the size of the second group, n= the size of the chosen sample., n ≤ r+b, The mean is: µ =, 9 This, , nr, r +b, , content is available online at <http://cnx.org/content/m16833/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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184, , CHAPTER 4. DISCRETE RANDOM VARIABLES, r, The standard deviation is: σ =, , rbn(r +b−n), (r + b )2 (r + b −1), , Formula 4.4: Poisson, X ∼ P(µ), X = the number of occurrences in the interval of interest, X takes on the values x = 0, 1, 2, 3, ..., The mean µ is typically given. (λ is often used as the mean instead of µ.) When the Poisson is, used to approximate the binomial, we use the binomial mean µ = np. n is the binomial number, of trials. p = the probability of a success for each trial. This formula is valid when n is "large" and, p "small" (a general rule is that n should be greater than or equal to 20 and p should be less than, or equal to 0.05). If n is large enough and p is small enough then the Poisson approximates the, √, binomial very well. The variance is σ2 = µ and the standard deviation is σ = µ, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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185, , 4.10 Practice 1: Discrete Distribution10, 4.10.1 Student Learning Outcomes, • The student will analyze the properties of a discrete distribution., , 4.10.2 Given:, A ballet instructor is interested in knowing what percent of each year’s class will continue on to the next,, so that she can plan what classes to offer. Over the years, she has established the following probability, distribution., , • Let X = the number of years a student will study ballet with the teacher., • Let P ( x ) = the probability that a student will study ballet x years., , 4.10.3 Organize the Data, Complete the table below using the data provided., x, , P(x), , 1, , 0.10, , 2, , 0.05, , 3, , 0.10, , x*P(x), , 4, 5, , 0.30, , 6, , 0.20, , 7, , 0.10, Table 4.9, , Exercise 4.10.1, In words, define the Random Variable X., Exercise 4.10.2, P ( x = 4) =, Exercise 4.10.3, P ( x < 4) =, Exercise 4.10.4, On average, how many years would you expect a child to study ballet with this teacher?, , 4.10.4 Discussion Question, Exercise 4.10.5, What does the column "P(x)" sum to and why?, Exercise 4.10.6, What does the column "x ∗ P(x)" sum to and why?, 10 This, , content is available online at <http://cnx.org/content/m16830/1.14/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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186, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.11 Practice 2: Binomial Distribution11, 4.11.1 Student Learning Outcomes, • The student will construct the Binomial Distribution., , 4.11.2 Given, The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time,, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied, that, yes, they believe that same-sex couples should have the right to legal marital status. (Source:, http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf). ), Suppose that you randomly pick 8 first-time, full-time freshmen from the survey. You are interested, in the number that believes that same sex-couples should have the right to legal marital status, , 4.11.3 Interpret the Data, Exercise 4.11.1, In words, define the random Variable X., , (Solution on p. 214.), , Exercise 4.11.2, X ∼___________, , (Solution on p. 214.), , Exercise 4.11.3, What values does the random variable X take on?, , (Solution on p. 214.), , Exercise 4.11.4, Construct the probability distribution function (PDF)., x, , P(x), , Table 4.10, , Exercise 4.11.5, On average (u), how many would you expect to answer yes?, , (Solution on p. 214.), , Exercise 4.11.6, What is the standard deviation (σ ) ?, , (Solution on p. 214.), , 11 This, , content is available online at <http://cnx.org/content/m17107/1.18/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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187, Exercise 4.11.7, What is the probability that at most 5 of the freshmen reply “yes”?, , (Solution on p. 214.), , Exercise 4.11.8, What is the probability that at least 2 of the freshmen reply “yes”?, , (Solution on p. 214.), , Exercise 4.11.9, Construct a histogram or plot a line graph. Label the horizontal and vertical axes with words., Include numerical scaling., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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188, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.12 Practice 3: Poisson Distribution12, 4.12.1 Student Learning Outcomes, • The student will analyze the properties of a Poisson distribution., , 4.12.2 Given, On average, eight teens in the U.S. die from motor vehicle injuries per day., As, a result, states across the country are debating raising the driving age., (Source:, http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html) ), , 4.12.3 Interpret the Data, Exercise 4.12.1, Assume the event occurs independently in any given day. In words, define the Random Variable, X., Exercise 4.12.2, X ∼______________, , (Solution on p. 214.), , Exercise 4.12.3, What values does X take on?, , (Solution on p. 214.), , Exercise 4.12.4, For the given values of the random variable X, fill in the corresponding probabilities., x, , P(x), , 0, 4, 8, 10, 11, 15, Table 4.11, , Exercise 4.12.5, (Solution on p. 214.), Is it likely that there will be no teens killed in the U.S. from motor vehicle injuries on any given, day? Justify your answer numerically., Exercise 4.12.6, (Solution on p. 214.), Is it likely that there will be more than 20 teens killed in the U.S. from motor vehicle injuries on, any given day? Justify your answer numerically., , 12 This, , content is available online at <http://cnx.org/content/m17109/1.15/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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189, , 4.13 Practice 4: Geometric Distribution13, 4.13.1 Student Learning Outcomes, • The student will analyze the properties of a geometric distribution., , 4.13.2 Given:, Use the information from the Binomial Distribution Practice (Section 4.11) shown below., The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time,, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied, that, yes, they believe that same-sex couples should have the right to legal marital status. (Source:, http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf ), Suppose that you randomly select freshman from the study until you find one who replies “yes.”, You are interested in the number of freshmen you must ask., , 4.13.3 Interpret the Data, Exercise 4.13.1, In words, define the Random Variable X., Exercise 4.13.2, X∼, , (Solution on p. 215.), , Exercise 4.13.3, What values does the random variable X take on?, , (Solution on p. 215.), , Exercise 4.13.4, Construct the probability distribution function (PDF). Stop at x = 6., x, , P(x), , 1, 2, 3, 4, 5, 6, Table 4.12, , Exercise 4.13.5, (Solution on p. 215.), On average(µ), how many freshmen would you expect to have to ask until you found one who, replies "yes?", Exercise 4.13.6, What is the probability that you will need to ask fewer than 3 freshmen?, 13 This, , (Solution on p. 215.), , content is available online at <http://cnx.org/content/m17108/1.17/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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190, , CHAPTER 4. DISCRETE RANDOM VARIABLES, Exercise 4.13.7, Construct a histogram or plot a line graph. Label the horizontal and vertical axes with words., Include numerical scaling., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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191, , 4.14 Practice 5: Hypergeometric Distribution14, 4.14.1 Student Learning Outcomes, • The student will analyze the properties of a hypergeometric distribution., , 4.14.2 Given, Suppose that a group of statistics students is divided into two groups: business majors and non-business, majors. There are 16 business majors in the group and 7 non-business majors in the group. A random, sample of 9 students is taken. We are interested in the number of business majors in the group., , 4.14.3 Interpret the Data, Exercise 4.14.1, In words, define the Random Variable X., Exercise 4.14.2, X∼, , (Solution on p. 215.), , Exercise 4.14.3, What values does X take on?, , (Solution on p. 215.), , Exercise 4.14.4, Construct the probability distribution function (PDF) for X., x, , P(x), , Table 4.13, , Exercise 4.14.5, On average(µ), how many would you expect to be business majors?, , 14 This, , (Solution on p. 215.), , content is available online at <http://cnx.org/content/m17106/1.13/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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192, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.15 Homework15, Exercise 4.15.1, 1. Complete the PDF and answer the questions., x, , P (X = x), , 0, , 0.3, , 1, , 0.2, , (Solution on p. 215.), , x · P (X = x), , 2, 3, , 0.4, Table 4.14, , a. Find the probability that x = 2., b. Find the expected value., Exercise 4.15.2, Suppose that you are offered the following “deal.” You roll a die. If you roll a 6, you win $10. If, you roll a 4 or 5, you win $5. If you roll a 1, 2, or 3, you pay $6., a. What are you ultimately interested in here (the value of the roll or the money you win)?, b. In words, define the Random Variable X., c. List the values that X may take on., d. Construct a PDF., e. Over the long run of playing this game, what are your expected average winnings per game?, f. Based on numerical values, should you take the deal? Explain your decision in complete sentences., Exercise 4.15.3, (Solution on p. 215.), A venture capitalist, willing to invest $1,000,000, has three investments to choose from. The first, investment, a software company, has a 10% chance of returning $5,000,000 profit, a 30% chance of, returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company,, a hardware company, has a 20% chance of returning $3,000,000 profit, a 40% chance of returning, $1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech, firm, has a 10% chance of returning $6,000,000 profit, a 70% of no profit or loss, and a 20% chance, of losing the million dollars., a., b., c., d., e., , Construct a PDF for each investment., Find the expected value for each investment., Which is the safest investment? Why do you think so?, Which is the riskiest investment? Why do you think so?, Which investment has the highest expected return, on average?, , Exercise 4.15.4, A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase, 4 tickets. The prize is 2 passes to a Broadway show, worth a total of $150., a. What are you interested in here?, 15 This, , content is available online at <http://cnx.org/content/m16823/1.21/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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193, b., c., d., e., , In words, define the Random Variable X., List the values that X may take on., Construct a PDF., If this fund-raiser is repeated often and you always purchase 4 tickets, what would be your, expected average winnings per raffle?, , Exercise 4.15.5, (Solution on p. 215.), Suppose that 20,000 married adults in the United States were randomly surveyed as to the number, of children they have. The results are compiled and are used as theoretical probabilities. Let X =, the number of children, x, , P (X = x), , 0, , 0.10, , 1, , 0.20, , 2, , 0.30, , x · P (X = x), , 3, 4, , 0.10, , 5, , 0.05, , 6 (or more), , 0.05, Table 4.15, , a., b., c., d., , Find the probability that a married adult has 3 children., In words, what does the expected value in this example represent?, Find the expected value., Is it more likely that a married adult will have 2 – 3 children or 4 – 6 children? How do you, know?, , Exercise 4.15.6, Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree, is given below., x, , P (X = x), , 3, , 0.05, , 4, , 0.40, , 5, , 0.30, , 6, , 0.15, , 7, , 0.10, Table 4.16, , a. In words, define the Random Variable X., b. What does it mean that the values 0, 1, and 2 are not included for x in the PDF?, c. On average, how many years do you expect it to take for an individual to earn a B.S.?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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194, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.15.1 For each problem:, a. In words, define the Random Variable X., b. List the values that X may take on., c. Give the distribution of X. X ∼, Then, answer the questions specific to each individual problem., Exercise 4.15.7, (Solution on p. 215.), Six different colored dice are rolled. Of interest is the number of dice that show a “1.”, d. On average, how many dice would you expect to show a “1”?, e. Find the probability that all six dice show a “1.”, f. Is it more likely that 3 or that 4 dice will show a “1”? Use numbers to justify your answer, numerically., Exercise 4.15.8, More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. (Source:, http://en.wikipedia.org/wiki/Distance_education), d. On average, how many schools would you expect to offer such courses?, e. Find the probability that at most 6 offer such courses., f. Is it more likely that 0 or that 13 will offer such courses? Use numbers to justify your answer, numerically and answer in a complete sentence., Exercise 4.15.9, (Solution on p. 215.), A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet, (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students, attend Tet festivities. We are interested in the number of students who will attend the festivities., d. How many of the 12 students do we expect to attend the festivities?, e. Find the probability that at most 4 students will attend., f. Find the probability that more than 2 students will attend., Exercise 4.15.10, Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating, students is randomly chosen., d. How many are expected to attend their graduation?, e. Find the probability that 17 or 18 attend., f. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your, answer numerically., Exercise 4.15.11, (Solution on p. 216.), At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey, 25 fencers at The Fencing Center. We are interested in the numbers that do not use the foil as their, main weapon., d. How many are expected to not use the foil as their main weapon?, e. Find the probability that six do not use the foil as their main weapon., f. Based on numerical values, would you be surprised if all 25 did not use foil as their main, weapon? Justify your answer numerically., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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195, Exercise 4.15.12, Approximately 8% of students at a local high school participate in after-school sports all four, years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that, participated in after-school sports all four years of high school., d. How many seniors are expected to have participated in after-school sports all four years of high, school?, e. Based on numerical values, would you be surprised if none of the seniors participated in afterschool sports all four years of high school? Justify your answer numerically., f. Based upon numerical values, is it more likely that 4 or that 5 of the seniors participated in, after-school sports all four years of high school? Justify your answer numerically., Exercise 4.15.13, (Solution on p. 216.), The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune, cookies, we are interested in the number of cookies with an extra fortune. Two distributions may, be used to solve this problem. Use one distribution to solve the problem., d. How many cookies do we expect to have an extra fortune?, e. Find the probability that none of the cookies have an extra fortune., f. Find the probability that more than 3 have an extra fortune., g. As n increases, what happens involving the probabilities using the two distributions? Explain, in complete sentences., Exercise 4.15.14, There are two games played for Chinese New Year and Vietnamese New Year. They are almost, identical. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with, a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish,, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play, with bets being $1. The player places a bet on a number or object. The “house” rolls three dice. If, none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the, dice shows the number or object bet (and the other two do not show it), the player gets back his, $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not, show it), the player gets back his $1 bet, plus $2 profit. If all three dice show the number or object, bet, the player gets back his $1 bet, plus $3 profit., Let X = number of matches and Y= profit per game., d. List the values that Y may take on. Then, construct one PDF table that includes both X & Y and, their probabilities., e. Calculate the average expected matches over the long run of playing this game for the player., f. Calculate the average expected earnings over the long run of playing this game for the player., g. Determine who has the advantage, the player or the house., Exercise 4.15.15, (Solution on p. 216.), According to the South Carolina Department of Mental Health web site, for, every 200 U.S. women, the average number who suffer from anorexia is one, ( http://www.state.sc.us/dmh/anorexia/statistics.htm 16 ). Out of a randomly chosen group of, 600 U.S. women:, d. How many are expected to suffer from anorexia?, e. Find the probability that no one suffers from anorexia., 16 http://www.state.sc.us/dmh/anorexia/statistics.htm, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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196, , CHAPTER 4. DISCRETE RANDOM VARIABLES, f. Find the probability that more than four suffer from anorexia., Exercise 4.15.16, The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one, Japanese woman is randomly chosen., 17, (, http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html, MHLW’s Pamphlet), d. Find the probability that she has no children., e. Find the probability that she has fewer children than the Japanese average., f. Find the probability that she has more children than the Japanese average., Exercise 4.15.17, (Solution on p. 216.), The average number of children a Spanish woman has in her lifetime is 1.47., Suppose that one Spanish woman is randomly chosen., ( http://www.typicallyspanish.com/news/publish/article_4897.shtml 18 )., d. Find the probability that she has no children., e. Find the probability that she has fewer children than the Spanish average., f. Find the probability that she has more children than the Spanish average ., Exercise 4.15.18, Fertile (female) cats produce an average of 3 litters per year. (Source: The Humane Society of, the United States). Suppose that one fertile, female cat is randomly chosen. In one year, find the, probability she produces:, d. No litters., e. At least 2 litters., f. Exactly 3 litters., Exercise 4.15.19, (Solution on p. 216.), A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership, that carries the car. She estimates the probability that any independent dealership will have the, car will be 28%. We are interested in the number of dealerships she must call., d. On average, how many dealerships would we expect her to have to call until she finds one that, has the car?, e. Find the probability that she must call at most 4 dealerships., f. Find the probability that she must call 3 or 4 dealerships., Exercise 4.15.20, Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each, person is considered independent. We are interested in the number of adults in America we must, survey until we find one who will watch the Super Bowl., d. How many adults in America do you expect to survey until you find one who will watch the, Super Bowl?, e. Find the probability that you must ask 7 people., f. Find the probability that you must ask 3 or 4 people., , 17 http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html, 18 http://www.typicallyspanish.com/news/publish/article_4897.shtml, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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197, Exercise 4.15.21, (Solution on p. 216.), A group of Martial Arts students is planning on participating in an upcoming demonstration., 6 are students of Tae Kwon Do; 7 are students of Shotokan Karate. Suppose that 8 students are, randomly picked to be in the first demonstration. We are interested in the number of Shotokan, Karate students in that first demonstration. Hint: Use the Hypergeometric distribution. Look in, the Formulas section of 4: Discrete Distributions and in the Appendix Formulas., d. How many Shotokan Karate students do we expect to be in that first demonstration?, e. Find the probability that 4 students of Shotokan Karate are picked for the first demonstration., f. Suppose that we are interested in the Tae Kwan Do students that are picked for the first demonstration. Find the probability that all 6 students of Tae Kwan Do are picked for the first, demonstration., Exercise 4.15.22, The chance of a IRS audit for a tax return with over $25,000 in income is about 2% per year. We, are interested in the expected number of audits a person with that income has in a 20 year period., Assume each year is independent., d. How many audits are expected in a 20 year period?, e. Find the probability that a person is not audited at all., f. Find the probability that a person is audited more than twice., Exercise 4.15.23, (Solution on p. 216.), Refer to the previous problem. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in 1 year. One way to solve this, problem is by using the Binomial Distribution. Since n is large and p is small, another discrete, distribution could be used to solve the following problems. Solve the following questions (d-f), using that distribution., d. How many are expected to be audited?, e. Find the probability that no one was audited., f. Find the probability that more than 2 were audited., Exercise 4.15.24, Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that 10 people will be randomly chosen to be on the committee from a group, of 28 volunteers, 20 who are technically proficient and 8 who are not. We are interested in the, number on the committee who are not technically proficient., d. How many instructors do you expect on the committee who are not technically proficient?, e. Find the probability that at least 5 on the committee are not technically proficient., f. Find the probability that at most 3 on the committee are not technically proficient., Exercise 4.15.25, (Solution on p. 217.), Refer back to Exercise 4.15.12. Solve this problem again, using a different, though still acceptable,, distribution., Exercise 4.15.26, Suppose that 9 Massachusetts athletes are scheduled to appear at a charity benefit. The 9 are randomly chosen from 8 volunteers from the Boston Celtics and 4 volunteers from the New England, Patriots. We are interested in the number of Patriots picked., d. Is it more likely that there will be 2 Patriots or 3 Patriots picked?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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198, , CHAPTER 4. DISCRETE RANDOM VARIABLES, Exercise 4.15.27, (Solution on p. 217.), On average, Pierre, an amateur chef, drops 3 pieces of egg shell into every 2 batters of cake he, makes. Suppose that you buy one of his cakes., d. On average, how many pieces of egg shell do you expect to be in the cake?, e. What is the probability that there will not be any pieces of egg shell in the cake?, f. Let’s say that you buy one of Pierre’s cakes each week for 6 weeks. What is the probability that, there will not be any egg shell in any of the cakes?, g. Based upon the average given for Pierre, is it possible for there to be 7 pieces of shell in the, cake? Why?, Exercise 4.15.28, It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we, find a resident who does not have adequate earthquake supplies., d. What is the probability that we must survey just 1 or 2 residents until we find a California, resident who does not have adequate earthquake supplies?, e. What is the probability that we must survey at least 3 California residents until we find a California resident who does not have adequate earthquake supplies?, f. How many California residents do you expect to need to survey until you find a California, resident who does not have adequate earthquake supplies?, g. How many California residents do you expect to need to survey until you find a California, resident who does have adequate earthquake supplies?, Exercise 4.15.29, (Solution on p. 217.), Refer to the above problem. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies., d. What is the probability that at least 8 have adequate earthquake supplies?, e. Is it more likely that none or that all of the residents surveyed will have adequate earthquake, supplies? Why?, f. How many residents do you expect will have adequate earthquake supplies?, , The next 2 questions refer to the following: In one of its Spring catalogs, L.L. Bean® advertised footwear on, 29 of its 192 catalog pages., Exercise 4.15.30, Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise, footwear. Each page may be picked at most once., d. How many pages do you expect to advertise footwear on them?, e. Is it probable that all 20 will advertise footwear on them? Why or why not?, f. What is the probability that less than 10 will advertise footwear on them?, Exercise 4.15.31, (Solution on p. 217.), Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise, footwear. This time, each page may be picked more than once., d. How many pages do you expect to advertise footwear on them?, e. Is it probable that all 20 will advertise footwear on them? Why or why not?, f. What is the probability that less than 10 will advertise footwear on them?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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199, g. Reminder: A page may be picked more than once. We are interested in the number of pages, that we must randomly survey until we find one that has footwear advertised on it. Define, the random variable X and give its distribution., h. What is the probability that you only need to survey at most 3 pages in order to find one that, advertises footwear on it?, i. How many pages do you expect to need to survey in order to find one that advertises footwear?, Exercise 4.15.32, Suppose that you roll a fair die until each face has appeared at least once. It does not matter in, what order the numbers appear. Find the expected number of rolls you must make until each face, has appeared at least once., , 4.15.2 Try these multiple choice problems., For the next three problems: The probability that the San Jose Sharks will win any given game is 0.3694, based on a 13 year win history of 382 wins out of 1034 games played (as of a certain date). An upcoming, monthly schedule contains 12 games., Let X = the number of games won in that upcoming month., Exercise 4.15.33, The expected number of wins for that upcoming month is:, , (Solution on p. 217.), , A. 1.67, B. 12, 382, C. 1043, D. 4.43, Exercise 4.15.34, (Solution on p. 217.), What is the probability that the San Jose Sharks win 6 games in that upcoming month?, A., B., C., D., , 0.1476, 0.2336, 0.7664, 0.8903, , Exercise 4.15.35, (Solution on p. 217.), What is the probability that the San Jose Sharks win at least 5 games in that upcoming month, A., B., C., D., , 0.3694, 0.5266, 0.4734, 0.2305, , For the next two questions: The average number of times per week that Mrs. Plum’s cats wake her up at, night because they want to play is 10. We are interested in the number of times her cats wake her up each, week., Exercise 4.15.36, In words, the random variable X =, , (Solution on p. 217.), , A. The number of times Mrs. Plum’s cats wake her up each week, B. The number of times Mrs. Plum’s cats wake her up each hour, C. The number of times Mrs. Plum’s cats wake her up each night, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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200, , CHAPTER 4. DISCRETE RANDOM VARIABLES, D. The number of times Mrs. Plum’s cats wake her up, Exercise 4.15.37, (Solution on p. 217.), Find the probability that her cats will wake her up no more than 5 times next week., A., B., C., D., , 0.5000, 0.9329, 0.0378, 0.0671, , Exercise 4.15.38, (Solution on p. 217.), People visiting video rental stores often rent more than one DVD at a time. The probability, distribution for DVD rentals per customer at Video To Go is given below. There is 5 video limit, per customer at this store, so nobody ever rents more than 5 DVDs., x, , 0, , 1, , 2, , 3, , 4, , 5, , P(X=x), , 0.03, , 0.50, , 0.24, , ?, , 0.07, , 0.04, , Table 4.17, , A., B., C., D., , Describe the random variable X in words., Find the probability that a customer rents three DVDs., Find the probability that a customer rents at least 4 DVDs., Find the probability that a customer rents at most 2 DVDs., , Another shop, Entertainment Headquarters, rents DVDs and videogames. The probability distribution for DVD rentals per customer at this shop is given below. They also have a 5 DVD limit per, customer., x, , 0, , 1, , 2, , 3, , 4, , 5, , P(X=x), , 0.35, , 0.25, , 0.20, , 0.10, , 0.05, , 0.05, , Table 4.18, , E. At which store is the expected number of DVDs rented per customer higher?, F. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they, expect to rent next week? Answer in sentence form., G. If Video to Go expects 300 customers next week and Entertainment HQ projects that they will, have 420 customers, for which store is the expected number of DVD rentals for next week, higher? Explain., H. Which of the two video stores experiences more variation in the number of DVD rentals per, customer? How do you know that?, Exercise 4.15.39, (Solution on p. 217.), A game involves selecting a card from a deck of cards and tossing a coin. The deck has 52 cards, and 12 cards are "face cards" (Jack, Queen, or King) The coin is a fair coin and is equally likely to, land on Heads or Tails, , • If the card is a face card and the coin lands on Heads, you win $6, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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201, , • If the card is a face card and the coin lands on Tails, you win $2, • If the card is not a face card, you lose $2, no matter what the coin shows., A. Find the expected value for this game (expected net gain or loss)., B. Explain what your calculations indicate about your long-term average profits and losses on, this game., C. Should you play this game to win money?, Exercise 4.15.40, (Solution on p. 218.), You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available, be sold in this lottery. In this lottery there is one $500 prize, 2 $100 prizes and 4 $25 prizes. Find, your expected gain or loss., Exercise 4.15.41, (Solution on p. 218.), A student takes a 10 question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the, questions correct., Exercise 4.15.42, (Solution on p. 218.), A student takes a 32 question multiple choice exam, but did not study and randomly guesses each, answer. Each question has 3 possible choices for the answer. Find the probability that the student, guesses more than 75% of the questions correctly., Exercise 4.15.43, (Solution on p. 219.), Suppose that you are perfoming the probability experiment of rolling one fair six-sided die. Let F, be the event of rolling a "4" or a "5". You are interested in how many times you need to roll the die, in order to obtain the first “4 or 5” as the outcome., , • p = probability of success (event F occurs), • q = probability of failure (event F does not occur), A. Write the description of the random variable X. What are the values that X can take on? Find, the values of p and q., B. Find the probability that the first occurrence of event F (rolling a “4” or “5”) is on the second, trial., C. How many trials would you expect until you roll a “4” or “5”?, **Exercises 38 - 43 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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202, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.16 Review19, The next two questions refer to the following:, A recent poll concerning credit cards found that 35 percent of respondents use a credit card that gives them, a mile of air travel for every dollar they charge. Thirty percent of the respondents charge more than $2000, per month. Of those respondents who charge more than $2000, 80 percent use a credit card that gives them, a mile of air travel for every dollar they charge., Exercise 4.16.1, (Solution on p. 219.), What is the probability that a randomly selected respondent will spend more than $2000 AND, use a credit card that gives them a mile of air travel for every dollar they charge?, A., B., C., D., , (0.30) (0.35), (0.80) (0.35), (0.80) (0.30), (0.80), , Exercise 4.16.2, (Solution on p. 219.), Based upon the above information, are using a credit card that gives a mile of air travel for each, dollar spent AND charging more than $2000 per month independent events?, A., B., C., D., , Yes, No, and they are not mutually exclusive either, No, but they are mutually exclusive, Not enough information given to determine the answer, , Exercise 4.16.3, (Solution on p. 219.), A sociologist wants to know the opinions of employed adult women about government funding, for day care. She obtains a list of 520 members of a local business and professional women’s, club and mails a questionnaire to 100 of these women selected at random. 68 questionnaires are, returned. What is the population in this study?, A., B., C., D., , All employed adult women, All the members of a local business and professional women’s club, The 100 women who received the questionnaire, All employed women with children, , The next two questions refer to the following: An article from The San Jose Mercury News was concerned, with the racial mix of the 1500 students at Prospect High School in Saratoga, CA. The table summarizes the, results. (Male and female values are approximate.) Suppose one Prospect High School student is randomly, selected., Ethnic Group, Gender, , White, , Asian, , Hispanic, , Black, , American Indian, , Male, , 400, , 168, , 115, , 35, , 16, , Female, , 440, , 132, , 140, , 40, , 14, , Table 4.19, 19 This, , content is available online at <http://cnx.org/content/m16832/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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203, Exercise 4.16.4, Find the probability that a student is Asian or Male., , (Solution on p. 219.), , Exercise 4.16.5, (Solution on p. 219.), Find the probability that a student is Black given that the student is Female., Exercise 4.16.6, (Solution on p. 219.), A sample of pounds lost, in a certain month, by individual members of a weight reducing clinic, produced the following statistics:, , •, •, •, •, •, •, , Mean = 5 lbs., Median = 4.5 lbs., Mode = 4 lbs., Standard deviation = 3.8 lbs., First quartile = 2 lbs., Third quartile = 8.5 lbs., , The correct statement is:, A., B., C., D., , One fourth of the members lost exactly 2 pounds., The middle fifty percent of the members lost from 2 to 8.5 lbs., Most people lost 3.5 to 4.5 lbs., All of the choices above are correct., , Exercise 4.16.7, What does it mean when a data set has a standard deviation equal to zero?, A., B., C., D., , (Solution on p. 219.), , All values of the data appear with the same frequency., The mean of the data is also zero., All of the data have the same value., There are no data to begin with., , Exercise 4.16.8, The statement that best describes the illustration below is:, , (Solution on p. 219.), , Figure 4.1, , A., B., C., D., , The mean is equal to the median., There is no first quartile., The lowest data value is the median., ( Q1+ Q3), The median equals, 2, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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204, , CHAPTER 4. DISCRETE RANDOM VARIABLES, Exercise 4.16.9, (Solution on p. 219.), According to a recent article (San Jose Mercury News) the average number of babies born with, significant hearing loss (deafness) is approximately 2 per 1000 babies in a healthy baby nursery., The number climbs to an average of 30 per 1000 babies in an intensive care nursery., Suppose that 1000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly 2 babies were born deaf., Exercise 4.16.10, (Solution on p. 219.), A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box, containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a, free gift., , •, •, •, •, , 10 of the coupons are for a free gift worth $6., 80 of the coupons are for a free gift worth $8., 6 of the coupons are for a free gift worth $12., 4 of the coupons are for a free gift worth $40., , Based upon the financial gain or loss over the long run, should you play the game?, A. Yes, I expect to come out ahead in money., B. No, I expect to come out behind in money., C. It doesn’t matter. I expect to break even., The next four questions refer to the following: Recently, a nurse commented that when a patient calls the, medical advice line claiming to have the flu, the chance that he/she truly has the flu (and not just a nasty, cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how, many actually have the flu., Exercise 4.16.11, Define the Random Variable and list its possible values., , (Solution on p. 219.), , Exercise 4.16.12, State the distribution of X ., , (Solution on p. 219.), , Exercise 4.16.13, Find the probability that at least 4 of the 25 patients actually have the flu., , (Solution on p. 219.), , Exercise 4.16.14, (Solution on p. 219.), On average, for every 25 patients calling in, how many do you expect to have the flu?, The next two questions refer to the following: Different types of writing can sometimes be distinguished, by the number of letters in the words used. A student interested in this fact wants to study the number of, letters of words used by Tom Clancy in his novels. She opens a Clancy novel at random and records the, number of letters of the first 250 words on the page., Exercise 4.16.15, What kind of data was collected?, , (Solution on p. 219.), , A. qualitative, B. quantitative - continuous, C. quantitative – discrete, Exercise 4.16.16, What is the population under study?, , (Solution on p. 219.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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205, , 4.17 Lab 1: Discrete Distribution (Playing Card Experiment)20, Class Time:, Names:, , 4.17.1 Student Learning Outcomes:, • The student will compare empirical data and a theoretical distribution to determine if everyday experiment fits a discrete distribution., • The student will demonstrate an understanding of long-term probabilities., , 4.17.2 Supplies:, • One full deck of playing cards, , 4.17.3 Procedure, The experiment procedure is to pick one card from a deck of shuffled cards., 1., 2., 3., 4., 5., 6., 7., 8., , The theorectical probability of picking a diamond from a deck is: _________, Shuffle a deck of cards., Pick one card from it., Record whether it was a diamond or not a diamond., Put the card back and reshuffle., Do this a total of 10 times, Record the number of diamonds picked., Let X = number of diamonds. Theoretically, X ∼ B (_____, _____), , 4.17.4 Organize the Data, 1. Record the number of diamonds picked for your class in the chart below. Then calculate the relative, frequency., 20 This, , content is available online at <http://cnx.org/content/m16827/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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207, , 4.17.5 Theoretical Distribution, 1. Build the theoretical PDF chart based on the distribution in the Procedure section above., x, , P (x), , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Table 4.21, , 2. Calculate the following:, a. µ =____________, b. σ =____________, 3. Construct a histogram of the theoretical distribution., , Figure 4.3, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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208, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 4.17.6 Using the Data, Calculate the following, rounding to 4 decimal places:, NOTE :, , RF = relative frequency, , Use the table from the section titled "Theoretical Distribution" here:, , • P ( x = 3) =, • P (1 < x < 4) =, • P ( x ≥ 8) =, Use the data from the section titled "Organize the Data" here:, , • RF ( x = 3) =, • RF (1 < x < 4) =, • RF ( x ≥ 8) =, , 4.17.7 Discussion Questions, For questions 1. and 2., think about the shapes of the two graphs, the probabilities and the relative frequencies, the means, and the standard deviations., 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. (Note: These answers may vary and still, be correct.), 2. Describe the three most significant differences between the graphs or distributions of the theoretical, and empirical distributions. (Note: These answers may vary and still be correct.), 3. Using your answers from the two previous questions, does it appear that the data fit the theoretical, distribution? In 1 - 3 complete sentences, explain why or why not., 4. Suppose that the experiment had been repeated 500 times. Which table (from "Organize the data", and "Theoretical Distributions") would you expect to change (and how would it change)? Why? Why, wouldn’t the other table change?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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209, , 4.18 Lab 2: Discrete Distribution (Lucky Dice Experiment)21, Class Time:, Names:, , 4.18.1 Student Learning Outcomes:, • The student will compare empirical data and a theoretical distribution to determine if a Tet gambling, game fits a discrete distribution., • The student will demonstrate an understanding of long-term probabilities., , 4.18.2 Supplies:, • 1 game “Lucky Dice” or 3 regular dice, NOTE : For a detailed game description, refer here. (The link goes to the beginning of Discrete, Random Variables Homework. Please refer to Problem #14.), , NOTE :, , Round relative frequencies and probabilities to four decimal places., , 4.18.3 The Procedure, 1. The experiment procedure is to bet on one object. Then, roll 3 Lucky Dice and count the number of, matches. The number of matches will decide your profit., 2. What is the theoretical probability of 1 die matching the object? _________, 3. Choose one object to place a bet on. Roll the 3 Lucky Dice. Count the number of matches., 4. Let X = number of matches. Theoretically, X ∼ B (______, ______), 5. Let Y = profit per game., , 4.18.4 Organize the Data, In the chart below, fill in the Y value that corresponds to each X value. Next, record the number of matches, picked for your class. Then, calculate the relative frequency., 1. Complete the table., x, , y, , Frequency, , Relative Frequency, , 0, 1, 2, 3, Table 4.22, 21 This, , content is available online at <http://cnx.org/content/m16826/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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210, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , 2. Calculate the Following:, a., b., c., d., , x=, sx =, y=, sy =, , 3. Explain what x represents., 4. Explain what y represents., 5. Based upon the experiment:, a. What was the average profit per game?, b. Did this represent an average win or loss per game?, c. How do you know? Answer in complete sentences., 6. Construct a histogram of the empirical data, , Figure 4.4, , 4.18.5 Theoretical Distribution, Build the theoretical PDF chart for X and Y based on the distribution from the section titled "The Procedure"., 1., x, , y, , P ( x ) = P (y), , 0, 1, 2, 3, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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211, Table 4.23, , 2. Calculate the following, a. µ x =, b. σx =, c. µy =, 3. Explain what µ x represents., 4. Explain what µy represents., 5. Based upon theory:, a. What was the expected profit per game?, b. Did the expected profit represent an average win or loss per game?, c. How do you know? Answer in complete sentences., 6. Construct a histogram of the theoretical distribution., , Figure 4.5, , 4.18.6 Use the Data, Calculate the following (rounded to 4 decimal places):, NOTE :, , RF = relative frequency, , Use the data from the section titled "Theoretical Distribution" here:, 1. P ( x = 3) =____________, 2. P (0 < x < 3) =____________, 3. P ( x ≥ 2) =____________, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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212, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , Use the data from the section titled "Organize the Data" here:, 1. RF ( x = 3) =____________, 2. RF (0 < x < 3) =____________, 3. RF ( x ≥ 2) = ____________, , 4.18.7 Discussion Question, For questions 1. and 2., consider the graphs, the probabilities and relative frequencies, the means and the, standard deviations., 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. (Note: these answers may vary and still, be correct.), 2. Describe the three most significant differences between the graphs or distributions of the theoretical, and empirical distributions. (Note: these answers may vary and still be correct.), 3. Thinking about your answers to 1. and 2.,does it appear that the data fit the theoretical distribution?, In 1 - 3 complete sentences, explain why or why not., 4. Suppose that the experiment had been repeated 500 times. Which table (from "Organize the Data" or, "Theoretical Distribution") would you expect to change? Why? How might the table change?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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213, , Solutions to Exercises in Chapter 4, Solution to Example 4.2, Problem 1 (p. 169), Let X = the number of days Nancy attends class per week., Solution to Example 4.2, Problem 2 (p. 169), 0, 1, 2, and 3, Solution to Example 4.2, Problem 3 (p. 169), , x, , P (x), , 0, , 0.01, , 1, , 0.04, , 2, , 0.15, , 3, , 0.80, , Table 4.24, , Solution to Example 4.5, Problem 1 (p. 171), X = amount of profit, Solution to Example 4.5, Problem 2 (p. 171), , x, , P (x), , xP ( x ), , WIN, , 10, , LOSE, , -6, , 1, 3, 2, 3, , 10, 3, −12, 3, , Table 4.25, , Solution to Example 4.5, Problem 3 (p. 172), Add the last column of the table. The expected value µ = −32 . You lose, on average, about 67 cents each, time you play the game so you do not come out ahead., Solution to Example 4.9, Problem 1 (p. 173), failure, Solution to Example 4.9, Problem 2 (p. 173), X = the number of statistics students who do their homework on time, Solution to Example 4.9, Problem 3 (p. 173), 0, 1, 2, . . ., 50, Solution to Example 4.9, Problem 4 (p. 173), Failure is a student who does not do his or her homework on time., Solution to Example 4.9, Problem 5 (p. 173), q = 0.30, Solution to Example 4.9, Problem 6 (p. 174), greater than or equal to (≥), Solution to Example 4.14, Problem 2 (p. 176), 1, 2, 3, . . ., (total number of students), Solution to Example 4.14, Problem 3 (p. 176), , • p = 0.55, • q = 0.45, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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214, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , Solution to Example 4.14, Problem 4 (p. 176), P ( x = 4), Solution to Example 4.18, Problem 1 (p. 178), Without, Solution to Example 4.18, Problem 2 (p. 178), The men, Solution to Example 4.18, Problem 3 (p. 179), 15 men, Solution to Example 4.18, Problem 4 (p. 179), 18 women, Solution to Example 4.18, Problem 5 (p. 179), Let X = the number of men on the committee. x = 0, 1, 2, . . ., 7., Solution to Example 4.18, Problem 6 (p. 179), P(x>4), Solution to Example 4.22, Problem 1 (p. 181), One broadcast, Solution to Example 4.22, Problem 2 (p. 181), 2, Solution to Example 4.22, Problem 3 (p. 181), Let X = the number of times the news reporter says "uh" during one broadcast., x = 0, 1, 2, 3, ..., Solution to Example 4.22, Problem 4 (p. 181), P(x > 2), , Solutions to Practice 2: Binomial Distribution, Solution to Exercise 4.11.1 (p. 186), X= the number that reply “yes”, Solution to Exercise 4.11.2 (p. 186), B(8,0.713), Solution to Exercise 4.11.3 (p. 186), 0,1,2,3,4,5,6,7,8, Solution to Exercise 4.11.5 (p. 186), 5.7, Solution to Exercise 4.11.6 (p. 186), 1.28, Solution to Exercise 4.11.7 (p. 187), 0.4151, Solution to Exercise 4.11.8 (p. 187), 0.9990, , Solutions to Practice 3: Poisson Distribution, Solution to Exercise 4.12.2 (p., P(8), Solution to Exercise 4.12.3 (p., 0,1,2,3,4,..., Solution to Exercise 4.12.5 (p., No, Solution to Exercise 4.12.6 (p., No, , 188), 188), 188), 188), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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215, , Solutions to Practice 4: Geometric Distribution, Solution to Exercise 4.13.2 (p., G(0.713), Solution to Exercise 4.13.3 (p., 1,2,. . ., Solution to Exercise 4.13.5 (p., 1.4, Solution to Exercise 4.13.6 (p., 0.9176, , 189), 189), 189), 189), , Solutions to Practice 5: Hypergeometric Distribution, Solution to Exercise 4.14.2 (p. 191), H(16,7,9), Solution to Exercise 4.14.3 (p. 191), 2,3,4,5,6,7,8,9, Solution to Exercise 4.14.5 (p. 191), 6.26, , Solutions to Homework, Solution to Exercise 4.15.1 (p. 192), a. 0.1, b. 1.6, Solution to Exercise 4.15.3 (p. 192), b., c., d., e., , $200,000;$600,000;$400,000, third investment, first investment, second investment, , Solution to Exercise 4.15.5 (p. 193), a. 0.2, c. 2.35, d. 2-3 children, Solution to Exercise 4.15.7 (p. 194), a. X = the number of dice that show a 1, b. 0,1,2,3,4,5,6, , , c. X ∼ B 6, 16, d. 1, e. 0.00002, f. 3 dice, Solution to Exercise 4.15.9 (p. 194), a., b., c., d., , X = the number of students that will attend Tet., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, X ∼B(12,0.18), 2.16, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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216, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , e. 0.9511, f. 0.3702, Solution to Exercise 4.15.11 (p. 194), a. X = the number of fencers that do not use foil as their main weapon, b. 0, 1, 2, 3,... 25, c. X ∼B(25,0.40), d. 10, e. 0.0442, f. Yes, Solution to Exercise 4.15.13 (p. 195), a. X = the number of fortune cookies that have an extra fortune, b. 0, 1, 2, 3,... 144, c. X ∼B(144, 0.03) or P(4.32), d. 4.32, e. 0.0124 or 0.0133, f. 0.6300 or 0.6264, Solution to Exercise 4.15.15 (p. 195), a. X = the number of women that suffer from anorexia, b. 0, 1, 2, 3,... 600 (can leave off 600), c. X ∼P(3), d. 3, e. 0.0498, f. 0.1847, Solution to Exercise 4.15.17 (p. 196), a. X = the number of children for a Spanish woman, b. 0, 1, 2, 3,..., c. X ∼P(1.47), d. 0.2299, e. 0.5679, f. 0.4321, Solution to Exercise 4.15.19 (p. 196), a. X = the number of dealers she calls until she finds one with a used red Miata, b. 1, 2, 3,..., c. X ∼G(0.28), d. 3.57, e. 0.7313, f. 0.2497, Solution to Exercise 4.15.21 (p. 197), d. 4.31, e. 0.4079, f. 0.0163, Solution to Exercise 4.15.23 (p. 197), d. 2, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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217, e. 0.1353, f. 0.3233, Solution to Exercise 4.15.25 (p. 197), a. X = the number of seniors that participated in after-school sports all 4 years of high school, b. 0, 1, 2, 3,... 60, c. X~P (4.8), d. 4.8, e. Yes, f. 4, Solution to Exercise 4.15.27 (p. 198), a. X = the number of shell pieces in one cake, b. 0, 1, 2, 3,..., c. X~P (1.5), d. 1.5, e. 0.2231, f. 0.0001, g. Yes, Solution to Exercise 4.15.29 (p. 198), d. 0.0043, e. none, f. 3.3, Solution to Exercise 4.15.31 (p. 198), d. 3.02, e. No, f. 0.9997, h. 0.3881, i. 6.6207 pages, Solution to Exercise 4.15.33 (p. 199), D: 4.43, Solution to Exercise 4.15.34 (p. 199), A: 0.1476, Solution to Exercise 4.15.35 (p. 199), C: 0.4734, Solution to Exercise 4.15.36 (p. 199), A: The number of times Mrs. Plum’s cats wake her up each week, Solution to Exercise 4.15.37 (p. 200), D: 0.0671, Solution to Exercise 4.15.38 (p. 200), Partial Answer:, A: X = the number of DVDs a Video to Go customer rents, B: 0.12, C: 0.11, D: 0.77, Solution to Exercise 4.15.39 (p. 200), The variable of interest is X = net gain or loss, in dollars, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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218, , CHAPTER 4. DISCRETE RANDOM VARIABLES, , The face cards J, Q, K (Jack, Queen, King). There are(3)(4) = 12 face cards and 52 – 12 = 40 cards that are not, face cards., We first need to construct the probability distribution for X. We use the card and coin events to determine, the probability for each outcome, but we use the monetary value of X to determine the expected value., Card Event, , $X net gain or loss, , P(X), , Face Card and Heads, , 6, , (12/52)(1/2) = 6/52, , Face Card and Tails, , 2, , (12/52)(1/2) = 6/52, , (Not Face Card) and (H or T), , –2, , (40/52)(1) = 40/52, , Table 4.26, , • Expected value = (6)(6/52) + (2)(6/52) + (–2) (40/52) = –32/52, • Expected value = –$0.62, rounded to the nearest cent, • If you play this game repeatedly, over a long number of games, you would expect to lost 62 cents per, game, on average., • You should not play this game to win money because the expected value indicates an expected average loss., Solution to Exercise 4.15.40 (p. 201), Start by writing the probability distribution. X is net gain or loss = prize (if any) less $10 cost of ticket, X = $ net gain or loss, , P(X), , $500–$10=$490, , 1/100, , $100–$10=$90, , 2/100, , $25–$10=$15, , 4/100, , $0–$10=$–10, , 93/100), , Table 4.27, , Expected Value = (490)(1/100) + (90)(2/100) + (15)(4/100) + (–10) (93/100) = –$2. There is an expected loss, of $2 per ticket, on average., Solution to Exercise 4.15.41 (p. 201), , • X = number of questions answered correctly, • X∼B(10, 0.5), • We are interested in AT LEAST 70% of 10 questions correct. 70% of 10 is 7. We want to find the, probability that X is greater than or equal to 7. The event "at least 7" is the complement of "less than, or equal to 6"., • Using your calculator’s distribution menu: 1 – binomcdf(10, .5, 6) gives 0.171875, • The probability of getting at least 70% of the 10 questions correct when randomly guessing is approximately 0.172, Solution to Exercise 4.15.42 (p. 201), , • X = number of questions answered correctly, • X∼B(32, 1/3), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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219, , • We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find, P(x>24). The event "more than 24" is the complement of "less than or equal to 24"., • Using your calculator’s distribution menu: 1 - binomcdf(32, 1/3, 24), • P(x>24) = 0.00000026761, • The probability of getting more than 75% of the 32 questions correct when randomly guessing is very, small and practically zero., Solution to Exercise 4.15.43 (p. 201), A: X can take on the values 1, 2, 3, .... p = 2/6, q = 4/6, B: 0.2222, C: 3, , Solutions to Review, Solution to Exercise 4.16.1 (p. 202), C, Solution to Exercise 4.16.2 (p. 202), B, Solution to Exercise 4.16.3 (p. 202), A, Solution to Exercise 4.16.4 (p. 203), 0.5773, Solution to Exercise 4.16.5 (p. 203), 0.0522, Solution to Exercise 4.16.6 (p. 203), B, Solution to Exercise 4.16.7 (p. 203), C, Solution to Exercise 4.16.8 (p. 203), C, Solution to Exercise 4.16.9 (p. 204), 0.2709, Solution to Exercise 4.16.10 (p. 204), B, Solution to Exercise 4.16.11 (p. 204), X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, ...25, Solution to Exercise 4.16.12 (p. 204), B (25, 0.04), Solution to Exercise 4.16.13 (p. 204), 0.0165, Solution to Exercise 4.16.14 (p. 204), 1, Solution to Exercise 4.16.15 (p. 204), C, Solution to Exercise 4.16.16 (p. 204), All words used by Tom Clancy in his novels, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 5, , Continuous Random Variables, 5.1 Continuous Random Variables1, 5.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Recognize and understand continuous probability density functions in general., • Recognize the uniform probability distribution and apply it appropriately., • Recognize the exponential probability distribution and apply it appropriately., , 5.1.2 Introduction, Continuous random variables have many applications. Baseball batting averages, IQ scores, the length, of time a long distance telephone call lasts, the amount of money a person carries, the length of time a, computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous, random variables., This chapter gives an introduction to continuous random variables and the many continuous distributions., We will be studying these continuous distributions for several chapters., NOTE: The values of discrete and continuous random variables can be ambiguous. For example,, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete, random variable. You count the miles. If X is the distance you drive to work, then you measure, values of X and X is a continuous random variable. How the random variable is defined is very, important., , 5.1.3 Properties of Continuous Probability Distributions, The graph of a continuous probability distribution is a curve. Probability is represented by area under the, curve., The curve is called the probability density function (abbreviated: pdf). We use the symbol f ( x ) to represent the curve. f ( x ) is the function that corresponds to the graph; we use the density function f ( x ) to, draw the graph of the probability distribution., 1 This, , content is available online at <http://cnx.org/content/m16808/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 221
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222, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , Area under the curve is given by a different function called the cumulative distribution function (abbreviated: cdf). The cumulative distribution function is used to evaluate probability as area., , •, •, •, •, , The outcomes are measured, not counted., The entire area under the curve and above the x-axis is equal to 1., Probability is found for intervals of x values rather than for individual x values., P (c < x < d) is the probability that the random variable X is in the interval between the values c and, d. P (c < x < d) is the area under the curve, above the x-axis, to the right of c and the left of d., • P ( x = c) = 0 The probability that x takes on any single individual value is 0. The area below the, curve, above the x-axis, and between x=c and x=c has no width, and therefore no area (area = 0)., Since the probability is equal to the area, the probability is also 0., , We will find the area that represents probability by using geometry, formulas, technology, or probability, tables. In general, calculus is needed to find the area under the curve for many probability density functions., When we use formulas to find the area in this textbook, the formulas were found by using the techniques, of integral calculus. However, because most students taking this course have not studied calculus, we will, not be using calculus in this textbook., There are many continuous probability distributions. When using a continuous probability distribution to, model probability, the distribution used is selected to best model and fit the particular situation., In this chapter and the next chapter, we will study the uniform distribution, the exponential distribution,, and the normal distribution. The following graphs illustrate these distributions., , Figure 5.1: The graph shows a Uniform Distribution with the area between x=3 and x=6 shaded to represent the probability that the value of the random variable X is in the interval between 3 and 6., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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223, , Figure 5.2: The graph shows an Exponential Distribution with the area between x=2 and x=4 shaded to, represent the probability that the value of the random variable X is in the interval between 2 and 4., , Figure 5.3: The graph shows the Standard Normal Distribution with the area between x=1 and x=2 shaded, to represent the probability that the value of the random variable X is in the interval between 1 and 2., , **With contributions from Roberta Bloom, , 5.2 Continuous Probability Functions2, We begin by defining a continuous probability density function. We use the function notation f ( x ). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the, functions we study are special. We define the function f ( x ) so that the area between it and the x-axis is, equal to a probability. Since the maximum probability is one, the maximum area is also one., For continuous probability distributions, PROBABILITY = AREA., 2 This, , content is available online at <http://cnx.org/content/m16805/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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224, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, Example 5.1, 1, 1, for 0 ≤ x ≤ 20. x = a real number. The graph of f ( x ) = 20, is a, Consider the function f ( x ) = 20, horizontal line. However, since 0 ≤ x ≤ 20 , f ( x ) is restricted to the portion between x = 0 and, x = 20, inclusive ., , f (x) =, , 1, 20, , for 0 ≤ x ≤ 20., , The graph of f ( x ) =, , 1, 20, , is a horizontal line segment when 0 ≤ x ≤ 20., , The area between f ( x ) =, 1, = 20 and height = 20, ., AREA = 20 ·, , 1, 20, , 1, 20, , where 0 ≤ x ≤ 20 and the x-axis is the area of a rectangle with base, , =1, , This particular function, where we have restricted x so that the area between the function and, the x-axis is 1, is an example of a continuous probability density function. It is used as a tool to, calculate probabilities., Suppose we want to find the area between f ( x ) =, , AREA = (2 − 0) ·, , 1, 20, , 1, 20, , and the x-axis where 0 < x < 2 ., , = 0.1, , (2 − 0) = 2 = base of a rectangle, 1, 20, , = the height., , The area corresponds to a probability. The probability that x is between 0 and 2 is 0.1, which can, be written mathematically as P(0<x<2) = P(x<2) = 0.1., Suppose we want to find the area between f ( x ) =, , 1, 20, , and the x-axis where 4 < x < 15 ., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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225, , AREA = (15 − 4) ·, , 1, 20, , = 0.55, , (15 − 4) = 11 = the base of a rectangle, 1, 20, , = the height., , The area corresponds to the probability P (4 < x < 15) = 0.55., Suppose we want to find P ( x = 15). On an x-y graph, x = 15 is a vertical, line. A vertical line has, , no width (or 0 width). Therefore, P ( x = 15) = (base)(height) = (0), , 1, 20, , = 0., , P ( X ≤ x ) (can be written as P ( X < x ) for continuous distributions) is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can use the CDF to, calculate P ( X > x ) . The CDF gives "area to the left" and P ( X > x ) gives "area to the right." We, calculate P ( X > x ) for continuous distributions as follows: P ( X > x ) = 1 − P ( X < x )., , Label the graph with f ( x ) and x. Scale the x and y axes with the maximum x and y values., 1, f ( x ) = 20, , 0 ≤ x ≤ 20., , P (2.3 < x < 12.7) = (base) (height) = (12.7 − 2.3), , , , 1, 20, , , , = 0.52, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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226, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , 5.3 The Uniform Distribution3, Example 5.2, The previous problem is an example of the uniform probability distribution., Illustrate the uniform distribution. The data that follows are 55 smiling times, in seconds, of an, eight-week old baby., 10.4, , 19.6, , 18.8, , 13.9, , 17.8, , 16.8, , 21.6, , 17.9, , 12.5, , 11.1, , 4.9, , 12.8, , 14.8, , 22.8, , 20.0, , 15.9, , 16.3, , 13.4, , 17.1, , 14.5, , 19.0, , 22.8, , 1.3, , 0.7, , 8.9, , 11.9, , 10.9, , 7.3, , 5.9, , 3.7, , 17.9, , 19.2, , 9.8, , 5.8, , 6.9, , 2.6, , 5.8, , 21.7, , 11.8, , 3.4, , 2.1, , 4.5, , 6.3, , 10.7, , 8.9, , 9.4, , 9.4, , 7.6, , 10.0, , 3.3, , 6.7, , 7.8, , 11.6, , 13.8, , 18.6, , Table 5.1, , sample mean = 11.49 and sample standard deviation = 6.23, We will assume that the smiling times, in seconds, follow a uniform distribution between 0 and 23, seconds, inclusive. This means that any smiling time from 0 to and including 23 seconds is equally, likely. The histogram that could be constructed from the sample is an empirical distribution that, closely matches the theoretical uniform distribution., Let X = length, in seconds, of an eight-week old baby’s smile., The notation for the uniform distribution is, X ∼ U ( a,b) where a = the lowest value of x and b = the highest value of x., The probability density function is f ( x ) =, , 1, b− a, , For this example, x ∼ U (0, 23) and f ( x ) =, , 1, 23−0, , for a ≤ x ≤ b., for 0 ≤ x ≤ 23., , Formulas for the theoretical mean and standard deviation are, q, ( b − a )2, µ = a+2 b and σ =, 12, For this problem, the theoretical mean and standard deviation are, q, (23−0)2, µ = 0+223 = 11.50 seconds and σ =, = 6.64 seconds, 12, Notice that the theoretical mean and standard deviation are close to the sample mean and standard, deviation., Example 5.3, Problem 1, What is the probability that a randomly chosen eight-week old baby smiles between 2 and 18, seconds?, Solution, Find P (2 < x < 18)., 3 This, , content is available online at <http://cnx.org/content/m16819/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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227, P (2 < x < 18) = (base) (height) = (18 − 2) ·, , 1, 23, , =, , 16, 23 ., , Problem 2, Find the 90th percentile for an eight week old baby’s smiling time., Solution, Ninety percent of the smiling times fall below the 90th percentile, k, so P ( x < k) = 0.90, P ( x < k ) = 0.90, , (base) (height) = 0.90, ( k − 0) ·, , 1, 23, , = 0.90, , k = 23 · 0.90 = 20.7, , Problem 3, Find the probability that a random eight week old baby smiles more than 12 seconds KNOWING, that the baby smiles MORE THAN 8 SECONDS., Solution, Find P ( x > 12| x > 8) There are two ways to do the problem. For the first way, use the fact that, this is a conditional and changes the sample space. The graph illustrates the new sample space., You already know the baby smiled more than 8 seconds., Write a new f ( x ): f ( x ) =, , 1, 23−8, , =, , 1, 15, , for 8 < x < 23, P ( x > 12| x > 8) = (23 − 12) ·, , 1, 15, , =, , 11, 15, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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228, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , For the second way, use the conditional formula from Probability Topics with the original distribution X ∼ U (0, 23):, P ( A| B) =, , P( A AND B), P( B), , For this problem, A is ( x > 12) and B is ( x > 8)., , So, P ( x > 12| x > 8) =, , ( x >12 AND x >8), P ( x >8), , =, , P( x >12), P ( x >8), , =, , 11, 23, 15, 23, , = 0.733, , Example 5.4, Uniform: The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive., Problem 1, What is the probability that a person waits fewer than 12.5 minutes?, Solution, Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. x ∼ U (0, 15). Write, 1, the probability density function. f ( x ) = 151−0 = 15, for 0 ≤ x ≤ 15., Find P ( x < 12.5). Draw a graph., P ( x < k ) = (base) (height) = (12.5 − 0) ·, , 1, 15, , = 0.8333, , The probability a person waits less than 12.5 minutes is 0.8333., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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229, , Problem 2, On the average, how long must a person wait?, Find the mean, µ, and the standard deviation, σ., Solution, µ = a+2 b = 152+0 = 7.5. On the average, a person must wait 7.5 minutes., q, q, ( b − a )2, (15−0)2, σ=, =, = 4.3. The Standard deviation is 4.3 minutes., 12, 12, Problem 3, Ninety percent of the time, the time a person must wait falls below what value?, N OTE : This asks for the 90th percentile., Solution, Find the 90th percentile. Draw a graph. Let k = the 90th percentile., , 1, P ( x < k ) = (base) (height) = (k − 0) · 15, 0.90 = k ·, , 1, 15, , k = (0.90) (15) = 13.5, k is sometimes called a critical value., The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5, minutes., , Example 5.5, Uniform: Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes,, inclusive. Let X = the time, in minutes, it takes a nine-year old child to eat a donut. Then X ∼, U (0.5, 4)., Problem 1, (Solution on p. 257.), The probability that a randomly selected nine-year old child eats a donut in at least two minutes, is _______., Problem 2, (Solution on p. 257.), Find the probability that a different nine-year old child eats a donut in more than 2 minutes given, that the child has already been eating the donut for more than 1.5 minutes., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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230, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, The second probability question has a conditional (refer to "Probability Topics (Section 3.1)"). You, are asked to find the probability that a nine-year old child eats a donut in more than 2 minutes, given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see the first example (Example 5.2)). You must reduce the sample space., First way: Since you already know the child has already been eating the donut for more than 1.5, minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes., Write a new f(x):, f (x) =, , 1, 4−1.5, , =, , 2, 5, , for 1.5 ≤ x ≤ 4., , Find P ( x > 2| x > 1.5). Draw a graph., , P ( x > 2| x > 1.5) = (base) (new height) = (4 − 2) (2/5) =?, The probability that a nine-year old child eats a donut in more than 2 minutes given that the child, has already been eating the donut for more than 1.5 minutes is 45 ., Second way: Draw the original graph for x ∼ U (0.5, 4). Use the conditional formula, P ( x > 2| x > 1.5) =, , P( x >2 AND x >1.5), P( x >1.5), , =, , P ( x >2), P( x >1.5), , =, , 2, 3.5, 2.5, 3.5, , = 0.8 =, , 4, 5, , NOTE : See "Summary of the Uniform and Exponential Probability Distributions (Section 5.5)" for, a full summary., , Example 5.6, Uniform: Ace Heating and Air Conditioning Service finds that the amount of time a repairman, needs to fix a furnace is uniformly distributed between 1.5 and 4 hours. Let x = the time needed, to fix a furnace. Then x ∼ U (1.5, 4)., 1., 2., 3., 4., , Find the problem that a randomly selected furnace repair requires more than 2 hours., Find the probability that a randomly selected furnace repair requires less than 3 hours., Find the 30th percentile of furnace repair times., The longest 25% of repair furnace repairs take at least how long? (In other words: Find the, minimum time for the longest 25% of repair times.) What percentile does this represent?, 5. Find the mean and standard deviation, Problem 1, Find the probability that a randomly selected furnace repair requires longer than 2 hours., Solution, To find f ( x ): f ( x ) =, , 1, 4−1.5, , =, , 1, 2.5, , so f ( x ) = 0.4, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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231, P(x>2) = (base)(height) = (4 − 2)(0.4) = 0.8, Example 4 Figure 1, , Figure 5.4: Uniform Distribution between 1.5 and 4 with shaded area between 2 and 4 representing the, probability that the repair time x is greater than 2, , Problem 2, Find the probability that a randomly selected furnace repair requires less than 3 hours. Describe, how the graph differs from the graph in the first part of this example., Solution, P ( x < 3) = (base)(height) = (3 − 1.5)(0.4) = 0.6, The graph of the rectangle showing the entire distribution would remain the same. However the, graph should be shaded between x=1.5 and x=3. Note that the shaded area starts at x=1.5 rather, than at x=0; since X∼U(1.5,4), x can not be less than 1.5., Example 4 Figure 2, , Figure 5.5: Uniform Distribution between 1.5 and 4 with shaded area between 1.5 and 3 representing the, probability that the repair time x is less than 3, , Problem 3, Find the 30th percentile of furnace repair times., Solution, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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232, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, Example 4 Figure 3, , Figure 5.6: Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the, shortest 30% of repair times., , P ( x < k ) = 0.30, P ( x < k ) = (base) (height) = (k − 1.5) · (0.4), 0.3 = (k − 1.5) (0.4) ; Solve to find k:, 0.75 = k − 1.5 , obtained by dividing both sides by 0.4, k = 2.25 , obtained by adding 1.5 to both sides, The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less., , Problem 4, The longest 25% of furnace repair times take at least how long? (Find the minimum time for, the longest 25% of repairs.), Solution, Example 4 Figure 4, , Figure 5.7: Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing, the longest 25% of repair times., , P ( x > k ) = 0.25, P ( x > k ) = (base) (height) = (4 − k) · (0.4), 0.25 = (4 − k)(0.4) ; Solve for k:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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233, 0.625 = 4 − k , obtained by dividing both sides by 0.4, −3.375 = −k , obtained by subtracting 4 from both sides, k=3.375, The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer)., Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are, 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times., , Problem 5, Find the mean and standard deviation, Solution, µ=, µ=, , a+b, 2, 1.5+4, 2, , q, and σ =, , ( b − a )2, 12, , q, , = 2.75 hours and σ =, , (4−1.5)2, 12, , = 0.7217 hours, , NOTE : See "Summary of the Uniform and Exponential Probability Distributions (Section 5.5)" for, a full summary., , **Example 5 contributed by Roberta Bloom, , 5.4 The Exponential Distribution4, The exponential distribution is often concerned with the amount of time until some specific event occurs., For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the, amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you, have in your pocket or purse approximately follows an exponential distribution., Values for an exponential random variable occur in the following way. There are fewer large values and, more small values. For example, the amount of money customers spend in one trip to the supermarket, follows an exponential distribution. There are more people that spend less money and fewer people that, spend large amounts of money., The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of, time a product lasts., Example 5.7, Illustrates the exponential distribution: Let X = amount of time (in minutes) a postal clerk, spends with his/her customer. The time is known to have an exponential distribution with the, average amount of time equal to 4 minutes., X is a continuous random variable since time is measured. It is given that µ = 4 minutes. To do, any calculations, you must know m, the decay parameter., m = µ1 . Therefore, m =, 4 This, , 1, 4, , = 0.25, , content is available online at <http://cnx.org/content/m16816/1.15/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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234, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, The standard deviation, σ, is the same as the mean. µ = σ, The distribution notation is X~Exp (m). Therefore, X~Exp (0.25)., The probability density function is f ( x ) = m · e−m· x The number e = 2.71828182846... It is a number, that is used often in mathematics. Scientific calculators have the key "e x ." If you enter 1 for x, the, calculator will display the value e., The curve is:, f ( x ) = 0.25 · e− 0.25· x where x is at least 0 and m = 0.25., For example, f (5) = 0.25 · e− 0.25·5 = 0.072, The graph is as follows:, , Notice the graph is a declining curve. When x = 0,, f ( x ) = 0.25 · e− 0.25·0 = 0.25 · 1 = 0.25 = m, Example 5.8, Problem 1, Find the probability that a clerk spends four to five minutes with a randomly selected customer., Solution, Find P (4 < x < 5)., The cumulative distribution function (CDF) gives the area to the left., P ( x < x ) = 1 − e−m· x, P ( x < 5) = 1 − e−0.25·5 = 0.7135 and P ( x < 4) = 1 − e−0.25·4 = 0.6321, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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235, , NOTE :, , You can do these calculations easily on a calculator., , The probability that a postal clerk spends four to five minutes with a randomly selected customer, is, P (4 < x < 5) = P ( x < 5) − P ( x < 4) = 0.7135 − 0.6321 = 0.0814, NOTE : TI-83+ and TI-84: On the home screen, enter (1-e^(-.25*5))-(1-e^(-.25*4)) or enter e^(-.25*4)e^(-.25*5)., , Problem 2, Half of all customers are finished within how long? (Find the 50th percentile), Solution, Find the 50th percentile., , P ( x < k ) = 0.50, k = 2.8 minutes (calculator or computer), Half of all customers are finished within 2.8 minutes., You can also do the calculation as follows:, P ( x < k ) = 0.50 and P ( x < k) = 1 − e−0.25·k, Therefore, 0.50 = 1 − e−0.25·k and e−0.25·k = 1 − 0.50 = 0.5, , , Take natural logs: ln e−0.25·k = ln (0.50). So, −0.25 · k = ln (0.50), Solve for k: k =, , ln(.50), −0.25, , = 2.8 minutes, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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236, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, LN (1− AreaToTheLe f t), −m, , NOTE :, , A formula for the percentile k is k =, , NOTE :, , TI-83+ and TI-84: On the home screen, enter LN(1-.50)/-.25. Press the (-) for the negative., , where LN is the natural log., , Problem 3, Which is larger, the mean or the median?, Solution, Is the mean or median larger?, From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is 4 minutes. The, mean is larger., , 5.4.1 Optional Collaborative Classroom Activity, Have each class member count the change he/she has in his/her pocket or purse. Your instructor will, record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use 5, intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then, calculate the mean., Let X = the amount of money a student in your class has in his/her pocket or purse., The distribution for X is approximately exponential with mean, µ = _______ and m = _______. The standard, deviation, σ = ________., Draw the appropriate exponential graph. You should label the x and y axes, the decay rate, and the mean., Shade the area that represents the probability that one student has less than $.40 in his/her pocket or purse., (Shade P ( x < 0.40))., Example 5.9, On the average, a certain computer part lasts 10 years. The length of time the computer part lasts, is exponentially distributed., Problem 1, What is the probability that a computer part lasts more than 7 years?, Solution, Let x = the amount of time (in years) a computer part lasts., µ = 10 so m =, , 1, µ, , =, , 1, 10, , = 0.1, , Find P ( x > 7). Draw a graph., P ( x > 7) = 1 − P ( x < 7)., Since P ( X < x ) = 1 − e−mx then P ( X > x ) = 1 − (1 − e−m· x ) = e−m· x, P ( x > 7) = e−0.1·7 = 0.4966. The probability that a computer part lasts more than 7 years is 0.4966., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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237, NOTE :, , TI-83+ and TI-84: On the home screen, enter e^(-.1*7)., , Problem 2, On the average, how long would 5 computer parts last if they are used one after another?, Solution, On the average, 1 computer part lasts 10 years. Therefore, 5 computer parts, if they are used one, right after the other would last, on the average,, , (5) (10) = 50 years., Problem 3, Eighty percent of computer parts last at most how long?, Solution, Find the 80th percentile. Draw a graph. Let k = the 80th percentile., , Solve for k: k =, , ln(1−.80), −0.1, , = 16.1 years, , Eighty percent of the computer parts last at most 16.1 years., NOTE :, , TI-83+ and TI-84: On the home screen, enter LN(1 - .80)/-.1, , Problem 4, What is the probability that a computer part lasts between 9 and 11 years?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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238, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, Solution, Find P (9 < x < 11). Draw a graph., , P (9 < x < 11) = P ( x < 11) − P ( x < 9) =, 0.0737. (calculator or computer), , , , 1 − e−0.1·11 − 1 − e−0.1·9 = 0.6671 − 0.5934 =, , The probability that a computer part lasts between 9 and 11 years is 0.0737., NOTE :, , TI-83+ and TI-84: On the home screen, enter e^(-.1*9) - e^(-.1*11)., , Example 5.10, Suppose that the length of a phone call, in minutes, is an exponential random variable with decay, 1, parameter = 12, . If another person arrives at a public telephone just before you, find the probability, that you will have to wait more than 5 minutes. Let X = the length of a phone call, in minutes., Problem, (Solution on p. 257.), What is m, µ, and σ? The probability that you must wait more than 5 minutes is _______ ., , NOTE : A summary for exponential distribution is available in "Summary of The Uniform and, Exponential Probability Distributions (Section 5.5)"., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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239, , 5.5 Summary of the Uniform and Exponential Probability Distributions5, Formula 5.1: Uniform, X = a real number between a and b (in some instances, X can take on the values a and b). a =, smallest X ; b = largest X, X ∼ U ( a,b), The mean is µ =, , a+b, 2, , q, The standard deviation is σ =, , ( b − a )2, 12, , Probability density function: f ( X ) =, , 1, b− a, , for a ≤ X ≤ b, , Area to the Left of x: P ( X < x ) = (base)(height), Area to the Right of x: P ( X > x ) = (base)(height), Area Between c and d: P (c < X < d) = (base) (height) = (d − c) (height)., Formula 5.2: Exponential, X ∼ Exp (m), X = a real number, 0 or larger. m = the parameter that controls the rate of decay or decline, The mean and standard deviation are the same., µ=σ=, , 1, m, , and m =, , 1, µ, , =, , 1, σ, , The probability density function: f ( X ) = m · e−m· X , X ≥ 0, Area to the Left of x: P ( X < x ) = 1 − e−m· x, Area to the Right of x: P ( X > x ) = e−m· x, , , Area Between c and d: P (c < X < d) = P ( X < d) − P ( X < c) = 1 − e− m·d − (1 − e− m·c ) =, e− m·c − e− m·d, Percentile, k: k =, , 5 This, , LN(1-AreaToTheLeft), −m, , content is available online at <http://cnx.org/content/m16813/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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240, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , 5.6 Practice 1: Uniform Distribution6, 5.6.1 Student Learning Outcomes, • The student will analyze data following a uniform distribution., , 5.6.2 Given, The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5, years) to 9.5 years., , 5.6.3 Describe the Data, Exercise 5.6.1, What is being measured here?, , (Solution on p. 257.), , Exercise 5.6.2, In words, define the Random Variable X., , (Solution on p. 257.), , Exercise 5.6.3, Are the data discrete or continuous?, , (Solution on p. 257.), , Exercise 5.6.4, The interval of values for x is:, , (Solution on p. 257.), , Exercise 5.6.5, The distribution for X is:, , (Solution on p. 257.), , 5.6.4 Probability Distribution, Exercise 5.6.6, Write the probability density function., , (Solution on p. 257.), , Exercise 5.6.7, Graph the probability distribution., , (Solution on p. 257.), , a. Sketch the graph of the probability distribution., , Figure 5.8, , 6 This, , content is available online at <http://cnx.org/content/m16812/1.14/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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241, b. Identify the following values:, i. Lowest value for x:, ii. Highest value for x:, iii. Height of the rectangle:, iv. Label for x-axis (words):, v. Label for y-axis (words):, , 5.6.5 Random Probability, Exercise 5.6.8, (Solution on p. 257.), Find the probability that a randomly chosen car in the lot was less than 4 years old., a. Sketch the graph. Shade the area of interest., , Figure 5.9, , b. Find the probability. P ( x < 4) =, Exercise 5.6.9, (Solution on p. 257.), Out of just the cars less than 7.5 years old, find the probability that a randomly chosen car in the, lot was less than 4 years old., a. Sketch the graph. Shade the area of interest., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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242, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , Figure 5.10, , b. Find the probability. P ( x < 4 | x < 7.5) =, Exercise 5.6.10: Discussion Question, What has changed in the previous two problems that made the solutions different?, , 5.6.6 Quartiles, Exercise 5.6.11, Find the average age of the cars in the lot., , (Solution on p. 257.), , Exercise 5.6.12, (Solution on p. 257.), Find the third quartile of ages of cars in the lot. This means you will have to find the value such, that 34 , or 75%, of the cars are at most (less than or equal to) that age., a. Sketch the graph. Shade the area of interest., , Figure 5.11, , b. Find the value k such that P ( x < k) = 0.75., c. The third quartile is:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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243, , 5.7 Practice 2: Exponential Distribution7, 5.7.1 Student Learning Outcomes, • The student will analyze data following the exponential distribution., , 5.7.2 Given, Carbon-14 is a radioactive element with a half-life of about 5730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121 . We start with 1 gram of carbon-14. We are interested in the time (years), it takes to decay carbon-14., , 5.7.3 Describe the Data, Exercise 5.7.1, What is being measured here?, Exercise 5.7.2, Are the data discrete or continuous?, , (Solution on p. 258.), , Exercise 5.7.3, In words, define the Random Variable X., , (Solution on p. 258.), , Exercise 5.7.4, What is the decay rate (m)?, , (Solution on p. 258.), , Exercise 5.7.5, The distribution for X is:, , (Solution on p. 258.), , 5.7.4 Probability, Exercise 5.7.6, (Solution on p. 258.), Find the amount (percent of 1 gram) of carbon-14 lasting less than 5730 years. This means, find, P ( x < 5730)., a. Sketch the graph. Shade the area of interest., , Figure 5.12, , 7 This, , content is available online at <http://cnx.org/content/m16811/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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244, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, b. Find the probability. P ( x < 5730) =, Exercise 5.7.7, Find the percentage of carbon-14 lasting longer than 10,000 years., , (Solution on p. 258.), , a. Sketch the graph. Shade the area of interest., , Figure 5.13, , b. Find the probability. P ( x > 10000) =, Exercise 5.7.8, Thirty percent (30%) of carbon-14 will decay within how many years?, , (Solution on p. 258.), , a. Sketch the graph. Shade the area of interest., , Figure 5.14, , b. Find the value k such that P ( x < k) = 0.30., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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245, , 5.8 Homework8, For each probability and percentile problem, DRAW THE PICTURE!, Exercise 5.8.1, Consider the following experiment. You are one of 100 people enlisted to take part in a study to, determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses, if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage, of nurses with an R.N. degree. You give that percentage to your supervisor., a. What part of the experiment will yield discrete data?, b. What part of the experiment will yield continuous data?, Exercise 5.8.2, When age is rounded to the nearest year, do the data stay continuous, or do they become discrete?, Why?, Exercise 5.8.3, (Solution on p. 258.), Births are approximately uniformly distributed between the 52 weeks of the year. They can be, said to follow a Uniform Distribution from 1 – 53 (spread of 52 weeks)., a. X ∼, b. Graph the probability distribution., c. f ( x ) =, d. µ =, e. σ =, f. Find the probability that a person is born at the exact moment week 19 starts. That is, find, P ( x = 19) =, g. P (2 < x < 31) =, h. Find the probability that a person is born after week 40., i. P (12 < x | x < 28) =, j. Find the 70th percentile., k. Find the minimum for the upper quarter., Exercise 5.8.4, A random number generator picks a number from 1 to 9 in a uniform manner., a. X~, b. Graph the probability distribution., c. f ( x ) =, d. µ =, e. σ =, f. P (3.5 < x < 7.25) =, g. P ( x > 5.67) =, h. P ( x > 5 | x > 3) =, i. Find the 90th percentile., Exercise 5.8.5, (Solution on p. 258.), The time (in minutes) until the next bus departs a major bus depot follows a distribution with, 1, where x goes from 25 to 45 minutes., f ( x ) = 20, a. Define the random variable. X =, 8 This, , content is available online at <http://cnx.org/content/m16807/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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246, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, b. X~, c. Graph the probability distribution., d. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous)., e. µ =, f. σ =, g. Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement., h. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of, the distribution. Shade the area of interest. Write the answer in a probability statement., i. P (25 < x < 55) = _________. State this in a probability statement (similar to g and h ), draw, the picture, and find the probability., j. Find the 90th percentile. This means that 90% of the time, the time is less than _____ minutes., k. Find the 75th percentile. In a complete sentence, state what this means. (See j.), l. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least, 30 minutes., Exercise 5.8.6, According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena, Hospital, the people who follow his program lose between 6 and 15 pounds a month until they, approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are, interested in the weight loss of a randomly selected individual following the program for one, month. (Source: The McDougall Program for Maximum Weight Loss by John A. McDougall,, M.D.), a. Define the random variable. X =, b. X~, c. Graph the probability distribution., d. f ( x ) =, e. µ =, f. σ =, g. Find the probability that the individual lost more than 10 pounds in a month., h. Suppose it is known that the individual lost more than 10 pounds in a month. Find the probability that he lost less than 12 pounds in the month., i. P (7 < x < 13 | x > 9) = __________. State this in a probability question (similar to g and h),, draw the picture, and find the probability., Exercise 5.8.7, (Solution on p. 258.), A subway train on the Red Line arrives every 8 minutes during rush hour. We are interested in the, length of time a commuter must wait for a train to arrive. The time follows a uniform distribution., a. Define the random variable. X =, b. X~, c. Graph the probability distribution., d. f ( x ) =, e. µ =, f. σ =, g. Find the probability that the commuter waits less than one minute., h. Find the probability that the commuter waits between three and four minutes., i. 60% of commuters wait more than how long for the train? State this in a probability question, (similar to g and h), draw the picture, and find the probability., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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247, Exercise 5.8.8, The age of a first grader on September 1 at Garden Elementary School is uniformly distributed, from 5.8 to 6.8 years. We randomly select one first grader from the class., a. Define the random variable. X =, b. X~, c. Graph the probability distribution., d. f ( x ) =, e. µ =, f. σ =, g. Find the probability that she is over 6.5 years., h. Find the probability that she is between 4 and 6 years., i. Find the 70th percentile for the age of first graders on September 1 at Garden Elementary School., Exercise 5.8.9, Let X~Exp(0.1), , (Solution on p. 259.), , a. decay rate=, b. µ =, c. Graph the probability distribution function., d. On the above graph, shade the area corresponding to P ( x < 6) and find the probability., e. Sketch a new graph, shade the area corresponding to P (3 < x < 6) and find the probability., f. Sketch a new graph, shade the area corresponding to P ( x > 7) and find the probability., g. Sketch a new graph, shade the area corresponding to the 40th percentile and find the value., h. Find the average value of x., Exercise 5.8.10, Suppose that the length of long distance phone calls, measured in minutes, is known to have an, exponential distribution with the average length of a call equal to 8 minutes., a. Define the random variable. X =, b. Is X continuous or discrete?, c. X~, d. µ =, e. σ =, f. Draw a graph of the probability distribution. Label the axes., g. Find the probability that a phone call lasts less than 9 minutes., h. Find the probability that a phone call lasts more than 9 minutes., i. Find the probability that a phone call lasts between 7 and 9 minutes., j. If 25 phone calls are made one after another, on average, what would you expect the total to be?, Why?, Exercise 5.8.11, (Solution on p. 259.), Suppose that the useful life of a particular car battery, measured in months, decays with parameter, 0.025. We are interested in the life of the battery., a. Define the random variable. X =, b. Is X continuous or discrete?, c. X~, d. On average, how long would you expect 1 car battery to last?, e. On average, how long would you expect 9 car batteries to last, if they are used one after another?, f. Find the probability that a car battery lasts more than 36 months., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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248, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, g. 70% of the batteries last at least how long?, Exercise 5.8.12, The percent of persons (ages 5 and older) in each state who speak a language at home other than, English is approximately exponentially distributed with a mean of 9.848 . Suppose we randomly, pick a state. (Source: Bureau of the Census, U.S. Dept. of Commerce), a. Define the random variable. X =, b. Is X continuous or discrete?, c. X~, d. µ =, e. σ =, f. Draw a graph of the probability distribution. Label the axes., g. Find the probability that the percent is less than 12., h. Find the probability that the percent is between 8 and 14., i. The percent of all individuals living in the United States who speak a language at home other, than English is 13.8 ., i. Why is this number different from 9.848%?, ii. What would make this number higher than 9.848%?, Exercise 5.8.13, (Solution on p. 259.), The time (in years) after reaching age 60 that it takes an individual to retire is approximately, exponentially distributed with a mean of about 5 years. Suppose we randomly pick one retired, individual. We are interested in the time after age 60 to retirement., a. Define the random variable. X =, b. Is X continuous or discrete?, c. X~, d. µ =, e. σ =, f. Draw a graph of the probability distribution. Label the axes., g. Find the probability that the person retired after age 70., h. Do more people retire before age 65 or after age 65?, i. In a room of 1000 people over age 80, how many do you expect will NOT have retired yet?, Exercise 5.8.14, The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150., a. Define the random variable. X =, b. X~, c. µ =, d. σ =, e. Draw a graph of the probability distribution. Label the axes., f. Find the probability that a car required over $300 for maintenance during its first year., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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249, , 5.8.1 Try these multiple choice problems, The next three questions refer to the following information. The average lifetime of a certain new cell, phone is 3 years. The manufacturer will replace any cell phone failing within 2 years of the date of purchase., The lifetime of these cell phones is known to follow an exponential distribution., Exercise 5.8.15, The decay rate is, A., B., C., D., , (Solution on p. 259.), , 0.3333, 0.5000, 2.0000, 3.0000, , Exercise 5.8.16, (Solution on p. 259.), What is the probability that a phone will fail within 2 years of the date of purchase?, A., B., C., d., , 0.8647, 0.4866, 0.2212, 0.9997, , Exercise 5.8.17, What is the median lifetime of these phones (in years)?, A., B., C., D., , (Solution on p. 259.), , 0.1941, 1.3863, 2.0794, 5.5452, , The next three questions refer to the following information. The Sky Train from the terminal to the rental, car and long term parking center is supposed to arrive every 8 minutes. The waiting times for the train are, known to follow a uniform distribution., Exercise 5.8.18, What is the average waiting time (in minutes)?, A., B., C., D., , 0.0000, 2.0000, 3.0000, 4.0000, , Exercise 5.8.19, Find the 30th percentile for the waiting times (in minutes)., A., B., C., D., , (Solution on p. 259.), , (Solution on p. 259.), , 2.0000, 2.4000, 2.750, 3.000, , Exercise 5.8.20, (Solution on p. 259.), The probability of waiting more than 7 minutes given a person has waited more than 4 minutes, is?, A. 0.1250, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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250, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, B. 0.2500, C. 0.5000, D. 0.7500, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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251, , 5.9 Review9, Exercise 5.9.1 – Exercise 5.9.7 refer to the following study: A recent study of mothers of junior high school, children in Santa Clara County reported that 76% of the mothers are employed in paid positions. Of those, mothers who are employed, 64% work full-time (over 35 hours per week), and 36% work part-time. However, out of all of the mothers in the population, 49% work full-time. The population under study is made, up of mothers of junior high school children in Santa Clara County., Let E =employed, Let F =full-time employment, Exercise 5.9.1, , (Solution on p. 259.), , a. Find the percent of all mothers in the population that NOT employed., b. Find the percent of mothers in the population that are employed part-time., Exercise 5.9.2, The type of employment is considered to be what type of data?, , (Solution on p. 259.), , Exercise 5.9.3, (Solution on p. 259.), Find the probability that a randomly selected mother works part-time given that she is employed., Exercise 5.9.4, (Solution on p. 260.), Find the probability that a randomly selected person from the population will be employed OR, work full-time., Exercise 5.9.5, (Solution on p. 260.), Based upon the above information, are being employed AND working part-time:, a. mutually exclusive events? Why or why not?, b. independent events? Why or why not?, Exercise 5.9.6 - Exercise 5.9.7 refer to the following: We randomly pick 10 mothers from the above population. We are interested in the number of the mothers that are employed. Let X =number of mothers that, are employed., Exercise 5.9.6, State the distribution for X., , (Solution on p. 260.), , Exercise 5.9.7, Find the probability that at least 6 are employed., , (Solution on p. 260.), , Exercise 5.9.8, (Solution on p. 260.), We expect the Statistics Discussion Board to have, on average, 14 questions posted to it per week., We are interested in the number of questions posted to it per day., a., b., c., d., , Define X., What are the values that the random variable may take on?, State the distribution for X., Find the probability that from 10 to 14 (inclusive) questions are posted to the Listserv on a, randomly picked day., , Exercise 5.9.9, (Solution on p. 260.), A person invests $1000 in stock of a company that hopes to go public in 1 year., , • The probability that the person will lose all his money after 1 year (i.e. his stock will be, worthless) is 35%., 9 This, , content is available online at <http://cnx.org/content/m16810/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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252, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , • The probability that the person’s stock will still have a value of $1000 after 1 year (i.e. no, profit and no loss) is 60%., • The probability that the person’s stock will increase in value by $10,000 after 1 year (i.e. will, be worth $11,000) is 5%., Find the expected PROFIT after 1 year., Exercise 5.9.10, (Solution on p. 260.), Rachel’s piano cost $3000. The average cost for a piano is $4000 with a standard deviation of, $2500. Becca’s guitar cost $550. The average cost for a guitar is $500 with a standard deviation, of $200. Matt’s drums cost $600. The average cost for drums is $700 with a standard deviation of, $100. Whose cost was lowest when compared to his or her own instrument? Justify your answer., Exercise 5.9.11, , (Solution on p. 260.), , For each statement below, explain why each is either true or false., a., b., c., d., , 25% of the data are at most 5., There is the same amount of data from 4 – 5 as there is from 5 – 7., There are no data values of 3., 50% of the data are 4., , Exercise 5.9.12 – Exercise 5.9.13 refer to the following: 64 faculty members were asked the number of, cars they owned (including spouse and children’s cars). The results are given in the following graph:, , Exercise 5.9.12, Find the approximate number of responses that were “3.”, , (Solution on p. 260.), , Exercise 5.9.13, (Solution on p. 260.), Find the first, second and third quartiles. Use them to construct a box plot of the data., Exercise 5.9.14 – Exercise 5.9.15 refer to the following study done of the Girls soccer team “Snow Leopards”:, Hair Style, , Hair Color, blond, , brown, , black, , ponytail, , 3, , 2, , 5, , plain, , 2, , 2, , 1, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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253, Table 5.2, , Suppose that one girl from the Snow Leopards is randomly selected., Exercise 5.9.14, (Solution on p. 260.), Find the probability that the girl has black hair GIVEN that she wears a ponytail., Exercise 5.9.15, Find the probability that the girl wears her hair plain OR has brown hair., , (Solution on p. 260.), , Exercise 5.9.16, (Solution on p. 260.), Find the probability that the girl has blond hair AND that she wears her hair plain., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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255, , Figure 5.15, , 2. Construct a histogram of the empirical data. Make 5 bars., , Figure 5.16, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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256, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , 5.10.4 Describe the Data, 1. Describe the shape of each graph. Use 2 – 3 complete sentences. (Keep it simple. Does the graph go, straight across, does it have a V shape, does it have a hump in the middle or at either end, etc.? One, way to help you determine a shape, is to roughly draw a smooth curve through the top of the bars.), 2. Describe how changing the number of bars might change the shape., , 5.10.5 Theoretical Distribution, 1. In words, X =, 2. The theoretical distribution of X is X ∼ U (0, 1). Use it for this part., 3. In theory, based upon the distribution X ∼ U (0, 1), complete the following., a., b., c., d., e., , µ=, σ=, 1st quartile =, 3rd quartile =, median = __________, , 4. Are the empirical values (the data) in the section titled "Collect the Data" close to the corresponding, theoretical values above? Why or why not?, , 5.10.6 Plot the Data, 1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges., 2. Do you notice any potential outliers? If so, which values are they? Either way, numerically justify your, answer. (Recall that any DATA are less than Q1 – 1.5*IQR or more than Q3 + 1.5*IQR are potential, outliers. IQR means interquartile range.), , 5.10.7 Compare the Data, 1. For each part below, use a complete sentence to comment on how the value obtained from the data, compares to the theoretical value you expected from the distribution in the section titled "Theoretical, Distribution.", a. minimum value:, b. 1st quartile:, c. median:, d. third quartile:, e. maximum value:, f. width of IQR:, g. overall shape:, 2. Based on your comments in the section titled "Collect the Data", how does the box plot fit or not fit, what you would expect of the distribution in the section titled "Theoretical Distribution?", , 5.10.8 Discussion Question, 1. Suppose that the number of values generated was 500, not 50. How would that affect what you would, expect the empirical data to be and the shape of its graph to look like?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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257, , Solutions to Exercises in Chapter 5, Solution to Example 5.5, Problem 1 (p. 229), 0.5714, Solution to Example 5.5, Problem 2 (p. 229), 4, 5, , Solution to Example 5.10, Problem (p. 238), 1, • m = 12, • µ = 12, • σ = 12, , P ( x > 5) = 0.6592, , Solutions to Practice 1: Uniform Distribution, Solution to Exercise 5.6.1 (p. 240), The age of cars in the staff parking lot, Solution to Exercise 5.6.2 (p. 240), X = The age (in years) of cars in the staff parking lot, Solution to Exercise 5.6.3 (p. 240), Continuous, Solution to Exercise 5.6.4 (p. 240), 0.5 - 9.5, Solution to Exercise 5.6.5 (p. 240), X ∼ U (0.5, 9.5), Solution to Exercise 5.6.6 (p. 240), f ( x ) = 19, Solution to Exercise 5.6.7 (p. 240), b.i. 0.5, b.ii. 9.5, b.iii. 19, b.iv. Age of Cars, b.v. f ( x ), Solution to Exercise 5.6.8 (p. 241), b.:, , 3.5, 9, , Solution to Exercise 5.6.9 (p. 241), b:, , 3.5, 7, , Solution to Exercise 5.6.11 (p. 242), µ=5, Solution to Exercise 5.6.12 (p. 242), b. k = 7.25, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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258, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , Solutions to Practice 2: Exponential Distribution, Solution to Exercise 5.7.2 (p. 243), Continuous, Solution to Exercise 5.7.3 (p. 243), X = Time (years) to decay carbon-14, Solution to Exercise 5.7.4 (p. 243), m = 0.000121, Solution to Exercise 5.7.5 (p. 243), X ∼ Exp(0.000121), Solution to Exercise 5.7.6 (p. 243), b. P ( x < 5730) = 0.5001, Solution to Exercise 5.7.7 (p. 244), b. P ( x > 10000) = 0.2982, Solution to Exercise 5.7.8 (p. 244), b. k = 2947.73, , Solutions to Homework, Solution to Exercise 5.8.3 (p. 245), a. X~U (1, 53), 1, where 1 ≤ x ≤ 53, c. f ( x ) = 52, d. 27, e. 15.01, f. 0, 29, g. 52, 13, h. 52, i. 16, 27, j. 37.4, k. 40, Solution to Exercise 5.8.5 (p. 245), b. X~U (25, 45), d. uniform; continuous, e. 35 minutes, f. 5.8 minutes, g. 0.25, h. 0.5, i. 1, j. 43 minutes, k. 40 minutes, l. 0.3333, Solution to Exercise 5.8.7 (p. 246), b. X~U (0, 8), d. f ( x ) = 81 where 0 ≤ x ≤ 8, e. 4, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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259, f. 2.31, g. 18, h. 81, i. 3.2, Solution to Exercise 5.8.9 (p. 247), a. 0.1, b. 10, d. 0.4512, e. 0.1920, f. 0.4966, g. 5.11, h. 10, Solution to Exercise 5.8.11 (p. 247), c. X~Exp (0.025), d. 40 months, e. 360 months, f. 0.4066, g. 14.27, Solution to Exercise 5.8.13 (p. 248), , c. X~Exp 51, d. 5, e. 5, g. 0.1353, h. Before, i. 18.3, Solution to Exercise 5.8.15 (p., A, Solution to Exercise 5.8.16 (p., B, Solution to Exercise 5.8.17 (p., C, Solution to Exercise 5.8.18 (p., D, Solution to Exercise 5.8.19 (p., B, Solution to Exercise 5.8.20 (p., B, , 249), 249), 249), 249), 249), 249), , Solutions to Review, Solution to Exercise 5.9.1 (p. 251), a. 24%, b. 27%, Solution to Exercise 5.9.2 (p. 251), Qualitative, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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260, , CHAPTER 5. CONTINUOUS RANDOM VARIABLES, , Solution to Exercise 5.9.3 (p. 251), 0.36, Solution to Exercise 5.9.4 (p. 251), 0.7636, Solution to Exercise 5.9.5 (p. 251), a. No,, b. No,, Solution to Exercise 5.9.6 (p. 251), B (10, 0.76), Solution to Exercise 5.9.7 (p. 251), 0.9330, Solution to Exercise 5.9.8 (p. 251), a., b., c., d., , X = the number of questions posted to the Statistics Listserv per day, x = 0, 1, 2, ..., X~P (2), 0, , Solution to Exercise 5.9.9 (p. 251), $150, Solution to Exercise 5.9.10 (p. 252), Matt, Solution to Exercise 5.9.11 (p. 252), a., b., c., d., , False, True, False, False, , Solution to Exercise 5.9.12 (p., 16, Solution to Exercise 5.9.13 (p., 2, 2, 3, Solution to Exercise 5.9.14 (p., 5, 10 = 0.5, Solution to Exercise 5.9.15 (p., 7, 15, , 252), 252), 253), 253), , Solution to Exercise 5.9.16 (p. 253), 2, 15, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 6, , The Normal Distribution, 6.1 The Normal Distribution1, 6.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Recognize the normal probability distribution and apply it appropriately., • Recognize the standard normal probability distribution and apply it appropriately., • Compare normal probabilities by converting to the standard normal distribution., , 6.1.2 Introduction, The normal, a continuous distribution, is the most important of all the distributions. It is widely used, and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines., Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics., Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are, normally distributed. Often real estate prices fit a normal distribution. The normal distribution is extremely, important but it cannot be applied to everything in the real world., In this chapter, you will study the normal distribution, the standard normal, and applications associated, with them., , 6.1.3 Optional Collaborative Classroom Activity, Your instructor will record the heights of both men and women in your class, separately. Draw histograms, of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do, you think that if you had recorded 200 data values for men and 200 for women that the curves would look, bell-shaped? Calculate the mean for each data set. Write the means on the x-axis of the appropriate graph, below the peak. Shade the approximate area that represents the probability that one randomly chosen, male is taller than 72 inches. Shade the approximate area that represents the probability that one randomly, chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability, appear to be more than 0.5?, 1 This, , content is available online at <http://cnx.org/content/m16979/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 261
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262, , CHAPTER 6. THE NORMAL DISTRIBUTION, , The normal distribution has two parameters (two numerical descriptive measures), the mean (µ) and the, standard deviation (σ). If X is a quantity to be measured that has a normal distribution with mean (µ) and, the standard deviation (σ), we designate this by writing, NORMAL:X ∼ N (µ, σ ), , The probability density function is a rather complicated function. Do not memorize it. It is not necessary., f (x) =, , √1, σ · 2· π, , 1, , · e− 2 ·(, , x −µ 2, σ, , ), , The cumulative distribution function is P ( X < x ) . It is calculated either by a calculator or a computer or, it is looked up in a table. Technology has made the tables basically obsolete. For that reason, as well as, the fact that there are various table formats, we are not including table instructions in this chapter. See the, NOTE in this chapter in Calculation of Probabilities., The curve is symmetrical about a vertical line drawn through the mean, µ. In theory, the mean is the same, as the median since the graph is symmetric about µ. As the notation indicates, the normal distribution, depends only on the mean and the standard deviation. Since the area under the curve must equal one, a, change in the standard deviation, σ, causes a change in the shape of the curve; the curve becomes fatter or, skinnier depending on σ. A change in µ causes the graph to shift to the left or right. This means there are an, infinite number of normal probability distributions. One of special interest is called the standard normal, distribution., , 6.2 The Standard Normal Distribution2, The standard normal distribution is a normal distribution of standardized values called z-scores. A zscore is measured in units of the standard deviation. For example, if the mean of a normal distribution is, 5 and the standard deviation is 2, the value 11 is 3 standard deviations above (or to the right of) the mean., The calculation is:, x = µ + (z) σ = 5 + (3) (2) = 11, , (6.0), , The z-score is 3., The mean for the standard normal distribution is 0 and the standard deviation is 1. The transformation, x −µ, , produces the distribution Z ∼ N (0, 1), z= σ, mean µ and standard deviation σ., 2 This, , . The value x comes from a normal distribution with, , content is available online at <http://cnx.org/content/m16986/1.7/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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263, , 6.3 Z-scores3, If X is a normally distributed random variable and X ∼ N (µ, σ), then the z-score is:, x−µ, (6.0), σ, The z-score tells you how many standard deviations that the value x is above (to the right of) or below, (to the left of) the mean, µ. Values of x that are larger than the mean have positive z-scores and values of x, that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of 0., z=, , Example 6.1, Suppose X ∼ N (5, 6). This says that X is a normally distributed random variable with mean, µ = 5 and standard deviation σ = 6. Suppose x = 17. Then:, 17 − 5, x−µ, =, =2, (6.0), σ, 6, This means that x = 17 is 2 standard deviations (2σ ) above or to the right of the mean µ = 5., The standard deviation is σ = 6., z=, , Notice that:, 5 + 2 · 6 = 17, , (The pattern is µ + zσ = x.), , (6.0), , Now suppose x = 1. Then:, 1−5, x−µ, =, = −0.67, (rounded to two decimal places), (6.0), σ, 6, This means that x = 1 is 0.67 standard deviations (− 0.67σ ) below or to the left of the mean, µ = 5. Notice that:, z=, , 5 + (−0.67) (6) is approximately equal to 1, , (This has the pattern µ + (−0.67) σ = 1 ), , Summarizing, when z is positive, x is above or to the right of µ and when z is negative, x is to the, left of or below µ., Example 6.2, Some doctors believe that a person can lose 5 pounds, on the average, in a month by reducing, his/her fat intake and by exercising consistently. Suppose weight loss has a normal distribution., Let X = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation, of 2 pounds. X ∼ N (5, 2). Fill in the blanks., Problem 1, (Solution on p. 285.), Suppose a person lost 10 pounds in a month. The z-score when x = 10 pounds is z = 2.5, (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right, or left) of the mean _____ (What is the mean?)., Problem 2, (Solution on p. 285.), Suppose a person gained 3 pounds (a negative weight loss). Then z = __________. This z-score, tells you that x = −3 is ________ standard deviations to the __________ (right or left) of the mean., Suppose the random variables X and Y have the following normal distributions: X ∼ N (5, 6) and, Y ∼ N (2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z?, z=, 3 This, , 4−2, y−µ, =, =2, σ, 1, , where µ=2 and σ=1., , content is available online at <http://cnx.org/content/m16991/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (6.0)
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264, , CHAPTER 6. THE NORMAL DISTRIBUTION, The z-score for y = 4 is z = 2. This means that 4 is z = 2 standard deviations to the right of, the mean. Therefore, x = 17 and y = 4 are both 2 (of their) standard deviations to the right of, their respective means., The z-score allows us to compare data that are scaled differently. To understand the concept,, suppose X ∼ N (5, 6) represents weight gains for one group of people who are trying to gain, weight in a 6 week period and Y ∼ N (2, 1) measures the same weight gain for a second group, of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each 2, standard deviations to the right of their means, they represent the same weight gain relative to, their means., , The Empirical Rule, If X is a random variable and has a normal distribution with mean µ and standard deviation σ then the, Empirical Rule says (See the figure below), , • About 68.27% of the x values lie between -1σ and +1σ of the mean µ (within 1 standard deviation of, the mean)., • About 95.45% of the x values lie between -2σ and +2σ of the mean µ (within 2 standard deviations of, the mean)., • About 99.73% of the x values lie between -3σ and +3σ of the mean µ (within 3 standard deviations of, the mean). Notice that almost all the x values lie within 3 standard deviations of the mean., • The z-scores for +1σ and –1σ are +1 and -1, respectively., • The z-scores for +2σ and –2σ are +2 and -2, respectively., • The z-scores for +3σ and –3σ are +3 and -3 respectively., , The Empirical Rule is also known as the 68-95-99.7 Rule., Example 6.3, Suppose X has a normal distribution with mean 50 and standard deviation 6., , • About 68.27% of the x values lie between -1σ = (-1)(6) = -6 and 1σ = (1)(6) = 6 of the mean 50., The values 50 - 6 = 44 and 50 + 6 = 56 are within 1 standard deviation of the mean 50. The, z-scores are -1 and +1 for 44 and 56, respectively., • About 95.45% of the x values lie between -2σ = (-2)(6) = -12 and 2σ = (2)(6) = 12 of the mean, 50. The values 50 - 12 = 38 and 50 + 12 = 62 are within 2 standard deviations of the mean 50., The z-scores are -2 and 2 for 38 and 62, respectively., • About 99.73% of the x values lie between -3σ = (-3)(6) = -18 and 3σ = (3)(6) = 18 of the mean, 50. The values 50 - 18 = 32 and 50 + 18 = 68 are within 3 standard deviations of the mean 50., The z-scores are -3 and +3 for 32 and 68, respectively., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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265, , 6.4 Areas to the Left and Right of x4, The arrow in the graph below points to the area to the left of x. This area is represented by the probability, P ( X < x ). Normal tables, computers, and calculators provide or calculate the probability P ( X < x )., , The area to the right is then P ( X > x ) = 1 − P ( X < x )., Remember, P ( X < x ) = Area to the left of the vertical line through x., P ( X > x ) = 1 − P ( X < x ) =. Area to the right of the vertical line through x, P ( X < x ) is the same as P ( X ≤ x ) and P ( X > x ) is the same as P ( X ≥ x ) for continuous distributions., , 6.5 Calculations of Probabilities5, Probabilities are calculated by using technology. There are instructions in the chapter for the TI-83+ and, TI-84 calculators., NOTE : In the Table of Contents for Collaborative Statistics, entry 15. Tables has a link to a table, of normal probabilities. Use the probability tables if so desired, instead of a calculator. The tables, include instructions for how to use then., , Example 6.4, If the area to the left is 0.0228, then the area to the right is 1 − 0.0228 = 0.9772., Example 6.5, The final exam scores in a statistics class were normally distributed with a mean of 63 and a, standard deviation of 5., Problem 1, Find the probability that a randomly selected student scored more than 65 on the exam., Solution, Let X = a score on the final exam. X ∼ N (63, 5), where µ = 63 and σ = 5, Draw a graph., Then, find P ( x > 65)., P ( x > 65) = 0.3446 (calculator or computer), 4 This, 5 This, , content is available online at <http://cnx.org/content/m16976/1.5/>., content is available online at <http://cnx.org/content/m16977/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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266, , CHAPTER 6. THE NORMAL DISTRIBUTION, , The probability that one student scores more than 65 is 0.3446., Using the TI-83+ or the TI-84 calculators, the calculation is as follows. Go into 2nd DISTR., After pressing 2nd DISTR, press 2:normalcdf., The syntax for the instructions are shown below., normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 ( = 1099 ) by pressing 1, the EE key (a 2nd key) and then 99., Or, you can enter 10^99 instead. The number 1099 is way out in the right tail of the normal curve., We are calculating the area between 65 and 1099 . In some instances, the lower number of the area, might be -1E99 ( = −1099 ). The number −1099 is way out in the left tail of the normal curve., H ISTORICAL N OTE : The TI probability program calculates a z-score and then the probability from, the z-score. Before technology, the z-score was looked up in a standard normal probability table, (because the math involved is too cumbersome) to find the probability. In this example, a standard, normal table with area to the left of the z-score was used. You calculate the z-score and look up, the area to the left. The probability is the area to the right., z=, , 65−63, 5, , = 0.4, , . Area to the left is 0.6554. P ( x > 65) = P (z > 0.4) = 1 − 0.6554 = 0.3446, , Problem 2, Find the probability that a randomly selected student scored less than 85., Solution, Draw a graph., Then find P ( x < 85). Shade the graph. P ( x < 85) = 1 (calculator or computer), The probability that one student scores less than 85 is approximately 1 (or 100%)., The TI-instructions and answer are as follows:, normalcdf(0,85,63,5) = 1 (rounds to 1), , Problem 3, Find the 90th percentile (that is, find the score k that has 90 % of the scores below k and 10% of, the scores above k)., Solution, Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the, x-axis. Shade the area that corresponds to the 90th percentile., Let k = the 90th percentile. k is located on the x-axis. P ( x < k) is the area to the left of k. The 90th, percentile k separates the exam scores into those that are the same or lower than k and those that, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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267, are the same or higher. Ninety percent of the test scores are the same or lower than k and 10% are, the same or higher. k is often called a critical value., k = 69.4 (calculator or computer), , The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall, at or above. For the TI-83+ or TI-84 calculators, use invNorm in 2nd DISTR. invNorm(area to the, left, mean, standard deviation) For this problem, invNorm(0.90,63,5) = 69.4, , Problem 4, Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of, the scores are above k)., Solution, Find the 70th percentile., Draw a new graph and label it appropriately. k = 65.6, The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall, at or above., invNorm(0.70,63,5) = 65.6, , Example 6.6, A computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking and a myriad of other things. Suppose that the average, number of hours a household personal computer is used for entertainment is 2 hours per day., Assume the times for entertainment are normally distributed and the standard deviation for the, times is half an hour., Problem 1, Find the probability that a household personal computer is used between 1.8 and 2.75 hours per, day., Solution, Let X = the amount of time (in hours) a household personal computer is used for entertainment., x ∼ N (2, 0.5) where µ = 2 and σ = 0.5., Find P (1.8 < x < 2.75)., The probability for which you are looking is the area between x, 2.75., P (1.8 < x < 2.75) = 0.5886, , =, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 1.8 and x, , =
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268, , CHAPTER 6. THE NORMAL DISTRIBUTION, , normalcdf(1.8,2.75,2,0.5) = 0.5886, The probability that a household personal computer is used between 1.8 and 2.75 hours per day, for entertainment is 0.5886., , Problem 2, Find the maximum number of hours per day that the bottom quartile of households use a personal, computer for entertainment., Solution, To find the maximum number of hours per day that the bottom quartile of households uses a, personal computer for entertainment, find the 25th percentile, k, where P ( x < k ) = 0.25., , invNorm(0.25,2,.5) = 1.66, The maximum number of hours per day that the bottom quartile of households uses a personal, computer for entertainment is 1.66 hours., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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269, , 6.6 Summary of Formulas6, Formula 6.1: Normal Probability Distribution, X ∼ N (µ, σ ), µ = the mean, , σ = the standard deviation, , Formula 6.2: Standard Normal Probability Distribution, Z ∼ N (0, 1), z = a standardized value (z-score), mean = 0, , standard deviation = 1, , Formula 6.3: Finding the kth Percentile, To find the kth percentile when the z-score is known: k = µ + (z) σ, Formula 6.4: z-score, x −µ, z= σ, Formula 6.5: Finding the area to the left, The area to the left: P ( X < x ), Formula 6.6: Finding the area to the right, The area to the right: P ( X > x ) = 1 − P ( X < x ), , 6 This, , content is available online at <http://cnx.org/content/m16987/1.5/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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270, , CHAPTER 6. THE NORMAL DISTRIBUTION, , 6.7 Practice: The Normal Distribution7, 6.7.1 Student Learning Outcomes, • The student will analyze data following a normal distribution., , 6.7.2 Given, The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation, of 1.3 years. A CD player is guaranteed for 3 years. We are interested in the length of time a CD player, lasts., , 6.7.3 Normal Distribution, Exercise 6.7.1, Define the Random Variable X in words. X =, Exercise 6.7.2, X∼, Exercise 6.7.3, (Solution on p. 285.), Find the probability that a CD player will break down during the guarantee period., a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability., , Figure 6.1, , b. P (0 < x < _________) = _________ (Use zero (0) for the minimum value of x.), Exercise 6.7.4, Find the probability that a CD player will last between 2.8 and 6 years., , (Solution on p. 285.), , a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability., 7 This, , content is available online at <http://cnx.org/content/m16983/1.10/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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271, , Figure 6.2, , b. P (_______ < x < _______) = _________, Exercise 6.7.5, Find the 70th percentile of the distribution for the time a CD player lasts., , (Solution on p. 285.), , a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower, 70%., , Figure 6.3, , b. P ( x < k ) = _________. Therefore, k = __________., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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272, , CHAPTER 6. THE NORMAL DISTRIBUTION, , 6.8 Homework8, Exercise 6.8.1, (Solution on p. 285.), According to a study done by De Anza students, the height for Asian adult males is normally, distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one, Asian adult male is randomly chosen. Let X =height of the individual., a. X ∼_______(_______,_______), b. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement., c. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not,, and justify your answer numerically., d. The middle 40% of heights fall between what two values? Sketch the graph and write the, probability statement., Exercise 6.8.2, IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one, individual is randomly chosen. Let X =IQ of an individual., a. X ∼_______(_______,_______), b. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement., c. Mensa is an organization whose members have the top 2% of all IQs. Find the minimum IQ, needed to qualify for the Mensa organization. Sketch the graph and write the probability, statement., d. The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement., Exercise 6.8.3, (Solution on p. 285.), The percent of fat calories that a person in America consumes each day is normally distributed, with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly, chosen. Let X =percent of fat calories., a. X ∼_______(_______,_______), b. Find the probability that the percent of fat calories a person consumes is more than 40. Graph, the situation. Shade in the area to be determined., c. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph, and write the probability statement., Exercise 6.8.4, Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with, a mean of 250 feet and a standard deviation of 50 feet., a. If X = distance in feet for a fly ball, then X ∼_______(_______,_______), b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball, traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region, corresponding to the probability. Find the probability., c. Find the 80th percentile of the distribution of fly balls. Sketch the graph and write the probability statement., 8 This, , content is available online at <http://cnx.org/content/m16978/1.20/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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273, Exercise 6.8.5, (Solution on p. 285.), In China, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live, in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount, of time spent alone is normally distributed. We randomly survey one Chinese 4-year-old living in, a rural area. We are interested in the amount of time the child spends alone per day. (Source: San, Jose Mercury News), a. In words, define the random variable X. X =, b. X ∼, c. Find the probability that the child spends less than 1 hour per day unsupervised. Sketch the, graph and write the probability statement., d. What percent of the children spend over 10 hours per day unsupervised?, e. 70% of the children spend at least how long per day unsupervised?, Exercise 6.8.6, In the 1992 presidential election, Alaska’s 40 election districts averaged 1956.8 votes per district, for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in, Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X =, number of votes for President Clinton for an election district. (Source: The World Almanac and, Book of Facts), a. State the approximate distribution of X. X ∼, b. Is 1956.8 a population mean or a sample mean? How do you know?, c. Find the probability that a randomly selected district had fewer than 1600 votes for President, Clinton. Sketch the graph and write the probability statement., d. Find the probability that a randomly selected district had between 1800 and 2000 votes for, President Clinton., e. Find the third quartile for votes for President Clinton., Exercise 6.8.7, (Solution on p. 285.), Suppose that the duration of a particular type of criminal trial is known to be normally distributed, with a mean of 21 days and a standard deviation of 7 days., a. In words, define the random variable X. X =, b. X ∼, c. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch, the graph and write the probability statement., d. 60% of all of these types of trials are completed within how many days?, Exercise 6.8.8, Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap, race) with a standard deviation of 2.28 seconds . The distribution of her race times is normally, distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri, Vogel), a., b., c., d., e., , In words, define the random variable X. X =, X∼, Find the percent of her laps that are completed in less than 130 seconds., The fastest 3% of her laps are under _______ ., The middle 80% of her laps are from _______ seconds to _______ seconds., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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274, , CHAPTER 6. THE NORMAL DISTRIBUTION, Exercise 6.8.9, (Solution on p. 285.), Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at, Lucky claimed to wait in the checkout line until their turn. Let X =time in line. Below are the, ordered real data (in minutes):, 0.50, , 4.25, , 5, , 6, , 7.25, , 1.75, , 4.25, , 5.25, , 6, , 7.25, , 2, , 4.25, , 5.25, , 6.25, , 7.25, , 2.25, , 4.25, , 5.5, , 6.25, , 7.75, , 2.25, , 4.5, , 5.5, , 6.5, , 8, , 2.5, , 4.75, , 5.5, , 6.5, , 8.25, , 2.75, , 4.75, , 5.75, , 6.5, , 9.5, , 3.25, , 4.75, , 5.75, , 6.75, , 9.5, , 3.75, , 5, , 6, , 6.75, , 9.75, , 3.75, , 5, , 6, , 6.75, , 10.75, , Table 6.1, , Calculate the sample mean and the sample standard deviation., Construct a histogram. Start the x − axis at −0.375 and make bar widths of 2 minutes., Draw a smooth curve through the midpoints of the tops of the bars., In words, describe the shape of your histogram and smooth curve., Let the sample mean approximate µ and the sample standard deviation approximate σ. The, distribution of X can then be approximated by X ∼, f. Use the distribution in (e) to calculate the probability that a person will wait fewer than 6.1, minutes., g. Determine the cumulative relative frequency for waiting less than 6.1 minutes., h. Why aren’t the answers to (f) and (g) exactly the same?, i. Why are the answers to (f) and (g) as close as they are?, j. If only 10 customers were surveyed instead of 50, do you think the answers to (f) and (g) would, have been closer together or farther apart? Explain your conclusion., a., b., c., d., e., , Exercise 6.8.10, Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In, complete sentences, explain why each of the following statements may be false., a. Ricardo’s actual GPA is lower than Anita’s actual GPA., b. Ricardo is not passing since his z-score is zero., c. Anita is in the 70th percentile of students at her college., Exercise 6.8.11, (Solution on p. 286.), Below is a sample of the maximum capacity (maximum number of spectators) of sports, stadiums. The table does not include horse racing or motor racing stadiums. (Source:, http://en.wikipedia.org/wiki/List_of_stadiums_by_capacity), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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275, , 40,000, , 40,000, , 45,050, , 45,500, , 46,249, , 48,134, , 49,133, , 50,071, , 50,096, , 50,466, , 50,832, , 51,100, , 51,500, , 51,900, , 52,000, , 52,132, , 52,200, , 52,530, , 52,692, , 53,864, , 54,000, , 55,000, , 55,000, , 55,000, , 55,000, , 55,000, , 55,000, , 55,082, , 57,000, , 58,008, , 59,680, , 60,000, , 60,000, , 60,492, , 60,580, , 62,380, , 62,872, , 64,035, , 65,000, , 65,050, , 65,647, , 66,000, , 66,161, , 67,428, , 68,349, , 68,976, , 69,372, , 70,107, , 70,585, , 71,594, , 72,000, , 72,922, , 73,379, , 74,500, , 75,025, , 76,212, , 78,000, , 80,000, , 80,000, , 82,300, , Table 6.2, , a. Calculate the sample mean and the sample standard deviation for the maximum capacity of, sports stadiums (the data)., b. Construct a histogram of the data., c. Draw a smooth curve through the midpoints of the tops of the bars of the histogram., d. In words, describe the shape of your histogram and smooth curve., e. Let the sample mean approximate µ and the sample standard deviation approximate σ. The, distribution of X can then be approximated by X ∼, f. Use the distribution in (e) to calculate the probability that the maximum capacity of sports, stadiums is less than 67,000 spectators., g. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is, less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a, maximum capacity less than 67,000. Divide by the total number of sports stadiums in the, sample., h. Why aren’t the answers to (f) and (g) exactly the same?, , 6.8.1 Try These Multiple Choice Questions, The questions below refer to the following: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days., Exercise 6.8.12, What is the median recovery time?, A., B., C., D., , (Solution on p. 286.), , 2.7, 5.3, 7.4, 2.1, , Exercise 6.8.13, What is the z-score for a patient who takes 10 days to recover?, , (Solution on p. 286.), , A. 1.5, B. 0.2, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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276, , CHAPTER 6. THE NORMAL DISTRIBUTION, C. 2.2, D. 7.3, Exercise 6.8.14, What is the probability of spending more than 2 days in recovery?, A., B., C., D., , 0.0580, 0.8447, 0.0553, 0.9420, , Exercise 6.8.15, The 90th percentile for recovery times is?, A., B., C., D., , (Solution on p. 286.), , (Solution on p. 286.), , 8.89, 7.07, 7.99, 4.32, , The questions below refer to the following: The length of time to find a parking space at 9 A.M. follows a, normal distribution with a mean of 5 minutes and a standard deviation of 2 minutes., Exercise 6.8.16, (Solution on p. 286.), Based upon the above information and numerically justified, would you be surprised if it took, less than 1 minute to find a parking space?, A. Yes, B. No, C. Unable to determine, Exercise 6.8.17, Find the probability that it takes at least 8 minutes to find a parking space., A., B., C., D., , (Solution on p. 286.), , 0.0001, 0.9270, 0.1862, 0.0668, , Exercise 6.8.18, (Solution on p. 286.), Seventy percent of the time, it takes more than how many minutes to find a parking space?, A., B., C., D., , 1.24, 2.41, 3.95, 6.05, , Exercise 6.8.19, (Solution on p. 286.), If the mean is significantly greater than the standard deviation, which of the following statements, is true?, I . The data cannot follow the uniform distribution., II . The data cannot follow the exponential distribution.., III . The data cannot follow the normal distribution., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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277, A., B., C., D., , I only, II only, III only, I, II, and III, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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278, , CHAPTER 6. THE NORMAL DISTRIBUTION, , 6.9 Review9, The next two questions refer to: X ∼ U (3, 13), Exercise 6.9.1, Explain which of the following are false and which are true., a:, b:, c:, d:, , (Solution on p. 286.), , 1, , 3 ≤ x ≤ 13, f ( x ) = 10, There is no mode., The median is less than the mean., P ( x > 10) = P ( x ≤ 6), , Exercise 6.9.2, Calculate:, , (Solution on p. 286.), , a: Mean, b: Median, c: 65th percentile., , Exercise 6.9.3, Which of the following is true for the above box plot?, a:, b:, c:, d:, , 25% of the data are at most 5., There is about the same amount of data from 4 – 5 as there is from 5 – 7., There are no data values of 3., 50% of the data are 4., , Exercise 6.9.4, If P ( G | H ) = P ( G ), then which of the following is correct?, A:, B:, C:, D:, , (Solution on p. 286.), , (Solution on p. 286.), , G and H are mutually exclusive events., P (G) = P ( H ), Knowing that H has occurred will affect the chance that G will happen., G and H are independent events., , Exercise 6.9.5, (Solution on p. 286.), If P ( J ) = 0.3, P (K ) = 0.6, and J and K are independent events, then explain which are correct, and which are incorrect., A:, B:, C:, D:, 9 This, , P ( JandK ) = 0, P ( JorK ) = 0.9, P ( JorK ) = 0.72, P ( J ) 6= P ( J | K ), content is available online at <http://cnx.org/content/m16985/1.10/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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279, Exercise 6.9.6, (Solution on p. 287.), On average, 5 students from each high school class get full scholarships to 4-year colleges. Assume, that most high school classes have about 500 students., X = the number of students from a high school class that get full scholarships to 4-year school., Which of the following is the distribution of X?, A., B., C., D., , P(5), B(500,5), Exp(1/5), N(5, (0.01)(0.99)/500), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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281, , Figure 6.4, , 3. Calculate the following., a. x =, b. s =, 4. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it, have a V-shape, does it have a hump in the middle or at either end, etc.?), , 6.10.4 Analyze the Distribution, Using your sample mean, sample standard deviation, and histogram to help, what was the approximate, theoretical distribution of the data?, , • X∼, • How does the histogram help you arrive at the approximate distribution?, , 6.10.5 Describe the Data, Use the Data from the section titled "Collect the Data" to complete the following statements., , •, •, •, •, •, •, •, , The IQR goes from __________ to __________., IQR = __________. (IQR=Q3-Q1), The 15th percentile is:, The 85th percentile is:, The median is:, The empirical probability that a randomly chosen lap time is more than 130 seconds =, Explain the meaning of the 85th percentile of this data., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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282, , CHAPTER 6. THE NORMAL DISTRIBUTION, , 6.10.6 Theoretical Distribution, Using the theoretical distribution from the section titled "Analyse the Distribution" complete the following, statements:, , •, •, •, •, •, •, •, , The IQR goes from __________ to __________., IQR =, The 15th percentile is:, The 85th percentile is:, The median is:, The probability that a randomly chosen lap time is more than 130 seconds =, Explain the meaning of the 85th percentile of this distribution., , 6.10.7 Discussion Questions, • Do the data from the section titled "Collect the Data" give a close approximation to the theoretical, distibution in the section titled "Analyze the Distribution"? In complete sentences and comparing the, result in the sections titled "Describe the Data" and "Theoretical Distribution", explain why or why, not., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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284, , CHAPTER 6. THE NORMAL DISTRIBUTION, a. x =, b. s =, , 4. Draw a smooth curve through the top of the bars of the histogram. Use 1-2 complete sentences to, describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it, have a V-shape, does it have a hump in the middle or at either end, etc.?), , 6.11.3 Analyze the Distribution, Using your sample mean, sample standard deviation, and histogram to help, what was the approximate, theoretical distribution of the data from the section titled "Collect the Data"?, , • X∼, • How does the histogram help you arrive at the approximate distribution?, , 6.11.4 Describe the Data, Using the data in the section titled "Collect the Data" complete the following statements. (Hint: order the, data), R EMEMBER : ( IQR = Q3 − Q1), , •, •, •, •, •, •, , IQR =, 15th percentile is:, 85th percentile is:, Median is:, What is the empirical probability that a randomly chosen pinkie length is more than 6.5 cm?, Explain the meaning the 85th percentile of this data., , 6.11.5 Theoretical Distribution, Using the Theoretical Distribution in the section titled "Analyze the Distribution", , •, •, •, •, •, •, , IQR =, 15th percentile is:, 85th percentile is:, Median is:, What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm?, Explain the meaning of the 85th percentile of this data., , 6.11.6 Discussion Questions, • Do the data from the section entitled "Collect the Data" give a close approximation to the theoretical, distribution in "Analyze the Distribution." In complete sentences and comparing the results in the, sections titled "Describe the Data" and "Theoretical Distribution", explain why or why not., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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285, , Solutions to Exercises in Chapter 6, Solution to Example 6.2, Problem 1 (p. 263), This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean 5., Solution to Example 6.2, Problem 2 (p. 263), z = -4. This z-score tells you that x = −3 is 4 standard deviations to the left of the mean., , Solutions to Practice: The Normal Distribution, Solution to Exercise 6.7.3 (p. 270), b. 3, 0.1979, Solution to Exercise 6.7.4 (p. 270), b. 2.8, 6, 0.7694, Solution to Exercise 6.7.5 (p. 271), b. 0.70, 4.78years, , Solutions to Homework, Solution to Exercise 6.8.1 (p. 272), a., b., c., d., , N (66, 2.5), 0.5404, No, Between 64.7 and 67.3 inches, , Solution to Exercise 6.8.3 (p. 272), a. N (36,10), b. 0.3446, c. 29.3, Solution to Exercise 6.8.5 (p. 273), a., b., c., d., e., , the time (in hours) a 4-year-old in China spends unsupervised per day, N (3, 1.5), 0.0912, 0, 2.21 hours, , Solution to Exercise 6.8.7 (p. 273), a., b., c., d., , The duration of a criminal trial, N (21, 7), 0.3341, 22.77, , Solution to Exercise 6.8.9 (p. 274), a., e., f., g., , The sample mean is 5.51 and the sample standard deviation is 2.15, N (5.51, 2.15), 0.6081, 0.64, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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286, , CHAPTER 6. THE NORMAL DISTRIBUTION, , Solution to Exercise 6.8.11 (p. 274), a., e., f., g., , The sample mean is 60,136.4 and the sample standard deviation is 10,468.1., N (60136.4, 10468.1), 0.7440, 0.7167, , Solution to Exercise 6.8.12 (p., B, Solution to Exercise 6.8.13 (p., C, Solution to Exercise 6.8.14 (p., D, Solution to Exercise 6.8.15 (p., C, Solution to Exercise 6.8.16 (p., A, Solution to Exercise 6.8.17 (p., D, Solution to Exercise 6.8.18 (p., C, Solution to Exercise 6.8.19 (p., B, , 275), 275), 276), 276), 276), 276), 276), 276), , Solutions to Review, Solution to Exercise 6.9.1 (p. 278), a:, b:, c:, d:, , True, True, False – the median and the mean are the same for this symmetric distribution, True, , Solution to Exercise 6.9.2 (p. 278), a: 8, b: 8, , 1, . k = 9.5, c: P ( x < k ) = 0.65 = (k − 3) ∗ 10, Solution to Exercise 6.9.3 (p. 278), a:, b:, c:, d:, , False – 43 of the data are at most 5, True – each quartile has 25% of the data, False – that is unknown, False – 50% of the data are 4 or less, , Solution to Exercise 6.9.4 (p. 278), D, Solution to Exercise 6.9.5 (p. 278), A: False - J and K are independent so they are not mutually exclusive which would imply dependency, (meaning P(J and K) is not 0)., B: False - see answer C., C: True - P(J or K) = P(J) + P(K) - P(J and K) = P(J) + P(K) - P(J)P(K) = 0.3 + 0.6 - (0.3)(0.6) = 0.72. Note that, P(J and K) = P(J)P(K) because J and K are independent., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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287, D: False - J and K are independent so P(J) = P(J|K)., Solution to Exercise 6.9.6 (p. 279), A, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 7, , The Central Limit Theorem, 7.1 The Central Limit Theorem1, 7.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, , Recognize the Central Limit Theorem problems., Classify continuous word problems by their distributions., Apply and interpret the Central Limit Theorem for Means., Apply and interpret the Central Limit Theorem for Sums., , 7.1.2 Introduction, Why are we so concerned with means? Two reasons are that they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem., The Central Limit Theorem (CLT for short) is one of the most powerful and useful ideas in all of statistics., Both alternatives are concerned with drawing finite samples of size n from a population with a known, mean, µ, and a known standard deviation, σ. The first alternative says that if we collect samples of size, n and n is "large enough," calculate each sample’s mean, and create a histogram of those means, then the, resulting histogram will tend to have an approximate normal bell shape. The second alternative says that, if we again collect samples of size n that are "large enough," calculate the sum of each sample and create a, histogram, then the resulting histogram will again tend to have a normal bell-shape., In either case, it does not matter what the distribution of the original population is, or whether you even, need to know it. The important fact is that the sample means and the sums tend to follow the normal, distribution. And, the rest you will learn in this chapter., The size of the sample, n, that is required in order to be to be ’large enough’ depends on the original, population from which the samples are drawn. If the original population is far from normal then more, observations are needed for the sample means or the sample sums to be normal. Sampling is done with, replacement., Optional Collaborative Classroom Activity, , 1 This, , content is available online at <http://cnx.org/content/m16953/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 289
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290, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , Do the following example in class: Suppose 8 of you roll 1 fair die 10 times, 7 of you roll 2 fair dice 10, times, 9 of you roll 5 fair dice 10 times, and 11 of you roll 10 fair dice 10 times., Each time a person rolls more than one die, he/she calculates the sample mean of the faces showing. For, example, one person might roll 5 fair dice and get a 2, 2, 3, 4, 6 on one roll., The 3.4 is one mean when 5 fair dice are rolled. This same person would, The mean is 2+2+53+4+6 = 3.4., roll the 5 dice 9 more times and calculate 9 more means for a total of 10 means., Your instructor will pass out the dice to several people as described above. Roll your dice 10 times. For, each roll, record the faces and find the mean. Round to the nearest 0.5., Your instructor (and possibly you) will produce one graph (it might be a histogram) for 1 die, one graph for, 2 dice, one graph for 5 dice, and one graph for 10 dice. Since the "mean" when you roll one die, is just the, face on the die, what distribution do these means appear to be representing?, Draw the graph for the means using 2 dice. Do the sample means show any kind of pattern?, Draw the graph for the means using 5 dice. Do you see any pattern emerging?, Finally, draw the graph for the means using 10 dice. Do you see any pattern to the graph? What can you, conclude as you increase the number of dice?, As the number of dice rolled increases from 1 to 2 to 5 to 10, the following is happening:, 1. The mean of the sample means remains approximately the same., 2. The spread of the sample means (the standard deviation of the sample means) gets smaller., 3. The graph appears steeper and thinner., You have just demonstrated the Central Limit Theorem (CLT)., The Central Limit Theorem tells you that as you increase the number of dice, the sample means tend, toward a normal distribution (the sampling distribution)., , 7.2 The Central Limit Theorem for Sample Means (Averages)2, Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:, a. µ X = the mean of X, b. σX = the standard deviation of X, If you draw random samples of size n, then as n increases, the random variable X which consists of sample, means, tends to be normally distributed and, , , σX, X ∼ N µX , √, n, The Central Limit Theorem for Sample Means says that if you keep drawing larger and larger samples, (like rolling 1, 2, 5, and, finally, 10 dice) and calculating their means the sample means form their own, normal distribution (the sampling distribution). The normal distribution has the same mean as the, original distribution and a variance that equals the original variance divided by n, the sample size. n is the, number of values that are averaged together not the number of times the experiment is done., 2 This, , content is available online at <http://cnx.org/content/m16947/1.23/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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291, To put it more formally, if you draw random samples of size n,the distribution of the random variable X, which consists of sample means, is called the sampling distribution of the mean. The sampling, distribution of the mean approaches a normal distribution as n, the sample size, increases., The random variable X has a different z-score associated with it than the random variable X. x is the value, of X in one sample., x−µ, z = X, , (7.0), , σX, √, n, , µ X is both the average of X and of X., σX =, , σX, √, n, , = standard deviation of X and is called the standard error of the mean., , Example 7.1, An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25, are drawn randomly from the population., Problem 1, Find the probability that the sample mean is between 85 and 92., Solution, Let X = one value from the original unknown population. The probability question asks you to, find a probability for the sample mean., Let X = the mean of a sample of size 25. Since µ X = 90, σX = 15, and n = 25;, , , then X ∼ N 90, √15, 25, , Find P (85 < x < 92), , Draw a graph., , P (85 < x < 92) = 0.6997, The probability that the sample mean is between 85 and 92 is 0.6997., , TI-83 or 84: normalcdf(lower value, upper value, mean, standard error of the mean), The parameter list is abbreviated (lower value, upper value, µ,, , √σ ), n, , , , normalcdf 85, 92, 90, √1525 = 0.6997, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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292, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , Problem 2, Find the value that is 2 standard deviations above the expected value (it is 90) of the sample mean., Solution, To find the value that is 2 standard deviations above the expected value 90, use the formula, , σX, value = µ X + (#ofSTDEVs) √, n, value = 90 + 2 ·, , √15, 25, , = 96, , So, the value that is 2 standard deviations above the expected value is 96., , Example 7.2, The length of time, in hours, it takes an "over 40" group of people to play one soccer match is, normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours. A sample of, size n = 50 is drawn randomly from the population., Problem, Find the probability that the sample mean is between 1.8 hours and 2.3 hours., Solution, Let X = the time, in hours, it takes to play one soccer match., The probability question asks you to find a probability for the sample mean time, in hours, it, takes to play one soccer match., Let X = the mean time, in hours, it takes to play one soccer match., If µ X = _________, σX = __________, and n = ___________, then X ∼ N(______, ______), by the Central Limit Theorem for Means., , , µ X = 2, σX = 0.5, n = 50, and X ∼ N 2 , √0.5, 50, , Find P (1.8 < x < 2.3)., , Draw a graph., , P (1.8 < x < 2.3) = 0.9977, , , normalcdf 1.8, 2.3, 2, √.5, = 0.9977, 50, , The probability that the mean time is between 1.8 hours and 2.3 hours is ______., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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293, , 7.3 The Central Limit Theorem for Sums3, Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:, a. µ X = the mean of X, b. σX = the standard deviation of X, If you draw random samples of size n, then as n increases, the random variable ΣX which consists of sums, tends to be normally distributed and, , √, ΣX ∼ N n · µ X , n · σX, The Central Limit Theorem for Sums says that if you keep drawing larger and larger samples and taking, their sums, the sums form their own normal distribution (the sampling distribution) which approaches a, normal distribution as the sample size increases. The normal distribution has a mean equal to the original, mean multiplied by the sample size and a standard deviation equal to the original standard deviation, multiplied by the square root of the sample size., The random variable ΣX has the following z-score associated with it:, a. Σx is one sum., Σx −n·µ, b. z = √n·σ X, X, , a. n√· µ X = the mean of ΣX, b. n · σX = standard deviation of ΣX, Example 7.3, An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is, drawn randomly from the population., Problem, a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than, 7500., b. Find the sum that is 1.5 standard deviations above the mean of the sums., Solution, Let X = one value from the original unknown population. The probability question asks you to, find a probability for the sum (or total of) 80 values., ΣX = the sum or total of 80 values. Since µ X = 90, σX = 15, and n = 80, then, , , √, ΣX ∼ N 80 · 90, 80 · 15, •. mean of the sums = n · µ X = (80) (90) = 7200, √, √, •. standard deviation of the sums = n · σX = 80 · 15, •. sum of 80 values = Σx = 7500, a: Find P (Σx > 7500), 3 This, , content is available online at <http://cnx.org/content/m16948/1.16/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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294, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, P (Σx > 7500) = 0.0127, , normalcdf(lower value, upper value, mean of sums, stdev of sums), √, The parameter list is abbreviated (lower, upper, n · µ X , n · σX ), √, normalcdf(7500,1E99, 80 · 90, 80 · 15 = 0.0127, Reminder: 1E99 = 1099 . Press the EE key for E., , b: Find Σx where z = 1.5:, √, √, Σx = n · µ X + z · n · σX = (80)(90) + (1.5)( 80) (15) = 7401.2, , 7.4 Using the Central Limit Theorem4, It is important for you to understand when to use the CLT. If you are being asked to find the probability of, the mean, use the CLT for the mean. If you are being asked to find the probability of a sum or total, use the, CLT for sums. This also applies to percentiles for means and sums., NOTE : If you are being asked to find the probability of an individual value, do not use the CLT., Use the distribution of its random variable., , 7.4.1 Examples of the Central Limit Theorem, Law of Large Numbers, The Law of Large Numbers says that if you take samples of larger and larger size from any population,, then the mean x of the sample tends to get closer and closer to µ. From the Central Limit Theorem, we, know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the, smaller the standard deviation gets. (Remember that the standard deviation for X is √σn .) This means that, the sample mean x must be close to the population mean µ. We can say that µ is the value that the sample, means approach as n gets larger. The Central Limit Theorem illustrates the Law of Large Numbers., Central Limit Theorem for the Mean and Sum Examples, 4 This, , content is available online at <http://cnx.org/content/m16958/1.21/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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295, Example 7.4, A study involving stress is done on a college campus among the students. The stress scores follow, a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a, sample of 75 students, find:, 1., 2., 3., 4., , The probability that the mean stress score for the 75 students is less than 2., The 90th percentile for the mean stress score for the 75 students., The probability that the total of the 75 stress scores is less than 200., The 90th percentile for the total stress score for the 75 students., , Let X = one stress score., Problems 1. and 2. ask you to find a probability or a percentile for a mean. Problems 3 and 4 ask, you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75., Since the individual stress scores follow a uniform distribution, X ∼ U (1, 5) where a = 1 and, b = 5 (See Continuous Random Variables (Section 5.1) for the uniform)., µX =, , a+b, 2, , q, σX =, , =, , 1+5, 2, , ( b − a )2, 12, , =, , =3, q, , (5−1)2, 12, , = 1.15, , For problems 1. and 2., let X = the mean stress score for the 75 students. Then,, , , 1.15, X ∼ N 3, √, where n = 75., 75, , Problem 1, Find P ( x < 2)., , Draw the graph., , Solution, P ( x < 2) = 0, The probability that the mean stress score is less than 2 is about 0., , , , 1.15, normalcdf 1, 2, 3, √, =0, 75, R EMINDER : The smallest stress score is 1. Therefore, the smallest mean for 75 stress scores is 1., , Problem 2, Find the 90th percentile for the mean of 75 stress scores. Draw a graph., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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296, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, Solution, Let k = the 90th precentile., Find k where P ( x < k) = 0.90., k = 3.2, , The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of, 75 stress scores are at most 3.2 and 10% are at least 3.2., , , 1.15, invNorm .90, 3, √, = 3.2, 75, , Forh problems c and id, let ΣX = the sum of the 75 stress scores., √, N (75) · (3) , 75 · 1.15, Problem 3, Find P (Σx < 200)., , Draw the graph., , Solution, The mean of the sum of 75 stress scores is 75 · 3 = 225, The standard deviation of the sum of 75 stress scores is, , √, , 75 · 1.15 = 9.96, , P (Σx < 200) = 0, , The probability that the total of 75 scores is less than 200 is about 0., , , √, normalcdf 75, 200, 75 · 3, 75 · 1.15 = 0., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , Then, ΣX ∼
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297, R EMINDER : The smallest total of 75 stress scores is 75 since the smallest single score is 1., , Problem 4, Find the 90th percentile for the total of 75 stress scores. Draw a graph., Solution, Let k = the 90th percentile., Find k where P (Σx < k) = 0.90., k = 237.8, , The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of, 75 scores are no more than 237.8 and 10% are no less than 237.8., , , √, invNorm .90, 75 · 3, 75 · 1.15 = 237.8, , Example 7.5, Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst, finds that for those people who exceed the time included in their basic contract, the excess time, used follows an exponential distribution with a mean of 22 minutes., Consider a random sample of 80 customers who exceed the time allowance included in their basic, cell phone contract., Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted, time allowance., , 1, X ∼ Exp 22, From Chapter 5, we know that µ = 22 and σ = 22., Let X = the mean excess time used by a sample of n = 80 customers who exceed their contracted, time allowance., , , X ∼ N 22, √22 by the CLT for Sample Means, 80, , Problem 1, Using the CLT to find Probability:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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298, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, a. Find the probability that the mean excess time used by the 80 customers in the sample is longer, Draw the graph., than 20 minutes. This is asking us to find P ( x > 20), b. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly, selected. Find the probability that this individual customer’s excess time is longer than 20, minutes. This is asking us to find P ( x > 20), c. Explain why the probabilities in (a) and (b) are different., Solution, Part a., Find: P ( x > 20), , , P ( x > 20) = 0.7919 using normalcdf 20, 1E99, 22, √22, 80, , The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of, 80 customers who exceed their contracted time allowance., , R EMINDER : 1E99 = 1099 and−1E99 = −1099 . Press the EE key for E. Or just use 10^99 instead of, 1E99., Part b., Find P(x>20) . Remember to use the exponential distribution for an individual: X∼Exp(1/22)., P(X>20) = e^(–(1/22)*20) or e^(–.04545*20) = 0.4029, Part c. Explain why the probabilities in (a) and (b) are different., P ( x > 20) = 0.4029 but P ( x > 20) = 0.7919, The probabilities are not equal because we use different distributions to calculate the probability, for individuals and for means., When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the CLT. Use the CLT with the normal distribution when you are, being asked to find the probability for an mean., , Problem 2, Using the CLT to find Percentiles:, Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed, their basic contract time allowances. Draw a graph., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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299, Solution, Let k = the 95th percentile. Find k where P ( x < k) = 0.95, , , k = 26.0 using invNorm .95, 22, √22 = 26.0, 80, , The 95th percentile for the sample mean excess time used is about 26.0 minutes for random, samples of 80 customers who exceed their contractual allowed time., 95% of such samples would have means under 26 minutes; only 5% of such samples would have, means above 26 minutes., , NOTE :, , (HISTORICAL): Normal Approximation to the Binomial, , Historically, being able to compute binomial probabilities was one of the most important applications of the, Central Limit Theorem. Binomial probabilities were displayed in a table in a book with a small value for n, (say, 20). To calculate the probabilities with large values of n, you had to use the binomial formula which, could be very complicated. Using the Normal Approximation to the Binomial simplified the process. To, compute the Normal Approximation to the Binomial, take a simple random sample from a population. You, must meet the conditions for a binomial distribution:, •. there are a certain number n of independent trials, •. the outcomes of any trial are success or failure, •. each trial has the same probability of a success p, Recall that if X is the binomial random variable, then X ∼ B (n, p). The shape of the binomial distribution, needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must, both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or, equal to 10). Then the binomial can be approximated by the normal distribution with mean µ = np and, √, standard deviation σ = npq. Remember that q = 1 − p. In order to get the best approximation, add 0.5 to, x or subtract 0.5 from x ( use x + 0.5 or x − 0.5. The number 0.5 is called the continuity correction factor., Example 7.6, Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the, population favor a charter school for grades K - 5. A simple random sample of 300 is surveyed., 1. Find the probability that at least 150 favor a charter school., 2. Find the probability that at most 160 favor a charter school., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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300, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, 3. Find the probability that more than 155 favor a charter school., 4. Find the probability that less than 147 favor a charter school., 5. Find the probability that exactly 175 favor a charter school., Let X = the number that favor a charter school for grades K - 5. X ∼ B (n, p) where n = 300 and, p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas, √, for the mean and standard deviation are µ = np and σ = npq. The mean is 159 and the standard, deviation is 8.6447. The random variable for the normal distribution is Y. Y ∼ N (159, 8.6447). See, The Normal Distribution for help with calculator instructions., For Problem 1., you include 150 so P ( x ≥ 150) has normal approximation P (Y ≥ 149.5) = 0.8641., , normalcdf (149.5, 10^99, 159, 8.6447) = 0.8641., For Problem 2., you include 160 so P ( x ≤ 160) has normal approximation P (Y ≤ 160.5) = 0.5689., , normalcdf (0, 160.5, 159, 8.6447) = 0.5689, For Problem 3., you exclude 155 so P ( x > 155) has normal approximation P (y > 155.5) = 0.6572., , normalcdf (155.5, 10^99, 159, 8.6447) = 0.6572, For Problem 4., you exclude 147 so P ( x < 147) has normal approximation P (Y < 146.5) = 0.0741., , normalcdf (0, 146.5, 159, 8.6447) = 0.0741, For Problem 5., P ( x = 175) has normal approximation P (174.5 < y < 175.5) = 0.0083., , normalcdf (174.5, 175.5, 159, 8.6447) = 0.0083, Because of calculators and computer software that easily let you calculate binomial probabilities, for large values of n, it is not necessary to use the the Normal Approximation to the Binomial, provided you have access to these technology tools. Most school labs have Microsoft Excel, an, example of computer software that calculates binomial probabilities. Many students have access, to the TI-83 or 84 series calculators and they easily calculate probabilities for the binomial. In an, Internet browser, if you type in "binomial probability distribution calculation," you can find at, least one online calculator for the binomial., For Example 3, the probabilities are calculated using the binomial (n = 300 and p = 0.53) below., Compare the binomial and normal distribution answers. See Discrete Random Variables for help, with calculator instructions for the binomial., P ( x ≥ 150): 1 - binomialcdf (300, 0.53, 149) = 0.8641, P ( x ≤ 160): binomialcdf (300, 0.53, 160) = 0.5684, P ( x > 155): 1 - binomialcdf (300, 0.53, 155) = 0.6576, P ( x < 147): binomialcdf (300, 0.53, 146) = 0.0742, P ( x = 175): (You use the binomial pdf.) binomialpdf (175, 0.53, 146) = 0.0083, **Contributions made to Example 2 by Roberta Bloom, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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301, , 7.5 Summary of Formulas5, Formula, Limit Theorem for Sample Means, 7.1: Central, , , σ, X, X ∼ N µ X , √n, The Mean X :, µX, Formula 7.2: Central Limit Theorem for Sample Means Z-Score and Standard Error of the Mean, , x −µ, σX, z = σX X, Standard Error of the Mean (Standard Deviation X ): √, n, √, , n, , Formula7.3: Central, √ Limit Theorem for Sums, Mean for Sums (ΣX ):, ΣX ∼ N (n) · µ X , n · σX, , n · µX, , Formula 7.4: Central Limit Theorem for Sums Z-Score and Standard, Deviation for Sums, √, Σx −n·µ, z = √n·σ X, Standard Deviation for Sums (ΣX ):, n · σX, X, , 5 This, , content is available online at <http://cnx.org/content/m16956/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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302, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , 7.6 Practice: The Central Limit Theorem6, 7.6.1 Student Learning Outcomes, • The student will calculate probabilities using the Central Limit Theorem., , 7.6.2 Given, Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees., From past experience, she has found that the reviews take her approximately 4 hours each to do with a, population standard deviation of 1.2 hours. Let X be the random variable representing the time it takes, her to complete one review. Assume X is normally distributed. Let X be the random variable representing, the mean time to complete the 16 reviews. Let ΣX be the total time it takes Yoonie to complete all of the, month’s reviews. Assume that the 16 reviews represent a random set of reviews., , 7.6.3 Distribution, Complete the distributions., 1. X ∼, 2. X ∼, 3. ΣX ∼, , 7.6.4 Graphing Probability, For each problem below:, a. Sketch the graph. Label and scale the horizontal axis. Shade the region corresponding to the probability., b. Calculate the value., Exercise 7.6.1, (Solution on p. 323.), Find the probability that one review will take Yoonie from 3.5 to 4.25 hours., , a., b. P ( ________ <x< ________ = _______, Exercise 7.6.2, (Solution on p. 323.), Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs., 6 This, , content is available online at <http://cnx.org/content/m16954/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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303, , a., b. P(________________ ) =_______, Exercise 7.6.3, (Solution on p. 323.), Find the 95th percentile for the mean time to complete one month’s reviews., , a., b. The 95th Percentile=, Exercise 7.6.4, (Solution on p. 323.), Find the probability that the sum of the month’s reviews takes Yoonie from 60 to 65 hours., , a., b. The Probability=, Exercise 7.6.5, Find the 95th percentile for the sum of the month’s reviews., , (Solution on p. 323.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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304, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , a., b. The 95th percentile=, , 7.6.5 Discussion Question, Exercise 7.6.6, What causes the probabilities in Exercise 7.6.1 and Exercise 7.6.2 to differ?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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305, , 7.7 Homework7, Exercise 7.7.1, (Solution on p. 323.), X ∼ N (60, 9). Suppose that you form random samples of 25 from this distribution. Let X be the, random variable of averages. Let ΣX be the random variable of sums. For c - f, sketch the graph,, shade the region, label and scale the horizontal axis for X, and find the probability., a. Sketch the distributions of X and X on the same graph., b. X ∼, c. P ( x < 60) =, d. Find the 30th percentile for the mean., e. P (56 < x < 62) =, f. P (18 < x < 58) =, g. Σx ∼, h. Find the minimum value for the upper quartile for the sum., i. P (1400 < Σx < 1550) =, Exercise 7.7.2, Determine which of the following are true and which are false. Then, in complete sentences,, justify your answers., a. When the sample size is large, the mean of X is approximately equal to the mean of X., b. When the sample size is large, X is approximately normally distributed., c. When the sample size is large, the standard deviation of X is approximately the same as the, standard deviation of X., Exercise 7.7.3, (Solution on p. 323.), The percent of fat calories that a person in America consumes each day is normally distributed, with a mean of about 36 and a standard deviation of about 10. Suppose that 16 individuals are, randomly chosen., Let X =average percent of fat calories., a. X~______ ( ______ , ______ ), b. For the group of 16, find the probability that the average percent of fat calories consumed is, more than 5. Graph the situation and shade in the area to be determined., c. Find the first quartile for the average percent of fat calories., Exercise 7.7.4, Previously, De Anza statistics students estimated that the amount of change daytime statistics, students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick, 25 daytime statistics students., In words, X =, X~, In words, X =, X~ ______ ( ______ , ______ ), Find the probability that an individual had between $0.80 and $1.00. Graph the situation and, shade in the area to be determined., f. Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the, situation and shade in the area to be determined., a., b., c., d., e., , 7 This, , content is available online at <http://cnx.org/content/m16952/1.24/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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306, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, g. Explain the why there is a difference in (e) and (f)., Exercise 7.7.5, (Solution on p. 323.), Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with, a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls., a. If X = average distance in feet for 49 fly balls, then X~_______ ( _______ , _______ ), b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the, graph. Scale the horizontal axis for X. Shade the region corresponding to the probability., Find the probability., c. Find the 80th percentile of the distribution of the average of 49 fly balls., Exercise 7.7.6, Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed, from 2 to 6 pounds. We randomly survey 64 homes with children., a. In words, X =, b. X~, c. µ X =, d. σX =, e. In words, ΣX =, f. ΣX~, g. Find the probability that the total weight of open boxes is less than 250 pounds., h. Find the 35th percentile for the total weight of open boxes of cereal., Exercise 7.7.7, (Solution on p. 323.), Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days, and a standard deviation of 7 days. We randomly sample 9 trials., a., b., c., d., , In words, ΣX =, ΣX~, Find the probability that the total length of the 9 trials is at least 225 days., 90 percent of the total of 9 of these types of trials will last at least how long?, , Exercise 7.7.8, According to the Internal Revenue Service, the average length of time for an individual to complete (record keep, learn, prepare, copy, assemble and send) IRS Form 1040 is 10.53 hours (without, any attached schedules). The distribution is unknown. Let us assume that the standard deviation, is 2 hours. Suppose we randomly sample 36 taxpayers., In words, X =, In words, X =, X~, Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more, than 12 hours? Explain why or why not in complete sentences., e. Would you be surprised if one taxpayer finished his Form 1040 in more than 12 hours? In a, complete sentence, explain why., , a., b., c., d., , Exercise 7.7.9, (Solution on p. 323.), Suppose that a category of world class runners are known to run a marathon (26 miles) in an, average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races., Let X = the average of the 49 races., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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307, a. X~, b. Find the probability that the runner will average between 142 and 146 minutes in these 49, marathons., c. Find the 80th percentile for the average of these 49 marathons., d. Find the median of the average running times., Exercise 7.7.10, The attention span of a two year-old is exponentially distributed with a mean of about 8 minutes., Suppose we randomly survey 60 two year-olds., a., b., c., d., e., , In words, X =, X~, In words, X =, X~, Before doing any calculations, which do you think will be higher? Explain why., i. the probability that an individual attention span is less than 10 minutes; or, ii. the probability that the average attention span for the 60 children is less than 10 minutes?, Why?, , f. Calculate the probabilities in part (e)., g. Explain why the distribution for X is not exponential., Exercise 7.7.11, (Solution on p. 324.), Suppose that the length of research papers is uniformly distributed from 10 to 25 pages. We, survey a class in which 55 research papers were turned in to a professor. The 55 research papers, are considered a random collection of all papers. We are interested in the average length of the, research papers., a. In words, X =, b. X~, c. µ X =, d. σX =, e. In words, X =, f. X~, g. In words, ΣX =, h. ΣX~, i. Without doing any calculations, do you think that it’s likely that the professor will need to read, a total of more than 1050 pages? Why?, j. Calculate the probability that the professor will need to read a total of more than 1050 pages., k. Why is it so unlikely that the average length of the papers will be less than 12 pages?, Exercise 7.7.12, The length of songs in a collector’s CD collection is uniformly distributed from 2 to 3.5 minutes., Suppose we randomly pick 5 CDs from the collection. There is a total of 43 songs on the 5 CDs., a. In words, X =, b. X~, c. In words, X=, d. X~, e. Find the first quartile for the average song length., f. The IQR (interquartile range) for the average song length is from _______ to _______., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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308, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, Exercise 7.7.13, (Solution on p. 324.), Salaries for teachers in a particular elementary school district are normally distributed with a, mean of $44,000 and a standard deviation of $6500. We randomly survey 10 teachers from that, district., a. In words, X =, b. In words, X =, c. X~, d. In words, ΣX =, e. ΣX~, f. Find the probability that the teachers earn a total of over $400,000., g. Find the 90th percentile for an individual teacher’s salary., h. Find the 90th percentile for the average teachers’ salary., i. If we surveyed 70 teachers instead of 10, graphically, how would that change the distribution, for X?, j. If each of the 70 teachers received a $3000 raise, graphically, how would that change the distribution for X?, Exercise 7.7.14, The distribution of income in some Third World countries is considered wedge shaped (many, very poor people, very few middle income people, and few to many wealthy people). Suppose we, pick a country with a wedge distribution. Let the average salary be $2000 per year with a standard, deviation of $8000. We randomly survey 1000 residents of that country., a., b., c., d., e., , In words, X =, In words, X =, X~, How is it possible for the standard deviation to be greater than the average?, Why is it more likely that the average of the 1000 residents will be from $2000 to $2100 than, from $2100 to $2200?, , Exercise 7.7.15, (Solution on p. 324.), The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital., a. In words, X =, b. In words, X =, c. X~, d. In words, ΣX =, e. ΣX~, f. Is it likely that an individual stayed more than 5 days in the hospital? Why or why not?, g. Is it likely that the average stay for the 80 women was more than 5 days? Why or why not?, h. Which is more likely:, i. an individual stayed more than 5 days; or, ii. the average stay of 80 women was more than 5 days?, i. If we were to sum up the women’s stays, is it likely that, collectively they spent more than a, year in the hospital? Why or why not?, Exercise 7.7.16, In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55, acres. Suppose we randomly survey 38 farmers from 1940. (Source: U.S. Dept. of Agriculture), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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309, a., b., c., d., , In words, X =, In words, X =, X~, The IQR for X is from _______ acres to _______ acres., , Exercise 7.7.17, (Solution on p. 324.), The stock closing prices of 35 U.S. semiconductor manufacturers are given below. (Source: Wall, Street Journal), 8.625; 30.25; 27.625; 46.75; 32.875; 18.25; 5; 0.125; 2.9375; 6.875; 28.25; 24.25; 21; 1.5; 30.25; 71; 43.5;, 49.25; 2.5625; 31; 16.5; 9.5; 18.5; 18; 9; 10.5; 16.625; 1.25; 18; 12.875; 7; 12.875; 2.875; 60.25; 29.25, a. In words, X =, b. i. x =, ii. s x =, iii. n =, c. Construct a histogram of the distribution of the averages. Start at x = −0.0005. Make bar, widths of 10., d. In words, describe the distribution of stock prices., e. Randomly average 5 stock prices together. (Use a random number generator.) Continue averaging 5 pieces together until you have 10 averages. List those 10 averages., f. Use the 10 averages from (e) to calculate:, i. x =, ii. s x =, g. Construct a histogram of the distribution of the averages. Start at x = −0.0005. Make bar, widths of 10., h. Does this histogram look like the graph in (c)?, i. In 1 - 2 complete sentences, explain why the graphs either look the same or look different?, j. Based upon the theory of the Central Limit Theorem, X~, Exercise 7.7.18, Use the Initial Public Offering data (Section 14.3.2: Stock Prices) (see “Table of Contents) to do this, problem., a. In words, X =, b. i. µ X =, ii. σX =, iii. n =, c. Construct a histogram of the distribution. Start at x = −0.50. Make bar widths of $5., d. In words, describe the distribution of stock prices., e. Randomly average 5 stock prices together. (Use a random number generator.) Continue averaging 5 pieces together until you have 15 averages. List those 15 averages., f. Use the 15 averages from (e) to calculate the following:, i. x =, ii. s x =, g. Construct a histogram of the distribution of the averages. Start at x = −0.50. Make bar widths, of $5., h. Does this histogram look like the graph in (c)? Explain any differences., i. In 1 - 2 complete sentences, explain why the graphs either look the same or look different?, j. Based upon the theory of the Central Limit Theorem, X~, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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310, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , 7.7.1 Try these multiple choice questions (Exercises19 - 23)., The next two questions refer to the following information: The time to wait for a particular rural bus, is distributed uniformly from 0 to 75 minutes. 100 riders are randomly sampled to learn how long they, waited., Exercise 7.7.19, (Solution on p. 324.), The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is:, A., B., C., D., , 315.0, 40.3, 38.5, 65.2, , Exercise 7.7.20, (Solution on p. 324.), Would you be surprised, based upon numerical calculations, if the sample average wait time (in, minutes) for 100 riders was less than 30 minutes?, A. Yes, B. No, C. There is not enough information., Exercise 7.7.21, Which of the following is NOT TRUE about the distribution for averages?, A., B., C., D., , (Solution on p. 324.), , The mean, median and mode are equal, The area under the curve is one, The curve never touches the x-axis, The curve is skewed to the right, , The next three questions refer to the following information: The cost of unleaded gasoline in the Bay Area, once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas, stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the, 16 gas stations., Exercise 7.7.22, (Solution on p. 324.), The distribution to use for the average cost of gasoline for the 16 gas stations is, A. X ∼ N (4.59, 0.10), 0.10, B. X ∼ N 4.59, √, 16 , , 0.10, C. X ∼ N 4.59, 16, , , 16, D. X ∼ N 4.59, 0.10, Exercise 7.7.23, (Solution on p. 324.), What is the probability that the average price for 16 gas stations is over $4.69?, A., B., C., D., , Almost zero, 0.1587, 0.0943, Unknown, , Exercise 7.7.24, (Solution on p. 324.), Find the probability that the average price for 30 gas stations is less than $4.55., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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311, A. 0.6554, B. 0.3446, C. 0.0142, D. 0.9858, E. 0, Exercise 7.7.25, (Solution on p. 324.), For the Charter School Problem (Example 6) in Central Limit Theorem: Using the Central Limit, Theorem, calculate the following using the normal approximation to the binomial., A. Find the probability that less than 100 favor a charter school for grades K - 5., B. Find the probability that 170 or more favor a charter school for grades K - 5., C. Find the probability that no more than 140 favor a charter school for grades K - 5., D. Find the probability that there are fewer than 130 that favor a charter school for grades K - 5., E. Find the probability that exactly 150 favor a charter school for grades K - 5., If you either have access to an appropriate calculator or computer software, try calculating these, probabilities using the technology. Try also using the suggestion that is at the bottom of Central, Limit Theorem: Using the Central Limit Theorem for finding a website that calculates binomial, probabilities., Exercise 7.7.26, (Solution on p. 324.), Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school., Each day the driver would be chosen by randomly selecting one of the four names. They carpool, to school for 96 days. Use the normal approximation to the binomial to calculate the following, probabilities. Round the standard deviation to 4 decimal places., A. Find the probability that Janice is the driver at most 20 days., B. Find the probability that Roberta is the driver more than 16 days., C. Find the probability that Barbara drives exactly 24 of those 96 days., If you either have access to an appropriate calculator or computer software, try calculating these, probabilities using the technology. Try also using the suggestion that is at the bottom of Central, Limit Theorem: Using the Central Limit Theorem for finding a website that calculates binomial, probabilities., **Exercise 24 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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312, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , 7.8 Review8, The next three questions refer to the following information: Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past, the 10 A.M. start time that individuals wait for their delivery., Exercise 7.8.1, X∼, A., B., C., D., , (Solution on p. 325.), , U (0, 4), U (10, 2), Exp (2), N (2, 1), , Exercise 7.8.2, The average wait time is:, A., B., C., D., , (Solution on p. 325.), , 1 hour, 2 hour, 2.5 hour, 4 hour, , Exercise 7.8.3, (Solution on p. 325.), Suppose that it is now past noon on a delivery day. The probability that a person must wait at, least 1 21 more hours is:, A., B., C., D., , 1, 4, 1, 2, 3, 4, 3, 8, , Exercise 7.8.4 , Given: X~Exp 31 ., , (Solution on p. 325.), , a. Find P ( x > 1), b. Calculate, value for the upper quartile., the minimum, , c. Find P x =, , 1, 3, , Exercise 7.8.5, , •, •, •, •, , (Solution on p. 325.), , 40% of full-time students took 4 years to graduate, 30% of full-time students took 5 years to graduate, 20% of full-time students took 6 years to graduate, 10% of full-time students took 7 years to graduate, , The expected time for full-time students to graduate is:, A., B., C., D., 8 This, , 4 years, 4.5 years, 5 years, 5.5 years, content is available online at <http://cnx.org/content/m16955/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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313, Exercise 7.8.6, (Solution on p. 325.), Which of the following distributions is described by the following example?, Many people can run a short distance of under 2 miles, but as the distance increases, fewer people, can run that far., A., B., C., D., , Binomial, Uniform, Exponential, Normal, , Exercise 7.8.7, (Solution on p. 325.), The length of time to brush one’s teeth is generally thought to be exponentially distributed with, a mean of 43 minutes. Find the probability that a randomly selected person brushes his/her teeth, less than 34 minutes., A. 0.5, B. 43, C. 0.43, D. 0.63, Exercise 7.8.8, Which distribution accurately describes the following situation?, , (Solution on p. 325.), , The chance that a teenage boy regularly gives his mother a kiss goodnight (and he should!!) is, about 20%. Fourteen teenage boys are randomly surveyed., X =the number of teenage boys that regularly give their mother a kiss goodnight, A. B (14, 0.20), B. P (2.8), C. N (2.8,, 2.24, ), 1, D. Exp 0.20, Exercise 7.8.9, Which distribution accurately describes the following situation?, , (Solution on p. 325.), , A 2008 report on technology use states that approximately 20 percent of U.S. households have, never sent an e-mail. (source: http://www.webguild.org/2008/05/20-percent-of-americanshave-never-used-email.php) Suppose that we select a random sample of fourteen U.S. households, ., X =the number of households in a 2008 sample of 14 households that have never sent an email, A. B (14, 0.20), B. P (2.8), C. N (2.8,, 2.24, ), 1, D. Exp 0.20, **Exercise 9 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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315, , Figure 7.1, , 4. Calculate the following (n = 1; surveying one person at a time):, a. x =, b. s =, 5. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve., , 7.9.3 Collecting Averages of Pairs, Repeat steps 1 - 5 (of the section above titled "Collect the Data") with one exception. Instead of recording, the change of 30 classmates, record the average change of 30 pairs., 1. Randomly survey 30 pairs of classmates. Record the values of the average of their change., __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , Table 7.2, , 2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collecting, the Data". Sketch the graph using a ruler and a pencil., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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317, 2. Construct a histogram. Scale the axes using the same scaling you did for the section titled "Collect the, Data". Sketch the graph using a ruler and a pencil., , Figure 7.3, , 3. Calculate the following (n = 5; surveying five people at a time):, a. x =, b. s =, 4. Draw a smooth curve through tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve., , 7.9.5 Discussion Questions, 1. As n changed, why did the shape of the distribution of the data change? Use 1 – 2 complete sentences, to explain what happened., 2. In the section titled "Collect the Data", what was the approximate distribution of the data? X ∼, 3. In the section titled "Collecting Averages of Groups of Five", what was the approximate distribution, of the averages? X ∼, 4. In 1 – 2 complete sentences, explain any differences in your answers to the previous two questions., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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318, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , 7.10 Lab 2: Central Limit Theorem (Cookie Recipes)10, Class Time:, Names:, , 7.10.1 Student Learning Outcomes:, • The student will demonstrate and compare properties of the Central Limit Theorem., , 7.10.2 Given:, X = length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of, the different recipes makes the same quantity of cookies.), Recipe #, , X, , Recipe #, , X, , Recipe #, , X, , Recipe #, , X, , 1, , 1, , 16, , 2, , 31, , 3, , 46, , 2, , 2, , 5, , 17, , 2, , 32, , 4, , 47, , 2, , 3, , 2, , 18, , 4, , 33, , 5, , 48, , 11, , 4, , 5, , 19, , 6, , 34, , 6, , 49, , 5, , 5, , 6, , 20, , 1, , 35, , 6, , 50, , 5, , 6, , 1, , 21, , 6, , 36, , 1, , 51, , 4, , 7, , 2, , 22, , 5, , 37, , 1, , 52, , 6, , 8, , 6, , 23, , 2, , 38, , 2, , 53, , 5, , 9, , 5, , 24, , 5, , 39, , 1, , 54, , 1, , 10, , 2, , 25, , 1, , 40, , 6, , 55, , 1, , 11, , 5, , 26, , 6, , 41, , 1, , 56, , 2, , 12, , 1, , 27, , 4, , 42, , 6, , 57, , 4, , 13, , 1, , 28, , 1, , 43, , 2, , 58, , 3, , 14, , 3, , 29, , 6, , 44, , 6, , 59, , 6, , 15, , 2, , 30, , 2, , 45, , 2, , 60, , 5, , Table 7.4, , Calculate the following:, a. µ x =, b. σx =, 10 This, , content is available online at <http://cnx.org/content/m16945/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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319, , 7.10.3 Collect the Data, Use a random number generator to randomly select 4 samples of size n = 5 from the given population., Record your samples below. Then, for each sample, calculate the mean to the nearest tenth. Record them in, the spaces provided. Record the sample means for the rest of the class., 1. Complete the table:, , Means:, , Sample 1, , Sample 2, , Sample 3, , Sample 4, , x=, , x=, , x=, , x=, , Sample means from other groups:, , Table 7.5, , 2. Calculate the following:, a. x =, b. s x =, 3. Again, use a random number generator to randomly select 4 samples from the population. This time,, make the samples of size n = 10. Record the samples below. As before, for each sample, calculate the, mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest, of the class., , Means:, , Sample 1, , Sample 2, , Sample 3, , Sample 4, , x=, , x=, , x=, , x=, , Sample means from other groups:, , Table 7.6, , 4. Calculate the following:, a. x =, b. s x =, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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320, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , 5. For the original population, construct a histogram. Make intervals with bar width = 1 day. Sketch the, graph using a ruler and pencil. Scale the axes., , Figure 7.4, , 6. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve., , 7.10.4 Repeat the Procedure for n=5, 1. For the sample of n = 5 days averaged together, construct a histogram of the averages (your means, together with the means of the other groups). Make intervals with bar widths = 21 day. Sketch the, graph using a ruler and pencil. Scale the axes., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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321, , Figure 7.5, , 2. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve., , 7.10.5 Repeat the Procedure for n=10, 1. For the sample of n = 10 days averaged together, construct a histogram of the averages (your means, together with the means of the other groups). Make intervals with bar widths = 21 day. Sketch the, graph using a ruler and pencil. Scale the axes., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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322, , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , Figure 7.6, , 2. Draw a smooth curve through the tops of the bars of the histogram. Use 1 – 2 complete sentences to, describe the general shape of the curve., , 7.10.6 Discussion Questions, 1. Compare the three histograms you have made, the one for the population and the two for the sample, means. In three to five sentences, describe the similarities and differences., 2. State the theoretical (according to the CLT) distributions for the sample means., a. n = 5: X ∼, b. n = 10: X ∼, 3. Are the sample means for n = 5 and n = 10 “close” to the theoretical mean, µ x ? Explain why or why, not., 4. Which of the two distributions of sample means has the smaller standard deviation? Why?, 5. As n changed, why did the shape of the distribution of the data change? Use 1 – 2 complete sentences, to explain what happened., NOTE :, , This lab was designed and contributed by Carol Olmstead., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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323, , Solutions to Exercises in Chapter 7, Solutions to Practice: The Central Limit Theorem, Solution to Exercise 7.6.1 (p. 302), b. 3.5, 4.25, 0.2441, Solution to Exercise 7.6.2 (p. 302), b. 0.7499, Solution to Exercise 7.6.3 (p. 303), b. 4.49 hours, Solution to Exercise 7.6.4 (p. 303), b. 0.3802, Solution to Exercise 7.6.5 (p. 303), b: 71.90, , Solutions to Homework, Solution to Exercise 7.7.1 (p. 305), , , b. Xbar~N 60, √9, 25, c. 0.5000, d. 59.06, e. 0.8536, f. 0.1333, h. 1530.35, i. 0.8536, Solution to Exercise 7.7.3 (p. 305), , , a. N 36, √10, 16, b. 1, c. 34.31, Solution to Exercise 7.7.5 (p. 306), , , a. N 250, √50, 49, b. 0.0808, c. 256.01 feet, Solution to Exercise 7.7.7 (p. 306), a., b., c., d., , The total length of time for 9 criminal trials, N (189, 21), 0.0432, 162.09, , Solution to Exercise 7.7.9 (p. 306), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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324, a., b., c., d., , CHAPTER 7. THE CENTRAL LIMIT THEOREM, , , , N 145, √14, 49, 0.6247, 146.68, 145 minutes, , Solution to Exercise 7.7.11 (p. 307), b. U (10, 25), c. 17.5, q, 225, d., 12 = 4.3301, f. N (17.5, 0.5839), h. N (962.5, 32.11), j. 0.0032, Solution to Exercise 7.7.13 (p. 308), , , √, c. N 44, 000, 6500, , , 10, √ , 10 (6500), e. N 440,000,, f. 0.9742, g. $52,330, h. $46,634, Solution to Exercise 7.7.15 (p. 308), , , c. N 2.4, √0.9, 80, , e. N (192,8.05), h. Individual, Solution to Exercise 7.7.17 (p. 309), b. $20.71; $17.31; 35, d. Exponential distribution, X ∼ Exp (1/20.71), f. $20.71;, $11.14 , , √, j. N 20.71, 17.31, 5, , Solution to Exercise 7.7.19 (p., B, Solution to Exercise 7.7.20 (p., A, Solution to Exercise 7.7.21 (p., D, Solution to Exercise 7.7.22 (p., B, Solution to Exercise 7.7.23 (p., A, Solution to Exercise 7.7.24 (p., C, Solution to Exercise 7.7.25 (p., , 310), 310), 310), 310), 310), 310), 311), , C. 0.0162, E. 0.0268, Solution to Exercise 7.7.26 (p. 311), A. 0.2047, B. 0.9615, C. 0.0938, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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325, , Solutions to Review, Solution to Exercise 7.8.1 (p., A, Solution to Exercise 7.8.2 (p., B, Solution to Exercise 7.8.3 (p., A, Solution to Exercise 7.8.4 (p., , 312), 312), 312), 312), , a. 0.7165, b. 4.16, c. 0, Solution to Exercise 7.8.5 (p., C, Solution to Exercise 7.8.6 (p., C, Solution to Exercise 7.8.7 (p., D, Solution to Exercise 7.8.8 (p., A, Solution to Exercise 7.8.9 (p., A, , 312), 313), 313), 313), 313), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 8, , Confidence Intervals, 8.1 Confidence Intervals1, 8.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Calculate and interpret confidence intervals for one population mean and one population proportion., • Interpret the student-t probability distribution as the sample size changes., • Discriminate between problems applying the normal and the student-t distributions., , 8.1.2 Introduction, Suppose you are trying to determine the mean rent of a two-bedroom apartment in your town. You might, look in the classified section of the newspaper, write down several rents listed, and average them together., You would have obtained a point estimate of the true mean. If you are trying to determine the percent of, times you make a basket when shooting a basketball, you might count the number of shots you make and, divide that by the number of shots you attempted. In this case, you would have obtained a point estimate, for the true proportion., We use sample data to make generalizations about an unknown population. This part of statistics is called, inferential statistics. The sample data help us to make an estimate of a population parameter. We realize, that the point estimate is most likely not the exact value of the population parameter, but close to it. After, calculating point estimates, we construct confidence intervals in which we believe the parameter lies., In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new, distribution, the Student’s-t, and how it is used with these intervals. Throughout the chapter, it is important, to keep in mind that the confidence interval is a random variable. It is the parameter that is fixed., If you worked in the marketing department of an entertainment company, you might be interested in the, mean number of compact discs (CD’s) a consumer buys per month. If so, you could conduct a survey, and calculate the sample mean, x, and the sample standard deviation, s. You would use x to estimate, the population mean and s to estimate the population standard deviation. The sample mean, x, is the, point estimate for the population mean, µ. The sample standard deviation, s, is the point estimate for the, population standard deviation, σ., 1 This, , content is available online at <http://cnx.org/content/m16967/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 327
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328, , CHAPTER 8. CONFIDENCE INTERVALS, , Each of x and s is also called a statistic., A confidence interval is another type of estimate but, instead of being just one number, it is an interval, of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The, confidence interval is likely to include an unknown population parameter., Suppose for the CD example we do not know the population mean µ but we do know that the population, standard deviation is σ = 1 and our sample size is 100. Then by the Central Limit Theorem, the standard, deviation for the sample mean is, √σ, n, , =, , √1, 100, , = 0.1., , The Empirical Rule, which applies to bell-shaped distributions, says that in approximately 95% of the, samples, the sample mean, x, will be within two standard deviations of the population mean µ. For our CD, example, two standard deviations is (2) (0.1) = 0.2. The sample mean x is likely to be within 0.2 units of, µ., Because x is within 0.2 units of µ, which is unknown, then µ is likely to be within 0.2 units of x in 95%, of the samples. The population mean µ is contained in an interval whose lower number is calculated by, taking the sample mean and subtracting two standard deviations ((2) (0.1)) and whose upper number is, calculated by taking the sample mean and adding two standard deviations. In other words, µ is between, x − 0.2 and x + 0.2 in 95% of all the samples., For the CD example, suppose that a sample produced a sample mean x = 2. Then the unknown population, mean µ is between, x − 0.2 = 2 − 0.2 = 1.8 and x + 0.2 = 2 + 0.2 = 2.2, We say that we are 95% confident that the unknown population mean number of CDs is between 1.8 and, 2.2. The 95% confidence interval is (1.8, 2.2)., The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean µ, or our sample produced an x that is not within 0.2 units of the true mean µ. The second possibility happens, for only 5% of all the samples (100% - 95%)., Remember that a confidence interval is created for an unknown population parameter like the population, mean, µ. Confidence intervals for some parameters have the form, (point estimate - margin of error, point estimate + margin of error), The margin of error depends on the confidence level or percentage of confidence., When you read newspapers and journals, some reports will use the phrase "margin of error." Other reports, will not use that phrase, but include a confidence interval as the point estimate + or - the margin of error., These are two ways of expressing the same concept., NOTE : Although the text only covers symmetric confidence intervals, there are non-symmetric, confidence intervals (for example, a confidence interval for the standard deviation)., , 8.1.3 Optional Collaborative Classroom Activity, Have your instructor record the number of meals each student in your class eats out in a week. Assume, that the standard deviation is known to be 3 meals. Construct an approximate 95% confidence interval for, the true mean number of meals students eat out each week., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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329, 1. Calculate the sample mean., 2. σ = 3 and n = the number, , of students surveyed., 3. Construct the interval x − 2 ·, , √σ , x + 2 · √σ, n, n, , We say we are approximately 95% confident that the true average number of meals that students eat out in, a week is between __________ and ___________., , 8.2 Confidence Interval, Single Population Mean, Population Standard, Deviation Known, Normal2, 8.2.1 Calculating the Confidence Interval, To construct a confidence interval for a single unknown population mean µ , where the population standard deviation is known, we need x as an estimate for µ and we need the margin of error. Here, the, margin of error is called the error bound for a population mean (abbreviated EBM). The sample mean x is, the point estimate of the unknown population mean µ, The confidence interval estimate will have the form:, (point estimate - error bound, point estimate + error bound) or, in symbols,( x − EBM, x + EBM), The margin of error depends on the confidence level (abbreviated CL). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population, parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the, choice of the person constructing the confidence interval to choose a confidence level of 90% or higher, because that person wants to be reasonably certain of his or her conclusions., There is another probability called alpha (α). α is related to the confidence level CL. α is the probability that, the interval does not contain the unknown population parameter., Mathematically, α + CL = 1., Example 8.1, Suppose we have collected data from a sample. We know the sample mean but we do not know, the mean for the entire population., The sample mean is 7 and the error bound for the mean is 2.5., x = 7 and EBM = 2.5., The confidence interval is (7 − 2.5, 7 + 2.5); calculating the values gives (4.5, 9.5)., If the confidence level (CL) is 95%, then we say that "We estimate with 95% confidence that the, true value of the population mean is between 4.5 and 9.5.", A confidence interval for a population mean with a known standard deviation is based on the fact that the, sample means follow an approximately normal distribution. Suppose that our sample has a mean of x = 10, and we have constructed the 90% confidence interval (5, 15) where EBM = 5., To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the, normal distribution., 2 This, , content is available online at <http://cnx.org/content/m16962/1.23/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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330, , CHAPTER 8. CONFIDENCE INTERVALS, , To capture the central 90%, we must go out 1.645 "standard deviations" on either side of the calculated, sample mean. 1.645 is the z-score from a Standard Normal probability distribution that puts an area of 0.90, in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail., It is important that the "standard deviation" used must be appropriate for the parameter we are estimating., So in this section, we need to use the standard deviation that applies to sample means, which is √σn . √σn is, commonly called the "standard error of the mean" in order to clearly distinguish the standard deviation for, a mean from the population standard deviation σ., In summary, as a result of the Central Limit Theorem:, , , • X is normally distributed, that is, X ∼ N µ X , √σn ., • When the population standard deviation σ is known, we use a Normal distribution to calculate, the error bound., Calculating the Confidence Interval:, To construct a confidence interval estimate for an unknown population mean, we need data from a random, sample. The steps to construct and interpret the confidence interval are:, , • Calculate the sample mean x from the sample data. Remember, in this section, we already know the, population standard deviation σ., • Find the Z-score that corresponds to the confidence level., • Calculate the error bound EBM, • Construct the confidence interval, • Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain, what the confidence interval means, in the words of the problem.), We will first examine each step in more detail, and then illustrate the process with some examples., Finding z for the stated Confidence Level, When we know the population standard deviation σ, we use a standard normal distribution to calculate, the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area, equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z∼N(0,1)., The confidence level, CL, is the area in the middle of the standard normal distribution. CL = 1 − α. So α is, the area that is split equally between the two tails. Each of the tails contains an area equal to α2 ., The z-score that has an area to the right of, , α, 2, , is denoted by z α, 2, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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331, For example, when CL = 0.95 then α = 0.05 and, , α, 2, , = 0.025 ; we write z α2 = z.025, , The area to the right of z.025 is 0.025 and the area to the left of z.025 is 1-0.025 = 0.975, z α = z0.025 = 1.96 , using a calculator, computer or a Standard Normal probability table., 2, , Using the TI83, TI83+ or TI84+ calculator: invNorm(0.975, 0, 1) = 1.96, CALCULATOR NOTE: Remember to use area to the LEFT of z α ; in this chapter the last two inputs in the, 2, invNorm command are 0,1 because you are using a Standard Normal Distribution Z∼N(0,1), EBM: Error Bound, The error bound formula for an unknown population mean µ when the population standard deviation σ is, known is, , • EBM = z α2 ·, , √σ, n, , Constructing the Confidence Interval, , • The confidence interval estimate has the format ( x − EBM, x + EBM)., The graph gives a picture of the entire situation., CL +, , α, 2, , +, , α, 2, , = CL + α = 1., , Writing the Interpretation, The interpretation should clearly state the confidence level (CL), explain what population parameter is, being estimated (here, a population mean), and should state the confidence interval (both endpoints). "We, estimate with ___% confidence that the true population mean (include context of the problem) is between, ___ and ___ (include appropriate units).", Example 8.2, Suppose scores on exams in statistics are normally distributed with an unknown population mean, and a population standard deviation of 3 points. A random sample of 36 scores is taken and gives, a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population, mean exam score (the mean score on all exams)., Problem, Find a 90% confidence interval for the true (population) mean of statistics exam scores., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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332, , CHAPTER 8. CONFIDENCE INTERVALS, Solution, , • You can use technology to directly calculate the confidence interval, • The first solution is shown step-by-step (Solution A)., • The second solution uses the TI-83, 83+ and 84+ calculators (Solution B)., Solution A, To find the confidence interval, you need the sample mean, x, and the EBM., x = 68, , EBM = z α · √σn, 2, σ = 3 ; n = 36 ; The confidence level is 90% (CL=0.90), CL = 0.90 so α = 1 − CL = 1 − 0.90 = 0.10, α, 2, , = 0.05, , z α = z.05, 2, , The area to the right of z.05 is 0.05 and the area to the left of z.05 is 1−0.05=0.95, z α = z.05 = 1.645, 2, , using invNorm(0.95,0,1) on the TI-83,83+,84+ calculators. This can also be found using appropriate, commands on other calculators, using a computer, or using a probability table for the Standard, Normal distribution., , , EBM = 1.645 · √3, = 0.8225, 36, , x − EBM = 68 − 0.8225 = 67.1775, x + EBM = 68 + 0.8225 = 68.8225, The 90% confidence interval is (67.1775, 68.8225)., Solution B, Using a function of the TI-83, TI-83+ or TI-84 calculators:, Press STAT and arrow over to TESTS., Arrow down to 7:ZInterval., Press ENTER., Arrow to Stats and press ENTER., Arrow down and enter 3 for σ, 68 for x , 36 for n, and .90 for C-level., Arrow down to Calculate and press ENTER., The confidence interval is (to 3 decimal places) (67.178, 68.822)., Interpretation, We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82., Explanation of 90% Confidence Level, 90% of all confidence intervals constructed in this way contain the true mean statistics exam score., For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to, contain the true population mean exam score., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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333, , 8.2.2 Changing the Confidence Level or Sample Size, Example 8.3: Changing the Confidence Level, Suppose we change the original problem by using a 95% confidence level. Find a 95% confidence, interval for the true (population) mean statistics exam score., Solution, To find the confidence interval, you need the sample mean, x, and the EBM., x = 68, , EBM = z α · √σn, 2, σ = 3 ; n = 36 ; The confidence level is 95% (CL=0.95), CL = 0.95 so α = 1 − CL = 1 − 0.95 = 0.05, α, 2, , = 0.025, , z α = z.025, 2, , The area to the right of z.025 is 0.025 and the area to the left of z.025 is 1−0.025=0.975, z α = z.025 = 1.96, 2, , using invnorm(.975,0,1) on the TI-83,83+,84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard, Normal distribution.), , , EBM = 1.96 · √3, = 0.98, 36, , x − EBM = 68 − 0.98 = 67.02, x + EBM = 68 + 0.98 = 68.98, Interpretation, We estimate with 95 % confidence that the true population mean for all statistics exam scores is, between 67.02 and 68.98., Explanation of 95% Confidence Level, 95% of all confidence intervals constructed in this way contain the true value of the population, mean statistics exam score., Comparing the results, The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The, 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the, area 0.90, it makes sense that the 95% confidence interval is wider., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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334, , CHAPTER 8. CONFIDENCE INTERVALS, , (a), , (b), , Figure 8.1, , Summary: Effect of Changing the Confidence Level, , • Increasing the confidence level increases the error bound, making the confidence interval, wider., • Decreasing the confidence level decreases the error bound, making the confidence interval, narrower., Example 8.4: Changing the Sample Size:, Suppose we change the original problem to see what happens to the error bound if the sample size, is changed., Problem, Leave everything the same except the sample size. Use the original 90% confidence level. What, happens to the error bound and the confidence interval if we increase the sample size and use, n=100 instead of n=36? What happens if we decrease the sample size to n=25 instead of n=36?, • x = 68, , • EBM = z α · √σn, 2, • σ = 3 ; The confidence level is 90% (CL=0.90) ; z α = z.05 = 1.645, 2, , Solution A, If we increase the sample size n to 100, we decrease the error bound., , , , When n = 100 : EBM = z α · √σn = 1.645 · √ 3, = 0.4935, 2, , 100, , Solution B, If we decrease the sample size n to 25, we increase the error bound., , , , When n = 25 : EBM = z α · √σn = 1.645 · √3, = 0.987, 2, , 25, , Summary: Effect of Changing the Sample Size, , • Increasing the sample size causes the error bound to decrease, making the confidence interval narrower., • Decreasing the sample size causes the error bound to increase, making the confidence interval wider., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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335, , 8.2.3 Working Backwards to Find the Error Bound or Sample Mean, Working Bacwards to find the Error Bound or the Sample Mean, When we calculate a confidence interval, we find the sample mean and calculate the error bound and use, them to calculate the confidence interval. But sometimes when we read statistical studies, the study may, state the confidence interval only. If we know the confidence interval, we can work backwards to find both, the error bound and the sample mean., Finding the Error Bound, , • From the upper value for the interval, subtract the sample mean, • OR, From the upper value for the interval, subtract the lower value. Then divide the difference by 2., Finding the Sample Mean, , • Subtract the error bound from the upper value of the confidence interval, • OR, Average the upper and lower endpoints of the confidence interval, Notice that there are two methods to perform each calculation. You can choose the method that is easier to, use with the information you know., Example 8.5, Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound., We may know that the sample mean is 68. Or perhaps our source only gave the confidence interval, and did not tell us the value of the the sample mean., Calculate the Error Bound:, , • If we know that the sample mean is 68: EBM = 68.82 − 68 = 0.82, • If we don’t know the sample mean: EBM = (68.82−2 67.18) = 0.82, Calculate the Sample Mean:, , • If we know the error bound: x = 68.82 − 0.82 = 68, • If we don’t know the error bound: x = (67.18+2 68.82) = 68, , 8.2.4 Calculating the Sample Size n, If researchers desire a specific margin of error, then they can use the error bound formula to calculate the, required sample size., The error bound, formula for a population mean when the population standard deviation is known is, α, EBM = z · √σn, 2, , The formula for sample size is n =, , z2 σ 2, ,, EBM2, , found by solving the error bound formula for n, , In this formula, z is z α , corresponding to the desired confidence level. A researcher planning a study who, 2, wants a specified confidence level and error bound can use this formula to calculate the size of the sample, needed for the study., Example 8.6, The population standard deviation for the age of Foothill College students is 15 years. If we, want to be 95% confident that the sample mean age is within 2 years of the true population mean, age of Foothill College students , how many randomly selected Foothill College students must be, surveyed?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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336, , CHAPTER 8. CONFIDENCE INTERVALS, From the problem, we know that σ = 15 and EBM=2, z = z.025 = 1.96, because the confidence level is 95%., 2 2, , 2, , 2, , n = z σ 2 = 1.962215 =216.09 using the sample size equation., EBM, Use n = 217: Always round the answer UP to the next higher integer to ensure that the sample, size is large enough., Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we, are within 2 years of the true population mean age of Foothill College students., **With contributions from Roberta Bloom, , 8.3 Confidence Interval, Single Population Mean, Standard Deviation, Unknown, Student’s-t3, In practice, we rarely know the population standard deviation. In the past, when the sample size was large,, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate, for σ and proceeded as before to calculate a confidence interval with close enough results. However,, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in, the confidence interval., William S. Gossett (1876-1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate, results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem, led him to "discover" what is called the Student’s-t distribution. The name comes from the fact that Gosset, wrote under the pen name "Student.", Up until the mid 1970s, some statisticians used the normal distribution approximation for large sample, sizes and only used the Student’s-t distribution for sample sizes of at most 30. With the common use of, graphing calculators and computers, the practice is to use the Student’s-t distribution whenever s is used, as an estimate for σ., If you draw a simple random sample of size n from a population that has approximately a normal distrix −µ, bution with mean µ and unknown population standard deviation σ and calculate the t-score t = s ,, √, , n, , then the t-scores follow a Student’s-t distribution with n − 1 degrees of freedom. The t-score has the same, interpretation as the z-score. It measures how far x is from its mean µ. For each sample size n, there is a, different Student’s-t distribution., The degrees of freedom, n − 1, come from the calculation of the sample standard deviation s. In Chapter, 2, we used n deviations ( x − x values) to calculate s. Because the sum of the deviations is 0, we can find, the last deviation once we know the other n − 1 deviations. The other n − 1 deviations can change or vary, freely. We call the number n − 1 the degrees of freedom (df)., Properties of the Student’s-t Distribution, , • The graph for the Student’s-t distribution is similar to the Standard Normal curve., • The mean for the Student’s-t distribution is 0 and the distribution is symmetric about 0., 3 This, , content is available online at <http://cnx.org/content/m16959/1.24/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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337, , • The Student’s-t distribution has more probability in its tails than the Standard Normal distribution, because the spread of the t distribution is greater than the spread of the Standard Normal. So the, graph of the Student’s-t distribution will be thicker in the tails and shorter in the center than the, graph of the Standard Normal distribution., • The exact shape of the Student’s-t distribution depends on the "degrees of freedom". As the degrees, of freedom increases, the graph Student’s-t distribution becomes more like the graph of the Standard, Normal distribution., • The underlying population of individual observations is assumed to be normally distributed with, unknown population mean µ and unknown population standard deviation σ. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the, assumption is met and doesn’t need discussion. Random sampling is assumed but it is a completely, separate assumption from normality., Calculators and computers can easily calculate any Student’s-t probabilities. The TI-83,83+,84+ have a tcdf, function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower, bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse, probability to find the value of t when we know the probability., For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works, similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of, freedom) The output is the t-score that corresponds to the area we specified., The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.), A probability table for the Student’s-t distribution can also be used. The table gives t-scores that correspond, to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program, or command, so if you are using that calculator, you need to use a probability table for the Student’s-t distribution.) When using t-table, note that some tables are formatted to show the confidence level in the column, headings, while the column headings in some tables may show only corresponding area in one or both tails., A Student’s-t table (See the Table of Contents 15. Tables) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily, calculate any Student’s-t probabilities., The notation for the Student’s-t distribution is (using T as the random variable) is, , • T ∼ tdf where df = n − 1., • For example, if we have a sample of size n=20 items, then we calculate the degrees of freedom as, df=n−1=20−1=19 and we write the distribution as T ∼ t19, If the population standard deviation is not known, the error bound for a population mean is:, , • EBM = t α · √sn, 2, , • t α2 is the t-score with area to the right equal to α2, • use df = n − 1 degrees of freedom, • s = sample standard deviation, The format for the confidence interval is:, , ( x − EBM, x + EBM)., The TI-83, 83+ and 84 calculators have a function that calculates the confidence interval directly. To get to, it,, Press STAT, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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338, , CHAPTER 8. CONFIDENCE INTERVALS, , Arrow over to TESTS., Arrow down to 8:TInterval and press ENTER (or just press 8)., Example 8.7, Suppose you do a study of acupuncture to determine how effective it is in relieving pain., You measure sensory rates for 15 subjects with the results given below. Use the sample data, to construct a 95% confidence interval for the mean sensory rate for the population (assumed, normal) from which you took the data., The solution is shown step-by-step and by using the TI-83, 83+ and 84+ calculators., 8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9, Solution, , • You can use technology to directly calculate the confidence interval., • The first solution is step-by-step (Solution A)., • The second solution uses the Ti-83+ and Ti-84 calculators (Solution B)., Solution A, To find the confidence interval, you need the sample mean, x, and the EBM., x = 8.2267, , s = 1.6722, , n = 15, , df = 15 − 1 = 14, CL = 0.95 so α = 1 − CL = 1 − 0.95 = 0.05, α, 2, , = 0.025, , t α = t.025, 2, , The area to the right of t.025 is 0.025 and the area to the left of t.025 is 1−0.025=0.975, t α = t.025 = 2.14 using invT(.975,14) on the TI-84+ calculator., 2, , EBM = t α ·, , , , 2, , EBM = 2.14 ·, , √s, n, , , , , , 1.6722, √, 15, , , , = 0.924, , x − EBM = 8.2267 − 0.9240 = 7.3, x + EBM = 8.2267 + 0.9240 = 9.15, The 95% confidence interval is (7.30, 9.15)., We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and, 9.15., Solution B, Using a function of the TI-83, TI-83+ or TI-84 calculators:, Press STAT and arrow over to TESTS., Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press, ENTER., Arrow down to List and enter the list name where you put the data., Arrow down to Freq and enter 1., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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339, Arrow down to C-level and enter .95, Arrow down to Calculate and press ENTER., The 95% confidence interval is (7.3006, 9.1527), NOTE : When calculating the error bound, a probability table for the Student’s-t distribution can, also be used to find the value of t. The table gives t-scores that correspond to the confidence level, (column) and degrees of freedom (row); the t-score is found where the row and column intersect, in the table., , **With contributions from Roberta Bloom, , 8.4 Confidence Interval for a Population Proportion4, During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show, that the candidate has 40% of the vote within 3 percentage points. Often, election polls are calculated with, 95% confidence. So, the pollsters would be 95% confident that the true proportion of voters who favored, the candidate would be between 0.37 and 0.43 : (0.40 − 0.03, 0.40 + 0.03)., Investors in the stock market are interested in the true proportion of stocks that go up and down each week., Businesses that sell personal computers are interested in the proportion of households in the United States, that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that, go up or down each week and for the true proportion of households in the United States that own personal, computers., The procedure to find the confidence interval, the sample size, the error bound, and the confidence level, for a proportion is similar to that for the population mean. The formulas are different., How do you know you are dealing with a proportion problem? First, the underlying distribution is, binomial. (There is no mention of a mean or average.) If X is a binomial random variable, then X ∼ B (n, p), where n = the number of trials and p = the probability of a success. To form a proportion, take X, the, random variable for the number of successes and divide it by n, the number of trials (or the sample size)., The random variable P’ (read "P prime") is that proportion,, P’ =, , X, n, , (Sometimes the random variable is denoted as P̂, read "P hat".), When n is large and p is not close to 0 or 1, we can use the normal distribution to approximate the binomial., , √, X ∼ N n · p, n · p · q, If we divide the random variable by n, the mean by n, and the standard deviation by n, we get a normal, distribution of proportions with P’, called the estimated proportion, as the random variable. (Recall that a, proportion = the number of successes divided by n.), , , √, n· p·q, n· p, X, =, P’, ∼, N, ,, n, n, n, √, , Using algebra to simplify :, 4 This, , n· p·q, n, , =, , q, , p·q, n, , content is available online at <http://cnx.org/content/m16963/1.20/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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340, , CHAPTER 8. CONFIDENCE INTERVALS, , , P’ follows a normal distribution for proportions: P’ ∼ N, , p,, , q, , p·q, n, , , , The confidence interval has the form ( p’ − EBP, p’ + EBP)., p’ =, , x, n, , p’ = the estimated proportion of successes (p’ is a point estimate for p, the true proportion), x = the number of successes., n = the size of the sample, The error bound for a proportion is, q, p’·q’, whereq’ = 1 − p’, EBP = z α ·, n, 2, , This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, theq, appropriate standard, deviation that we use is, , √σ ., n, , For a proportion, the appropriate standard deviation is, , However, in the error bound formula, we use, , q, , p’·q’, n, , p·q, n ., , as the standard deviation, instead of, , However, in the error bound formula, the standard deviation is, , q, , q, , p·q, n, , p’·q’, n ., , In the error bound formula, the sample proportions p’ and q’ are estimates of the unknown population, proportions p and q. The estimated proportions p’ and q’ are used because p and q are not known. p’ and, q’ are calculated from the data. p’ is the estimated proportion of successes. q’ is the estimated proportion of, failures., The confidence interval can only be used if the number of successes np’ and the number of failures nq’ are, both larger than 5., NOTE :, , For the normal distribution of proportions, the z-score formula is as follows., q , p·q, p’− p, If P’ ∼ N p,, then the z-score formula is z = √ p·q, n, n, , Example 8.8, Suppose that a market research firm is hired to estimate the percent of adults living in a large, city who have cell phones. 500 randomly selected adult residents in this city are surveyed to, determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they, own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the, true proportion of adults residents of this city who have cell phones., Solution, , • You can use technology to directly calculate the confidence interval., • The first solution is step-by-step (Solution A)., • The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B)., Solution A, Let, X = the, number of people in the sample who have cell phones. X is binomial. X ∼, B 500, 421, 500 ., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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341, To calculate the confidence interval, you must find p’, q’, and EBP., n = 500, p’ =, , x, n, , x = the number of successes = 421, , =, , 421, 500, , = 0.842, , p’ = 0.842 is the sample proportion; this is the point estimate of the population proportion., q’ = 1 − p’ = 1 − 0.842 = 0.158, Since CL = 0.95, then α = 1 − CL = 1 − 0.95 = 0.05, , α, 2, , = 0.025., , Then z α = z.025 = 1.96, 2, , Use the TI-83, 83+ or 84+ calculator command invNorm(0.975,0,1) to find z.025 . Remember that the, area to the right of z.025 is 0.025 and the area to the left of z0.025 is 0.975. This can also be found, using appropriate commands on other calculators, using a computer, or using a Standard Normal, probability table., q, q, p’·q’, (0.842)·(0.158), EBP = z α ·, =, 1.96, ·, = 0.032, n, 500, 2, , p’ − EBP = 0.842 − 0.032 = 0.81, p’ + EBP = 0.842 + 0.032 = 0.874, The, confidence, interval, for, ( p’ − EBP, p’ + EBP) =(0.810, 0.874)., , the, , true, , binomial, , population, , proportion, , is, , Interpretation, We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city, have cell phones., Explanation of 95% Confidence Level, 95% of the confidence intervals constructed in this way would contain the true value for the, population proportion of all adult residents of this city who have cell phones., Solution B, Using a function of the TI-83, 83+ or 84 calculators:, Press STAT and arrow over to TESTS., Arrow down to A:1-PropZint. Press ENTER., Arrow down to x and enter 421., Arrow down to n and enter 500., Arrow down to C-Level and enter .95., Arrow down to Calculate and press ENTER., The confidence interval is (0.81003, 0.87397)., , Example 8.9, For a class project, a political science student at a large university wants to estimate the percent, of students that are registered voters. He surveys 500 students and finds that 300 are registered, voters. Compute a 90% confidence interval for the true percent of students that are registered, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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342, , CHAPTER 8. CONFIDENCE INTERVALS, voters and interpret the confidence interval., , Solution, , • You can use technology to directly calculate the confidence interval., • The first solution is step-by-step (Solution A)., • The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B)., Solution A, x = 300 and n = 500., p’ =, , x, n, , =, , 300, 500, , = 0.600, , q’ = 1 − p’ = 1 − 0.600 = 0.400, Since CL = 0.90, then α = 1 − CL = 1 − 0.90 = 0.10, , α, 2, , = 0.05., , z α = z.05 = 1.645, 2, , Use the TI-83, 83+ or 84+ calculator command invNorm(0.95,0,1) to find z.05 . Remember that, the area to the right of z.05 is 0.05 and the area to the left of z.05 is 0.95. This can also be found, using appropriate commands on other calculators, using a computer, or using a Standard Normal, probability table., q, q, p’·q’, (0.60)·(0.40), EBP = z α ·, =, 1.645, ·, = 0.036, n, 500, 2, , p’ − EBP = 0.60 − 0.036 = 0.564, p’ + EBP = 0.60 + 0.036 = 0.636, The, confidence, interval, for, ( p’ − EBP, p’ + EBP) =(0.564, 0.636)., , the, , true, , binomial, , population, , proportion, , is, , Interpretation:, , • We estimate with 90% confidence that the true percent of all students that are registered, voters is between 56.4% and 63.6%., • Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL, students are registered voters., Explanation of 90% Confidence Level, 90% of all confidence intervals constructed in this way contain the true value for the population, percent of students that are registered voters., Solution B, Using a function of the TI-83, 83+ or 84 calculators:, Press STAT and arrow over to TESTS., Arrow down to A:1-PropZint. Press ENTER., Arrow down to x and enter 300., Arrow down to n and enter 500., Arrow down to C-Level and enter .90., Arrow down to Calculate and press ENTER., The confidence interval is (0.564, 0.636)., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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343, , 8.4.1 Calculating the Sample Size n, If researchers desire a specific margin of error, then they can use the error bound formula to calculate the, required sample size., The error bound formula for a population proportion is, q, • EBP = z α2 · p’q’, n, • Solving for n gives you an equation for the sample size., , • n=, , z α 2 · p’q’, 2, , EBP2, , Example 8.10, Suppose a mobile phone company wants to determine the current percentage of customers, aged 50+ that use text messaging on their cell phone. How many customers aged 50+ should, the company survey in order to be 90% confident that the estimated (sample) proportion is, within 3 percentage points of the true population proportion of customers aged 50+ that use text, messaging on their cell phone., , Solution, From the problem, we know that EBP=0.03 (3%=0.03) and, z α = z.05 = 1.645 because the confidence level is 90%, 2, , However, in order to find n , we need to know the estimated (sample) proportion p’. Remember, that q’=1-p’. But, we do not know p’ yet. Since we multiply p’ and q’ together, we make them both, equal to 0.5 because p’q’= (.5)(.5)=.25 results in the largest possible product. (Try other products:, (.6)(.4)=.24; (.3)(.7)=.21; (.2)(.8)=.16 and so on). The largest possible product gives us the largest n., This gives us a large enough sample so that we can be 90% confident that we are within 3 percentage points of the true population proportion. To calculate the sample size n, use the formula and, make the substitutions., n=, , z2 p’q’, EBP2, , gives n =, , 1.6452 (.5)(.5), =751.7, .032, , Round the answer to the next higher value. The sample size should be 752 cell phone customers, aged 50+ in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers aged 50+ that use text messaging on, their cell phone., **With contributions from Roberta Bloom., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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344, , CHAPTER 8. CONFIDENCE INTERVALS, , 8.5 Summary of Formulas5, Formula 8.1: General form of a confidence interval, (lower value, upper value) = (point estimate − error bound, point estimate + error bound), Formula 8.2: To find the error bound when you know the confidence interval, upper value−lower value, error bound = upper value − point estimate, OR, error bound =, 2, Formula 8.3: Single Population Mean, Known Standard Deviation, Normal Distribution, Use the Normal Distribution for Means (Section 7.2), EBM = z α · √σn, 2, , The confidence interval has the format ( x − EBM, x + EBM)., Formula 8.4: Single Population Mean, Unknown Standard Deviation, Student’s-t Distribution, Use the Student’s-t Distribution with degrees of freedom df = n − 1. EBM = t α · √sn, 2, , Formula 8.5: Single Population Proportion, Normal Distribution, Use the Normal Distribution for a single population proportion p’ =, q, p’·q’, EBP = z α ·, p’ + q’ = 1, n, 2, The confidence interval has the format ( p’ − EBP, p’ + EBP)., Formula 8.6: Point Estimates, x is a point estimate for µ, p’ is a point estimate for ρ, , x, n, , s is a point estimate for σ, , 5 This, , content is available online at <http://cnx.org/content/m16973/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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345, , 8.6 Practice 1: Confidence Intervals for Means, Known Population Standard Deviation6, 8.6.1 Student Learning Outcomes, • The student will calculate confidence intervals for means when the population standard deviation is, known., , 8.6.2 Given, The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected., The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College, students. (http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm7, Let X = the age of a Winter Foothill College student, , 8.6.3 Calculating the Confidence Interval, Exercise 8.6.1, x=, , (Solution on p. 371.), , Exercise 8.6.2, n=, , (Solution on p. 371.), , Exercise 8.6.3, 15=(insert symbol here), , (Solution on p. 371.), , Exercise 8.6.4, Define the Random Variable, X, in words., , (Solution on p. 371.), , X=, Exercise 8.6.5, What is x estimating?, , (Solution on p. 371.), , Exercise 8.6.6, Is σx known?, , (Solution on p. 371.), , Exercise 8.6.7, (Solution on p. 371.), As a result of your answer to (4), state the exact distribution to use when calculating the Confidence Interval., , 8.6.4 Explaining the Confidence Interval, Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students., Exercise 8.6.8, How much area is in both tails (combined)? α = ________, , (Solution on p. 371.), , Exercise 8.6.9, How much area is in each tail?, , (Solution on p. 371.), α, 2, , = ________, , Exercise 8.6.10, Identify the following specifications:, 6 This, , (Solution on p. 371.), , content is available online at <http://cnx.org/content/m16970/1.13/>., , 7 http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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346, , CHAPTER 8. CONFIDENCE INTERVALS, a. lower limit =, b. upper limit =, c. error bound =, Exercise 8.6.11, The 95% Confidence Interval is:__________________, , (Solution on p. 371.), , Exercise 8.6.12, Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval, and, the sample mean., , Figure 8.2, , Exercise 8.6.13, In one complete sentence, explain what the interval means., , 8.6.5 Discussion Questions, Exercise 8.6.14, Using the same mean, standard deviation and level of confidence, suppose that n were 69 instead, of 25. Would the error bound become larger or smaller? How do you know?, Exercise 8.6.15, Using the same mean, standard deviation and sample size, how would the error bound change if, the confidence level were reduced to 90%? Why?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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347, , 8.7 Practice 2: Confidence Intervals for Means, Unknown Population, Standard Deviation8, 8.7.1 Student Learning Outcomes, • The student will calculate confidence intervals for means when the population standard deviation is, unknown., , 8.7.2 Given, The following real data are the result of a random survey of 39 national flags (with replacement between, picks) from various countries. We are interested in finding a confidence interval for the true mean number, of colors on a national flag. Let X = the number of colors on a national flag., X, , Freq., , 1, , 1, , 2, , 7, , 3, , 18, , 4, , 7, , 5, , 6, , Table 8.1, , 8.7.3 Calculating the Confidence Interval, Exercise 8.7.1, Calculate the following:, , (Solution on p. 371.), , a. x =, b. s x =, c. n =, Exercise 8.7.2, (Solution on p. 371.), Define the Random Variable, X, in words. X = __________________________, Exercise 8.7.3, What is x estimating?, , (Solution on p. 371.), , Exercise 8.7.4, Is σx known?, , (Solution on p. 371.), , Exercise 8.7.5, (Solution on p. 371.), As a result of your answer to (4), state the exact distribution to use when calculating the Confidence Interval., 8 This, , content is available online at <http://cnx.org/content/m16971/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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348, , CHAPTER 8. CONFIDENCE INTERVALS, , 8.7.4 Confidence Interval for the True Mean Number, Construct a 95% Confidence Interval for the true mean number of colors on national flags., Exercise 8.7.6, How much area is in both tails (combined)? α =, , (Solution on p. 371.), , Exercise 8.7.7, How much area is in each tail?, , (Solution on p. 371.), α, 2, , =, , Exercise 8.7.8, Calculate the following:, , (Solution on p. 371.), , a. lower limit =, b. upper limit =, c. error bound =, Exercise 8.7.9, The 95% Confidence Interval is:, , (Solution on p. 372.), , Exercise 8.7.10, Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval, and the sample mean., , Figure 8.3, , Exercise 8.7.11, In one complete sentence, explain what the interval means., , 8.7.5 Discussion Questions, Exercise 8.7.12, Using the same x, s x , and level of confidence, suppose that n were 69 instead of 39. Would the, error bound become larger or smaller? How do you know?, Exercise 8.7.13, Using the same x, s x , and n = 39, how would the error bound change if the confidence level were, reduced to 90%? Why?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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349, , 8.8 Practice 3: Confidence Intervals for Proportions9, 8.8.1 Student Learning Outcomes, • The student will calculate confidence intervals for proportions., , 8.8.2 Given, The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a, bucket. The 5 P.M., Monday night, ages 8 - 12, beginning ice-skating class was picked. In that class were 64, girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 - 12, in all beginning, ice-skating classes at the Ice Chalet. Assume that the children in the selected class is a random sample of, the population., , 8.8.3 Estimated Distribution, Exercise 8.8.1, What is being counted?, Exercise 8.8.2, In words, define the Random Variable X. X =, , (Solution on p. 372.), , Exercise 8.8.3, Calculate the following:, , (Solution on p. 372.), , a. x =, b. n =, c. p’ =, Exercise 8.8.4, State the estimated distribution of X. X ∼, , (Solution on p. 372.), , Exercise 8.8.5, Define a new Random Variable P’. What is p’ estimating?, , (Solution on p. 372.), , Exercise 8.8.6, In words, define the Random Variable P’ . P’ =, , (Solution on p. 372.), , Exercise 8.8.7, State the estimated distribution of P’. P’ ∼, , 8.8.4 Explaining the Confidence Interval, Construct a 92% Confidence Interval for the true proportion of girls in the age 8 - 12 beginning ice-skating, classes at the Ice Chalet., Exercise 8.8.8, How much area is in both tails (combined)? α =, , (Solution on p. 372.), , Exercise 8.8.9, How much area is in each tail?, , (Solution on p. 372.), α, 2, , =, , Exercise 8.8.10, Calculate the following:, 9 This, , (Solution on p. 372.), , content is available online at <http://cnx.org/content/m16968/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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350, , CHAPTER 8. CONFIDENCE INTERVALS, a. lower limit =, b. upper limit =, c. error bound =, Exercise 8.8.11, The 92% Confidence Interval is:, , (Solution on p. 372.), , Exercise 8.8.12, Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval, and, the sample proportion., , Figure 8.4, , Exercise 8.8.13, In one complete sentence, explain what the interval means., , 8.8.5 Discussion Questions, Exercise 8.8.14, Using the same p’ and level of confidence, suppose that n were increased to 100. Would the error, bound become larger or smaller? How do you know?, Exercise 8.8.15, Using the same p’ and n = 80, how would the error bound change if the confidence level were, increased to 98%? Why?, Exercise 8.8.16, If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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351, , 8.9 Homework10, NOTE : If you are using a student’s-t distribution for a homework problem below, you may assume, that the underlying population is normally distributed. (In general, you must first prove that, assumption, though.), , Exercise 8.9.1, (Solution on p. 372.), Among various ethnic groups, the standard deviation of heights is known to be approximately 3, inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. 48, male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8, inches., a. i. x =________, ii. σ = ________, iii. s x =________, iv. n =________, v. n − 1 =________, b. Define the Random Variables X and X, in words., c. Which distribution should you use for this problem? Explain your choice., d. Construct a 95% confidence interval for the population mean height of male Swedes., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. What will happen to the level of confidence obtained if 1000 male Swedes are surveyed instead, of 48? Why?, Exercise 8.9.2, In six packages of “The Flintstones® Real Fruit Snacks” there were 5 Bam-Bam snack pieces. The, total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval, for the population proportion of Bam-Bam snack pieces., a., b., c., d., , Define the Random Variables X and P’, in words., Which distribution should you use for this problem? Explain your choice, Calculate p’., Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces, per bag., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., , e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why, or why not?, Exercise 8.9.3, (Solution on p. 372.), A random survey of enrollment at 35 community colleges across the United States yielded the, following figures (source: Microsoft Bookshelf ): 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722;, 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314;, 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622. Assume the underlying, population is normal., a. i. x =, 10 This, , content is available online at <http://cnx.org/content/m16966/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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352, , CHAPTER 8. CONFIDENCE INTERVALS, ii. s x = ________, iii. n = ________, iv. n − 1 =________, b. Define the Random Variables X and X, in words., c. Which distribution should you use for this problem? Explain your choice., d. Construct a 95% confidence interval for the population mean enrollment at community colleges, in the United States., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. What will happen to the error bound and confidence interval if 500 community colleges were, surveyed? Why?, Exercise 8.9.4, From a stack of IEEE Spectrum magazines, announcements for 84 upcoming engineering conferences were randomly picked. The mean length of the conferences was 3.94 days, with a standard, deviation of 1.28 days. Assume the underlying population is normal., a. Define the Random Variables X and X, in words., b. Which distribution should you use for this problem? Explain your choice., c. Construct a 95% confidence interval for the population mean length of engineering conferences., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., Exercise 8.9.5, (Solution on p. 373.), Suppose that a committee is studying whether or not there is waste of time in our judicial system., It is interested in the mean amount of time individuals waste at the courthouse waiting to be called, for service. The committee randomly surveyed 81 people. The sample mean was 8 hours with a, sample standard deviation of 4 hours., a. i. x =________, ii. s x = ________, iii. n =________, iv. n − 1 =________, b. Define the Random Variables X and X, in words., c. Which distribution should you use for this problem? Explain your choice., d. Construct a 95% confidence interval for the population mean time wasted., a. State the confidence interval., b. Sketch the graph., c. Calculate the error bound., e. Explain in a complete sentence what the confidence interval means., Exercise 8.9.6, Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known, standard deviation of 7.0 hours. The population distribution is assumed to be normal., a. i. x = ________, ii. σ =________, iii. s x =________, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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353, iv. n = ________, v. n − 1 =________, b. Define the Random Variables X and X, in words., c. Which distribution should you use for this problem? Explain your choice., d. Construct a 90% confidence interval for the population mean time to complete the tax forms., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. If the firm wished to increase its level of confidence and keep the error bound the same by, taking another survey, what changes should it make?, f. If the firm did another survey, kept the error bound the same, and only surveyed 49 people,, what would happen to the level of confidence? Why?, g. Suppose that the firm decided that it needed to be at least 96% confident of the population, mean length of time to within 1 hour. How would the number of people the firm surveys, change? Why?, Exercise 8.9.7, (Solution on p. 373.), A sample of 16 small bags of the same brand of candies was selected. Assume that the population, distribution of bag weights is normal. The weight of each bag was then recorded. The mean, weight was 2 ounces with a standard deviation of 0.12 ounces. The population standard deviation, is known to be 0.1 ounce., a. i. x = ________, ii. σ = ________, iii. s x =________, iv. n = ________, v. n − 1 = ________, b. Define the Random Variable X, in words., c. Define the Random Variable X, in words., d. Which distribution should you use for this problem? Explain your choice., e. Construct a 90% confidence interval for the population mean weight of the candies., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., f. Construct a 98% confidence interval for the population mean weight of the candies., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., g. In complete sentences, explain why the confidence interval in (f) is larger than the confidence, interval in (e)., h. In complete sentences, give an interpretation of what the interval in (f) means., Exercise 8.9.8, A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length, of time they last is approximately normal. Researchers in a hospital used the drug on a random, sample of 9 patients. The effective period of the tranquilizer for each patient (in hours) was as, follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4 ., a. i. x = ________, ii. s x = ________, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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354, , CHAPTER 8. CONFIDENCE INTERVALS, , b., c., d., e., , iii. n = ________, iv. n − 1 = ________, Define the Random Variable X, in words., Define the Random Variable X, in words., Which distribution should you use for this problem? Explain your choice., Construct a 95% confidence interval for the population mean length of time., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., , f. What does it mean to be “95% confident” in this problem?, Exercise 8.9.9, (Solution on p. 373.), Suppose that 14 children were surveyed to determine how long they had to use training wheels., It was revealed that they used them an average of 6 months with a sample standard deviation of, 3 months. Assume that the underlying population distribution is normal., a. i. x = ________, ii. s x = ________, iii. n = ________, iv. n − 1 = ________, b. Define the Random Variable X, in words., c. Define the Random Variable X, in words., d. Which distribution should you use for this problem? Explain your choice., e. Construct a 99% confidence interval for the population mean length of time using training, wheels., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., f. Why would the error bound change if the confidence level was lowered to 90%?, Exercise 8.9.10, Insurance companies are interested in knowing the population percent of drivers who always, buckle up before riding in a car., a. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03?, b. If it was later determined that it was important to be more than 95% confident and a new survey, was commissioned, how would that affect the minimum number you would need to survey?, Why?, Exercise 8.9.11, (Solution on p. 373.), Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and, found that 320 claimed to always buckle up. We are interested in the population proportion of, drivers who claim to always buckle up., a. i. x = ________, ii. n = ________, iii. p’ = ________, b. Define the Random Variables X and P’, in words., c. Which distribution should you use for this problem? Explain your choice., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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355, d. Construct a 95% confidence interval for the population proportion that claim to always buckle, up., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. If this survey were done by telephone, list 3 difficulties the companies might have in obtaining, random results., Exercise 8.9.12, Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the, records of 225 flights are randomly selected and the number of unoccupied seats is noted for each, of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1, seats., a. i. x = ________, ii. s x = ________, iii. n = ________, iv. n − 1 = ________, b. Define the Random Variables X and X, in words., c. Which distribution should you use for this problem? Explain your choice., d. Construct a 92% confidence interval for the population mean number of unoccupied seats per, flight., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., Exercise 8.9.13, (Solution on p. 373.), According to a recent survey of 1200 people, 61% feel that the president is doing an acceptable, job. We are interested in the population proportion of people who feel the president is doing an, acceptable job., a. Define the Random Variables X and P’, in words., b. Which distribution should you use for this problem? Explain your choice., c. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., Exercise 8.9.14, A survey of the mean amount of cents off that coupons give was done by randomly surveying one, coupon per page from the coupon sections of a recent San Jose Mercury News. The following data, were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the, underlying distribution is approximately normal., a. i. x = ________, ii. s x = ________, iii. n = ________, iv. n − 1 = ________, b. Define the Random Variables X and X, in words., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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356, , CHAPTER 8. CONFIDENCE INTERVALS, c. Which distribution should you use for this problem? Explain your choice., d. Construct a 95% confidence interval for the population mean worth of coupons., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. If many random samples were taken of size 14, what percent of the confident intervals constructed should contain the population mean worth of coupons? Explain why., Exercise 8.9.15, (Solution on p. 374.), An article regarding interracial dating and marriage recently appeared in the Washington Post. Of, the 1709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves, as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this, survey, 86% of blacks said that their families would welcome a white person into their families., Among Asians, 77% would welcome a white person into their families, 71% would welcome a, Latino, and 66% would welcome a black person., a. We are interested in finding the 95% confidence interval for the percent of all black families that, would welcome a white person into their families. Define the Random Variables X and P’,, in words., b. Which distribution should you use for this problem? Explain your choice., c. Construct a 95% confidence interval, i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., Exercise 8.9.16, Refer to the problem above., a. Construct three 95% confidence intervals., i: Percent of all Asians that would welcome a white person into their families., ii: Percent of all Asians that would welcome a Latino into their families., iii: Percent of all Asians that would welcome a black person into their families., b. Even though the three point estimates are different, do any of the confidence intervals overlap?, Which?, c. For any intervals that do overlap, in words, what does this imply about the significance of the, differences in the true proportions?, d. For any intervals that do not overlap, in words, what does this imply about the significance of, the differences in the true proportions?, Exercise 8.9.17, (Solution on p. 374.), A camp director is interested in the mean number of letters each child sends during his/her camp, session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken., The mean from the sample is 7.9 with a sample standard deviation of 2.8., a. i. x = ________, ii. σ = ________, iii. s x = ________, iv. n = ________, v. n − 1 = ________, b. Define the Random Variables X and X, in words., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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357, c. Which distribution should you use for this problem? Explain your choice., d. Construct a 90% confidence interval for the population mean number of letters campers send, home., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., e. What will happen to the error bound and confidence interval if 500 campers are surveyed?, Why?, Exercise 8.9.18, Stanford University conducted a study of whether running is healthy for men and women over, age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness, Association died. We are interested in the proportion of people over 50 who ran and died in the, same eight–year period., a. Define the Random Variables X and P’, in words., b. Which distribution should you use for this problem? Explain your choice., c. Construct a 97% confidence interval for the population proportion of people over 50 who ran, and died in the same eight–year period., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., d. Explain what a “97% confidence interval” means for this study., Exercise 8.9.19, (Solution on p. 374.), In a recent sample of 84 used cars sales costs, the sample mean was $6425 with a standard deviation, of $3156. Assume the underlying distribution is approximately normal., a. Which distribution should you use for this problem? Explain your choice., b. Define the Random Variable X, in words., c. Construct a 95% confidence interval for the population mean cost of a used car., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., d. Explain what a “95% confidence interval” means for this study., Exercise 8.9.20, A telephone poll of 1000 adult Americans was reported in an issue of Time Magazine. One of the, questions asked was “What is the main problem facing the country?” 20% answered “crime”. We, are interested in the population proportion of adult Americans who feel that crime is the main, problem., a. Define the Random Variables X and P’, in words., b. Which distribution should you use for this problem? Explain your choice., c. Construct a 95% confidence interval for the population proportion of adult Americans who feel, that crime is the main problem., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., d. Suppose we want to lower the sampling error. What is one way to accomplish that?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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358, , CHAPTER 8. CONFIDENCE INTERVALS, e. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ± 3%. In, 1-3 complete sentences, explain what the ± 3% represents., Exercise 8.9.21, (Solution on p. 374.), Refer to the above problem. Another question in the poll was “[How much are] you worried, about the quality of education in our schools?” 63% responded “a lot”. We are interested in the, population proportion of adult Americans who are worried a lot about the quality of education in, our schools., 1. Define the Random Variables X and P’, in words., 2. Which distribution should you use for this problem? Explain your choice., 3. Construct a 95% confidence interval for the population proportion of adult Americans worried a lot about the quality of education in our schools., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., 4. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ± 3%., In 1-3 complete sentences, explain what the ± 3% represents., Exercise 8.9.22, Six different national brands of chocolate chip cookies were randomly selected at the supermarket., The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is, approximately normal., a. Calculate a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets., i. State the confidence interval., ii. Sketch the graph., iii. Calculate the error bound., b. If you wanted a smaller error bound while keeping the same level of confidence, what should, have been changed in the study before it was done?, c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies., d. Calculate the mean., e. Is the mean within the interval you calculated in part (a)? Did you expect it to be? Why or why, not?, Exercise 8.9.23, A confidence interval for a proportion is given to be (– 0.22, 0.34). Why doesn’t the lower limit of, the confidence interval make practical sense? How should it be changed? Why?, , 8.9.1 Try these multiple choice questions., The next three problems refer to the following: According to a Field Poll, 79% of California adults, (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California. (Source:, http://field.com/fieldpollonline/subscribers/), Exercise 8.9.24, A point estimate for the true population proportion is:, , (Solution on p. 374.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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359, A., B., C., D., , 0.90, 1.27, 0.79, 400, , Exercise 8.9.25, A 90% confidence interval for the population proportion is:, A., B., C., D., , (0.761, 0.820), (0.125, 0.188), (0.755, 0.826), (0.130, 0.183), , Exercise 8.9.26, The error bound is approximately, A., B., C., D., , (Solution on p. 374.), , (Solution on p. 374.), , 1.581, 0.791, 0.059, 0.030, , The next two problems refer to the following:, A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of, soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55., Assume the underlying population is normally distributed., Exercise 8.9.27, (Solution on p. 374.), Find the 95% Confidence Interval for the true population mean for the amount of soda served., A., B., C., D., , (12.42, 14.18), (12.32, 14.29), (12.50, 14.10), Impossible to determine, , Exercise 8.9.28, What is the error bound?, A., B., C., D., , (Solution on p. 374.), , 0.87, 1.98, 0.99, 1.74, , Exercise 8.9.29, (Solution on p. 374.), What is meant by the term “90% confident” when constructing a confidence interval for a mean?, A. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval., B. If we took repeated samples, approximately 90% of the confidence intervals calculated from, those samples would contain the sample mean., C. If we took repeated samples, approximately 90% of the confidence intervals calculated from, those samples would contain the true value of the population mean., D. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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360, , CHAPTER 8. CONFIDENCE INTERVALS, , The next two problems refer to the following:, Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed, to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three, (173) of the homes surveyed met the minimum recommendations for earthquake preparedness and 338 did, not., Exercise 8.9.30, (Solution on p. 374.), Find the Confidence Interval at the 90% Confidence Level for the true population proportion of, southern California community homes meeting at least the minimum recommendations for earthquake preparedness., A., B., C., D., , (0.2975, 0.3796), (0.6270, 6959), (0.3041, 0.3730), (0.6204, 0.7025), , Exercise 8.9.31, (Solution on p. 374.), The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is:, A., B., C., D., , 0.6614, 0.3386, 173, 338, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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361, , 8.10 Review11, The next three problems refer to the following situation: Suppose that a sample of 15 randomly chosen, people were put on a special weight loss diet. The amount of weight lost, in pounds, follows an unknown, distribution with mean equal to 12 pounds and standard deviation equal to 3 pounds. Assume that the, distribution for the weight loss is normal., Exercise 8.10.1, (Solution on p. 375.), To find the probability that the mean amount of weight lost by 15 people is no more than 14, pounds, the random variable should be:, A., B., C., D., , The number of people who lost weight on the special weight loss diet, The number of people who were on the diet, The mean amount of weight lost by 15 people on the special weight loss diet, The total amount of weight lost by 15 people on the special weight loss diet, , Exercise 8.10.2, Find the probability asked for in the previous problem., , (Solution on p. 375.), , Exercise 8.10.3, Find the 90th percentile for the mean amount of weight lost by 15 people., , (Solution on p. 375.), , The next three questions refer to the following situation: The time of occurrence of the first accident, during rush-hour traffic at a major intersection is uniformly distributed between the three hour interval 4, p.m. to 7 p.m. Let X = the amount of time (hours) it takes for the first accident to occur., , • So, if an accident occurs at 4 p.m., the amount of time, in hours, it took for the accident to occur is, _______., • µ = _______, • σ2 = _______, Exercise 8.10.4, (Solution on p. 375.), What is the probability that the time of occurrence is within the first half-hour or the last hour of, the period from 4 to 7 p.m.?, A. Cannot be determined from the information given, B. 16, C. 12, D. 13, Exercise 8.10.5, The 20th percentile occurs after how many hours?, A., B., C., D., , (Solution on p. 375.), , 0.20, 0.60, 0.50, 1, , Exercise 8.10.6, (Solution on p. 375.), Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let, C = the total cumulative time. Then C follows which distribution?, A. U (0, 3), 11 This, , content is available online at <http://cnx.org/content/m16972/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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362, , CHAPTER 8. CONFIDENCE INTERVALS, B. Exp, , , 1, 3, , C. N (60, 5.477), D. N (1.5, 0.01875), Exercise 8.10.7, (Solution on p. 375.), Using the information in question #6, find the probability that the total time for all first accidents, to occur is more than 43 hours., The next two questions refer to the following situation: The length of time a parent must wait for his, children to clean their rooms is uniformly distributed in the time interval from 1 to 15 days., Exercise 8.10.8, (Solution on p. 375.), How long must a parent expect to wait for his children to clean their rooms?, A., B., C., D., , 8 days, 3 days, 14 days, 6 days, , Exercise 8.10.9, (Solution on p. 375.), What is the probability that a parent will wait more than 6 days given that the parent has already, waited more than 3 days?, A., B., C., D., , 0.5174, 0.0174, 0.7500, 0.2143, , The next five problems refer to the following study: Twenty percent of the students at a local community, college live in within five miles of the campus. Thirty percent of the students at the same community college, receive some kind of financial aid. Of those who live within five miles of the campus, 75% receive some, kind of financial aid., Exercise 8.10.10, (Solution on p. 375.), Find the probability that a randomly chosen student at the local community college does not live, within five miles of the campus., A., B., C., D., , 80%, 20%, 30%, Cannot be determined, , Exercise 8.10.11, (Solution on p. 375.), Find the probability that a randomly chosen student at the local community college lives within, five miles of the campus or receives some kind of financial aid., A., B., C., D., , 50%, 35%, 27.5%, 75%, , Exercise 8.10.12, (Solution on p. 375.), Based upon the above information, are living in student housing within five miles of the campus, and receiving some kind of financial aid mutually exclusive?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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363, A. Yes, B. No, C. Cannot be determined, Exercise 8.10.13, The interest rate charged on the financial aid is _______ data., A., B., C., D., , (Solution on p. 375.), , quantitative discrete, quantitative continuous, qualitative discrete, qualitative, , Exercise 8.10.14, (Solution on p. 375.), What follows is information about the students who receive financial aid at the local community, college., , • 1st quartile = $250, • 2nd quartile = $700, • 3rd quartile = $1200, (These amounts are for the school year.) If a sample of 200 students is taken, how many are, expected to receive $250 or more?, A., B., C., D., , 50, 250, 150, Cannot be determined, , The next two problems refer to the following information:, independent events., Exercise 8.10.15, P ( A AND B) =, A., B., C., D., , (Solution on p. 375.), , 0.5, 0.6, 0, 0.06, , Exercise 8.10.16, P ( A OR B) =, A., B., C., D., , P ( A) = 0.2 , P ( B) = 0.3 , A and B are, , (Solution on p. 375.), , 0.56, 0.5, 0.44, 1, , Exercise 8.10.17, (Solution on p. 375.), If H and D are mutually exclusive events, P ( H ) = 0.25 , P ( D ) = 0.15 , then P ( H | D ), A., B., C., D., , 1, 0, 0.40, 0.0375, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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364, , CHAPTER 8. CONFIDENCE INTERVALS, , 8.11 Lab 1: Confidence Interval (Home Costs)12, Class Time:, Names:, , 8.11.1 Student Learning Outcomes:, • The student will calculate the 90% confidence interval for the mean cost of a home in the area in which, this school is located., • The student will interpret confidence intervals., • The student will determine the effects that changing conditions has on the confidence interval., , 8.11.2 Collect the Data, Check the Real Estate section in your local newspaper. (Note: many papers only list them one day per, week. Also, we will assume that homes come up for sale randomly.) Record the sales prices for 35 randomly, selected homes recently listed in the county., 1. Complete the table:, __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , Table 8.2, , 8.11.3 Describe the Data, 1. Compute the following:, a. x =, b. s x =, c. n =, 2. Define the Random Variable X, in words. X =, 3. State the estimated distribution to use. Use both words and symbols., 12 This, , content is available online at <http://cnx.org/content/m16960/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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365, , 8.11.4 Find the Confidence Interval, 1. Calculate the confidence interval and the error bound., a. Confidence Interval:, b. Error Bound:, 2. How much area is in both tails (combined)? α =, 3. How much area is in each tail? α2 =, 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the, upper and lower limits of the confidence interval and the sample mean., , Figure 8.5, , 5. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data on, the first page and count how many of the data values lie within the confidence interval. What percent, is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%., , 8.11.5 Describe the Confidence Interval, 1. In two to three complete sentences, explain what a Confidence Interval means (in general), as if you, were talking to someone who has not taken statistics., 2. In one to two complete sentences, explain what this Confidence Interval means for this particular, study., , 8.11.6 Use the Data to Construct Confidence Intervals, 1. Using the above information, construct a confidence interval for each confidence level, given., Confidence level, , EBM / Error Bound, , Confidence Interval, , 50%, 80%, 95%, 99%, Table 8.3, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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366, , CHAPTER 8. CONFIDENCE INTERVALS, , 2. What happens to the EBM as the confidence level increases? Does the width of the confidence interval, increase or decrease? Explain why this happens., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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367, , 8.12 Lab 2: Confidence Interval (Place of Birth)13, Class Time:, Names:, , 8.12.1 Student Learning Outcomes:, • The student will calculate the 90% confidence interval for proportion of students in this school that, were born in this state., • The student will interpret confidence intervals., • The student will determine the effects that changing conditions have on the confidence interval., , 8.12.2 Collect the Data, 1. Survey the students in your class, asking them if they were born in this state. Let X = the number that, were born in this state., a. n =____________, b. x =____________, 2. Define the Random Variable P’ in words., 3. State the estimated distribution to use., , 8.12.3 Find the Confidence Interval and Error Bound, 1. Calculate the confidence interval and the error bound., a. Confidence Interval:, b. Error Bound:, 2. How much area is in both tails (combined)? α=, 3. How much area is in each tail? α2 =, 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the, upper and lower limits of the confidence interval and the sample proportion., , Figure 8.6, , 13 This, , content is available online at <http://cnx.org/content/m16961/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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368, , CHAPTER 8. CONFIDENCE INTERVALS, , 8.12.4 Describe the Confidence Interval, 1. In two to three complete sentences, explain what a Confidence Interval means (in general), as if you, were talking to someone who has not taken statistics., 2. In one to two complete sentences, explain what this Confidence Interval means for this particular, study., 3. Using the above information, construct a confidence interval for each given confidence level, given., Confidence level, , EBP / Error Bound, , Confidence Interval, , 50%, 80%, 95%, 99%, Table 8.4, , 4. What happens to the EBP as the confidence level increases? Does the width of the confidence interval, increase or decrease? Explain why this happens., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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369, , 8.13 Lab 3: Confidence Interval (Womens’ Heights)14, Class Time:, Names:, , 8.13.1 Student Learning Outcomes:, • The student will calculate a 90% confidence interval using the given data., • The student will determine the relationship between the confidence level and the percent of constructed intervals that contain the population mean., , 8.13.2 Given:, 1. Heights of 100 Women (in Inches), 59.4, , 71.6, , 69.3, , 65.0, , 62.9, , 66.5, , 61.7, , 55.2, , 67.5, , 67.2, , 63.8, , 62.9, , 63.0, , 63.9, , 68.7, , 65.5, , 61.9, , 69.6, , 58.7, , 63.4, , 61.8, , 60.6, , 69.8, , 60.0, , 64.9, , 66.1, , 66.8, , 60.6, , 65.6, , 63.8, , 61.3, , 59.2, , 64.1, , 59.3, , 64.9, , 62.4, , 63.5, , 60.9, , 63.3, , 66.3, , 61.5, , 64.3, , 62.9, , 60.6, , 63.8, , 58.8, , 64.9, , 65.7, , 62.5, , 70.9, , 62.9, , 63.1, , 62.2, , 58.7, , 64.7, , 66.0, , 60.5, , 64.7, , 65.4, , 60.2, , 65.0, , 64.1, , 61.1, , 65.3, , 64.6, , 59.2, , 61.4, , 62.0, , 63.5, , 61.4, , 65.5, , 62.3, , 65.5, , 64.7, , 58.8, , 66.1, , 64.9, , 66.9, , 57.9, , 69.8, , 58.5, , 63.4, , 69.2, , 65.9, , 62.2, , 60.0, , 58.1, , 62.5, , 62.4, , 59.1, , 66.4, , 61.2, , 60.4, , 58.7, , 66.7, , 67.5, , 63.2, , 56.6, , 67.7, , 62.5, , Table 8.5, , Listed above are the heights of 100 women. Use a random number generator to randomly select 10, data values., 14 This, , content is available online at <http://cnx.org/content/m16964/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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370, , CHAPTER 8. CONFIDENCE INTERVALS, , 2. Calculate the sample mean and sample standard deviation. Assume that the population standard, deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your, sample of 10 values. Write the confidence interval you obtained in the first space of the table below., 3. Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into the table below:, 90% Confidence Intervals, __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , __________, , Table 8.6, , 8.13.3 Discussion Questions, 1. The actual population mean for the 100 heights given above is µ = 63.4. Using the class listing of, confidence intervals, count how many of them contain the population mean µ; i.e., for how many, intervals does the value of µ lie between the endpoints of the confidence interval?, 2. Divide this number by the total number of confidence intervals generated by the class to determine, the percent of confidence intervals that contains the mean µ. Write this percent below., 3. Is the percent of confidence intervals that contain the population mean µ close to 90%?, 4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percent, of confidence intervals that contained the population mean?, 5. When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean, by this phrase., 6. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given, (the heights of women) and count how many of the data values lie within the confidence interval that, you generated on that page. How many of the 100 data values lie within your confidence interval?, What percent is this? Is this percent close to 90%?, 7. Explain why it does not make sense to count data values that lie in a confidence interval. Think about, the random variable that is being used in the problem., 8. Suppose you obtained the heights of 10 women and calculated a confidence interval from this information. Without knowing the population mean µ, would you have any way of knowing for certain, if your interval actually contained the value of µ? Explain., NOTE :, , This lab was designed and contributed by Diane Mathios., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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371, , Solutions to Exercises in Chapter 8, Solutions to Practice 1: Confidence Intervals for Means, Known Population Standard, Deviation, Solution to Exercise 8.6.1 (p. 345), 30.4, Solution to Exercise 8.6.2 (p. 345), 25, Solution to Exercise 8.6.3 (p. 345), σ, Solution to Exercise 8.6.4 (p. 345), the mean age of 25 randomly selected Winter Foothill students, Solution to Exercise 8.6.5 (p. 345), µ, Solution to Exercise 8.6.6 (p. 345), yes, Solution to Exercise 8.6.7 (p. 345), Normal, Solution to Exercise 8.6.8 (p. 345), 0.05, Solution to Exercise 8.6.9 (p. 345), 0.025, Solution to Exercise 8.6.10 (p. 345), a. 24.52, b. 36.28, c. 5.88, Solution to Exercise 8.6.11 (p. 346), (24.52, 36.28), , Solutions to Practice 2: Confidence Intervals for Means, Unknown Population Standard Deviation, Solution to Exercise 8.7.1 (p. 347), a. 3.26, b. 1.02, c. 39, Solution to Exercise 8.7.2 (p. 347), the mean number of colors of 39 flags, Solution to Exercise 8.7.3 (p. 347), µ, Solution to Exercise 8.7.4 (p. 347), No, Solution to Exercise 8.7.5 (p. 347), t38, Solution to Exercise 8.7.6 (p. 348), 0.05, Solution to Exercise 8.7.7 (p. 348), 0.025, Solution to Exercise 8.7.8 (p. 348), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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372, , CHAPTER 8. CONFIDENCE INTERVALS, , a. 2.93, b. 3.59, c. 0.33, Solution to Exercise 8.7.9 (p. 348), 2.93; 3.59, , Solutions to Practice 3: Confidence Intervals for Proportions, Solution to Exercise 8.8.2 (p. 349), The number of girls, age 8-12, in the beginning ice skating class, Solution to Exercise 8.8.3 (p. 349), a. 64, b. 80, c. 0.8, Solution to Exercise 8.8.4 (p. 349), B (80, 0.80), Solution to Exercise 8.8.5 (p. 349), p, Solution to Exercise 8.8.6 (p. 349), The proportion of girls, age 8-12, in the beginning ice skating class., Solution to Exercise 8.8.8 (p. 349), 1 - 0.92 = 0.08, Solution to Exercise 8.8.9 (p. 349), 0.04, Solution to Exercise 8.8.10 (p. 349), a. 0.72, b. 0.88, c. 0.08, Solution to Exercise 8.8.11 (p. 350), (0.72; 0.88), , Solutions to Homework, Solution to Exercise 8.9.1 (p. 351), a. i. 71, ii. 3, iii. 2.8, iv. 48, v. 47 , c. N 71, √3, 48, d. i. CI: (70.15,71.85), iii. EB = 0.85, Solution to Exercise 8.9.3 (p. 351), a. i. 8629, ii. 6944, iii. 35, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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373, iv. 34, c. t34, d. i. CI: (6244, 11,014), iii. EB = 2385, e. It will become smaller, Solution to Exercise 8.9.5 (p. 352), a. i. 8, ii. 4, iii. 81, iv. 80, c. t80, d. i. CI: (7.12, 8.88), iii. EB = 0.88, Solution to Exercise 8.9.7 (p. 353), a. i. 2, ii. 0.1, iii . 0.12, iv. 16, v. 15, b. the weight of 1 small bag of candies, c. themean weight, of 16 small bags of candies, , d. N 2, √0.1, 16, e. i. CI: (1.96, 2.04), iii. EB = 0.04, f. i. CI: (1.94, 2.06), iii. EB = 0.06, , Solution to Exercise 8.9.9 (p. 354), a. i. 6, ii. 3, iii. 14, iv. 13, b. the time for a child to remove his training wheels, c. the mean time for 14 children to remove their training wheels., d. t13, e. i. CI: (3.58, 8.42), iii. EB = 2.42, Solution to Exercise 8.9.11 (p. 354), a. i. 320, ii . 400, iii., , 0.80, q, (0.80)(0.20), c. N 0.80,, 400, d. i. CI: (0.76, 0.84), iii. EB = 0.04, Solution to Exercise 8.9.13 (p. 355), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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374, , CHAPTER 8. CONFIDENCE INTERVALS, , , b. N 0.61,, , q, , (0.61)(0.39), 1200, , , , c. i. CI: (0.59, 0.63), iii. EB = 0.02, Solution to Exercise 8.9.15 (p. 356), , , q, (0.86)(0.14), b. N 0.86,, 323, c. i. CI: (0.823, 0.898), iii. EB = 0.038, Solution to Exercise 8.9.17 (p. 356), a. i. 7.9, ii. 2.5, iii. 2.8, iv. 20, v. 19, , c. N 7.9, √2.5, 20, d. i. CI: (6.98, 8.82), iii. EB: 0.92, Solution to Exercise 8.9.19 (p. 357), a. t83, b. mean cost of 84 used cars, c. i. CI: (5740.10, 7109.90), iii. EB = 684.90, Solution to Exercise 8.9.21 (p. 358), , , q, (0.63)(0.37), b. N 0.63,, 1000, c. i. CI: (0.60, 0.66), iii. EB = 0.03, Solution to Exercise 8.9.24 (p., C, Solution to Exercise 8.9.25 (p., A, Solution to Exercise 8.9.26 (p., D, Solution to Exercise 8.9.27 (p., B, Solution to Exercise 8.9.28 (p., C, Solution to Exercise 8.9.29 (p., C, Solution to Exercise 8.9.30 (p., C, Solution to Exercise 8.9.31 (p., A, , 358), 359), 359), 359), 359), 359), 360), 360), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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375, , Solutions to Review, Solution to Exercise 8.10.1 (p. 361), C, Solution to Exercise 8.10.2 (p. 361), 0.9951, Solution to Exercise 8.10.3 (p. 361), 12.99, Solution to Exercise 8.10.4 (p. 361), C, Solution to Exercise 8.10.5 (p. 361), B, Solution to Exercise 8.10.6 (p. 361), C, Solution to Exercise 8.10.7 (p. 362), 0.9990, Solution to Exercise 8.10.8 (p. 362), A, Solution to Exercise 8.10.9 (p. 362), C, Solution to Exercise 8.10.10 (p. 362), A, Solution to Exercise 8.10.11 (p. 362), B, Solution to Exercise 8.10.12 (p. 362), B, Solution to Exercise 8.10.13 (p. 363), B, Solution to Exercise 8.10.14 (p. 363), C. 150, Solution to Exercise 8.10.15 (p. 363), D, Solution to Exercise 8.10.16 (p. 363), C, Solution to Exercise 8.10.17 (p. 363), B, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 9, , Hypothesis Testing: Single Mean and, Single Proportion, 9.1 Hypothesis Testing: Single Mean and Single Proportion1, 9.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Differentiate between Type I and Type II Errors, • Describe hypothesis testing in general and in practice, • Conduct and interpret hypothesis tests for a single population mean, population standard deviation, known., • Conduct and interpret hypothesis tests for a single population mean, population standard deviation, unknown., • Conduct and interpret hypothesis tests for a single population proportion., , 9.1.2 Introduction, One job of a statistician is to make statistical inferences about populations based on samples taken from the, population. Confidence intervals are one way to estimate a population parameter. Another way to make, a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that, its new small truck gets 35 miles per gallon, on the average. A tutoring service claims that its method of, tutoring helps 90% of its students get an A or a B. A company says that women managers in their company, earn an average of $60,000 per year., A statistician will make a decision about these claims. This process is called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a, decision as to whether or not there is sufficient evidence based upon analyses of the data, to reject the null, hypothesis., In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also, learn about the errors associated with these tests., Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data,, and a conclusion. To perform a hypothesis test, a statistician will:, 1 This, , content is available online at <http://cnx.org/content/m16997/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 377
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 378, 1., 2., 3., 4., , Set up two contradictory hypotheses., Collect sample data (in homework problems, the data or summary statistics will be given to you)., Determine the correct distribution to perform the hypothesis test., Analyze sample data by performing the calculations that ultimately will allow you to reject or fail to, reject the null hypothesis., 5. Make a decision and write a meaningful conclusion., NOTE : To do the hypothesis test homework problems for this chapter and later chapters, make, copies of the appropriate special solution sheets. See the Table of Contents topic "Solution Sheets"., , 9.2 Null and Alternate Hypotheses2, The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternate, hypothesis. These hypotheses contain opposing viewpoints., Ho : The null hypothesis: It is a statement about the population that will be assumed to be true unless it, can be shown to be incorrect beyond a reasonable doubt., Ha : The alternate hypothesis: It is a claim about the population that is contradictory to Ho and what we, conclude when we reject Ho ., Example 9.1, Ho : No more than 30% of the registered voters in Santa Clara County voted in the primary election., Ha : More than 30% of the registered voters in Santa Clara County voted in the primary election., Example 9.2, We want to test whether the mean grade point average in American colleges is different from 2.0, (out of 4.0)., Ho : µ = 2.0, , Ha : µ 6= 2.0, , Example 9.3, We want to test if college students take less than five years to graduate from college, on the average., Ho : µ ≥ 5, , Ha : µ < 5, , Example 9.4, In an issue of U. S. News and World Report, an article on school standards stated that about half, of all students in France, Germany, and Israel take advanced placement exams and a third pass., The same article stated that 6.6% of U. S. students take advanced placement exams and 4.4 % pass., Test if the percentage of U. S. students who take advanced placement exams is more than 6.6%., Ho : p= 0.066, , Ha : p > 0.066, , Since the null and alternate hypotheses are contradictory, you must examine evidence to decide if you have, enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data., After you have determined which hypothesis the sample supports, you make a decision. There are two, options for a decision. They are "reject Ho " if the sample information favors the alternate hypothesis or "do, not reject Ho " or "fail to reject Ho " if the sample information is insufficient to reject the null hypothesis., 2 This, , content is available online at <http://cnx.org/content/m16998/1.14/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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379, Mathematical Symbols Used in Ho and Ha :, Ho, , Ha, , equal (=), , not equal (6=) or greater than (> ) or less than (<), , greater than or equal to (≥), , less than (<), , less than or equal to (≤), , more than (> ), Table 9.1, , NOTE : Ho always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The, choice of symbol depends on the wording of the hypothesis test. However, be aware that many, researchers (including one of the co-authors in research work) use = in the Null Hypothesis, even, with > or < as the symbol in the Alternate Hypothesis. This practice is acceptable because we, only make the decision to reject or not reject the Null Hypothesis., , 9.2.1 Optional Collaborative Classroom Activity, Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from, which your group can write a null and alternate hypotheses. Discuss your hypotheses with the rest of the, class., , 9.3 Outcomes and the Type I and Type II Errors3, When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or, falseness) of the null hypothesis Ho and the decision to reject or not. The outcomes are summarized in the, following table:, Ho IS ACTUALLY, , ..., , True, , False, , Do not reject Ho, , Correct Outcome, , Type II error, , Reject Ho, , Type I Error, , Correct Outcome, , ACTION, , Table 9.2, , The four possible outcomes in the table are:, , •, •, •, •, , The decision is to not reject Ho when, in fact, Ho is true (correct decision)., The decision is to reject Ho when, in fact, Ho is true (incorrect decision known as a Type I error)., The decision is to not reject Ho when, in fact, Ho is false (incorrect decision known as a Type II error)., The decision is to reject Ho when, in fact, Ho is false (correct decision whose probability is called the, Power of the Test)., , Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities., α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the, null hypothesis is true., 3 This, , content is available online at <http://cnx.org/content/m17006/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 380, , β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when, the null hypothesis is false., α and β should be as small as possible because they are probabilities of errors. They are rarely 0., The Power of the Test is 1 − β. Ideally, we want a high power that is as close to 1 as possible. Increasing the, sample size can increase the Power of the Test., The following are examples of Type I and Type II errors., Example 9.5, Suppose the null hypothesis, Ho , is: Frank’s rock climbing equipment is safe., Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really, is safe. Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it, is not safe., α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it, really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in, fact, it is not safe., Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks, his rock climbing equipment is safe, he will go ahead and use it.), Example 9.6, Suppose the null hypothesis, Ho , is: The victim of an automobile accident is alive when he arrives, at the emergency room of a hospital., Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive., Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is, dead., α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive, = P(Type I error). β = probability that the emergency crew does not know if the victim is alive, when, in fact, the victim is dead = P(Type II error)., The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim, is dead, they will not treat him.), , 9.4 Distribution Needed for Hypothesis Testing4, Earlier in the course, we discussed sampling distributions. Particular distributions are associated with, hypothesis testing. Perform tests of a population mean using a normal distribution or a student’s-t distribution. (Remember, use a student’s-t distribution when the population standard deviation is unknown, and the distribution of the sample mean is approximately normal.) In this chapter we perform tests of a, population proportion using a normal distribution (usually n is large or the sample size is large)., If you are testing a single population mean, the distribution for the test is for means:, , , σX, X ∼ N µX , √, or, tdf, n, 4 This, , content is available online at <http://cnx.org/content/m17017/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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381, The population parameter is µ. The estimated value (point estimate) for µ is x, the sample mean., If you are testing a single population proportion, the distribution for the test is for proportions or percentages:, q , p·q, P’ ∼ N p,, n, The population parameter is p. The estimated value (point estimate) for p is p’. p’ =, number of successes and n is the sample size., , x, n, , where x is the, , 9.5 Assumption5, When you perform a hypothesis test of a single population mean µ using a Student’s-t distribution (often, called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately, normally distributed. You use the sample standard deviation to approximate the population standard, deviation. (Note that if the sample size is sufficiently large, a t-test will work even if the population is not, approximately normally distributed)., When you perform a hypothesis test of a single population mean µ using a normal distribution (often, called a z-test), you take a simple random sample from the population. The population you are testing, is normally distributed or your sample size is sufficiently large. You know the value of the population, standard deviation., When you perform a hypothesis test of a single population proportion p, you take a simple random, sample from the population. You must meet the conditions for a binomial distribution which are there are, a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has, the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape, of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5, and nq > 5). Then the binomial distribution, q of sample (estimated) proportion can be approximated by the, normal distribution with µ = p and σ =, , p·q, n ., , Remember that q = 1 − p., , 9.6 Rare Events6, Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely, to occur if the assumption is true, then you would conclude that your assumption about the population is, probably incorrect. (Remember that your assumption is just an assumption - it is not a fact and it may or, may not be true. But your sample data are real and the data are showing you a fact that seems to contradict, your assumption.), For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line, to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are, 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill., Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The, 1, probability of this happening is 200, = 0.005. Because this is so unlikely, Ali is hoping that what the two, of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi, getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket., 5 This, 6 This, , content is available online at <http://cnx.org/content/m17002/1.16/>., content is available online at <http://cnx.org/content/m16994/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 382, , 9.7 Using the Sample to Test the Null Hypothesis7, Use the sample data to calculate the actual probability of getting the test result, called the p-value. The, p-value is the probability that, if the null hypothesis is true, the results from another randomly selected, sample will be as extreme or more extreme as the results obtained from the given sample., A large p-value calculated from the data indicates that we should fail to reject the null hypothesis. The, smaller the p-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it., Draw a graph that shows the p-value. The hypothesis test is easier to perform if you use a graph because, you see the problem more clearly., Example 9.7: (to illustrate the p-value), Suppose a baker claims that his bread height is more than 15 cm, on the average. Several of his, customers do not believe him. To persuade his customers that he is right, the baker decides to do a, hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The, baker knows from baking hundreds of loaves of bread that the standard deviation for the height, is 0.5 cm. and the distribution of heights is normal., The null hypothesis could be Ho : µ ≤ 15 The alternate hypothesis is Ha : µ > 15, The words "is more than" translates as a "> " so "µ > 15" goes into the alternate hypothesis. The, null hypothesis must contradict the alternate hypothesis., Since σ is known (σ = 0.5 cm.), the distribution for the population is known to be normal with, mean µ= 15 and standard deviation √σn = √0.5 = 0.16., 10, , Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then, is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test, works by asking the question how unlikely the sample mean would be if the null hypothesis, were true. The graph shows how far out the sample mean is on the normal curve. The p-value is, the probability that, if we were to take other samples, any other sample mean would fall at least, as far out as 17 cm., The p-value, then, is the probability that a sample mean is the same or greater than 17 cm., when the population mean is, in fact, 15 cm. We can calculate this probability using the normal, distribution for means from Chapter 7., , p-value = P ( x > 17) which is approximately 0., 7 This, , content is available online at <http://cnx.org/content/m16995/1.17/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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383, A p-value of approximately 0 tells us that it is highly unlikely that a loaf of bread rises no more, than 15 cm, on the average. That is, almost 0% of all loaves of bread would be at least as high, as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the, outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone), we conclude, that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There, is sufficient evidence that the true mean height for the population of the baker’s loaves of bread is, greater than 15 cm., , 9.8 Decision and Conclusion8, A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the, p-value and a preset or preconceived α (also called a "significance level"). A preset α is the probability of, a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given, to you at the beginning of the problem., When you make a decision to reject or not reject Ho , do as follows:, , • If α > p-value, reject Ho . The results of the sample data are significant. There is sufficient evidence to, conclude that Ho is an incorrect belief and that the alternative hypothesis, Ha , may be correct., • If α ≤ p-value, do not reject Ho . The results of the sample data are not significant. There is not, sufficient evidence to conclude that the alternative hypothesis, Ha , may be correct., • When you "do not reject Ho ", it does not mean that you should believe that Ho is true. It simply, means that the sample data have failed to provide sufficient evidence to cast serious doubt about the, truthfulness of Ho ., Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms, of the given problem., , 9.9 Additional Information9, • In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is, the preconceived or preset α., • The statistician setting up the hypothesis test selects the value of α to use before collecting the sample, data., • If no level of significance is given, the accepted standard is to use α = 0.05., • When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right, tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two, tailed., • The alternate hypothesis, Ha , tells you if the test is left, right, or two-tailed. It is the key to conducting, the appropriate test., • Ha never has a symbol that contains an equal sign., • Thinking about the meaning of the p-value: A data analyst (and anyone else) should have more, confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for, example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value, like 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should, have more confidence that she made the correct decision in failing to reject the null hypothesis. This, makes the data analyst use judgment rather than mindlessly applying rules., 8 This, 9 This, , content is available online at <http://cnx.org/content/m16992/1.11/>., content is available online at <http://cnx.org/content/m16999/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 384, , The following examples illustrate a left, right, and two-tailed test., Example 9.8, Ho : µ = 5, , Ha : µ < 5, , Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as, follows:, , Example 9.9, Ho : p ≤ 0.2, , Ha : p > 0.2, , This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of, the p-value is as follows:, , Example 9.10, Ho : µ = 50, , Ha : µ 6= 50, , This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the, p-value is as follows., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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385, , 9.10 Summary of the Hypothesis Test10, The hypothesis test itself has an established process. This can be summarized as follows:, Determine Ho and Ha . Remember, they are contradictory., Determine the random variable., Determine the distribution for the test., Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score, and a t-score are examples of test statistics.), 5. Compare the preconceived α with the p-value, make a decision (reject or do not reject Ho ), and write, a clear conclusion using English sentences., 1., 2., 3., 4., , Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the, sample size of the data that is used in calculating the p-value. Remember that the quantity 1 − β is called, the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the, sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when, it should be., , 9.11 Examples11, Example 9.11, Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard, freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could, swim the 25-yard freestyle faster by using goggles. Frank bought Jeffrey a new pair of expensive, goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey’s mean time, was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43, seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the, 25-yard freestyle are normal., Solution, Set up the Hypothesis Test:, Since the problem is about a mean, this is a test of a single population mean., Ho : µ = 16.43, , Ha : µ < 16.43, , For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is lefttailed., Determine the distribution needed:, Random variable: X = the mean time to swim the 25-yard freestyle., Distribution for the test: X is normal (population standard deviation is known: σ = 0.8), , , , , σX, 0.8, √, X ∼ N µ, √, Therefore,, X, ∼, N, 16.43,, n, 15, , µ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15., Calculate the p-value using the normal distribution for a mean:, 10 This, 11 This, , content is available online at <http://cnx.org/content/m16993/1.6/>., content is available online at <http://cnx.org/content/m17005/1.25/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 386, , p-value = P, , −, , , , x < 16, , = 0.0187 where the sample mean in the problem is given as 16., , p-value = 0.0187 (This is called the actual level of significance.) The p-value is the area to the left, of the sample mean is given as 16., Graph:, , Figure 9.1, , µ = 16.43 comes from Ho . Our assumption is µ = 16.43., Interpretation of the p-value: If Ho is true, there is a 0.0187 probability (1.87%) that Jeffrey’s mean, time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean, time of 16 seconds or less is unlikely to have happened randomly. It is a rare event., Compare α and the p-value:, α = 0.05, , p-value = 0.0187, , α > p-value, , Make a decision: Since α > p-value, reject Ho ., This means that you reject µ = 16.43. In other words, you do not think Jeffrey swims the 25-yard, freestyle in 16.43 seconds but faster with the new goggles., Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new, goggles. The sample data show there is sufficient evidence that Jeffrey’s mean time to swim the, 25-yard freestyle is less than 16.43 seconds., The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:, Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow, down and enter 16.43 for µ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow, down to µ: (alternate hypothesis) and arrow over to <µ0 . Press ENTER. Arrow down to Calculate, and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates, the test statistic (z-score) for the sample mean. µ < 16.43 is the alternate hypothesis. Do this set, of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph, appears with z = −2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw, that no other equations are highlighted in Y = and the plots are turned off., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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387, When the calculator does a Z-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the Central Limit Theorem:, , √ , P ( x < 16 ) = 2nd DISTR normcdf −10^99, 16, 16.43, 0.8/ 15 ., The Type I and Type II errors for this problem are as follows:, The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than, 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds., (Reject the null hypothesis when the null hypothesis is true.), The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style,, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style,, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis, is false.), , Historical Note: The traditional way to compare the two probabilities, α and the p-value, is to compare, the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the, x −µ, p-value is −2.08. (From the Central Limit Theorem, the test statistic formula is z = σX X . For this problem,, √, , n, , x = 16, µ X = 16.43 from the null hypothesis, σX = 0.8, and n = 15.) You can find the critical value for, α = 0.05 in the normal table (see 15.Tables in the Table of Contents). The z-score for an area to the left, equal to 0.05 is midway between -1.65 and -1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is, -1.645. Since −1.645 > − 2.08 (which demonstrates that α > p-value), reject Ho . Traditionally, the decision, to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very, common. For this problem, the p-value, 0.0187 is considerably smaller than α, 0.05. You can be confident, about your decision to reject. The graph shows α, the p-value, and the test statistics and the critical value., , Figure 9.2, , Example 9.12, A college football coach thought that his players could bench press a mean weight of 275 pounds., It is known that the standard deviation is 55 pounds. Three of his players thought that the mean, weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 388, , different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2);, 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). (Source: data from Reuben Davis, Kraig, Evans, and Scott Gunderson.), Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean, is more than 275 pounds., Solution, Set up the Hypothesis Test:, Since the problem is about a mean weight, this is a test of a single population mean., Ho : µ = 275, , Ha : µ > 275, , This is a right-tailed test., , Calculating the distribution needed:, Random variable: X = the mean weight, in pounds, lifted by the football players., Distribution for the test: It is normal because σ is known., , , X ∼ N 275, √55, 30, , x = 286.2 pounds (from the data)., σ = 55 pounds (Always use σ if you know it.) We assume µ = 275 pounds unless our data shows, us otherwise., Calculate the p-value using the normal distribution for a mean and using the sample mean as, input (see the calculator instructions below for using the data as input):, p-value = P ( x > 286.2) = 0.1323., Interpretation of the p-value: If Ho is true, then there is a 0.1331 probability (13.23%) that the, football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large, enough, a mean weight lift of 286.2 pounds or more is not a rare event., , Figure 9.3, , Compare α and the p-value:, α = 0.025, , p-value = 0.1323, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
Page 399 : 389, Make a decision: Since α<p-value, do not reject Ho ., Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence, to conclude that the true mean weight lifted is more than 275 pounds., The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:, Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow, over to Data and press ENTER. Arrow down and enter 275 for µ0 , 55 for σ, the name of the list where, you put the data, and the name of the list where you put the frequencies. Arrow down to µ : and, arrow over to > µ0 . Press ENTER. Arrow down to Calculate and press ENTER. The calculator not, only calculates the p-value (p = 0.1331, a little different from the above calculation - in it we, used the sample mean rounded to one decimal place instead of the data) but it also calculates the, test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation., µ > 275 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead, of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331, (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the, plots are turned off., , Example 9.13, Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor, thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores, 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance., The data are from a normal distribution., Solution, Set up the Hypothesis Test:, A 5% level of significance means that α = 0.05. This is a test of a single population mean., Ho : µ = 65, , Ha : µ > 65, , Since the instructor thinks the average score is higher, use a "> ". The "> " means the test is, right-tailed., Determine the distribution needed:, Random variable: X = average score on the first statistics test., Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also, that the data come from a normal distribution. This means that the distribution for the test is a, student’s-t., Use tdf . Therefore, the distribution for the test is t9 where n = 10 and df = 10 − 1 = 9., Calculate the p-value using the Student’s-t distribution:, p-value = P ( x > 67 )= 0.0396 where the sample mean and sample standard deviation are, calculated as 67 and 3.1972 from the data., Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability, (3.96%) that the sample mean is 67 or more., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 390, , Figure 9.4, , Compare α and the p-value:, Since α = .05 and p-value = 0.0396. Therefore, α > p-value., Make a decision: Since α > p-value, reject Ho ., This means you reject µ = 65. In other words, you believe the average test score is more than 65., Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean, (average) test score is more than 65, just as the math instructor thinks., The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:, Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to, Data and press ENTER. Arrow down and enter 65 for µ0 , the name of the list where you put the, data, and 1 for Freq:. Arrow down to µ : and arrow over to > µ0 . Press ENTER. Arrow down, to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it, also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample, standard deviation. µ > 65 is the alternate hypothesis. Do this set of instructions again except, arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test, statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are, highlighted in Y = and the plots are turned off., , Example 9.14, Joon believes that 50% of first-time brides in the United States are younger than their grooms., She performs a hypothesis test to determine if the percentage is the same or different from 50%., Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the, hypothesis test, she uses a 1% level of significance., Solution, Set up the Hypothesis Test:, The 1% level of significance means that α = 0.01. This is a test of a single population proportion., Ho : p = 0.50, , Ha : p 6= 0.50, , The words "is the same or different from" tell you this is a two-tailed test., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
Page 401 : 391, Calculate the distribution needed:, Random variable: P’ = the percent of of first-time brides who are younger than their grooms., Distribution for the test: The problem contains no mention of a mean. The information is given, in terms of percentages. Use the distribution for P’, the estimated proportion., , , q , q, p·q, 0.5·0.5, Therefore, P’ ∼ N 0.5,, where p = 0.50, q = 1 − p = 0.50, and, P’ ∼ N p,, n, 100, n = 100., Calculate the p-value using the normal distribution for proportions:, p-value = P ( p’ < 0.47 or p’ > 0.53 ) = 0.5485, where x = 53, p’ =, , x, n, , =, , 53, 100, , = 0.53., , Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%), that the sample (estimated) proportion p’ is 0.53 or more OR 0.47 or less (see the graph below)., , Figure 9.5, , µ = p = 0.50 comes from Ho , the null hypothesis., p’= 0.53. Since the curve is symmetrical and the test is two-tailed, the p’ for the left tail is equal to, 0.50 − 0.03 = 0.47 where µ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.), Compare α and the p-value:, Since α = 0.01 and p-value = 0.5485. Therefore, α< p-value., Make a decision: Since α<p-value, you cannot reject Ho ., Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that, the percentage of first-time brides that are younger than their grooms is different from 50%., The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:, Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for p0 , 53 for x and 100 for, n. Arrow down to Prop and arrow to not equals p0 . Press ENTER. Arrow down to Calculate, and press ENTER. The calculator calculates the p-value (p = 0.5485) and the test statistic (z-score)., Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to, Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 0.6 (test statistic) and, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 392, , p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in, Y = and the plots are turned off., The Type I and Type II errors are as follows:, The Type I error is to conclude that the proportion of first-time brides that are younger than their, grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true)., The Type II error is there is not enough evidence to conclude that the proportion of first time brides, that are younger than their grooms differs from 50% when, in fact, the proportion does differ from, 50%. (Do not reject the null hypothesis when the null hypothesis is false.), , Example 9.15, Problem 1, Suppose a consumer group suspects that the proportion of households that have three cell phones, is 30%. A cell phone company has reason to believe that the proportion is 30%. Before they start, a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150, households with the result that 43 of the households have three cell phones., Solution, Set up the Hypothesis Test:, Ho : p = 0.30, , Ha : p6= 0.30, , Determine the distribution needed:, The random variable is P’ = proportion of households that have three cell phones., , , q, (0.30)·(0.70), The distribution for the hypothesis test is P’ ∼ N 0.30,, 150, Problem 2, The value that helps determine the p-value is p’. Calculate p’., Problem 3, What is a success for this problem?, Problem 4, What is the level of significance?, Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately., Problem 5, Calculate the p-value., Problem 6, Make a decision. _____________(Reject/Do not reject) H0 because____________., , The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem, follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the, null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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393, proportion p’ is the proportion of fleas killed to the total fleas found on Fido. This is sample information., The problem gives a preconceived α = 0.01, for comparison, and a 95% confidence interval computation., The poem is clever and humorous, so please enjoy it!, NOTE : Hypothesis testing problems consist of multiple steps. To help you do the problems, solution sheets are provided for your use. Look in the Table of Contents Appendix for the topic, "Solution Sheets." If you like, use copies of the appropriate solution sheet for homework problems., , Example 9.16, , My dog has so many fleas,, They do not come off with ease., As for shampoo, I have tried many types, Even one called Bubble Hype,, Which only killed 25% of the fleas,, Unfortunately I was not pleased., I've used all kinds of soap,, Until I had give up hope, Until one day I saw, An ad that put me in awe., A shampoo used for dogs, Called GOOD ENOUGH to Clean a Hog, Guaranteed to kill more fleas., I gave Fido a bath, And after doing the math, His number of fleas, Started dropping by 3's!, Before his shampoo, I counted 42., At the end of his bath,, I redid the math, And the new shampoo had killed 17 fleas., So now I was pleased., Now it is time for you to have some fun, With the level of significance being .01,, You must help me figure out, Use the new shampoo or go without?, Solution, Set up the Hypothesis Test:, Ho : p = 0.25, , Ha : p > 0.25, , Determine the distribution needed:, In words, CLEARLY state what your random variable X or P’ represents., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
Page 404 : CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 394, , P’ = The proportion of fleas that are killed by the new shampoo, State the distribution to use for the test., , , q, (0.25)(1−0.25), Normal: N 0.25,, 42, Test Statistic: z = 2.3163, Calculate the p-value using the normal distribution for proportions:, p-value =0.0103, In 1 – 2 complete sentences, explain what the p-value means for this problem., If the null hypothesis is true (the proportion, is 0.25), then there is a 0.0103 probability that the, sample (estimated) proportion is 0.4048 17, 42 or more., Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the, horizontal axis and shade the region(s) corresponding to the p-value., , Figure 9.6, , Compare α and the p-value:, Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and, write an appropriate conclusion, using COMPLETE SENTENCES., alpha, , decision, , reason for decision, , 0.01, , Do not reject Ho, , α<p-value, , Table 9.3, , Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that, the percentage of fleas that are killed by the new shampoo is more than 25%., Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the, graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence, Interval., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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395, , Figure 9.7, , Confidence Interval: (0.26, 0.55) We are 95% confident that the true population proportion p of, fleas that are killed by the new shampoo is between 26% and 55%., NOTE : This test result is not very definitive since the p-value is very close to alpha. In reality, one, would probably do more tests by giving the dog another bath after the fleas have had a chance to, return., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 396, , 9.12 Summary of Formulas12, Ho and Ha are contradictory., If Ho has:, , equal (=), , greater than or equal to, (≥), , less than or equal to, (≤), , then Ha has:, , not equal (6=) or greater, than (> ) or less than, (< ), , less than (< ), , greater than (> ), , Table 9.4, , If α ≤ p-value, then do not reject Ho ., If α > p-value, then reject Ho ., α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data., α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the, null hypothesis is true., β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when, the null hypothesis is false., If there is no given preconceived α, then use α = 0.05., Types of Hypothesis Tests, , • Single population mean, known population variance (or standard deviation): Normal test., • Single population mean, unknown population variance (or standard deviation): Student’s-t test., • Single population proportion: Normal test., , 12 This, , content is available online at <http://cnx.org/content/m16996/1.9/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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397, , 9.13 Practice 1: Single Mean, Known Population Standard Deviation13, 9.13.1 Student Learning Outcomes, • The student will conduct a hypothesis test of a single mean with known population standard deviation., , 9.13.2 Given, Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5, years. A study was then done to see if the mean time has increased in the new century. A random sample, of 26 first–time convicted burglars in a recent year was picked. The mean length of time in jail from the, survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the, population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time, has increased. The distribution of the population is normal., , 9.13.3 Hypothesis Testing: Single Mean, Exercise 9.13.1, Is this a test of means or proportions?, , (Solution on p. 423.), , Exercise 9.13.2, State the null and alternative hypotheses., , (Solution on p. 423.), , a. Ho :, b. Ha :, Exercise 9.13.3, Is this a right-tailed, left-tailed, or two-tailed test? How do you know?, , (Solution on p. 423.), , Exercise 9.13.4, What symbol represents the Random Variable for this test?, , (Solution on p. 423.), , Exercise 9.13.5, In words, define the Random Variable for this test., , (Solution on p. 423.), , Exercise 9.13.6, Is the population standard deviation known and, if so, what is it?, , (Solution on p. 423.), , Exercise 9.13.7, Calculate the following:, , (Solution on p. 423.), , a., b., c., d., , x=, σ=, sx =, n=, , Exercise 9.13.8, (Solution on p. 423.), Since both σ and s x are given, which should be used? In 1 -2 complete sentences, explain why., Exercise 9.13.9, State the distribution to use for the hypothesis test., , (Solution on p. 423.), , Exercise 9.13.10, Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized mean and the, sample mean x. Shade the area corresponding to the p-value., 13 This, , content is available online at <http://cnx.org/content/m17004/1.11/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 398, , Exercise 9.13.11, Find the p-value., , (Solution on p. 423.), , Exercise 9.13.12, At a pre-conceived α = 0.05, what is your:, , (Solution on p. 423.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , 9.13.4 Discussion Questions, Exercise 9.13.13, Does it appear that the mean jail time spent for first time convicted burglars has increased? Why, or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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399, , 9.14 Practice 2: Single Mean, Unknown Population Standard Deviation14, 9.14.1 Student Learning Outcomes, • The student will conduct a hypothesis test of a single mean with unknown population standard deviation., , 9.14.2 Given, A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years, with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time, on death row could likely be 15 years., , 9.14.3 Hypothesis Testing: Single Mean, Exercise 9.14.1, Is this a test of means or proportions?, , (Solution on p. 423.), , Exercise 9.14.2, State the null and alternative hypotheses., , (Solution on p. 423.), , a. Ho :, b. Ha :, Exercise 9.14.3, Is this a right-tailed, left-tailed, or two-tailed test? How do you know?, , (Solution on p. 423.), , Exercise 9.14.4, What symbol represents the Random Variable for this test?, , (Solution on p. 423.), , Exercise 9.14.5, In words, define the Random Variable for this test., , (Solution on p. 423.), , Exercise 9.14.6, Is the population standard deviation known and, if so, what is it?, , (Solution on p. 423.), , Exercise 9.14.7, Calculate the following:, , (Solution on p. 423.), , a. x =, b. 6.3 =, c. n =, Exercise 9.14.8, Which test should be used? In 1 -2 complete sentences, explain why., , (Solution on p. 424.), , Exercise 9.14.9, State the distribution to use for the hypothesis test., , (Solution on p. 424.), , Exercise 9.14.10, Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized mean and the, sample mean, x. Shade the area corresponding to the p-value., 14 This, , content is available online at <http://cnx.org/content/m17016/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 400, , Figure 9.8, , Exercise 9.14.11, Find the p-value., , (Solution on p. 424.), , Exercise 9.14.12, At a pre-conceived α = 0.05, what is your:, , (Solution on p. 424.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , 9.14.4 Discussion Question, Does it appear that the mean time on death row could be 15 years? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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401, , 9.15 Practice 3: Single Proportion15, 9.15.1 Student Learning Outcomes, • The student will conduct a hypothesis test of a single population proportion., , 9.15.2 Given, The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness., (http://www.nimh.nih.gov/publicat/depression.cfm) Suppose that in a survey of 100 people in a certain, town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower, than the percent in the general adult American population., , 9.15.3 Hypothesis Testing: Single Proportion, Exercise 9.15.1, Is this a test of means or proportions?, , (Solution on p. 424.), , Exercise 9.15.2, State the null and alternative hypotheses., , (Solution on p. 424.), , a. Ho :, b. Ha :, Exercise 9.15.3, Is this a right-tailed, left-tailed, or two-tailed test? How do you know?, , (Solution on p. 424.), , Exercise 9.15.4, What symbol represents the Random Variable for this test?, , (Solution on p. 424.), , Exercise 9.15.5, In words, define the Random Variable for this test., , (Solution on p. 424.), , Exercise 9.15.6, Calculate the following:, , (Solution on p. 424.), , a: x =, b: n =, c: p’ =, Exercise 9.15.7, Calculate σp’ . Make sure to show how you set up the formula., , (Solution on p. 424.), , Exercise 9.15.8, State the distribution to use for the hypothesis test., , (Solution on p. 424.), , Exercise 9.15.9, Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized mean and the, sample proportion, p-hat. Shade the area corresponding to the p-value., , 15 This, , content is available online at <http://cnx.org/content/m17003/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 402, Exercise 9.15.10, Find the p-value, , (Solution on p. 424.), , Exercise 9.15.11, At a pre-conceived α = 0.05, what is your:, , (Solution on p. 424.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , 9.15.4 Discusion Question, Exercise 9.15.12, Does it appear that the proportion of people in that town with depression or a depressive illness, is lower than general adult American population? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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403, , 9.16 Homework16, Exercise 9.16.1, (Solution on p. 424.), Some of the statements below refer to the null hypothesis, some to the alternate hypothesis., State the null hypothesis, Ho , and the alternative hypothesis, Ha , in terms of the appropriate parameter (µ or p)., a. The mean number of years Americans work before retiring is 34., b. At most 60% of Americans vote in presidential elections., c. The mean starting salary for San Jose State University graduates is at least $100,000 per year., d. 29% of high school seniors get drunk each month., e. Fewer than 5% of adults ride the bus to work in Los Angeles., f. The mean number of cars a person owns in her lifetime is not more than 10., g. About half of Americans prefer to live away from cities, given the choice., h. Europeans have a mean paid vacation each year of six weeks., i. The chance of developing breast cancer is under 11% for women., j. Private universities mean tuition cost is more than $20,000 per year., Exercise 9.16.2, For (a) - (j) above, state the Type I and Type II errors in complete sentences., , (Solution on p. 425.), , Exercise 9.16.3, For (a) - (j) above, in complete sentences:, a. State a consequence of committing a Type I error., b. State a consequence of committing a Type II error., , D IRECTIONS : For each of the word problems, use a solution sheet to do the hypothesis test. The, solution sheet is found in 14. Appendix (online book version: the link is "Solution Sheets"; PDF, book version: look under 14.5 Solution Sheets). Please feel free to make copies of the solution, sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files., NOTE : If you are using a student’s-t distribution for a homework problem below, you may assume, that the underlying population is normally distributed. (In general, you must first prove that, assumption, though.), , Exercise 9.16.4, A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs, to be replaced. From past studies of this tire, the standard deviation is known to be 8000. A survey, of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was, 46,500 miles with a standard deviation of 9800 miles. Do the data support the claim at the 5%, level?, Exercise 9.16.5, (Solution on p. 425.), From generation to generation, the mean age when smokers first start to smoke varies. However,, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers, of this generation was done to see if the mean starting age is at least 19. The sample mean was, 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?, 16 This, , content is available online at <http://cnx.org/content/m17001/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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404, , CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, Exercise 9.16.6, The cost of a daily newspaper varies from city to city. However, the variation among prices, remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean, cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation, of 18¢. Do the data support the claim at the 1% level?, Exercise 9.16.7, (Solution on p. 425.), An article in the San Jose Mercury News stated that students in the California state university, system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that, the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with, a sample standard deviation of 1.2. Do the data support your claim at the 1% level?, Exercise 9.16.8, The mean number of sick days an employee takes per year is believed to be about 10. Members, of a personnel department do not believe this figure. They randomly survey 8 employees. The, number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the, number of sick days they took for the past year. Should the personnel team believe that the mean, number is about 10?, Exercise 9.16.9, (Solution on p. 425.), In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80, hour week. Recently, many groups have been studying whether or not the women’s movement, has, in fact, resulted in an increase in the average work week for women (combining employment, and at-home work). Suppose a study was done to determine if the mean work week has increased., 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was 10. Does it appear that the mean work week has increased for women at the, 5% level?, Exercise 9.16.10, Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics, class go through life feeling more enriched. For some reason that she can’t quite figure out, most, people don’t believe her. You decide to check this out on your own. You randomly survey 64 of, her past Elementary Statistics students and find that 34 feel more enriched as a result of her class., Now, what do you think?, Exercise 9.16.11, (Solution on p. 425.), A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average, brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown, trout’s mean I.Q. is greater than 4. You catch 12 brown trout. A fish psychologist determines the, I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief., Exercise 9.16.12, Refer to the previous problem. Conduct a hypothesis test to see if your decision and conclusion, would change if your belief were that the brown trout’s mean I.Q. is not 4., Exercise 9.16.13, (Solution on p. 425.), According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the, birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent, of girls born in China. You conduct a study. In this study, you count the number of girls and boys, born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based, on your study, do you believe that the percent of girls born in China is 46.7?, Exercise 9.16.14, A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an, angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out, of 76 Americans surveyed, only 2 had seen or sensed the presence of an angel. As a result of the, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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405, contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give, three reasons why the two polls might give different results., Exercise 9.16.15, (Solution on p. 425.), The mean work week for engineers in a start-up company is believed to be about 60 hours. A, newly hired engineer hopes that it’s shorter. She asks 10 engineering friends in start-ups for the, lengths of their mean work weeks. Based on the results that follow, should she count on the mean, work week to be shorter than 60 hours?, Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55., Exercise 9.16.16, Use the “Lap time” data for Lap 4 (see Table of Contents) to test the claim that Terri finishes Lap, 4, on average, in less than 129 seconds. Use all twenty races given., Exercise 9.16.17, Use the “Initial Public Offering” data (see Table of Contents) to test the claim that the mean offer, price was $18 per share. Do not use all the data. Use your random number generator to randomly, survey 15 prices., , NOTE :, , The following questions were written by past students. They are excellent problems!, , Exercise 9.16.18, 18. "Asian Family Reunion" by Chau Nguyen, , Every two years it comes around, We all get together from different towns., In my honest opinion, It's not a typical family reunion, Not forty, or fifty, or sixty,, But how about seventy companions!, The kids would play, scream, and shout, One minute they're happy, another they'll pout., The teenagers would look, stare, and compare, From how they look to what they wear., The men would chat about their business, That they make more, but never less., Money is always their subject, And there's always talk of more new projects., The women get tired from all of the chats, They head to the kitchen to set out the mats., Some would sit and some would stand, Eating and talking with plates in their hands., Then come the games and the songs, And suddenly, everyone gets along!, With all that laughter, it's sad to say, That it always ends in the same old way., They hug and kiss and say "good-bye", And then they all begin to cry!, I say that 60 percent shed their tears, But my mom counted 35 people this year., She said that boys and men will always have their pride,, So we won't ever see them cry., I myself don't think she's correct,, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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406, , CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , So could you please try this problem to see if you object?, Exercise 9.16.19, "The Problem with Angels" by Cyndy Dowling, , (Solution on p. 425.), , Although this problem is wholly mine,, The catalyst came from the magazine, Time., On the magazine cover I did find, The realm of angels tickling my mind., Inside, 69% I found to be, In angels, Americans do believe., Then, it was time to rise to the task,, Ninety-five high school and college students I did ask., Viewing all as one group,, Random sampling to get the scoop., So, I asked each to be true,, "Do you believe in angels?" Tell me, do!, Hypothesizing at the start,, Totally believing in my heart, That the proportion who said yes, Would be equal on this test., Lo and behold, seventy-three did arrive,, Out of the sample of ninety-five., Now your job has just begun,, Solve this problem and have some fun., Exercise 9.16.20, "Blowing Bubbles" by Sondra Prull, , Studying stats just made me tense,, I had to find some sane defense., Some light and lifting simple play, To float my math anxiety away., Blowing bubbles lifts me high, Takes my troubles to the sky., POIK! They're gone, with all my stress, Bubble therapy is the best., The label said each time I blew, The average number of bubbles would be at least 22., I blew and blew and this I found, From 64 blows, they all are round!, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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407, , But the number of bubbles in 64 blows, Varied widely, this I know., 20 per blow became the mean, They deviated by 6, and not 16., From counting bubbles, I sure did relax, But now I give to you your task., Was 22 a reasonable guess?, Find the answer and pass this test!, Exercise 9.16.21, 21. "Dalmatian Darnation" by Kathy Sparling, , (Solution on p. 426.), , A greedy dog breeder named Spreckles, Bred puppies with numerous freckles, The Dalmatians he sought, Possessed spot upon spot, The more spots, he thought, the more shekels., His competitors did not agree, That freckles would increase the fee., They said, ``Spots are quite nice, But they don't affect price;, One should breed for improved pedigree.'', The breeders decided to prove, This strategy was a wrong move., Breeding only for spots, Would wreak havoc, they thought., His theory they want to disprove., They proposed a contest to Spreckles, Comparing dog prices to freckles., In records they looked up, One hundred one pups:, Dalmatians that fetched the most shekels., They asked Mr. Spreckles to name, An average spot count he'd claim, To bring in big bucks., Said Spreckles, ``Well, shucks,, It's for one hundred one that I aim.'', Said an amateur statistician, Who wanted to help with this mission., ``Twenty-one for the sample, Standard deviation's ample:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 408, , They examined one hundred and one, Dalmatians that fetched a good sum., They counted each spot,, Mark, freckle and dot, And tallied up every one., Instead of one hundred one spots, They averaged ninety six dots, Can they muzzle Spreckles', Obsession with freckles, Based on all the dog data they've got?, Exercise 9.16.22, "Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall, As a poor starving student I don’t have much money to spend for even the bare necessities. So, my favorite and main staple food is macaroni and cheese. It’s high in taste and low in cost and, nutritional value., One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so, important, food of my life. So I went down the street to Greatway to get a box of macaroni and, cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The, world is changing fast. I had thought that the mean cost of a box (the normal size, not some supergigantic-family-value-pack) was at most $1, but now I wasn’t so sure. However, I was determined, to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and, cheese. Here are the data I wrote in my notebook:, Price per box of Mac and Cheese:, , •, •, •, •, •, •, •, •, , 5 stores @ $2.02, 15 stores @ $0.25, 3 stores @ $1.29, 6 stores @ $0.35, 4 stores @ $2.27, 7 stores @ $1.50, 5 stores @ $1.89, 8 stores @ 0.75., , I could see that the costs varied but I had to sit down to figure out whether or not I was right. If, it does turn out that this mouth-watering dish is at most $1, then I’ll throw a big cheesy party in, our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving, student I can’t be expected to feed our class of animals!), Exercise 9.16.23, (Solution on p. 426.), "William Shakespeare: The Tragedy of Hamlet, Prince of Denmark" by Jacqueline Ghodsi, THE CHARACTERS (in order of appearance):, , • HAMLET, Prince of Denmark and student of Statistics, • POLONIUS, Hamlet’s tutor, • HOROTIO, friend to Hamlet and fellow student, Scene: The great library of the castle, in which Hamlet does his lessons, Act I, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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409, (The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is, reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table, at right stage.), POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of, your imagination!, HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin, damn’d, be their intents wicked or charitable., POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true, to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns, to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how, infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-andsixty and discover what the true proportion be. For my part, I will never succumb to this fantasy,, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one, hundred hold true., HORATIO (to Hamlet): What should we do, my Lord?, HAMLET: Go to thy purpose, Horatio., HORATIO: To what end, my Lord?, HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the, consonance of our youth, but the obligation of our ever-preserved love, be even and direct with, me, whether I am right or no., (Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.), Act II, (The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and, places some books upon the table just a moment before Horatio enters.), POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?, HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that, one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might, not this believe, without the sensible and true avouch of mine own eyes., POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look, to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and, Polonius leave together.), HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the, slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing,, end them. (Hamlet resignedly attends to his task.), (Curtain falls), Exercise 9.16.24, "Untitled" by Stephen Chen, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 410, , I’ve often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult, to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the, likelihood of generating an error is 1%., So, armed with this knowledge, I wrote a new test program Y that will generate the same error that, test program X creates, but more often. To find out if my test program is better than the original,, so that I can convince the management that I’m right, I ran my test program to find out how often, I can generate the same error. When I ran my test program 50 times, I generated the error twice., While this may not seem much better, I think that I can convince the management to use my test, program instead of the original test program. Am I right?, Exercise 9.16.25, Japanese Girls’ Names, , (Solution on p. 426.), , by Kumi Furuichi, It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have, started around my grandmothers’ generation and its peak might have been around my mother’s, generation.) “Ko” means “child” in Chinese character. Parents would name their daughters with, “ko” attaching to other Chinese characters which have meanings that they want their daughters, to become, such as Sachiko – a happy child, Yoshiko – a good child, Yasuko – a healthy child, and, so on., However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have, names which end with “ko.” More and more, parents seem to have become creative, modernized,, and, sometimes, westernized in naming their children., I have a feeling that, while 70 percent or more of my mother’s generation would have names with, “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese, friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Below are, the names. (Some are repeats.) Test to see if the proportion has dropped for this generation., Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko,, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki,, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko,, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko,, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko,, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko,, Yuko., Exercise 9.16.26, Phillip’s Wish by Suzanne Osorio, , My nephew likes to play, Chasing the girls makes his day., He asked his mother, If it is okay, To get his ear pierced., She said, ``No way!'', To poke a hole through your ear,, Is not what I want for you, dear., He argued his point quite well,, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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411, , Says even my macho pal, Mel,, Has gotten this done., It's all just for fun., C'mon please, mom, please, what the hell., Again Phillip complained to his mother,, Saying half his friends (including their brothers), Are piercing their ears, And they have no fears, He wants to be like the others., She said, ``I think it's much less., We must do a hypothesis test., And if you are right,, I won't put up a fight., But, if not, then my case will rest.'', We proceeded to call fifty guys, To see whose prediction would fly., Nineteen of the fifty, Said piercing was nifty, And earrings they'd occasionally buy., Then there's the other thirty-one,, Who said they'd never have this done., So now this poem's finished., Will his hopes be diminished,, Or will my nephew have his fun?, Exercise 9.16.27, The Craven by Mark Salangsang, , (Solution on p. 426.), , Once upon a morning dreary, In stats class I was weak and weary., Pondering over last night's homework, Whose answers were now on the board, This I did and nothing more., While I nodded nearly napping, Suddenly, there came a tapping., As someone gently rapping,, Rapping my head as I snore., Quoth the teacher, ``Sleep no more.'', ``In every class you fall asleep,'', The teacher said, his voice was deep., ``So a tally I've begun to keep, Of every class you nap and snore., The percentage being forty-four.'', ``My dear teacher I must confess,, While sleeping is what I do best., The percentage, I think, must be less,, A percentage less than forty-four.'', This I said and nothing more., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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412, , CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , ``We'll see,'' he said and walked away,, And fifty classes from that day, He counted till the month of May, The classes in which I napped and snored., The number he found was twenty-four., At a significance level of 0.05,, Please tell me am I still alive?, Or did my grade just take a dive, Plunging down beneath the floor?, Upon thee I hereby implore., Exercise 9.16.28, Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public, speaking. A student believes that less than 40% of students at her school fear public speaking., She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking., Conduct a hypothesis test to determine if the percent at her school is less than 40%. (Source:, http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1 17, ), Exercise 9.16.29, (Solution on p. 426.), 68% of online courses taught at community colleges nationwide were taught by full-time faculty., To test if 68% also represents California’s percent for full-time faculty teaching the online classes,, Long Beach City College (LBCC), CA, was randomly selected for comparison. In the same year, 34, of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test, to determine if 68% represents CA. NOTE: For more accurate results, use more CA community, colleges and this past year’s data. (Sources: Growing by Degrees by Allen and Seaman; Amit, Schitai, Director of Instructional Technology and Distance Learning, LBCC)., Exercise 9.16.30, According to an article in Bloomberg Businessweek, New York City’s most recent adult smoking, rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if, the rate is still 14% or if it has decreased. (Source: http://www.businessweek.com/news/201109-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html 18 ), Exercise 9.16.31, (Solution on p. 426.), The mean age of De Anza College students in a previous term was 26.6 years old. An instructor, thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. (Source: http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf 19 ), Exercise 9.16.32, Registered nurses earned an average annual salary of $69,110. For that same year, a survey, was conducted of 41 California registered nurses to determine if the annual salary is higher than, $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of, $7,489. Conduct a hypothesis test. (Source: http://www.bls.gov/oes/current/oes291111.htm 20, ), Exercise 9.16.33, (Solution on p. 426.), La Leche League International reports that the mean age of weaning a child from breastfeeding, is age 4 to 5 worldwide. In America, most nursing mothers wean their children much earlier., 17 http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1, 18 http://www.businessweek.com/news/2011-09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html, 19 http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf, 20 http://www.bls.gov/oes/current/oes291111.htm, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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413, Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children., The mean weaning age was 9 months (3/4 year) with a standard deviation of 4 months. Conduct, a hypothesis test to determine if the mean weaning age in the U.S. is less than 4 years old. (Source:, http://www.lalecheleague.org/Law/BAFeb01.html 21 ), , 9.16.1 Try these multiple choice questions., Exercise 9.16.34, (Solution on p. 426.), When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug., Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?, A., B., C., D., , To conclude the drug is safe when in, fact, it is unsafe, To not conclude the drug is safe when, in fact, it is safe., To conclude the drug is safe when, in fact, it is safe., To not conclude the drug is unsafe when, in fact, it is unsafe, , The next two questions refer to the following information: Over the past few decades, public health, officials have examined the link between weight concerns and teen girls smoking. Researchers surveyed a, group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four, years the girls were surveyed again. Sixty-three (63) said they smoked to stay thin. Is there good evidence, that more than thirty percent of the teen girls smoke to stay thin?, Exercise 9.16.35, The alternate hypothesis is, A., B., C., D., , (Solution on p. 426.), , p < 0.30, p ≤ 0.30, p ≥ 0.30, p > 0.30, , Exercise 9.16.36, After conducting the test, your decision and conclusion are, , (Solution on p. 426.), , A. Reject Ho : There is sufficient evidence to conclude that more than 30% of teen girls smoke to, stay thin., B. Do not reject Ho : There is not sufficient evidence to conclude that less than 30% of teen girls, smoke to stay thin., C. Do not reject Ho : There is not sufficient evidence to conclude that more than 30% of teen girls, smoke to stay thin., D. Reject Ho : There is sufficient evidence to conclude that less than 30% of teen girls smoke to, stay thin., The next three questions refer to the following information: A statistics instructor believes that fewer, than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of, the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of attended the midnight, showing., Exercise 9.16.37, An appropriate alternative hypothesis is, , (Solution on p. 427.), , 21 http://www.lalecheleague.org/Law/BAFeb01.html, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 414, A., B., C., D., , p = 0.20, p > 0.20, p < 0.20, p ≤ 0.20, , Exercise 9.16.38, At a 1% level of significance, an appropriate conclusion is:, , (Solution on p. 427.), , A. There is insufficient evidence to conclude that the percent of EVC students that attended, midnight showing of Harry Potter is less than 20%., B. There is sufficient evidence to conclude that the percent of EVC students that attended, midnight showing of Harry Potter is more than 20%., C. There is sufficient evidence to conclude that the percent of EVC students that attended, midnight showing of Harry Potter is less than 20%., D. There is insufficient evidence to conclude that the percent of EVC students that attended, midnight showing of Harry Potter is at least 20%., , the, the, the, the, , Exercise 9.16.39, (Solution on p. 427.), The Type I error is to conclude that the percent of EVC students who attended is, A., B., C., D., , at least 20%, when in fact, it is less than 20%., 20%, when in fact, it is 20%., less than 20%, when in fact, it is at least 20%., less than 20%, when in fact, it is less than 20%., , The next two questions refer to the following information:, It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than 7, hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a, mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC, Intermediate Algebra students get less than 7 hours of sleep per night, on average?, Exercise 9.16.40, The distribution to be used for this test is X ∼, , , 1.93, A. N 7.24, √, , (Solution on p. 427.), , 22, , B. N (7.24, 1.93), C. t22, D. t21, Exercise 9.16.41, (Solution on p. 427.), The Type II error is to not reject that the mean number of hours of sleep LTCC students get per, night is at least 7 when, in fact, the mean number of hours, A., B., C., D., , is more than 7 hours., is at most 7 hours., is at least 7 hours., is less than 7 hours., , The next three questions refer to the following information: Previously, an organization reported that, teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently,, the mean is higher. Fifteen (15) randomly chosen teenagers were asked how many hours per week they, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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415, spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a, hypothesis test., Exercise 9.16.42, The null and alternate hypotheses are:, A., B., C., D., , Ho : x = 4.5, Ha : x > 4.5, Ho : µ ≥ 4.5 Ha : µ < 4.5, Ho : µ = 4.75 Ha: µ > 4.75, Ho : µ = 4.5 Ha : µ > 4.5, , Exercise 9.16.43, At a significance level of a = 0.05, what is the correct conclusion?, A., B., C., D., , (Solution on p. 427.), , (Solution on p. 427.), , There is enough evidence to conclude that the mean number of hours is more than 4.75, There is enough evidence to conclude that the mean number of hours is more than 4.5, There is not enough evidence to conclude that the mean number of hours is more than 4.5, There is not enough evidence to conclude that the mean number of hours is more than 4.75, , Exercise 9.16.44, The Type I error is:, , (Solution on p. 427.), , A. To conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher., B. To conclude that the current mean hours per week is higher than 4.5, when in fact, it is the, same., C. To conclude that the mean hours per week currently is 4.5, when in fact, it is higher., D. To conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is, not higher., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 416, , 9.17 Review22, Exercise 9.17.1, (Solution on p. 427.), Rebecca and Matt are 14 year old twins. Matt’s height is 2 standard deviations below the mean, for 14 year old boys’ height. Rebecca’s height is 0.10 standard deviations above the mean for 14, year old girls’ height. Interpret this., A., B., C., D., , Matt is 2.1 inches shorter than Rebecca, Rebecca is very tall compared to other 14 year old girls., Rebecca is taller than Matt., Matt is shorter than the average 14 year old boy., , Exercise 9.17.2, (Solution on p. 427.), Construct a histogram of the IPO data (see Table of Contents, 14. Appendix, Data Sets). Use 5, intervals., The next three exercises refer to the following information: Ninety homeowners were asked the number, of estimates they obtained before having their homes fumigated. X = the number of estimates., x, , Rel. Freq., , 1, , 0.3, , 2, , 0.2, , 4, , 0.4, , 5, , 0.1, , Cumulative Rel. Freq., , Table 9.5, , Complete the cumulative relative frequency column., Exercise 9.17.3, (Solution on p. 427.), Calculate the sample mean (a), the sample standard deviation (b) and the percent of the estimates, that fall at or below 4 (c)., Exercise 9.17.4, (Solution on p. 427.), Calculate the median, M, the first quartile, Q1, the third quartile, Q3. Then construct a boxplot of, the data., Exercise 9.17.5, The middle 50% of the data are between _____ and _____., , (Solution on p. 427.), , The next three questions refer to the following table: Seventy 5th and 6th graders were asked their favorite, dinner., Pizza, , Hamburgers, , Spaghetti, , Fried shrimp, , 5th grader, , 15, , 6, , 9, , 0, , 6th grader, , 15, , 7, , 10, , 8, , Table 9.6, , Exercise 9.17.6, (Solution on p. 427.), Find the probability that one randomly chosen child is in the 6th grade and prefers fried shrimp., 22 This, , content is available online at <http://cnx.org/content/m17013/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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417, A., B., C., D., , 32, 70, 8, 32, 8, 8, 8, 70, , Exercise 9.17.7, Find the probability that a child does not prefer pizza., , (Solution on p. 427.), , 30, A. 70, 30, B. 40, C. 40, 70, D. 1, , Exercise 9.17.8, (Solution on p. 427.), Find the probability a child is in the 5th grade given that the child prefers spaghetti., A., B., C., D., , 9, 19, 9, 70, 9, 30, 19, 70, , Exercise 9.17.9, A sample of convenience is a random sample., , (Solution on p. 427.), , A. true, B. false, Exercise 9.17.10, A statistic is a number that is a property of the population., , (Solution on p. 427.), , A. true, B. false, Exercise 9.17.11, You should always throw out any data that are outliers., , (Solution on p. 427.), , A. true, B. false, Exercise 9.17.12, (Solution on p. 427.), Lee bakes pies for a small restaurant in Felton, CA. She generally bakes 20 pies in a day, on the, average. Of interest is the num.ber of pies she bakes each day, a. Define the Random Variable X., b. State the distribution for X., c. Find the probability that Lee bakes more than 25 pies in any given day., Exercise 9.17.13, (Solution on p. 428.), Six different brands of Italian salad dressing were randomly selected at a supermarket. The grams, of fat per serving are 7, 7, 9, 6, 8, 5. Assume that the underlying distribution is normal. Calculate a, 95% confidence interval for the population mean grams of fat per serving of Italian salad dressing, sold in supermarkets., Exercise 9.17.14, (Solution on p. 428.), Given: uniform, exponential, normal distributions. Match each to a statement below., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 418, a. mean = median 6= mode, b. mean > median > mode, c. mean = median = mode, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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419, , 9.18 Lab: Hypothesis Testing of a Single Mean and Single Proportion23, Class Time:, Names:, , 9.18.1 Student Learning Outcomes:, • The student will select the appropriate distributions to use in each case., • The student will conduct hypothesis tests and interpret the results., , 9.18.2 Television Survey, In a recent survey, it was stated that Americans watch television on average four hours per day. Assume, that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for, students at your school is lower., 1., 2., 3., 4., 5., 6., , Ho :, Ha :, In words, define the random variable. __________ =, The distribution to use for the test is:, Determine the test statistic using your data., Draw a graph and label it appropriately.Shade the actual level of significance., a. Graph:, , 23 This, , content is available online at <http://cnx.org/content/m17007/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 420, , Figure 9.9, , b. Determine the p-value:, 7. Do you or do you not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , 9.18.3 Language Survey, About 42.3% of Californians and 19.6% of all Americans over age 5 speak a language other than English, at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the, students at your school that speak a language other than English at home is different from 42.3%. (Source:, http://www.census.gov/hhes/socdemo/language/ 24 ), 1., 2., 3., 4., 5., 6., , Ho :, Ha :, In words, define the random variable. __________ =, The distribution to use for the test is:, Determine the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., , 24 http://cnx.org/content/m17007/latest/, , http://www.census.gov/hhes/socdemo/language/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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421, a. Graph:, , Figure 9.10, , b. Determine the p-value:, 7. Do you or do you not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , 9.18.4 Jeans Survey, Suppose that young adults own an average of 3 pairs of jeans. Survey 8 people from your class to determine, if the average is higher than 3., 1., 2., 3., 4., 5., 6., , Ho :, Ha :, In words, define the random variable. __________ =, The distribution to use for the test is:, Determine the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., a. Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 422, , Figure 9.11, , b. Determine the p-value:, 7. Do you or do you not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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423, , Solutions to Exercises in Chapter 9, Solutions to Practice 1: Single Mean, Known Population Standard Deviation, Solution to Exercise 9.13.1 (p. 397), Means, Solution to Exercise 9.13.2 (p. 397), a: Ho : µ = 2. 5 (or, Ho : µ ≤ 2.5), b: Ha : µ > 2.5, Solution to Exercise 9.13.3 (p. 397), right-tailed, Solution to Exercise 9.13.4 (p. 397), X, Solution to Exercise 9.13.5 (p. 397), The mean time spent in jail for 26 first time convicted burglars, Solution to Exercise 9.13.6 (p. 397), Yes, 1.5, Solution to Exercise 9.13.7 (p. 397), a., b., c., d., , 3, 1.5, 1.8, 26, , Solution to Exercise 9.13.8 (p. 397), σ, Solution, 9.13.9 (p. 397), to Exercise, , , X~N 2.5, √1.5, 26, Solution to Exercise 9.13.11 (p. 398), 0.0446, Solution to Exercise 9.13.12 (p. 398), a. Reject the null hypothesis, , Solutions to Practice 2: Single Mean, Unknown Population Standard Deviation, Solution to Exercise 9.14.1 (p. 399), averages, Solution to Exercise 9.14.2 (p. 399), a. Ho : µ = 15, b. Ha : µ 6= 15, Solution to Exercise 9.14.3 (p. 399), two-tailed, Solution to Exercise 9.14.4 (p. 399), X, Solution to Exercise 9.14.5 (p. 399), the mean time spent on death row for the 75 inmates, Solution to Exercise 9.14.6 (p. 399), No, Solution to Exercise 9.14.7 (p. 399), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 424, a. 17.4, b. s, c. 75, , Solution to Exercise 9.14.8 (p. 399), t−test, Solution to Exercise 9.14.9 (p. 399), t74, Solution to Exercise 9.14.11 (p. 400), 0.0015, Solution to Exercise 9.14.12 (p. 400), a. Reject the null hypothesis, , Solutions to Practice 3: Single Proportion, Solution to Exercise 9.15.1 (p. 401), Proportions, Solution to Exercise 9.15.2 (p. 401), a. Ho : p = 0.095, b. Ha : p < 0.095, Solution to Exercise 9.15.3 (p. 401), left-tailed, Solution to Exercise 9.15.4 (p. 401), P’, Solution to Exercise 9.15.5 (p. 401), the proportion of people in that town surveyed suffering from depression or a depressive illness, Solution to Exercise 9.15.6 (p. 401), a. 7, b. 100, c. 0.07, Solution to Exercise 9.15.7 (p. 401), 0.0293, Solution to Exercise 9.15.8 (p. 401), Normal, Solution to Exercise 9.15.10 (p. 402), 0.1969, Solution to Exercise 9.15.11 (p. 402), a. Do not reject the null hypothesis, , Solutions to Homework, Solution to Exercise 9.16.1 (p. 403), a. Ho : µ = 34 ; Ha : µ 6= 34, c. Ho : µ ≥ 100, 000 ; Ha : µ < 100, 000, d. Ho : p = 0.29 ; Ha : p 6= 0.29, g. Ho : p = 0.50 ; Ha : p 6= 0.50, i. Ho : p ≥ 0.11 ; Ha : p < 0.11, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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425, Solution to Exercise 9.16.2 (p. 403), a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We do, not conclude that the mean is not 34 years, when it is not really 34 years., c. Type I error: We conclude that the mean is less than $100,000, when it really is at least $100,000. Type II, error: We do not conclude that the mean is less than $100,000, when it is really less than $100,000., d. Type I error: We conclude that the proportion of h.s. seniors who get drunk each month is not 29%,, when it really is 29%. Type II error: We do not conclude that the proportion of h.s. seniors that get, drunk each month is not 29%, when it is really not 29%., i. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error:, We do not conclude that the proportion is less than 11%, when it really is less than 11%., Solution to Exercise 9.16.5 (p. 403), e. z = −2.71, f. 0.0034, h. Decision: Reject null; Conclusion: µ < 19, i. (17.449, 18.757), Solution to Exercise 9.16.7 (p. 404), e. 3.5, f. 0.0005, h. Decision: Reject null; Conclusion: µ > 4.5, i. (4.7553, 5.4447), Solution to Exercise 9.16.9 (p. 404), e. 2.7, f. 0.0042, h. Decision: Reject Null, i. (80.789, 85.211), Solution to Exercise 9.16.11 (p. 404), d. t11, e. 1.96, f. 0.0380, h. Decision: Reject null when a = 0.05 ; do not reject null when a = 0.01, i. (3.8865, 5.9468), Solution to Exercise 9.16.13 (p. 404), e. -1.64, f. 0.1000, h. Decision: Do not reject null, i. (0.3216, 0.4784), Solution to Exercise 9.16.15 (p. 405), d. t9, e. -1.33, f. 0.1086, h. Decision: Do not reject null, i. (51.886, 62.114), Solution to Exercise 9.16.19 (p. 406), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , 426, e. 1.65, f. 0.0984, h. Decision: Do not reject null, i. (0.6836, 0.8533), , Solution to Exercise 9.16.21 (p. 407), e. -2.39, f. 0.0093, h. Decision: Reject null, i. (91.854, 100.15), Solution to Exercise 9.16.23 (p. 408), e. -1.82, f. 0.0345, h. Decision: Do not reject null, i. (0.5331, 0.7685), Solution to Exercise 9.16.25 (p. 410), e. z = −2.99, f. 0.0014, h. Decision: Reject null; Conclusion: p < .70, i. (0.4529, 0.6582), Solution to Exercise 9.16.27 (p. 411), e. 0.57, f. 0.7156, h. Decision: Do not reject null, i. (0.3415, 0.6185), Solution to Exercise 9.16.29 (p. 412), e. 1.32, f. 0.1873, h. Decision: Do not reject null, i. (0.65, 0.90), Solution to Exercise 9.16.31 (p. 412), e. 9.98, f. 0.0000, h. Decision: Reject null, i. (28.8, 30.0), Solution to Exercise 9.16.33 (p. 412), e. -44.7, f. 0.0000, h. Decision: Reject null, i. (0.60, 0.90) - in years, Solution to Exercise 9.16.34 (p. 413), B, Solution to Exercise 9.16.35 (p. 413), D, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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427, Solution to Exercise 9.16.36 (p., C, Solution to Exercise 9.16.37 (p., C, Solution to Exercise 9.16.38 (p., A, Solution to Exercise 9.16.39 (p., C, Solution to Exercise 9.16.40 (p., D, Solution to Exercise 9.16.41 (p., D, Solution to Exercise 9.16.42 (p., D, Solution to Exercise 9.16.43 (p., C, Solution to Exercise 9.16.44 (p., B, , 413), 413), 414), 414), 414), 414), 415), 415), 415), , Solutions to Review, Solution to Exercise 9.17.1 (p. 416), D, Solution to Exercise 9.17.2 (p. 416), No solution provided. There are several ways in which the histogram could be constructed., Solution to Exercise 9.17.3 (p. 416), a. 2.8, b. 1.48, c. 90%, Solution to Exercise 9.17.4 (p. 416), M = 3 ; Q1 = 1 ; Q3 = 4, Solution to Exercise 9.17.5 (p. 416), 1 and 4, Solution to Exercise 9.17.6 (p. 416), D, Solution to Exercise 9.17.7 (p. 417), C, Solution to Exercise 9.17.8 (p. 417), A, Solution to Exercise 9.17.9 (p. 417), B, Solution to Exercise 9.17.10 (p. 417), B, Solution to Exercise 9.17.11 (p. 417), B, Solution to Exercise 9.17.12 (p. 417), b. P (20), c. 0.1122, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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428, , CHAPTER 9. HYPOTHESIS TESTING: SINGLE MEAN AND SINGLE, PROPORTION, , Solution to Exercise 9.17.13 (p. 417), CI: (5.52, 8.48), Solution to Exercise 9.17.14 (p. 417), a. uniform, b. exponential, c. normal, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 10, , Hypothesis Testing: Two Means, Paired, Data, Two Proportions, 10.1 Hypothesis Testing: Two Population Means and Two Population, Proportions1, 10.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , • Classify hypothesis tests by type., • Conduct and interpret hypothesis tests for two population means, population standard deviations, known., • Conduct and interpret hypothesis tests for two population means, population standard deviations, unknown., • Conduct and interpret hypothesis tests for two population proportions., • Conduct and interpret hypothesis tests for matched or paired samples., , 10.1.2 Introduction, Studies often compare two groups. For example, researchers are interested in the effect aspirin has in, preventing heart attacks. Over the last few years, newspapers and magazines have reported about various, aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a, placebo. Then, the heart attack rate is studied over several years., There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income, brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses, really help raise their scores., In the previous chapter, you learned to conduct hypothesis tests on single means and single proportions., You will expand upon that in this chapter. You will compare two means or two proportions to each other., The general procedure is still the same, just expanded., 1 This, , content is available online at <http://cnx.org/content/m17029/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 429
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 430, , To compare two means or two proportions, you work with two groups. The groups are classified either as, independent or matched pairs. Independent groups mean that the two samples taken are independent,, that is, sample values selected from one population are not related in any way to sample values selected, from the other population. Matched pairs consist of two samples that are dependent. The parameter tested, using matched pairs is the population mean. The parameters tested using independent groups are either, population means or population proportions., NOTE : This chapter relies on either a calculator or a computer to calculate the degrees of freedom,, the test statistics, and p-values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. When using the TI-83+/TI-84 calculators, we do not need to separate two population, means, independent groups, population variances unknown into large and small sample sizes., However, most statistical computer software has the ability to differentiate these tests., , This chapter deals with the following hypothesis tests:, Independent groups (samples are independent), • Test of two population means., • Test of two population proportions., Matched or paired samples (samples are dependent), • Becomes a test of one population mean., , 10.2 Comparing Two Independent Population Means with Unknown, Population Standard Deviations2, 1. The two independent samples are simple random samples from two distinct populations., 2. Both populations are normally distributed with the population means and standard deviations unknown unless the sample sizes are greater than 30. In that case, the populations need not be normally, distributed., NOTE : The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom, formula was developed by Aspin-Welch., , The comparison of two population means is very common. A difference between the two samples depends, on both the means and the standard deviations. Very different means can occur by chance if there is great, variation among the individual samples. In order to account for the variation, we take the difference of, the sample means, X1 - X2 , and divide by the standard error (shown below) in order to standardize the, difference. The result is a t-score test statistic (shown below)., Because we do not know the population standard deviations, we estimate them using the two sample, standard deviations from our independent samples. For the hypothesis test, we calculate the estimated, standard deviation, or standard error, of the difference in sample means, X1 - X2 ., The standard error is:, , s, , ( S1 ) 2 ( S2 ) 2, +, n1, n2, , The test statistic (t-score) is calculated as follows:, 2 This, , content is available online at <http://cnx.org/content/m17025/1.18/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (10.0)
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431, t-score, , ( x1 − x2 ) − ( µ1 − µ2 ), r, ( S1 ) 2, n1, , +, , (10.0), , ( S2 ) 2, n2, , where:, , • s1 and s2 , the sample standard deviations, are estimates of σ1 and σ2 , respectively., • σ1 and σ2 are the unknown population standard deviations., • x1 and x2 are the sample means. µ1 and µ2 are the population means., The degrees of freedom (df) is a somewhat complicated calculation. However, a computer or calculator calculates it easily. The dfs are not always a whole number. The test statistic calculated above is approximated, by the student’s-t distribution with dfs as follows:, Degrees of freedom, , df =, 1, n1 −1, , , , ·, , ( s1 )2, n1, , ( s1 )2, n1, , 2, , +, , ( s2 )2, n2, , +, , 1, n2 −1, , 2, , , , ·, , ( s2 )2, n2, , (10.0), , 2, , When both sample sizes n1 and n2 are five or larger, the student’s-t approximation is very good. Notice that, the sample variances s1 2 and s2 2 are not pooled. (If the question comes up, do not pool the variances.), NOTE :, , It is not necessary to compute this by hand. A calculator or computer easily computes it., , Example 10.1: Independent groups, The average amount of time boys and girls ages 7 through 11 spend playing sports each day is, believed to be the same. An experiment is done, data is collected, resulting in the table below., Both populations have a normal distribution., , Sample Size, , Average Number of, Hours Playing Sports, Per Day, , Girls, , 9, , 2 hours, , Boys, , 16, , 3.2 hours, , Sample Standard, Deviation, , √, , 0.75, , 1.00, , Table 10.1, , Problem, Is there a difference in the mean amount of time boys and girls ages 7 through 11 play sports each, day? Test at the 5% level of significance., Solution, The population standard deviations are not known. Let g be the subscript for girls and b be the, subscript for boys. Then, µ g is the population mean for girls and µb is the population mean for, boys. This is a test of two independent groups, two population means., Random variable: Xg − Xb = difference in the sample mean amount of time girls and boys play, sports each day., Ho : µ g = µb, , µ g − µb = 0, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 432, Ha : µ g 6 = µ b, , µ g − µb 6= 0, , The words "the same" tell you Ho has an "=". Since there are no other words to indicate Ha , then, assume "is different." This is a two-tailed test., Distribution for the test: Use td f where d f is calculated using the d f formula for independent, groups, two population means. Using a calculator, d f is approximately 18.8462. Do not pool the, variances., Calculate the p-value using a student’s-t distribution: p-value = 0.0054, Graph:, , Figure 10.1, , sg =, , √, , 0.75, , sb = 1, So, x g − xb = 2 − 3.2 = −1.2, Half the p-value is below -1.2 and half is above 1.2., Make a decision: Since α > p-value, reject Ho ., This means you reject µ g = µb . The means are different., Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to, conclude that the mean number of hours that girls and boys aged 7 through 11 play sports per day, is different (mean number of hours boys aged 7 through 11 play sports per day is greater than the, mean number of hours played by girls OR the mean number of hours girls aged 7 through 11 play, sports per day is greater than the mean number of hours played by boys)., NOTE : TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over, √, 0.75 for Sx1, 9, to Stats and press ENTER. Arrow down and enter 2 for the first sample mean,, for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to µ1: and arrow, to does not equal µ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to, Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the, test statistic is -3.14. Do the procedure again but instead of Calculate do Draw., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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433, , Example 10.2, A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math, classes with a standard deviation of 1.5 math classes. College B samples 9 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes, that a student who graduates from college A has taken more math classes, on the average. Both, populations have a normal distribution. Test at a 1% significance level. Answer the following, questions., Problem 1, Is this a test of two means or two proportions?, , (Solution on p. 466.), , Problem 2, Are the populations standard deviations known or unknown?, , (Solution on p. 466.), , Problem 3, Which distribution do you use to perform the test?, , (Solution on p. 466.), , Problem 4, What is the random variable?, , (Solution on p. 466.), , Problem 5, What are the null and alternate hypothesis?, , (Solution on p. 466.), , Problem 6, Is this test right, left, or two tailed?, , (Solution on p. 466.), , Problem 7, What is the p-value?, , (Solution on p. 466.), , Problem 8, Do you reject or not reject the null hypothesis?, , (Solution on p. 466.), , Conclusion:, At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude, that a student who graduates from college A has taken more math classes, on the average, than a, student who graduates from college B., , 10.3 Comparing Two Independent Population Means with Known Population Standard Deviations3, Even though this situation is not likely (knowing the population standard deviations is not likely), the, following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations, must be normal. The random variable is X1 − X2 . The normal distribution has the following format:, Normal distribution, , , X1 − X2 ∼ N u 1 − u 2 ,, , 3 This, , s, , , (σ1 )2 (σ2 )2 , +, n1, n2, , content is available online at <http://cnx.org/content/m17042/1.10/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (10.1)
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 434, The standard deviation is:, , s, , The test statistic (z-score) is:, z=, , (σ1 )2 (σ2 )2, +, n1, n2, , (10.1), , ( x1 − x2 ) − ( µ1 − µ2 ), r, (σ1 )2, n1, , +, , (10.1), , (σ2 )2, n2, , Example 10.3, independent groups, population standard deviations known: The mean lasting time of 2 competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both, populations have a normal distribution. The following table is the result., Wax, , Sample Mean Number of Months Floor Wax Last, , Population Standard Deviation, , 1, , 3, , 0.33, , 2, , 2.9, , 0.36, Table 10.2, , Problem, Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance., Solution, This is a test of two independent groups, two population means, population standard deviations, known., Random Variable: X1 − X2 = difference in the mean number of months the competing floor waxes, last., Ho : µ1 ≤ µ2, Ha : µ 1 > µ 2, The words "is more effective" says that wax 1 lasts longer than wax 2, on the average. "Longer", is a ” > ” symbol and goes into Ha . Therefore, this is a right-tailed test., Distribution for the test: The population standard deviations are known so the distribution is, normal. Using the formula above, the distribution is:, q, , 2, 0.362, X1 − X2 ∼ N 0, 0.33, +, 20, 20, Since µ1 ≤ µ2 then µ1 − µ2 ≤ 0 and the mean for the normal distribution is 0., Calculate the p-value using the normal distribution: p-value = 0.1799, Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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435, , Figure 10.2, , x1 − x2 = 3 − 2.9 = 0.1, Compare α and the p-value: α = 0.05 and p-value = 0.1799. Therefore, α < p-value., Make a decision: Since α < p-value, do not reject Ho ., Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence, to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time, wax 2 lasts., NOTE : TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over, to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first, sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to µ1: and, arrow to > µ2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1799, and the test statistic is 0.9157. Do the procedure again but instead of Calculate do Draw., , 10.4 Comparing Two Independent Population Proportions4, 1. The two independent samples are simple random samples that are independent., 2. The number of successes is at least five and the number of failures is at least five for each of the, samples., Comparing two proportions, like comparing two means, is common. If two estimated proportions are, different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can, help determine if a difference in the estimated proportions ( PA − PB ) reflects a difference in the population, proportions., The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, Ho : p A = p B . To conduct the test, we use a pooled, proportion, pc ., 4 This, , content is available online at <http://cnx.org/content/m17043/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 436, , The pooled proportion is calculated as follows:, pc =, , x A + xB, n A + nB, , (10.2), , The distribution for the differences is:, " s, P’ A − P’B ∼ N 0,, , The test statistic (z-score) is:, , , p c · (1 − p c ) ·, , 1, 1, +, nA, nB, , #, (10.2), , ( p’ − p’B ) − ( p A − p B ), z= r A, , , pc · (1 − pc ) · n1 + n1B, , (10.2), , A, , Example 10.4: Two population proportions, Two types of medication for hives are being tested to determine if there is a difference in the, proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random, sample of 200 adults given medication B still had hives 30 minutes after taking the medication., Test at a 1% level of significance., , 10.4.1 Determining the solution, This is a test of 2 population proportions., Problem, How do you know?, , (Solution on p. 466.), , Let A and B be the subscripts for medication A and medication B. Then p A and p B are the desired, population proportions., Random Variable:, P’ A − P’B = difference in the proportions of adult patients who did not react after 30 minutes to, medication A and medication B., Ho : p A = p B, , p A − pB = 0, , Ha : p A 6 = p B, , p A − p B 6= 0, , The words "is a difference" tell you the test is two-tailed., Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:, pc =, , x A +xB, n A +n B, , Therefore,, , =, , 20+12, 200+200, , = 0.08, , 1 − pc = 0.92, , r, , 1, P’ A − P’B ∼ N 0, (0.08) · (0.92) · 200, +, , 1, 200, , , , P’ A − P’B follows an approximate normal distribution., Calculate the p-value using the normal distribution: p-value = 0.1404., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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437, Estimated proportion for group A:, , p’ A =, , Estimated proportion for group B:, , p’B =, , xA, nA, xB, nB, , =, =, , 20, 200, 12, 200, , = 0.1, = 0.06, , Graph:, , Figure 10.3, , P’ A − P’B = 0.1 − 0.06 = 0.04., Half the p-value is below -0.04 and half is above 0.04., Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value., Make a decision: Since α < p-value, do not reject Ho ., Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to, conclude that there is a difference in the proportions of adult patients who did not react after 30, minutes to medication A and medication B., NOTE : TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down, and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not, equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404, and the test statistic is 1.47. Do the procedure again but instead of Calculate do Draw., , 10.5 Matched or Paired Samples5, 1., 2., 3., 4., 5., , Simple random sampling is used., Sample sizes are often small., Two measurements (samples) are drawn from the same pair of individuals or objects., Differences are calculated from the matched or paired samples., The differences form the sample that is used for the hypothesis test., , 5 This, , content is available online at <http://cnx.org/content/m17033/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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438, , CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 6. The matched pairs have differences that either come from a population that is normal or the number of, differences is sufficiently large so the distribution of the sample mean of differences is approximately, normal., In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, µd , is then tested using, a Student-t test for a single population mean with n − 1 degrees of freedom where n is the number of, differences., The test statistic (t-score) is:, , x −µ, t = d d, , (10.3), , sd, √, n, , Example 10.5: Matched or paired samples, A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results, for randomly selected subjects are shown in the table. The "before" value is matched to an "after", value and the differences are calculated. The differences have a normal distribution., Subject:, , A, , B, , C, , D, , E, , F, , G, , H, , Before, , 6.6, , 6.5, , 9.0, , 10.3, , 11.3, , 8.1, , 6.3, , 11.6, , After, , 6.8, , 2.4, , 7.4, , 8.5, , 8.1, , 6.1, , 3.4, , 2.0, , Table 10.3, , Problem, Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level., Solution, Corresponding "before" and "after" values form matched pairs. (Calculate "sfter" - "before")., After Data, , Before Data, , Difference, , 6.8, , 6.6, , 0.2, , 2.4, , 6.5, , -4.1, , 7.4, , 9, , -1.6, , 8.5, , 10.3, , -1.8, , 8.1, , 11.3, , -3.2, , 6.1, , 8.1, , -2, , 3.4, , 6.3, , -2.9, , 2, , 11.6, , -9.6, , Table 10.4, , The data for the test are the differences: {0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}, The sample mean and sample standard deviation of the differences are:, sd = 2.91 Verify these values., , xd = −3.13 and, , Let µd be the population mean for the differences. We use the subscript d to denote "differences.", , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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439, Random Variable: Xd = the mean difference of the sensory measurements, Ho : µd ≥ 0, , (10.3), , There is no improvement. (µd is the population mean of the differences.), Ha : µ d < 0, , (10.3), , There is improvement. The score should be lower after hypnotism so the difference ought to be, negative to indicate improvement., Distribution for the test: The distribution is a student-t with d f = n − 1 = 8 − 1 = 7. Use t7 ., (Notice that the test is for a single population mean.), Calculate the p-value using the Student-t distribution: p-value = 0.0095, Graph:, , Figure 10.4, , X d is the random variable for the differences., The sample mean and sample standard deviation of the differences are:, x d = −3.13, sd = 2.91, Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value., Make a decision: Since α > p-value, reject Ho ., This means that µd < 0 and there is improvement., Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears, to be effective in reducing pain., NOTE : For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time, (after - before) and put the differences into a list or you can put the after data into a first list and, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 440, , the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list, name - 2nd list name. The calculator will do the subtraction and you will have the differences in, the third list., NOTE : TI-83+ and TI-84: Use your list of differences as the data. Press STAT and arrow over to, TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for µ0 , the, name of the list where you put the data, and 1 for Freq:. Arrow down to µ: and arrow over to <, µ0 . Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094 and the test, statistic is -3.04. Do these instructions again except arrow to Draw (instead of Calculate). Press, ENTER., , Example 10.6, A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked 4 of his, players to participate in a study. The amount of weight they could each lift was recorded before, they took the strength development class. After completing the class, the amount of weight they, could each lift was again measured. The data are as follows:, Weight (in pounds), , Player 1, , Player 2, , Player 3, , Player 4, , Amount of weighted lifted prior to the class, , 205, , 241, , 338, , 368, , Amount of weight lifted after the class, , 295, , 252, , 330, , 360, , Table 10.5, , The coach wants to know if the strength development class makes his players stronger, on, average., Problem, (Solution on p. 466.), Record the differences data. Calculate the differences by subtracting the amount of weight lifted, prior to the class from the weight lifted after completing the class. The data for the differences are:, {90, 11, -8, -8}. The differences have a normal distribution., Using the differences data, calculate the sample mean and the sample standard deviation., x d = 21.3, , sd = 46.7, , Using the difference data, this becomes a test of a single __________ (fill in the blank)., Define the random variable: X d = mean difference in the maximum lift per player., The distribution for the hypothesis test is t3 ., Ho : µd ≤ 0, , Ha : µ d > 0, , Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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441, , Figure 10.5, , Calculate the p-value: The p-value is 0.2150, Decision: If the level of significance is 5%, the decision is to not reject the null hypothesis because, α < p-value., What is the conclusion?, Example 10.7, Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put, with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they, could push equal distances with either hand. The following data was collected., Distance, (in feet), using, , Student 1, , Student 2, , Student 3, , Student 4, , Student 5, , Student 6, , Student 7, , Dominant, Hand, , 30, , 26, , 34, , 17, , 19, , 26, , 20, , Weaker, Hand, , 28, , 14, , 27, , 18, , 17, , 26, , 16, , Table 10.6, , Problem, (Solution on p. 466.), Conduct a hypothesis test to determine whether the mean difference in distances between the, children’s dominant versus weaker hands is significant., H INT: use a t-test on the difference data. Assume the differences have a normal distribution. The, random variable is the mean difference., C HECK : The test statistic is 2.18 and the p-value is 0.0716., What is your conclusion?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 442, , 10.6 Summary of Types of Hypothesis Tests6, Two Population Means, , • Populations are independent and population standard deviations are unknown., • Populations are independent and population standard deviations are known (not likely)., Matched or Paired Samples, , • Two samples are drawn from the same set of objects., • Samples are dependent., Two Population Proportions, , • Populations are independent., , 6 This, , content is available online at <http://cnx.org/content/m17044/1.5/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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443, , 10.7 Practice 1: Hypothesis Testing for Two Proportions7, 10.7.1 Student Learning Outcomes, • The student will conduct a hypothesis test of two proportions., , 10.7.2 Given, In the recent Census, 3 percent of the U.S. population reported being two or more, races., However, the percent varies tremendously from state to state., (Source:, http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf ) Suppose that two random surveys, are conducted. In the first random survey, out of 1000 North Dakotans, only 9 people reported being of, two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two, or more races. Conduct a hypothesis test to determine if the population percents are the same for the two, states or if the percent for Nevada is statistically higher than for North Dakota., , 10.7.3 Hypothesis Testing: Two Proportions, Exercise 10.7.1, Is this a test of means or proportions?, , (Solution on p. 466.), , Exercise 10.7.2, State the null and alternative hypotheses., , (Solution on p. 466.), , a. H0 :, b. Ha :, Exercise 10.7.3, Is this a right-tailed, left-tailed, or two-tailed test? How do you know?, , (Solution on p. 466.), , Exercise 10.7.4, What is the Random Variable of interest for this test?, Exercise 10.7.5, In words, define the Random Variable for this test., Exercise 10.7.6, (Solution on p. 466.), Which distribution (Normal or student’s-t) would you use for this hypothesis test?, Exercise 10.7.7, Explain why you chose the distribution you did for the above question., Exercise 10.7.8, Calculate the test statistic., , (Solution on p. 466.), , Exercise 10.7.9, Sketch a graph of the situation. Mark the hypothesized difference and the sample difference., Shade the area corresponding to the p−value., 7 This, , content is available online at <http://cnx.org/content/m17027/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 444, , Figure 10.6, , Exercise 10.7.10, Find the p−value:, , (Solution on p. 466.), , Exercise 10.7.11, At a pre-conceived α = 0.05, what is your:, , (Solution on p. 466.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , 10.7.4 Discussion Question, Exercise 10.7.12, Does it appear that the proportion of Nevadans who are two or more races is higher than the, proportion of North Dakotans? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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445, , 10.8 Practice 2: Hypothesis Testing for Two Averages8, 10.8.1 Student Learning Outcome, • The student will conduct a hypothesis test of two means., , 10.8.2 Given, The U.S. Center for Disease Control reports that the mean life expectancy for whites born in 1900 was, 47.6 years and for nonwhites it was 33.0 years. (http://www.cdc.gov/nchs/data/dvs/nvsr53_06t12.pdf ), Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124, whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the, mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if, the mean life spans in the county were the same for whites and nonwhites., , 10.8.3 Hypothesis Testing: Two Means, Exercise 10.8.1, Is this a test of means or proportions?, , (Solution on p. 467.), , Exercise 10.8.2, State the null and alternative hypotheses., , (Solution on p. 467.), , a. H0 :, b. Ha :, Exercise 10.8.3, Is this a right-tailed, left-tailed, or two-tailed test? How do you know?, , (Solution on p. 467.), , Exercise 10.8.4, What is the Random Variable of interest for this test?, , (Solution on p. 467.), , Exercise 10.8.5, In words, define the Random Variable of interest for this test., , (Solution on p. 467.), , Exercise 10.8.6, Which distribution (Normal or student’s-t) would you use for this hypothesis test?, Exercise 10.8.7, Explain why you chose the distribution you did for the above question., Exercise 10.8.8, Calculate the test statistic., , (Solution on p. 467.), , Exercise 10.8.9, Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and, the sample difference. Shade the area corresponding to the p−value., 8 This, , content is available online at <http://cnx.org/content/m17039/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 446, , Figure 10.7, , Exercise 10.8.10, Find the p−value:, , (Solution on p. 467.), , Exercise 10.8.11, At a pre-conceived α = 0.05, what is your:, , (Solution on p. 467.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , 10.8.4 Discussion Question, Exercise 10.8.12, Does it appear that the means are the same? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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447, , 10.9 Homework9, For questions Exercise 10.9.1 - Exercise 10.9.10, indicate which of the following choices best identifies the, hypothesis test., A. Independent group means, population standard deviations and/or variances known, B. Independent group means, population standard deviations and/or variances unknown, C. Matched or paired samples, D. Single mean, E. 2 proportions, F. Single proportion, Exercise 10.9.1, (Solution on p. 467.), A powder diet is tested on 49 people and a liquid diet is tested on 36 different people. The population standard deviations are 2 pounds and 3 pounds, respectively. Of interest is whether the, liquid diet yields a higher mean weight loss than the powder diet., Exercise 10.9.2, A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children, that like the new chocolate bar is greater than the proportion of adults that like it., Exercise 10.9.3, (Solution on p. 467.), The mean number of English courses taken in a two–year time period by male and female college, students is believed to be about the same. An experiment is conducted and data are collected from, 9 males and 16 females., Exercise 10.9.4, A football league reported that the mean number of touchdowns per game was 5. A study is done, to determine if the mean number of touchdowns has decreased., Exercise 10.9.5, (Solution on p. 467.), A study is done to determine if students in the California state university system take longer to, graduate than students enrolled in private universities. 100 students from both the California state, university system and private universities are surveyed. From years of research, it is known that, the population standard deviations are 1.5811 years and 1 year, respectively., Exercise 10.9.6, According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A, study is done to verify this., Exercise 10.9.7, (Solution on p. 467.), According to a recent study, U.S. companies have an mean maternity-leave of six weeks., Exercise 10.9.8, A recent drug survey showed an increase in use of drugs and alcohol among local high school, students as compared to the national percent. Suppose that a survey of 100 local youths and 100, national youths is conducted to see if the proportion of drug and alcohol use is higher locally than, nationally., Exercise 10.9.9, (Solution on p. 467.), A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are, recorded. Of interest is the mean increase in SAT scores., Exercise 10.9.10, University of Michigan researchers reported in the Journal of the National Cancer Institute that, quitting smoking is especially beneficial for those under age 49. In this American Cancer Society, 9 This, , content is available online at <http://cnx.org/content/m17023/1.21/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 448, , study, the risk (probability) of dying of lung cancer was about the same as for those who had never, smoked., , 10.9.1, D IRECTIONS : For each of the word problems, use a solution sheet to do the hypothesis test. The, solution sheet is found in 14. Appendix (online book version: the link is "Solution Sheets"; PDF, book version: look under 14.5 Solution Sheets). Please feel free to make copies of the solution, sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files., NOTE : If you are using a student’s-t distribution for a homework problem below, including for, paired data, you may assume that the underlying population is normally distributed. (In general,, you must first prove that assumption, though.), , Exercise 10.9.11, (Solution on p. 467.), A powder diet is tested on 49 people and a liquid diet is tested on 36 different people. Of interest, is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet, group had an mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid, diet group had an mean weight loss of 45 pounds with a standard deviation of 14 pounds., Exercise 10.9.12, The mean number of English courses taken in a two–year time period by male and female college, students is believed to be about the same. An experiment is conducted and data are collected from, 29 males and 16 females. The males took an average of 3 English courses with a standard deviation, of 0.8. The females took an average of 4 English courses with a standard deviation of 1.0. Are the, means statistically the same?, Exercise 10.9.13, (Solution on p. 467.), A study is done to determine if students in the California state university system take longer, to graduate, on average, than students enrolled in private universities. 100 students from both, the California state university system and private universities are surveyed. Suppose that from, years of research, it is known that the population standard deviations are 1.5811 years and 1 year,, respectively. The following data are collected. The California state university system students, took on average 4.5 years with a standard deviation of 0.8. The private university students took, on average 4.1 years with a standard deviation of 0.3., Exercise 10.9.14, A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are, recorded. Of interest is the mean increase in SAT scores. The following data are collected:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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449, , Pre-course score, , Post-course score, , 1200, , 1300, , 960, , 920, , 1010, , 1100, , 840, , 880, , 1100, , 1070, , 1250, , 1320, , 860, , 860, , 1330, , 1370, , 790, , 770, , 990, , 1040, , 1110, , 1200, , 740, , 850, Table 10.7, , Exercise 10.9.15, (Solution on p. 467.), A recent drug survey showed an increase in use of drugs and alcohol among local high school, seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100, national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than, nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60, national seniors reported using them., Exercise 10.9.16, A student at a four-year college claims that mean enrollment at four–year colleges is higher than, at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges, surveyed, the mean enrollment was 5068 with a standard deviation of 4777. Of the 35 four-year, colleges surveyed, the mean enrollment was 5466 with a standard deviation of 8191. (Source:, Microsoft Bookshelf ), Exercise 10.9.17, (Solution on p. 467.), A study was conducted by the U.S. Army to see if applying antiperspirant to soldiers’ feet for a, few days before a major hike would help cut down on the number of blisters soldiers had on their, feet. In the experiment, for three nights before they went on a 13-mile hike, a group of 328 West, Point cadets put an alcohol-based antiperspirant on their feet. A “control group” of 339 soldiers, put on a similar, but inactive, preparation on their feet. On the day of the hike, the temperature, reached 83 ◦ F. At the end of the hike, 21% of the soldiers who had used the antiperspirant and 48%, of the control group had developed foot blisters. Conduct a hypothesis test to see if the proportion, of soldiers using the antiperspirant was significantly lower than the control group. (Source: U.S., Army study reported in Journal of the American Academy of Dermatologists), Exercise 10.9.18, We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the, same for the white and the black races in the United States. We randomly pick one year, 1992,, to compare the races. The number of suicides estimated in the United States in 1992 for white, females is 4930. 580 were aged 15 to 24. The estimate for black females is 330. 40 were aged 15 to, 24. We will let female suicide victims be our population. (Source: the National Center for Health, Statistics, U.S. Dept. of Health and Human Services), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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450, , CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, Exercise 10.9.19, (Solution on p. 468.), At Rachel’s 11th birthday party, 8 girls were timed to see how long (in seconds) they could hold, their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping., The girls thought that the mean difference between their jumping and relaxed times would be 0., Test their hypothesis., Relaxed time (seconds), , Jumping time (seconds), , 26, , 21, , 47, , 40, , 30, , 28, , 22, , 21, , 23, , 25, , 45, , 43, , 37, , 35, , 29, , 32, Table 10.8, , Exercise 10.9.20, Elizabeth Mjelde,, was interested in whether the value from the Golden, an art history professor,, , larger +smaller dimension, Ratio formula,, was the same in the Whitney Exhibit for works from 1900, larger dimension, – 1919 as for works from 1920 – 1942. 37 early works were sampled. They averaged 1.74 with, a standard deviation of 0.11. 65 of the later works were sampled. They averaged 1.746 with a, standard deviation of 0.1064. Do you think that there is a significant difference in the Golden, Ratio calculation? (Source: data from Whitney Exhibit on loan to San Jose Museum of Art ), Exercise 10.9.21, (Solution on p. 468.), One of the questions in a study of marital satisfaction of dual–career couples was to rate the statement, “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went, from 1 (strongly agree) to 5 (strongly disagree). Below are ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus, the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife)., Wife’s score, , 2, , 2, , 3, , 3, , 4, , 2, , 1, , 1, , 2, , 4, , Husband’s score, , 2, , 2, , 1, , 3, , 2, , 1, , 1, , 1, , 2, , 4, , Table 10.9, , Exercise 10.9.22, Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. Evaluate the data, below. Do you think that their cholesterol levels were significantly lowered?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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451, , Starting cholesterol level, , Ending cholesterol level, , 140, , 140, , 220, , 230, , 110, , 120, , 240, , 220, , 200, , 190, , 180, , 150, , 190, , 200, , 360, , 300, , 280, , 300, , 260, , 240, Table 10.10, , Exercise 10.9.23, (Solution on p. 468.), Mean entry level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. (Source: http://, www.graduatingengineer.com 10 ). A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their, mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3450 and, $4210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry level, mechanical engineering salary is lower than the mean entry level electrical engineering salary., Exercise 10.9.24, A recent year was randomly picked from 1985 to the present. In that year, there were 2051 Hispanic, students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were, 321 Hispanic students out of a total of 2441 students. In general, do you think that the percent, of Hispanic students at the two colleges is basically the same or different? (Source: Chancellor’s, Office, California Community Colleges, November 1994 ), Exercise 10.9.25, (Solution on p. 468.), Eight runners were convinced that the mean difference in their individual times for running one, mile versus race walking one mile was at most 2 minutes. Below are their times. Do you agree, that the mean difference is at most 2 minutes?, 10 http://www.graduatingengineer.com/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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452, , CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, Running time (minutes), , Race walking time (minutes), , 5.1, , 7.3, , 5.6, , 9.2, , 6.2, , 10.4, , 4.8, , 6.9, , 7.1, , 8.9, , 4.2, , 9.5, , 6.1, , 9.4, , 4.4, , 7.9, Table 10.11, , Exercise 10.9.26, Marketing companies have collected data implying that teenage girls use more ring tones on their, cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls, and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2, with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8., Conduct a hypothesis test to determine if the means are approximately the same or if the girls’, mean is higher than the boys’ mean., Exercise 10.9.27, (Solution on p. 468.), While her husband spent 2½ hours picking out new speakers, a statistician decided to determine, whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. 8 of the 24 women surveyed, claimed to enjoy the activity. Interpret the results of the survey., Exercise 10.9.28, We are interested in whether children’s educational computer software costs less, on average,, than children’s entertainment software. 36 educational software titles were randomly picked from, a catalog. The mean cost was $31.14 with a standard deviation of $4.69. 35 entertainment software, titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard, deviation of $10.87. Decide whether children’s educational software costs less, on average, than, children’s entertainment software. (Source: Educational Resources, December catalog), Exercise 10.9.29, (Solution on p. 468.), Parents of teenage boys often complain that auto insurance costs more, on average, for teenage, boys than for teenage girls. A group of concerned parents examines a random sample of insurance, bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From, past years, it is known that the population standard deviation for each group is $180. Determine, whether or not you believe that the mean cost for auto insurance for teenage boys is greater than, that for teenage girls., Exercise 10.9.30, A group of transfer bound students wondered if they will spend the same mean amount on texts, and supplies each year at their four-year university as they have at their community college. They, conducted a random survey of 54 students at their community college and 66 students at their, local four-year university. The sample means were $947 and $1011, respectively. The population, standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to, determine if the means are statistically the same., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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453, Exercise 10.9.31, (Solution on p. 468.), Joan Nguyen recently claimed that the proportion of college–age males with at least one pierced, ear is as high as the proportion of college–age females. She conducted a survey in her classes. Out, of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do, you believe that the proportion of males has reached the proportion of females?, Exercise 10.9.32, Some manufacturers claim that non-hybrid sedan cars have a lower mean miles per gallon (mpg), than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a, standard deviation of 7 mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard, deviation of 4 mpg. Suppose that the population standard deviations are known to be 6 and 3,, respectively. Conduct a hypothesis test to the manufacturers claim., Questions Exercise 10.9.33 – Exercise 10.9.37 refer to the Terri Vogel’s data set (see Table of Contents)., Exercise 10.9.33, (Solution on p. 468.), Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices., Exercise 10.9.34, Repeat the test in Exercise 10.9.33, but use Lap 5 data this time., Exercise 10.9.35, (Solution on p. 469.), Repeat the test in Exercise 10.9.33, but this time combine the data from Laps 1 and 5., Exercise 10.9.36, In 2 – 3 complete sentences, explain in detail how you might use Terri Vogel’s data to answer the, following question. “Does Terri Vogel drive faster in races than she does in practices?”, Exercise 10.9.37, (Solution on p. 469.), Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of, practice laps she completes slower than 135 seconds?, Exercise 10.9.38, "To Breakfast or Not to Breakfast?" by Richard Ayore, In the American society, birthdays are one of those days that everyone looks forward to. People of, different ages and peer groups gather to mark the 18th, 20th, . . . birthdays. During this time, one, looks back to see what he or she had achieved for the past year, and also focuses ahead for more, to come., If, by any chance, I am invited to one of these parties, my experience is always different. Instead, of dancing around with my friends while the music is booming, I get carried away by memories, of my family back home in Kenya. I remember the good times I had with my brothers and sister, while we did our daily routine., Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one, day arguing with my brother as to why he always remained behind just to join us an hour later. In, his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This, is why I always work more hours than you guys!”, And so, to prove his wrong or right, we decided to give it a try. One day we went to work as usual, without breakfast, and recorded the time we could work before getting tired and stopping. On, the next day, we all ate breakfast before going to work. We recorded how long we worked again, before getting tired and stopping. Of interest was our mean increase in work time. Though not, sure, my brother insisted that it is more than two hours. Using the data below, solve our problem., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 454, , Work hours with breakfast, , Work hours without breakfast, , 8, , 6, , 7, , 5, , 9, , 5, , 5, , 4, , 9, , 7, , 8, , 7, , 10, , 7, , 7, , 5, , 6, , 6, , 9, , 5, Table 10.12, , 10.9.2 Try these multiple choice questions., For questions Exercise 10.9.39 – Exercise 10.9.40, use the following information., A new AIDS prevention drugs was tried on a group of 224 HIV positive patients. Forty-five (45) patients, developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after, four years. We want to test whether the method of treatment reduces the proportion of patients that develop, AIDS after four years or if the proportions of the treated group and the untreated group stay the same., Let the subscript t= treated patient and ut= untreated patient., Exercise 10.9.39, The appropriate hypotheses are:, A., B., C., D., , (Solution on p. 469.), , Ho : pt < put and Ha : pt ≥ put, Ho : pt ≤ put and Ha : pt > put, Ho : pt = put and Ha : pt 6= put, Ho : pt = put and Ha : pt < put, , Exercise 10.9.40, If the p -value is 0.0062 what is the conclusion (use α = 0.05 )?, , (Solution on p. 469.), , A. The method has no effect., B. There is sufficient evidence to conclude that the method reduces the proportion of HIV positive, patients that develop AIDS after four years., C. There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients that develop AIDS after four years., D. There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients that develop AIDS after four years., Exercise 10.9.41, (Solution on p. 469.), Lesley E. Tan investigated the relationship between left-handedness and right-handedness and, motor competence in preschool children. Random samples of 41 left-handers and 41 right-handers, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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455, were given several tests of motor skills to determine if there is evidence of a difference between the, children based on this experiment. The experiment produced the means and standard deviations, shown below. Determine the appropriate test and best distribution to use for that test., Left-handed, , Right-handed, , Sample size, , 41, , 41, , Sample mean, , 97.5, , 98.1, , Sample standard deviation, , 17.5, , 19.2, , Table 10.13, , A., B., C., D., , Two independent means, normal distribution, Two independent means, student’s-t distribution, Matched or paired samples, student’s-t distribution, Two population proportions, normal distribution, , For questions Exercise 10.9.42 – Exercise 10.9.43, use the following information., An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a, “biofeedback exercise program.” Six (6) subjects were randomly selected and the blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated, (after − before) producing the following results: x d = −10.2 sd = 8.4. Using the data, test the hypothesis, that the blood pressure has decreased after the training,, Exercise 10.9.42, The distribution for the test is, , (Solution on p. 469.), , A. t5, B. t6, C. N (−, 10.2, 8.4), 8.4, D. N −10.2, √, 6, , Exercise 10.9.43, If α = 0.05, the p-value and the conclusion are, , (Solution on p. 469.), , A. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the, training, B. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training, C. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the, training, D. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the, training, For questions Exercise 10.9.44– Exercise 10.9.45, use the following information., The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new, players to develop their skills. Data for a randomly picked date showed the following annual goals., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 456, , Western, , Eastern, , Los Angeles 9, , D.C. United 9, , FC Dallas 3, , Chicago 8, , Chivas USA 4, , Columbus 7, , Real Salt Lake 3, , New England 6, , Colorado 4, , MetroStars 5, , San Jose 4, , Kansas City 3, Table 10.14, , Conduct a hypothesis test to determine if the Western Reserve Division teams score, on average, fewer goals, than the Eastern Reserve Division teams. Subscripts: 1 Western Reserve Division (W); 2 Eastern Reserve, Division (E), Exercise 10.9.44, The exact distribution for the hypothesis test is:, A., B., C., D., , (Solution on p. 469.), , The normal distribution., The student’s-t distribution., The uniform distribution., The exponential distribution., , Exercise 10.9.45, If the level of significance is 0.05, the conclusion is:, , (Solution on p. 469.), , A. There is sufficient evidence to conclude that the W Division teams score, on average, fewer, goals than the E teams., B. There is insufficient evidence to conclude that the W Division teams score, on average, more, goals than the E teams., C. There is insufficient evidence to conclude that the W teams score, on average, fewer goals than, the E teams score., D. Unable to determine., Questions Exercise 10.9.46 – Exercise 10.9.48 refer to the following., Neuroinvasive West Nile virus refers to a severe disease that affects a person’s nervous system . It, is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported, cases of neuroinvasive West Nile virus out of a total of 1021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of, neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus, cases? Using a 1% level of significance, conduct an appropriate hypothesis test. (Source: http://, http://www.cdc.gov/ncidod/dvbid/westnile/index.htm 11 ), , • “2011” subscript: 2011 group., • “2010” subscript: 2010 group, Exercise 10.9.46, This is:, , (Solution on p. 469.), , A. a test of two proportions, 11 http://cnx.org/content/m17023/latest/, , http://www.cdc.gov/ncidod/dvbid/westnile/index.htm, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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457, B. a test of two independent means, C. a test of a single mean, D. a test of matched pairs., Exercise 10.9.47, An appropriate null hypothesis is:, A., B., C., D., , (Solution on p. 469.), , p2011 ≤ p2010, p2011 ≥ p2010, µ2011 ≤ µ2010, p2011 > p2010, , Exercise 10.9.48, (Solution on p. 469.), The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is, A. There is sufficient evidence to conclude that the proportion of people in the United States in, 2011 that got neuroinvasive West Nile disease is less than the proportion of people in the, United States in 2010 that got neuroinvasive West Nile disease., B. There is insufficient evidence to conclude that the proportion of people in the United States in, 2011 that got neuroinvasive West Nile disease is more than the proportion of people in the, United States in 2010 that got neuroinvasive West Nile disease., C. There is insufficient evidence to conclude that the proportion of people in the United States, in 2011 that got neuroinvasive West Nile disease is less than the proportion of people in the, United States in 2010 that got neuroinvasive West Nile disease., D. There is sufficient evidence to conclude that the proportion of people in the United States in, 2011 that got neuroinvasive West Nile disease is more than the proportion of people in the, United States in 2010 that got neuroinvasive West Nile disease., Questions Exercise 10.9.49 and Exercise 10.9.50 refer to the following:, A golf instructor is interested in determining if her new technique for improving players’ golf scores is, effective. She takes four (4) new students. She records their 18-holes scores before learning the technique, and then after having taken her class. She conducts a hypothesis test. The data are as follows., Player 1, , Player 2, , Player 3, , Player 4, , Mean score before class, , 83, , 78, , 93, , 87, , Mean score after class, , 80, , 80, , 86, , 86, , Table 10.15, , Exercise 10.9.49, This is:, A., B., C., D., , (Solution on p. 469.), , a test of two independent means, a test of two proportions, a test of a single proportion, a test of matched pairs., , Exercise 10.9.50, The correct decision is:, , (Solution on p. 469.), , A. Reject Ho, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 458, B. Do not reject Ho, , Questions Exercise 10.9.51 and Exercise 10.9.52 refer to the following:, Suppose a statistics instructor believes that there is no significant difference between the mean class scores, of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples, from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86, and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day”, subscript refers to the statistics day students. The “night” subscript refers to the statistics night students., Exercise 10.9.51, An appropriate alternate hypothesis for the hypothesis test is:, A., B., C., D., , (Solution on p. 469.), , µday > µnight, µday < µnight, µday = µnight, µday 6= µnight, , Exercise 10.9.52, A concluding statement is:, , (Solution on p. 469.), , A. There is sufficient evidence to conclude that statistics night students mean on Exam 2 is better, than the statistics day students mean on Exam 2., B. There is insufficient evidence to conclude that the statistics day students mean on Exam 2 is, better than the statistics night students mean on Exam 2., C. There is insufficient evidence to conclude that there is a significant difference between the, means of the statistics day students and night students on Exam 2., D. There is sufficient evidence to conclude that there is a significant difference between the means, of the statistics day students and night students on Exam 2., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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459, , 10.10 Review12, The next three questions refer to the following information:, In a survey at Kirkwood Ski Resort the following information was recorded:, Sport Participation by Age, 0 – 10, , 11 - 20, , 21 - 40, , 40+, , Ski, , 10, , 12, , 30, , 8, , Snowboard, , 6, , 17, , 12, , 5, , Table 10.16, , Suppose that one person from of the above was randomly selected., Exercise 10.10.1, Find the probability that the person was a skier or was age 11 – 20., , (Solution on p. 469.), , Exercise 10.10.2, (Solution on p. 469.), Find the probability that the person was a snowboarder given he/she was age 21 – 40., Exercise 10.10.3, Explain which of the following are true and which are false., a., b., c., d., , (Solution on p. 469.), , Sport and Age are independent events., Ski and age 11 – 20 are mutually exclusive events., P (Ski and age 21 − 40) < P (Ski | age 21 − 40), P (Snowboard or age 0 − 10) < P (Snowboard | age 0 − 10), , Exercise 10.10.4, (Solution on p. 470.), The average length of time a person with a broken leg wears a cast is approximately 6 weeks., The standard deviation is about 3 weeks. Thirty people who had recently healed from broken, legs were interviewed. State the distribution that most accurately reflects total time to heal for the, thirty people., Exercise 10.10.5, (Solution on p. 470.), The distribution for X is Uniform. What can we say for certain about the distribution for X when, n = 1?, A. The distribution for X is still Uniform with the same mean and standard dev. as the distribution, for X., B. The distribution for Xis Normal with the different mean and a different standard deviation as, the distribution for X., C. The distribution for X is Normal with the same mean but a larger standard deviation than the, distribution for X., D. The distribution for X is Normal with the same mean but a smaller standard deviation than, the distribution for X., Exercise 10.10.6, (Solution on p. 470.), The distribution for X is uniform. What can we say for certain about the distribution for ∑ X, when n = 50?, 12 This, , content is available online at <http://cnx.org/content/m17021/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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460, , CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, A. The distribution for ∑ Xis still uniform with the same mean and standard deviation as the, distribution for X., B. The distribution for ∑ X is Normal with the same mean but a larger standard deviation as the, distribution for X., C. The distribution for ∑ X is Normal with a larger mean and a larger standard deviation than the, distribution for X., D. The distribution for ∑ X is Normal with the same mean but a smaller standard deviation than, the distribution for X., , The next three questions refer to the following information:, A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They, calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches., Suppose we randomly survey 16 five-pound bags of baby carrots., Exercise 10.10.7, (Solution on p. 470.), State the approximate distribution for X, the distribution for the average lengths of baby carrots, in 16 five-pound bags. X~, Exercise 10.10.8, Explain why we cannot find the probability that one individual randomly chosen carrot is greater, than 2.25 inches., Exercise 10.10.9, Find the probability that x is between 2 and 2.25 inches., , (Solution on p. 470.), , The next three questions refer to the following information:, At the beginning of the term, the amount of time a student waits in line at the campus store is normally, distributed with a mean of 5 minutes and a standard deviation of 2 minutes., Exercise 10.10.10, Find the 90th percentile of waiting time in minutes., , (Solution on p. 470.), , Exercise 10.10.11, Find the median waiting time for one student., , (Solution on p. 470.), , Exercise 10.10.12, (Solution on p. 470.), Find the probability that the average waiting time for 40 students is at least 4.5 minutes., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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461, , 10.11 Lab: Hypothesis Testing for Two Means and Two Proportions13, Class Time:, Names:, , 10.11.1 Student Learning Outcomes:, • The student will select the appropriate distributions to use in each case., • The student will conduct hypothesis tests and interpret the results., , 10.11.2 Supplies:, • The business section from two consecutive days’ newspapers, • 3 small packages of M&Ms®, • 5 small packages of Reese’s Pieces®, , 10.11.3 Increasing Stocks Survey, Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion, of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks, that increased. As randomly as possible, choose 40 NYSE stocks and 32 NASDAQ stocks and complete the, following statements., 1., 2., 3., 4., 5., 6., , Ho, Ha, In words, define the Random Variable. ____________=, The distribution to use for the test is:, Calculate the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., a. Graph:, , 13 This, , content is available online at <http://cnx.org/content/m17022/1.13/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 462, , Figure 10.8, , b. Calculate the p-value:, 7. Do you reject or not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , 10.11.4 Decreasing Stocks Survey, Randomly pick 8 stocks from the newspaper. Using two consecutive days’ business sections, test whether, the stocks went down, on average, for the second day., 1., 2., 3., 4., 5., 6., , Ho, Ha, In words, define the Random Variable. ____________=, The distribution to use for the test is:, Calculate the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., a. Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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463, , Figure 10.9, , b. Calculate the p-value:, 7. Do you reject or not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , 10.11.5 Candy Survey, Buy three small packages of M&Ms and 5 small packages of Reese’s Pieces (same net weight as the M&Ms)., Test whether or not the mean number of candy pieces per package is the same for the two brands., 1., 2., 3., 4., 5., 6., , Ho :, Ha :, In words, define the random variable. __________=, What distribution should be used for this test?, Calculate the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., a. Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 464, , Figure 10.10, , b. Calculate the p-value:, 7. Do you reject or not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , 10.11.6 Shoe Survey, Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes,, sandals, and boots. Use your class as the sample., 1., 2., 3., 4., 5., 6., , Ho, Ha, In words, define the Random Variable. ____________=, The distribution to use for the test is:, Calculate the test statistic using your data., Draw a graph and label it appropriately. Shade the actual level of significance., a. Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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465, , Figure 10.11, , b. Calculate the p-value:, 7. Do you reject or not reject the null hypothesis? Why?, 8. Write a clear conclusion using a complete sentence., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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466, , CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , Solutions to Exercises in Chapter 10, Solution to Example 10.2, Problem 1 (p., two means, Solution to Example 10.2, Problem 2 (p., unknown, Solution to Example 10.2, Problem 3 (p., student’s-t, Solution to Example 10.2, Problem 4 (p., X A − XB, Solution to Example 10.2, Problem 5 (p., , 433), 433), 433), 433), 433), , • Ho : µ A ≤ µ B, • Ha : µ A > µ B, Solution to Example 10.2, Problem 6 (p. 433), right, Solution to Example 10.2, Problem 7 (p. 433), 0.1928, Solution to Example 10.2, Problem 8 (p. 433), Do not reject., Solution to Example 10.4, Problem (p. 436), The problem asks for a difference in proportions., Solution to Example 10.6, Problem (p. 440), means; At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that, the strength development class helped to make the players stronger, on average., Solution to Example 10.7, Problem (p. 441), H0 : µd equals 0; Ha : µd does not equal 0; Do not reject the null; At a 5% significance level, from the, sample data, there is not sufficient evidence to conclude that the mean difference in distances between the, children’s dominant versus weaker hands is significant (there is not sufficient evidence to show that the, children could push the shot-put further with their dominant hand). Alpha and the p-value are close so the, test is not strong., , Solutions to Practice 1: Hypothesis Testing for Two Proportions, Solution to Exercise 10.7.1 (p. 443), Proportions, Solution to Exercise 10.7.2 (p. 443), a. H0 :PN =PND, a. Ha :PN > PND, Solution to Exercise 10.7.3 (p. 443), right-tailed, Solution to Exercise 10.7.6 (p. 443), Normal, Solution to Exercise 10.7.8 (p. 443), 3.50, Solution to Exercise 10.7.10 (p. 444), 0.0002, Solution to Exercise 10.7.11 (p. 444), a. Reject the null hypothesis, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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467, , Solutions to Practice 2: Hypothesis Testing for Two Averages, Solution to Exercise 10.8.1 (p. 445), Means, Solution to Exercise 10.8.2 (p. 445), a. H0 : µW = µNW, b. Ha : µW 6= µNW, Solution to Exercise 10.8.3 (p. 445), two-tailed, Solution to Exercise 10.8.4 (p. 445), XW − X NW, Solution to Exercise 10.8.5 (p. 445), The difference between the mean life spans of whites and nonwhites., Solution to Exercise 10.8.8 (p. 445), 5.42, Solution to Exercise 10.8.10 (p. 446), 0.0000, Solution to Exercise 10.8.11 (p. 446), a. Reject the null hypothesis, , Solutions to Homework, Solution to Exercise 10.9.1 (p. 447), A, Solution to Exercise 10.9.3 (p. 447), B, Solution to Exercise 10.9.5 (p. 447), A, Solution to Exercise 10.9.7 (p. 447), D, Solution to Exercise 10.9.9 (p. 447), C, Solution to Exercise 10.9.11 (p. 448), d. t68.44, e. -1.04, f. 0.1519, h. Decision: Do not reject null, Solution to Exercise 10.9.13 (p. 448), Standard Normal, e. z = 2.14, f. 0.0163, h. Decision: Reject null when α = 0.05; Do not reject null when α = 0.01, Solution to Exercise 10.9.15 (p. 449), e. 0.73, f. 0.2326, h. Decision: Do not reject null, Solution to Exercise 10.9.17 (p. 449), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 468, e. -7.33, f. 0, h. Decision: Reject null, , Solution to Exercise 10.9.19 (p. 450), d. t7, e. -1.51, f. 0.1755, h. Decision: Do not reject null, Solution to Exercise 10.9.21 (p. 450), d. t9, e. t = −1.86, f. 0.0479, h. Decision: Reject null, but run another test, Solution to Exercise 10.9.23 (p. 451), d. t108, e. t = −0.82, f. 0.2066, h. Decision: Do not reject null, Solution to Exercise 10.9.25 (p. 451), d. t7, e. t = 2.9850, f. 0.0102, h. Decision: Reject null; There is sufficient evidence to conclude that the mean difference is more than 2, minutes., Solution to Exercise 10.9.27 (p. 452), e. 0.22, f. 0.4133, h. Decision: Do not reject null, Solution to Exercise 10.9.29 (p. 452), e. z = 2.50, f. 0.0063, h. Decision: Reject null, Solution to Exercise 10.9.31 (p. 453), e. -4.82, f. 0, h. Decision: Reject null, Solution to Exercise 10.9.33 (p. 453), d. t20.32, e. -4.70, f. 0.0001, h. Decision: Reject null, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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469, Solution to Exercise 10.9.35 (p. 453), d. t40.94, e. -5.08, f. 0, h. Decision: Reject null, Solution to Exercise 10.9.37 (p. 453), e. -0.9223, f. 0.1782, h. Decision: Do not reject null, Solution to Exercise 10.9.39 (p., D, Solution to Exercise 10.9.40 (p., B, Solution to Exercise 10.9.41 (p., B, Solution to Exercise 10.9.42 (p., A, Solution to Exercise 10.9.43 (p., C, Solution to Exercise 10.9.44 (p., B, Solution to Exercise 10.9.45 (p., C, Solution to Exercise 10.9.46 (p., A, Solution to Exercise 10.9.47 (p., A, Solution to Exercise 10.9.48 (p., D, Solution to Exercise 10.9.49 (p., D, Solution to Exercise 10.9.50 (p., B, Solution to Exercise 10.9.51 (p., D, Solution to Exercise 10.9.52 (p., C, , 454), 454), 454), 455), 455), 456), 456), 456), 457), 457), 457), 457), 458), 458), , Solutions to Review, Solution to Exercise 10.10.1 (p. 459), 77, 100, , Solution to Exercise 10.10.2 (p. 459), 12, 42, , Solution to Exercise 10.10.3 (p. 459), a. False, b. False, c. True, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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CHAPTER 10. HYPOTHESIS TESTING: TWO MEANS, PAIRED DATA, TWO, PROPORTIONS, , 470, d. False, , Solution to Exercise 10.10.4 (p., N (180, 16.43), Solution to Exercise 10.10.5 (p., A, Solution to Exercise 10.10.6 (p., C, Solution, to, Exercise 10.10.7 (p., , , 459), 459), 459), 460), , 2, √.25, 16, , N, Solution to Exercise 10.10.9 (p. 460), 0.5000, Solution to Exercise 10.10.10 (p. 460), 7.6, Solution to Exercise 10.10.11 (p. 460), 5, Solution to Exercise 10.10.12 (p. 460), 0.9431, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 11, , The Chi-Square Distribution, 11.1 The Chi-Square Distribution1, 11.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, •, , Interpret the chi-square probability distribution as the sample size changes., Conduct and interpret chi-square goodness-of-fit hypothesis tests., Conduct and interpret chi-square test of independence hypothesis tests., Conduct and interpret chi-square homogeneity hypothesis tests., Conduct and interpret chi-square single variance hypothesis tests., , 11.1.2 Introduction, Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a, greater frequency? How about if the types of movies people preferred were different across different age, groups? What about if a coffee machine was dispensing approximately the same amount of coffee each, time? You could answer these questions by conducting a hypothesis test., You will now study a new distribution, one that is used to determine the answers to the above examples., This distribution is called the Chi-square distribution., In this chapter, you will learn the three major applications of the Chi-square distribution:, , • The goodness-of-fit test, which determines if data fit a particular distribution, such as with the lottery, example, • The test of independence, which determines if events are independent, such as with the movie example, • The test of a single variance, which tests variability, such as with the coffee example, NOTE : Though the Chi-square calculations depend on calculators or computers for most of the, calculations, there is a table available (see the Table of Contents 15. Tables). TI-83+ and TI-84, calculator instructions are included in the text., 1 This, , content is available online at <http://cnx.org/content/m17048/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 471
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472, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.1.3 Optional Collaborative Classroom Activity, Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, etc.). Plot a histogram and a boxplot, using your data. See if you can determine a probability distribution that your data fits. Have a discussion, with the class about your choice., , 11.2 Notation2, The notation for the chi-square distribution is:, χ2 ∼ χ2df, where d f = degrees of freedom depend on how chi-square is being used. (If you want to practice calculating chi-square probabilities then use d f = n − 1. The degrees of freedom for the three major uses are each, calculated differently.), 2, For, p the χ distribution, the population mean is µ = d f and the population standard deviation is σ =, 2 · df., , The random variable is shown as χ2 but may be any upper case letter., The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent,, squared standard normal variables., χ2 = ( Z1 )2 + ( Z2 )2 + ... + ( Zk )2, , 11.3 Facts About the Chi-Square Distribution3, 1. The curve is nonsymmetrical and skewed to the right., 2. There is a different chi-square curve for each d f ., 2 This, 3 This, , content is available online at <http://cnx.org/content/m17052/1.6/>., content is available online at <http://cnx.org/content/m17045/1.6/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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473, , (a), , (b), , Figure 11.1, , 3. The test statistic for any test is always greater than or equal to zero., 4. When d f > 90, the chi-square curve approximates the normal. For X ∼ χ21000 the mean, µ = d f = 1000, √, and the standard deviation, σ = 2 · 1000 = 44.7. Therefore, X ∼ N (1000, 44.7), approximately., 5. The mean, µ, is located just to the right of the peak., , Figure 11.2, , In the next sections, you will learn about four different applications of the Chi-Square Distribution. These, hypothesis tests are almost always right-tailed tests. In order to understand why the tests are mostly righttailed, you will need to look carefully at the actual definition of the test statistic. Think about the following, while you study the next four sections. If the expected and observed values are "far" apart, then the test, statistic will be "large" and we will reject in the right tail. The only way to obtain a test statistic very close to, zero, would be if the observed and expected values are very, very close to each other. A left-tailed test could, be used to determine if the fit were "too good." A "too good" fit might occur if data had been manipulated, or invented. Think about the implications of right-tailed versus left-tailed hypothesis tests as you learn the, applications of the Chi-Square Distribution., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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474, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.4 Goodness-of-Fit Test4, In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For, example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null, and the alternate hypotheses for this test may be written in sentences or may be stated as equations or, inequalities., The test statistic for a goodness-of-fit test is:, , (O − E)2, E, k, , Σ, , (11.2), , where:, , • O = observed values (data), • E = expected values (from theory), • k = the number of different data cells or categories, The observed values are the data values and the expected values are the values you would expect to get, if the null hypothesis were true. There are n terms of the form, , (O− E)2, ., E, , The degrees of freedom are df = (number of categories - 1)., The goodness-of-fit test is almost always right tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the, right tail of the chi-square curve., NOTE :, , The expected value for each cell needs to be at least 5 in order to use this test., , Example 11.1, Absenteeism of college students from math classes is a major concern to math instructors because, missing class appears to increase the drop rate. Suppose that a study was done to determine if the, actual student absenteeism follows faculty perception. The faculty expected that a group of 100, students would miss class according to the following chart., Number absences per term, , Expected number of students, , 0-2, , 50, , 3-5, , 30, , 6-8, , 12, , 9 - 11, , 6, , 12+, , 2, Table 11.1, , A random survey across all mathematics courses was then done to determine the actual number, (observed) of absences in a course. The next chart displays the result of that survey., 4 This, , content is available online at <http://cnx.org/content/m17192/1.8/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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475, , Number absences per term, , Actual number of students, , 0-2, , 35, , 3-5, , 40, , 6-8, , 20, , 9 - 11, , 1, , 12+, , 4, Table 11.2, , Determine the null and alternate hypotheses needed to conduct a goodness-of-fit test., Ho : Student absenteeism fits faculty perception., The alternate hypothesis is the opposite of the null hypothesis., Ha : Student absenteeism does not fit faculty perception., Problem 1, Can you use the information as it appears in the charts to conduct the goodness-of-fit test?, Solution, No. Notice that the expected number of absences for the "12+" entry is less than 5 (it is 2)., Combine that group with the "9 - 11" group to create new tables where the number of students for, each entry are at least 5. The new tables are below., , Number absences per term, , Expected number of students, , 0-2, , 50, , 3-5, , 30, , 6-8, , 12, , 9+, , 8, Table 11.3, , Number absences per term, , Actual number of students, , 0-2, , 35, , 3-5, , 40, , 6-8, , 20, , 9+, , 5, Table 11.4, , Problem 2, What are the degrees of freedom (d f )?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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476, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, Solution, There are 4 "cells" or categories in each of the new tables., d f = number o f cells − 1 = 4 − 1 = 3, , Example 11.2, Employers particularly want to know which days of the week employees are absent in a five, day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week, did they have the highest number of employee absences. The results were distributed as follows:, Day of the Week Employees were most Absent, , Number of Absences, , Monday, , Tuesday, , Wednesday, , Thursday, , Friday, , 15, , 12, , 9, , 9, , 15, , Table 11.5, , Problem, For the population of employees, do the days for the highest number of absences occur with equal, frequencies during a five day work week? Test at a 5% significance level., Solution, The null and alternate hypotheses are:, , • Ho : The absent days occur with equal frequencies, that is, they fit a uniform distribution., • Ha : The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution., If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample:, 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday,, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the, table are the observed (O) values or data., This time, calculate the χ2 test statistic by hand. Make a chart with the following headings and fill, in the columns:, , •, •, •, •, , Expected (E) values (12, 12, 12, 12, 12), Observed (O) values (15, 12, 9, 9, 15), (O − E), (O − E)2, , •, , (O − E)2, E, (O − E)2, , The last column ( E ) should have 0.75, 0, 0.75, 0.75, 0.75., Now add (sum) the last column. Verify that the sum is 3. This is the χ2 test statistic., , To find the p-value, calculate P χ2 > 3 . This test is right-tailed., (Use a computer or calculator to find the p-value. You should get p-value = 0.5578.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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477, The d f s are the number of cells − 1 = 5 − 1 = 4., TI-83+ and TI-84: Press 2nd DISTR. Arrow down to χ2 cdf. Press ENTER. Enter (3,10^99,4)., Rounded to 4 decimal places, you should see 0.5578 which is the p-value., Next, complete a graph like the one below with the proper labeling and shading. (You should, shade the right tail.), , The decision is to not reject the null hypothesis., Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to, conclude that the absent days do not occur with equal frequencies., NOTE : TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the, goodness-of-fit test. The next example (Example 11-3) has the calculator instructions. The newer, TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values, (the data) into a first list and the expected values (the values you expect if the null hypothesis is, true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list, and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you, clear any lists before you start. See below., NOTE : To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of, the particular list. Press CLEAR and then arrow down. The list will be cleared. Or, you can press, STAT and press 4 (for ClrList). Enter the list name and press ENTER., , Example 11.3, One study indicates that the number of televisions that American families have is distributed (this, is the given distribution for the American population) as follows:, Number of Televisions, , Percent, , 0, , 10, , 1, , 16, , 2, , 55, , 3, , 11, , over 3, , 8, Table 11.6, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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478, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, The table contains expected (E) percents., A random sample of 600 families in the far western United States resulted in the following data:, Number of Televisions, , Frequency, , 0, , 66, , 1, , 119, , 2, , 340, , 3, , 60, , over 3, , 15, Total = 600, Table 11.7, , The table contains observed (O) frequency values., Problem, At the 1% significance level, does it appear that the distribution "number of televisions" of far, western United States families is different from the distribution for the American population as a, whole?, Solution, This problem asks you to test whether the far western United States families distribution fits the, distribution of the American families. This test is always right-tailed., The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are:, Number of Televisions, , Percent, , Expected Frequency, , 0, , 10, , (0.10) · (600) = 60, , 1, , 16, , (0.16) · (600) = 96, , 2, , 55, , (0.55) · (600) = 330, , 3, , 11, , (0.11) · (600) = 66, , over 3, , 8, , (0.08) · (600) = 48, , Table 11.8, , Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the, calculator do the math. For example, instead of 60, enter .10*600., Ho : The "number of televisions" distribution of far western United States families is the same as, the "number of televisions" distribution of the American population., Ha : The "number of televisions" distribution of far western United States families is different from, the "number of televisions" distribution of the American population., Distribution for the test: χ24 where d f = (the number of cells) − 1 = 5 − 1 = 4., NOTE :, , d f 6= 600 − 1, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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479, Calculate the test statistic: χ2 = 29.65, Graph:, , , Probability statement: p-value = P χ2 > 29.65 = 0.000006., Compare α and the p-value:, , • α = 0.01, • p-value = 0.000006, So, α > p-value., Make a decision: Since α > p-value, reject Ho ., This means you reject the belief that the distribution for the far western states is the same as that, of the American population as a whole., Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude, that the "number of televisions" distribution for the far western United States is different from the, "number of televisions" distribution for the American population as a whole., NOTE : TI-83+ and some TI-84 calculators: Press STAT and ENTER. Make sure to clear lists L1,, L2, and L3 if they have data in them (see the note at the end of Example 11-2). Into L1, put, the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600,, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter, (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You, should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd, DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded, to 4 places, you should see 5.77E-6 = .000006 (rounded to 6 decimal places) which is the p-value., , The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the, observed values (the data) into a first list and the expected values (the values you expect if the, null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names, for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or, draw. Make sure you clear any lists before you start., , Example 11.4, Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the, coins fair? Test at a 5% significance level., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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480, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, Solution, This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair, coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT., This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does, the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?", Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the value, 0, 1, 2. (There are 0, 1, or 2 heads in the flip of 2 coins.) Therefore, the number of cells is 3. Since, X = the number of heads, the observed frequencies are 20 (for 2 heads), 57 (for 1 head), and 23 (for, 0 heads or both tails). The expected frequencies are 25 (for 2 heads), 50 (for 1 head), and 25 (for 0, heads or both tails). This test is right-tailed., Ho : The coins are fair., Ha : The coins are not fair., Distribution for the test: χ22 where d f = 3 − 1 = 2., Calculate the test statistic: χ2 = 2.14, Graph:, , , Probability statement: p-value = P χ2 > 2.14 = 0.3430, Compare α and the p-value:, , • α = 0.05, • p-value = 0.3430, So, α < p-value., Make a decision: Since α < p-value, do not reject Ho ., Conclusion: There is insufficient evidence to conclude that the coins are not fair., NOTE : TI-83+ and some TI- 84 calculators: Press STAT and ENTER. Make sure you clear lists L1, L2,, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put, the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area "L3". Enter, (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You, should see "sum".Enter L3. Rounded to 2 decimal places, you should see 2.14. Press 2nd DISTR., Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 2.14,1E99,2). Rounded to 4 places, you, should see .3430 which is the p-value., , The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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481, observed values (the data) into a first list and the expected values (the values you expect if the, null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names, for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or, draw. Make sure you clear any lists before you start., , 11.5 Test of Independence5, Tests of independence involve using a contingency table of observed (data) values. You first saw a contingency table when you studied probability in the Probability Topics (Section 3.1) chapter., The test statistic for a test of independence is similar to that of a goodness-of-fit test:, , (O − E)2, E, (i · j ), Σ, , (11.2), , where:, , •, •, •, •, , O = observed values, E = expected values, i = the number of rows in the table, j = the number of columns in the table, , There are i · j terms of the form, , (O− E)2, ., E, , A test of independence determines whether two factors are independent or not. You first encountered, the term independence in Chapter 3. As a review, consider the following example., NOTE :, , The expected value for each cell needs to be at least 5 in order to use this test., , Example 11.5, Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and, B are independent then P ( A AND B) = P ( A) P ( B). A AND B is the event that a driver received, a speeding violation last year and is also a cell phone user while driving. Suppose, in a study of, drivers who received speeding violations in the last year and who uses cell phones while driving,, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305, were cell phone users while driving and 450 were not., Let y = expected number of drivers that use a cell phone while driving and received speeding, violations., If A and B are independent, then P ( A AND B) = P ( A) P ( B). By substitution,, y, 755, , =, , 70, 755, , ·, , 305, 755, , Solve for y : y =, 5 This, , 70·305, 755, , = 28.3, , content is available online at <http://cnx.org/content/m17191/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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482, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, About 28 people from the sample are expected to be cell phone users while driving and to receive, speeding violations., In a test of independence, we state the null and alternate hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent, and the alternate hypothesis states that they are not independent (dependent). If we do a test of, independence using the example above, then the null hypothesis is:, Ho : Being a cell phone user while driving and receiving a speeding violation are independent, events., If the null hypothesis were true, we would expect about 28 people to be cell phone users while, driving and to receive a speeding violation., The test of independence is always right-tailed because of the calculation of the test statistic. If, the expected and observed values are not close together, then the test statistic is very large and, way out in the right tail of the chi-square curve, like goodness-of-fit., The degrees of freedom for the test of independence are:, df = (number of columns - 1)(number of rows - 1), The following formula calculates the expected number (E):, E=, , (row total)(column total), total number surveyed, , Example 11.6, In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend, time with a disabled senior citizen. The program recruits among community college students,, four-year college students, and nonstudents. The following table is a sample of the adult volunteers and the number of hours they volunteer per week., Number of Hours Worked Per Week by Volunteer Type (Observed), Type of Volunteer, , 1-3 Hours, , 4-6 Hours, , 7-9 Hours, , Row Total, , Community College Students, , 111, , 96, , 48, , 255, , Four-Year College Students, , 96, , 133, , 61, , 290, , Nonstudents, , 91, , 150, , 53, , 294, , Column Total, , 298, , 379, , 162, , 839, , Table 11.9: The table contains observed (O) values (data)., , Problem, Are the number of hours volunteered independent of the type of volunteer?, Solution, The observed table and the question at the end of the problem, "Are the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two, factors are number of hours volunteered and type of volunteer. This test is always right-tailed., Ho : The number of hours volunteered is independent of the type of volunteer., Ha : The number of hours volunteered is dependent on the type of volunteer., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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483, The expected table is:, Number of Hours Worked Per Week by Volunteer Type (Expected), Type of Volunteer, , 1-3 Hours, , 4-6 Hours, , 7-9 Hours, , Community College Students, , 90.57, , 115.19, , 49.24, , Four-Year College Students, , 103.00, , 131.00, , 56.00, , Nonstudents, , 104.42, , 132.81, , 56.77, , Table 11.10: The table contains expected (E) values (data)., , For example, the calculation for the expected frequency for the top left cell is, E=, , (row total)(column total), total number surveyed, , =, , 255·298, 839, , = 90.57, , Calculate the test statistic: χ2 = 12.99, , (calculator or computer), , Distribution for the test: χ24, df = (3 columns − 1) (3 rows − 1) = (2) (2) = 4, Graph:, , , Probability statement: p-value = P χ2 > 12.99 = 0.0113, Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0113. α > p-value., Make a decision: Since α > p-value, reject Ho . This means that the factors are not independent., Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude, that the number of hours volunteered and the type of volunteer are dependent on one another., For the above example, if there had been another type of volunteer, teenagers, what would the, degrees of freedom be?, NOTE :, , Calculator instructions follow., , TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3, ENTER 3 ENTER. Enter the table values by row from Example 11-6. Press ENTER after each. Press, 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You, should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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484, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time but arrow down to, Draw instead of calculate., , Example 11.7, De Anza College is interested in the relationship between anxiety level and the need to succeed, in school. A random sample of 400 students took a test that measured anxiety level and need to, succeed in school. The table shows the results. De Anza College wants to know if anxiety level, and need to succeed in school are independent events., Need to Succeed in School vs. Anxiety Level, Need, to, Succeed in, School, , High, Anxiety, , Med-high, Anxiety, , Medium, Anxiety, , Med-low, Anxiety, , Low, Anxiety, , Row Total, , High Need, , 35, , 42, , 53, , 15, , 10, , 155, , Medium, Need, , 18, , 48, , 63, , 33, , 31, , 193, , Low Need, , 4, , 5, , 11, , 15, , 17, , 52, , Column Total, , 57, , 95, , 127, , 63, , 58, , 400, , Table 11.11, , Problem 1, How many high anxiety level students are expected to have a high need to succeed in school?, Solution, The column total for a high anxiety level is 57. The row total for high need to succeed in school is, 155. The sample size or total surveyed is 400., E=, , (row total)(column total), total surveyed, , =, , 155·57, 400, , = 22.09, , The expected number of students who have a high anxiety level and a high need to succeed in, school is about 22., , Problem 2, If the two variables are independent, how many students do you expect to have a low need to, succeed in school and a med-low level of anxiety?, Solution, The column total for a med-low anxiety level is 63. The row total for a low need to succeed in, school is 52. The sample size or total surveyed is 400., Problem 3, total)(column total), a. E = (row total, =, surveyed, b. The expected number of students who have a med-low anxiety level and a low need to succeed, in school is about:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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485, , 11.6 The Chi-Square Distribution: Test for Homogeneity6, The Goodness of Fit test can be used to decide whether a population fits a given distribution, but the Goodness of Fit test will not suffice to compare whether two populations follow the same unknown distribution., A different test, called the Test for Homogeneity, can be used to make a conclusion about whether two, populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the, same procedure as with the Test of Independence., NOTE :, , The expected value for each cell needs to be at least 5 in order to use this test., , Hypotheses, Ho : The distributions of the two populations are the same., Ha : The distributions of the two populations are not the same., Test Statistic, Use a χ2 test statistic. It is computed in the same way as the test for independence., Degrees of Freedom (df), df = number of columns - 1, Requirements, All values in the table must be greater than or equal to 5., Common Uses, Comparing two populations. For example: men versus women, before vs. after, east vs. west. The variable, is categorical with more than two possible response values., Example 11.8, Do male and female college students have the same distribution of living conditions? Use a level of, significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly, selected female college students were asked about their living conditions: Dormitory, Apartment,, With Parents, Other. The results are shown in the table below., Distribution of Living Conditions for College Males and College Females, Dormitory, , Apartment, , With Parents, , Other, , Males, , 72, , 84, , 49, , 45, , Females, , 91, , 86, , 88, , 35, , Table 11.12, 6 This, , content is available online at <http://cnx.org/content/m43655/1.2/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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486, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, Problem, Do male and female college students have the same distribution of living conditions?, Solution, Ho : The distribution of living conditions for male college students is the same as the distribution, of living conditions for female college students., Ha : The distribution of living conditions for male college students is not the same as the, distribution of living conditions for female college students., Degrees of Freedom (df):, df = number of columns - 1 = 4 - 1 = 3, Distribution for the test: χ23, Calculate the test statistic: χ2 = 10.1287, , (calculator or computer), , , Probability statement: p-value = P χ2 > 10.1287 = 0.0175, TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press, 2 ENTER 4 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT., Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see, Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic, is 10.1287 and the p-value = 0.0175. Do the procedure a second time but arrow down to Draw, instead of calculate., Compare α and the p-value: Since no α is given, assume α = 0.05., α > p-value., , p-value = 0.0175., , Make a decision: Since α > p-value, reject Ho . This means that the distributions are not, the same., Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living conditions for male and female college students are not the, same., Notice that the conclusion is only that the distributions are not the same., the Test for Homogeneity to make any conclusions about how they differ., , We cannot use, , Example 11.9, Both before and after a recent earthquake, surveys were conducted asking voters which of the, three candidates they planned on voting for in the upcoming city council election. Has there been, a change since the earthquake? Use a level of significance of 0.05. The table below shows the, results of the survey., Perez, , Chung, , Stevens, , Before, , 167, , 128, , 135, , After, , 214, , 197, , 225, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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487, Table 11.13, , Problem, Has there been a change in the distribution of voter preferences since the earthquake?, Solution, Ho : The distribution of voter preferences was the same before and after the earthquake., Ha : The distribution of voter preferences was not the same before and after the earthquake., Degrees of Freedom (df):, df = number of columns - 1 = 3 - 1 = 2, Distribution for the test: χ22, Calculate the test statistic: χ2 = 3.2603, , (calculator or computer), , , Probability statement: p-value = P χ2 > 3.2603 = 0.1959, TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press, 2 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT., Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see, Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is, 3.2603 and the p-value = 0.1959. Do the procedure a second time but arrow down to Draw instead, of calculate., Compare α and the p-value: α = 0.05 and the p-value = 0.1959. α < p-value., Make a decision: Since α < p-value, do not reject Ho ., Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to, conclude that the distribution of voter preferences was not the same before and after the, earthquake., , ** Contributed by Dr. Larry Green, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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488, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.7 The Chi-Square Distribution: Comparison Summary of the ChiSquare Tests Goodness-of-Fit, Independence and Homogeneity7, Comparison Summary of the Chi-Square Tests: Goodness-of-Fit, Independence and Homogeneity, You have seen the χ2 test statistic used in three different circumstances. Below is a summary that will help, you decide which χ2 test is the appropriate one to use., , • Goodness-of-Fit: Use the Goodness-of-Fit Test to decide whether a population with unknown distribution "fits" a known distribution. In this case there will be a single qualitative survey question or a, single outcome of an experiment from a single population. Goodness-of-Fit is typically used to see, if the population is uniform (all outcomes occur with equal frequency), the population is normal, or, the population is the same as another population with known distribution. The null and alternative, hypotheses are:, Ho : The population fits the given distribution., Ha : The population does not fit the given distribution., • Independence: Use the Test for Independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a, contingency table will be constructed. The goal is to see if the two variables are unrelated (independent) or related (dependent). The null and alternative hypotheses are:, Ho :The two variables (factors) are independent., Ha :The two variables (factors) are dependent., • Homogeneity: Use the Test for Homogeneity to decide if two populations with unknown distribution, have the same distribution as each other. In this case there will be a single qualitative survey question, or experiment given to two different populations. The null and alternative hypotheses are:, Ho :The two populations follow the same distribution., Ha :The two populations have different distributions., **With contributions by Dr. Larry Green, , 11.8 Test of a Single Variance8, A test of a single variance assumes that the underlying distribution is normal. The null and alternate, hypotheses are stated in terms of the population variance (or population standard deviation). The test, statistic is:, , ( n − 1) · s2, σ2, where:, , • n = the total number of data, • s2 = sample variance, • σ2 = population variance, You may think of s as the random variable in this test. The degrees of freedom are df = n − 1., A test of a single variance may be right-tailed, left-tailed, or two-tailed., 7 This, 8 This, , content is available online at <http://cnx.org/content/m43654/1.2/>., content is available online at <http://cnx.org/content/m17059/1.7/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (11.2)
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489, The following example will show you how to set up the null and alternate hypotheses. The null and, alternate hypotheses contain statements about the population variance., Example 11.10, Math instructors are not only interested in how their students do on exams, on average, but how, the exam scores vary. To many instructors, the variance (or standard deviation) may be more, important than the average., Suppose a math instructor believes that the standard deviation for his final exam is 5 points. One, of his best students thinks otherwise. The student claims that the standard deviation is more than, 5 points. If the student were to conduct a hypothesis test, what would the null and alternate, hypotheses be?, Solution, Even though we are given the population standard deviation, we can set the test up using the, population variance as follows., , • Ho : σ2 = 52, • H a : σ 2 > 52, , Example 11.11, With individual lines at its various windows, a post office finds that the standard deviation for, normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post, office experiments with a single main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes., With a significance level of 5%, test the claim that a single line causes lower variation among, waiting times (shorter waiting times) for customers., Solution, Since the claim is that a single line causes lower variation, this is a test of a single variance. The, parameter is the population variance, σ2 , or the population standard deviation, σ., Random Variable: The sample standard deviation, s, is the random variable. Let s = standard, deviation for the waiting times., , • Ho : σ2 = 7.22, • Ha : σ2 <7.22, The word "lower" tells you this is a left-tailed test., Distribution for the test: χ224 , where:, , • n = the number of customers sampled, • df = n − 1 = 25 − 1 = 24, Calculate the test statistic:, χ2 =, , (n−1)·s2, σ2, , =, , (25−1)·3.52, 7.22, , = 5.67, , where n = 25, s = 3.5, and σ = 7.2., Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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490, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , Probability statement: p-value = P χ2 < 5.67, Compare α and the p-value: α = 0.05, , , , = 0.000042, , p-value = 0.000042, , α > p-value, , Make a decision: Since α > p-value, reject Ho ., This means that you reject σ2 = 7.22 . In other words, you do not think the variation in waiting, times is 7.2 minutes, but lower., Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude, that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes., TI-83+ and TI-84 calculators: In 2nd DISTR, use 7:χ2cdf. The syntax is (lower, upper, df) for, the parameter list. For Example 11-9, χ2cdf(-1E99,5.67,24). The p-value = 0.000042., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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491, , 11.9 Summary of Formulas9, The Chi-Square√, Probability Distribution, µ = df and σ = 2 · df, Goodness-of-Fit Hypothesis Test, , • Use goodness-of-fit to test whether a data set fits a particular probability distribution., • The degrees of freedom are number of cells or categories - 1., 2, , • The test statistic is Σ (O−EE) , where O = observed values (data), E = expected values (from theory),, k, , and k = the number of different data cells or categories., • The test is right-tailed., Test of Independence, , • Use the test of independence to test whether two factors are independent or not., • The degrees of freedom are equal to (number of columns - 1)(number of rows - 1)., • The test statistic is Σ, , (i · j ), , (O− E)2, E, , where O = observed values, E = expected values, i = the number of rows, , in the table, and j = the number of columns in the table., • The test is right-tailed., total)(column total), • If the null hypothesis is true, the expected number E = (row total, ., surveyed, Test of Homogeneity, , • Use the test for homogeneity to decide if two populations with unknown distributions have the same, distribution as each other., • The degrees of freedom are equal to number of columns - 1., • The test statistic is Σ, , (i · j ), , (O− E)2, E, , where O = observed values, E = expected values, i = the number of rows, , in the table, and j = the number of columns in the table., • The test is right-tailed., total)(column total), • If the null hypothesis is true, the expected number E = (row total, ., surveyed, NOTE : The expected value for each cell needs to be at least 5 in order to use the Goodness-of-Fit,, Independence and Homogeneity tests., , Test of a Single Variance, , • Use the test to determine variation., • The degrees of freedom are the number of samples - 1., 2, , • The test statistic is (n−σ12)·s , where n = the total number of data, s2 = sample variance, and σ2 =, population variance., • The test may be left, right, or two-tailed., , 9 This, , content is available online at <http://cnx.org/content/m17058/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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492, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.10 Practice 1: Goodness-of-Fit Test10, 11.10.1 Student Learning Outcomes, • The student will conduct a goodness-of-fit test., , 11.10.2 Given, The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is, broken down by ethnicity as follows: (Source: HIV/AIDS Epidemiology Santa Clara County, Santa Clara, County Public Health Department, May 2011 ), Ethnicity, , Number of Cases, , White, , 2229, , Hispanic, , 1157, , Black/African-American, , 457, , Asian, Pacific Islander, , 232, Total = 4075, , Table 11.14, , The percentage of each ethnic group in Santa Clara County is as follows:, Ethnicity, , Percentage of total county population, , Number expected (round to 2, decimal places), , White, , 42.9%, , 1748.18, , Hispanic, , 26.7%, , Black/African-American, , 2.6%, , Asian, Pacific Islander, , 27.8%, Total = 100%, Table 11.15, , 11.10.3 Expected Results, If the ethnicity of AIDS victims followed the ethnicity of the total county population, fill in the expected, number of cases per ethnic group., , 11.10.4 Goodness-of-Fit Test, Perform a goodness-of-fit test to determine whether the make-up of AIDS cases follows the ethnicity of the, general population of Santa Clara County., Exercise 11.10.1, Ho :, 10 This, , content is available online at <http://cnx.org/content/m17054/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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493, Exercise 11.10.2, Ha :, Exercise 11.10.3, Is this a right-tailed, left-tailed, or two-tailed test?, Exercise 11.10.4, degrees of freedom =, , (Solution on p. 518.), , Exercise 11.10.5, Chi2 test statistic =, , (Solution on p. 518.), , Exercise 11.10.6, p-value =, , (Solution on p. 518.), , Exercise 11.10.7, Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade, in the region corresponding to the p-value., , Let α = 0.05, Decision:, Reason for the Decision:, Conclusion (write out in complete sentences):, , 11.10.5 Discussion Question, Exercise 11.10.8, Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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494, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.11 Practice 2: Contingency Tables11, 11.11.1 Student Learning Outcomes, • The student will conduct a test for independence using contingency tables., Conduct a hypothesis test to determine if smoking level and ethnicity are independent., , 11.11.2 Collect the Data, Copy the data provided in Probability Topics Practice 1: Contingency Tables into the table below., Smoking Levels by Ethnicity (Observed), Smoking, Level Per, Day, , African, American, , Native, Hawaiian, , Latino, , Japanese, Americans, , White, , TOTALS, , 1-10, 11-20, 21-30, 31+, TOTALS, Table 11.16, , 11.11.3 Hypothesis, State the hypotheses., , • Ho :, • Ha :, , 11.11.4 Expected Values, Enter expected values in the above below. Round to two decimal places., , 11.11.5 Analyze the Data, Calculate the following values:, Exercise 11.11.1, Degrees of freedom =, , (Solution on p. 518.), , Exercise 11.11.2, Chi2 test statistic =, , (Solution on p. 518.), , Exercise 11.11.3, p-value =, Exercise 11.11.4, Is this a right-tailed, left-tailed, or two-tailed test? Explain why., , (Solution on p. 518.), , 11 This, , (Solution on p. 518.), , content is available online at <http://cnx.org/content/m17056/1.13/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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495, , 11.11.6 Graph the Data, Exercise 11.11.5, Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade, in the region corresponding to the p-value., , 11.11.7 Conclusions, State the decision and conclusion (in a complete sentence) for the following preconceived levels of α ., Exercise 11.11.6, α = 0.05, , (Solution on p. 518.), , a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, Exercise 11.11.7, α = 0.01, a. Decision:, b. Reason for the decision:, c. Conclusion (write out in a complete sentence):, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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496, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.12 Practice 3: Test of a Single Variance12, 11.12.1 Student Learning Outcomes, • The student will conduct a test of a single variance., , 11.12.2 Given, Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The, average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes., , 11.12.3 Sample Variance, Exercise 11.12.1, Is the traveler disputing the claim about the average or about the variance?, Exercise 11.12.2, (Solution on p. 518.), A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes., Exercise 11.12.3, Is this a right-tailed, left-tailed, or two-tailed test?, , 11.12.4 Hypothesis Test, Perform a hypothesis test on the consistency part of the claim., Exercise 11.12.4, Ho :, Exercise 11.12.5, Ha :, Exercise 11.12.6, Degrees of freedom =, , (Solution on p. 518.), , Exercise 11.12.7, Chi2 test statistic =, , (Solution on p. 518.), , Exercise 11.12.8, (Solution on p. 518.), p-value =, Exercise 11.12.9, Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade, the p-value., 12 This, , content is available online at <http://cnx.org/content/m17053/1.8/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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497, , Exercise 11.12.10, Let α = 0.05, Decision:, Conclusion (write out in a complete sentence):, , 11.12.5 Discussion Questions, Exercise 11.12.11, How did you know to test the variance instead of the mean?, Exercise 11.12.12, If an additional test were done on the claim of the average delay, which distribution would you, use?, Exercise 11.12.13, If an additional test was done on the claim of the average delay, but 45 flights were surveyed,, which distribution would you use?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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498, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.13 Homework13, Exercise 11.13.1, a. Explain why the “goodness of fit” test and the “test for independence” are generally right tailed, tests., b. If you did a left-tailed test, what would you be testing?, , 11.13.1 Word Problems, For each word problem, use a solution sheet to solve the hypothesis test problem. Go to The Table of, Contents 14. Appendix for the chi-square solution sheet. Round expected frequency to two decimal places., Exercise 11.13.2, A 6-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data below are the result of the 120 rolls., Face Value, , Frequency, , 1, , 15, , 2, , 29, , 3, , 16, , 4, , 15, , 5, , 30, , 6, , 15, , Expected Frequency, , Table 11.17, , Exercise 11.13.3, (Solution on p. 518.), The marital status distribution of the U.S. male population, age 15 and older, is as shown below., (Source: U.S. Census Bureau, Current Population Reports), Marital Status, , Percent, , never married, , 31.3, , married, , 56.1, , widowed, , 2.5, , divorced/separated, , 10.1, , Expected Frequency, , Table 11.18, , Suppose that a random sample of 400 U.S. young adult males, 18 – 24 years old, yielded the, following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying, 400 people. Fill in the above table, rounding to two decimal places., 13 This, , content is available online at <http://cnx.org/content/m17028/1.20/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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499, , Marital Status, , Frequency, , never married, , 140, , married, , 238, , widowed, , 2, , divorced/separated, , 20, , Table 11.19, , The next two questions refer to the following information. The columns in the chart below contain the, Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class and the Overall Student Population. (Source: http://www.collegeboard.com)., Suppose the right column contains the result of a survey of 1000 local students from that year who took an, AP Exam., , Race/Ethnicity, , AP Examinee Population, , Overall Student Population, , Survey Frequency, , Asian, Asian American, or Pacific Islander, , 10.2%, , 5.4%, , 113, , Black or African American, , 8.2%, , 14.5%, , 94, , Hispanic or Latino, , 15.5%, , 15.9%, , 136, , 0.6%, , 1.2%, , 10, , White, , 59.4%, , 61.6%, , 604, , Not reported/other, , 6.1%, , 1.4%, , 43, , American Indian, Alaska Native, , or, , Table 11.20, , Exercise 11.13.4, Perform a goodness-of-fit test to determine whether the local results follow the distribution of the, U. S. Overall Student Population based on ethnicity., Exercise 11.13.5, (Solution on p. 518.), Perform a goodness-of-fit test to determine whether the local results follow the distribution of U., S. AP Examinee Population, based on ethnicity., Exercise 11.13.6, The City of South Lake Tahoe, CA, has an Asian population of 1419 people, out of a total population of 23,609 (Source: U.S. Census Bureau). Suppose that a survey of 1419 self-reported Asians, in Manhattan, NY, area yielded the data in the table below. Conduct a goodness of fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe, area., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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500, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , Race, , Lake Tahoe Frequency, , Manhattan Frequency, , Asian Indian, , 131, , 174, , Chinese, , 118, , 557, , Filipino, , 1045, , 518, , Japanese, , 80, , 54, , Korean, , 12, , 29, , Vietnamese, , 9, , 21, , Other, , 24, , 66, Table 11.21, , The next two questions refer to the following information: UCLA conducted a survey of more than, 263,000 college freshmen from 385 colleges in fall 2005. The results of student expected majors by gender, were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5000 graduating, females and 5000 graduating males was done as a follow-up last year to determine what their actual major, was. The results are shown in the tables for Exercises 7 and 8. The second column in each table does not, add to 100% because of rounding., Exercise 11.13.7, (Solution on p. 519.), Conduct a hypothesis test to determine if the actual college major of graduating females fits the, distribution of their expected majors., Major, , Women - Expected Major, , Women - Actual Major, , Arts & Humanities, , 14.0%, , 670, , Biological Sciences, , 8.4%, , 410, , Business, , 13.1%, , 685, , Education, , 13.0%, , 650, , Engineering, , 2.6%, , 145, , Physical Sciences, , 2.6%, , 125, , Professional, , 18.9%, , 975, , Social Sciences, , 13.0%, , 605, , Technical, , 0.4%, , 15, , Other, , 5.8%, , 300, , Undecided, , 8.0%, , 420, Table 11.22, , Exercise 11.13.8, Conduct a hypothesis test to determine if the actual college major of graduating males fits the, distribution of their expected majors., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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501, , Major, , Men - Expected Major, , Men - Actual Major, , Arts & Humanities, , 11.0%, , 600, , Biological Sciences, , 6.7%, , 330, , Business, , 22.7%, , 1130, , Education, , 5.8%, , 305, , Engineering, , 15.6%, , 800, , Physical Sciences, , 3.6%, , 175, , Professional, , 9.3%, , 460, , Social Sciences, , 7.6%, , 370, , Technical, , 1.8%, , 90, , Other, , 8.2%, , 400, , Undecided, , 6.6%, , 340, Table 11.23, , Exercise 11.13.9, (Solution on p. 519.), A recent debate about where in the United States skiers believe the skiing is best prompted the, following survey. Test to see if the best ski area is independent of the level of the skier., U.S. Ski Area, , Beginner, , Intermediate, , Advanced, , Tahoe, , 20, , 30, , 40, , Utah, , 10, , 30, , 60, , Colorado, , 10, , 40, , 50, , Table 11.24, , Exercise 11.13.10, Car manufacturers are interested in whether there is a relationship between the size of car an, individual drives and the number of people in the driver’s family (that is, whether car size and, family size are independent). To test this, suppose that 800 car owners were randomly surveyed, with the following results. Conduct a test for independence., Family Size, , Sub & Compact, , Mid-size, , Full-size, , Van & Truck, , 1, , 20, , 35, , 40, , 35, , 2, , 20, , 50, , 70, , 80, , 3-4, , 20, , 50, , 100, , 90, , 5+, , 20, , 30, , 70, , 70, , Table 11.25, , Exercise 11.13.11, (Solution on p. 519.), College students may be interested in whether or not their majors have any effect on starting, salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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502, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, in college and their starting salaries after graduation. Below are the data. Conduct a test for, independence., Major, , < $50,000, , $50,000 - $68,999, , $69,000 +, , English, , 5, , 20, , 5, , Engineering, , 10, , 30, , 60, , Nursing, , 10, , 15, , 15, , Business, , 10, , 20, , 30, , Psychology, , 20, , 30, , 20, , Table 11.26, , Exercise 11.13.12, Some travel agents claim that honeymoon hot spots vary according to age of the bride and groom., Suppose that 280 East Coast recent brides were interviewed as to where they spent their honeymoons. The information is given below. Conduct a test for independence., Location, , 20 - 29, , 30 - 39, , 40 - 49, , 50 and over, , Niagara Falls, , 15, , 25, , 25, , 20, , Poconos, , 15, , 25, , 25, , 10, , Europe, , 10, , 25, , 15, , 5, , Virgin Islands, , 20, , 25, , 15, , 5, , Table 11.27, , Exercise 11.13.13, (Solution on p. 519.), A manager of a sports club keeps information concerning the main sport in which members, participate and their ages. To test whether there is a relationship between the age of a member, and his or her choice of sport, 643 members of the sports club are randomly selected. Conduct a, test for independence., Sport, , 18 - 25, , 26 - 30, , 31 - 40, , 41 and over, , racquetball, , 42, , 58, , 30, , 46, , tennis, , 58, , 76, , 38, , 65, , swimming, , 72, , 60, , 65, , 33, , Table 11.28, , Exercise 11.13.14, A major food manufacturer is concerned that the sales for its skinny French fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold, across the country to determine if the type of fries sold is independent of the area of the country., The results of the study are below. Conduct a test for independence., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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503, , Type of Fries, , Northeast, , South, , Central, , West, , skinny fries, , 70, , 50, , 20, , 25, , curly fries, , 100, , 60, , 15, , 30, , steak fries, , 20, , 40, , 10, , 10, , Table 11.29, , Exercise 11.13.15, (Solution on p. 519.), According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following, is a breakdown of the amount of life insurance purchased by males in the following age groups., He is interested in whether the age of the male and the amount of life insurance purchased are, independent events. Conduct a test for independence., Age of Males, , None, , < $200,000, , $200,000 - $400,000, , $401,001 - $1,000,000, , $1,000,000 +, , 20 - 29, , 40, , 15, , 40, , 0, , 5, , 30 - 39, , 35, , 5, , 20, , 20, , 10, , 40 - 49, , 20, , 0, , 30, , 0, , 30, , 50 +, , 40, , 30, , 15, , 15, , 10, , Table 11.30, , Exercise 11.13.16, Suppose that 600 thirty–year–olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test for independence., , Annual Salary, , Not a high school, graduate, , High school graduate, , College graduate, , Masters or doctorate, , < $30,000, , 15, , 25, , 10, , 5, , $30,000 - $40,000, , 20, , 40, , 70, , 30, , $40,000 - $50,000, , 10, , 20, , 40, , 55, , $50,000 - $60,000, , 5, , 10, , 20, , 60, , $60,000 +, , 0, , 5, , 10, , 150, , Table 11.31, , Exercise 11.13.17, (Solution on p. 519.), A Psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown below., Conduct a Test of Homogeneity. Test at a 5% level of significance., Open, , Conscientious, , Extrovert, , Agreeable Neurotic, , Business, , 41, , 52, , 46, , 61 58, , Social Science, , 72, , 75, , 63, , 80 65, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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504, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, Table 11.32, , Exercise 11.13.18, (Solution on p. 519.), Do men and women select different breakfasts? The breakfast ordered by randomly selected men, and women at a popular breakfast place is shown below. Conduct a test of homogeneity. Test at a, 5% level of significance, French Toast, , Pancakes, , Waffles, , Omelettes, , Men, , 47, , 35, , 28, , 53, , Women, , 65, , 59, , 55, , 60, , Table 11.33, , Exercise 11.13.19, (Solution on p. 520.), Is there a difference between the distribution of community college statistics students and the, distribution of university statistics students in what technology they use on their homework? Of, the randomly selected community college students 43 used a computer, 102 used a calculator, with built in statistics functions, and 65 used a table from the textbook. Of the randomly selected, university students 28 used a computer, 33 used a calculator with built in statistics functions, and, 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of, significance., Exercise 11.13.20, (Solution on p. 520.), A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the, same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in, Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish., Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other, trout, 67 were bass, and 53 were catfish. Perform the hypothesis test at a 5% level of significance., Exercise 11.13.21, (Solution on p. 520.), A plant manager is concerned her equipment may need recalibrating. It seems that the actual, weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be, at most 21 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected, boxes of cereal from the next day’s production were weighed. The standard deviation of the 84, boxes was 0.54. Does the machine need to be recalibrated?, Exercise 11.13.22, Consumers may be interested in whether the cost of a particular calculator varies from store to, store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard, deviation of $12, test the claim that the standard deviation is greater than $15., Exercise 11.13.23, (Solution on p. 520.), Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time, to run the 7 ½ mile race is at most 3 minutes. To test her claim, Rupinder looks up 5 of her race, times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes., Exercise 11.13.24, Airline companies are interested in the consistency of the number of babies on each flight, so that, they have adequate safety equipment. They are also interested in the variation of the number of, babies. Suppose that an airline executive believes the average number of babies on flights is 6 with, a variance of 9 at most. The airline conducts a survey. The results of the 18 flights surveyed give, a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the, airline executive’s belief., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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505, Exercise 11.13.25, (Solution on p. 520.), The number of births per woman in China is 1.6 down from 5.91 in 1966 (SourceWorld Bank,, 6/5/12 ). This fertility rate has been attributed to the law passed in 1979 restricting births to one, per woman. Suppose that a group of students studied whether or not the standard deviation of, births per woman was greater than 0.75. They asked 50 women across China the number of births, they had. Below are the results. Does the students’ survey indicate that the standard deviation is, greater than 0.75?, # of births, , Frequency, , 0, , 5, , 1, , 30, , 2, , 10, , 3, , 5, Table 11.34, , Exercise 11.13.26, According to an avid aquariest, the average number of fish in a 20–gallon tank is 10, with a, standard deviation of 2. His friend, also an aquariest, does not believe that the standard deviation, is 2. She counts the number of fish in 15 other 20–gallon tanks. Based on the results that follow, do, you think that the standard deviation is different from 2? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7;, 9; 11; 10; 11, Exercise 11.13.27, (Solution on p. 520.), The manager of "Frenchies" is concerned that patrons are not consistently receiving the same, amount of French fries with each order. The chef claims that the standard deviation for a 10–, ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly, weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of 2 oz., , 11.13.2 Try these true/false questions., Exercise 11.13.28, (Solution on p. 521.), As the degrees of freedom increase, the graph of the chi-square distribution looks more and more, symmetrical., Exercise 11.13.29, The standard deviation of the chi-square distribution is twice the mean., , (Solution on p. 521.), , Exercise 11.13.30, (Solution on p. 521.), The mean and the median of the chi-square distribution are the same if df = 24., Exercise 11.13.31, (Solution on p. 521.), In a Goodness-of-Fit test, the expected values are the values we would expect if the null hypothesis were true., Exercise 11.13.32, (Solution on p. 521.), In general, if the observed values and expected values of a Goodness-of-Fit test are not close, together, then the test statistic can get very large and on a graph will be way out in the right tail., Exercise 11.13.33, (Solution on p. 521.), The degrees of freedom for a Test for Independence are equal to the sample size minus 1., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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506, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, Exercise 11.13.34, (Solution on p. 521.), Use a Goodness-of-Fit test to determine if high school principals believe that students are absent, equally during the week or not., Exercise 11.13.35, (Solution on p. 521.), The Test for Independence uses tables of observed and expected data values., Exercise 11.13.36, (Solution on p. 521.), The test to use when determining if the college or university a student chooses to attend is related, to his/her socioeconomic status is a Test for Independence., Exercise 11.13.37, (Solution on p. 521.), The test to use to determine if a six-sided die is fair is a Goodness-of-Fit test., Exercise 11.13.38, (Solution on p. 521.), In a Test of Independence, the expected number is equal to the row total multiplied by the column, total divided by the total surveyed., Exercise 11.13.39, (Solution on p. 521.), In a Goodness-of Fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis., Exercise 11.13.40, (Solution on p. 521.), For a Chi-Square distribution with degrees of freedom of 17, the probability that a value is greater, than 20 is 0.7258., Exercise 11.13.41, (Solution on p. 521.), If df = 2, the chi-square distribution has a shape that reminds us of the exponential., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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507, , 11.14 Review14, The next two questions refer to the following real study:, A recent survey of U.S. teenage pregnancy was answered by 720 girls, age 12 - 19. 6% of the girls surveyed, said they have been pregnant. (Parade Magazine) We are interested in the true proportion of U.S. girls, age, 12 - 19, who have been pregnant., Exercise 11.14.1, (Solution on p. 521.), Find the 95% confidence interval for the true proportion of U.S. girls, age 12 - 19, who have been, pregnant., Exercise 11.14.2, (Solution on p. 521.), The report also stated that the results of the survey are accurate to within ± 3.7% at the 95%, confidence level. Suppose that a new study is to be done. It is desired to be accurate to within 2%, of the 95% confidence level. What is the minimum number that should be surveyed?, Exercise 11.14.3 , Given: X ∼ Exp 31 . Sketch the graph that depicts: P ( x > 1)., The next four questions refer to the following information:, Suppose that the time that owners keep their cars (purchased new) is normally distributed with a mean, of 7 years and a standard deviation of 2 years. We are interested in how long an individual keeps his car, (purchased new). Our population is people who buy their cars new., Exercise 11.14.4, 60% of individuals keep their cars at most how many years?, , (Solution on p. 521.), , Exercise 11.14.5, (Solution on p. 521.), Suppose that we randomly survey one person. Find the probability that person keeps his/her car, less than 2.5 years., Exercise 11.14.6, (Solution on p. 521.), If we are to pick individuals 10 at a time, find the distribution for the mean car length ownership., Exercise 11.14.7, (Solution on p. 521.), If we are to pick 10 individuals, find the probability that the sum of their ownership time is more, than 55 years., Exercise 11.14.8, For which distribution is the median not equal to the mean?, A., B., C., D., , (Solution on p. 521.), , Uniform, Exponential, Normal, Student-t, , Exercise 11.14.9, (Solution on p. 521.), Compare the standard normal distribution to the student-t distribution, centered at 0. Explain, which of the following are true and which are false., a. As the number surveyed increases, the area to the left of -1 for the student-t distribution approaches the area for the standard normal distribution., b. As the degrees of freedom decrease, the graph of the student-t distribution looks more like the, graph of the standard normal distribution., c. If the number surveyed is 15, the normal distribution should never be used., 14 This, , content is available online at <http://cnx.org/content/m17057/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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508, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , The next five questions refer to the following information:, We are interested in the checking account balance of a twenty-year-old college student. We randomly, survey 16 twenty-year-old college students. We obtain a sample mean of $640 and a sample standard, deviation of $150. Let X = checking account balance of an individual twenty year old college student., Exercise 11.14.10, Explain why we cannot determine the distribution of X., Exercise 11.14.11, (Solution on p. 522.), If you were to create a confidence interval or perform a hypothesis test for the population mean, checking account balance of 20-year old college students, what distribution would you use?, Exercise 11.14.12, (Solution on p. 522.), Find the 95% confidence interval for the true mean checking account balance of a twenty-year-old, college student., Exercise 11.14.13, What type of data is the balance of the checking account considered to be?, , (Solution on p. 522.), , Exercise 11.14.14, What type of data is the number of 20 year olds considered to be?, , (Solution on p. 522.), , Exercise 11.14.15, (Solution on p. 522.), On average, a busy emergency room gets a patient with a shotgun wound about once per week., We are interested in the number of patients with a shotgun wound the emergency room gets per, 28 days., a. Define the random variable X., b. State the distribution for X., c. Find the probability that the emergency room gets no patients with shotgun wounds in the next, 28 days., The next two questions refer to the following information:, The probability that a certain slot machine will pay back money when a quarter is inserted is 0.30 . Assume, that each play of the slot machine is independent from each other. A person puts in 15 quarters for 15 plays., Exercise 11.14.16, (Solution on p. 522.), Is the expected number of plays of the slot machine that will pay back money greater than, less, than or the same as the median? Explain your answer., Exercise 11.14.17, (Solution on p. 522.), Is it likely that exactly 8 of the 15 plays would pay back money? Justify your answer numerically., Exercise 11.14.18, A game is played with the following rules:, , •, •, •, •, , (Solution on p. 522.), , it costs $10 to enter, a fair coin is tossed 4 times, if you do not get 4 heads or 4 tails, you lose your $10, if you get 4 heads or 4 tails, you get back your $10, plus $30 more, , Over the long run of playing this game, what are your expected earnings?, Exercise 11.14.19, , (Solution on p. 522.), , • The mean grade on a math exam in Rachel’s class was 74, with a standard deviation of 5., Rachel earned an 80., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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509, , • The mean grade on a math exam in Becca’s class was 47, with a standard deviation of 2., Becca earned a 51., • The mean grade on a math exam in Matt’s class was 70, with a standard deviation of 8. Matt, earned an 83., Find whose score was the best, compared to his or her own class. Justify your answer numerically., The next two questions refer to the following information:, A random sample of 70 compulsive gamblers were asked the number of days they go to casinos per week., The results are given in the following graph:, , Figure 11.3, , Exercise 11.14.20, Find the number of responses that were “5"., , (Solution on p. 522.), , Exercise 11.14.21, (Solution on p. 522.), Find the mean, standard deviation, the median, the first quartile, the third quartile and the IQR., Exercise 11.14.22, (Solution on p. 522.), Based upon research at De Anza College, it is believed that about 19% of the student population, speaks a language other than English at home., Suppose that a study was done this year to see if that percent has decreased. Ninety-eight students, were randomly surveyed with the following results. Fourteen said that they speak a language, other than English at home., a., b., c., d., e., , State an appropriate null hypothesis., State an appropriate alternate hypothesis., Define the Random Variable, P’., Calculate the test statistic., Calculate the p-value., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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510, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, f. At the 5% level of decision, what is your decision about the null hypothesis?, g. What is the Type I error?, h. What is the Type II error?, Exercise 11.14.23, (Solution on p. 522.), Assume that you are an emergency paramedic called in to rescue victims of an accident. You, need to help a patient who is bleeding profusely. The patient is also considered to be a high risk, for contracting AIDS. Assume that the null hypothesis is that the patient does not have the HIV, virus. What is a Type I error?, Exercise 11.14.24, (Solution on p. 522.), It is often said that Californians are more casual than the rest of Americans. Suppose that a, survey was done to see if the proportion of Californian professionals that wear jeans to work is, greater than the proportion of non-Californian professionals. Fifty of each was surveyed with the, following results. 15 Californians wear jeans to work and 6 non-Californians wear jeans to work., , • C = Californian professional, • NC = non-Californian professional, a. State appropriate null and alternate hypotheses., b. Define the Random Variable., c. Calculate the test statistic and p-value., d. At the 5% significance level, what is your decision?, e. What is the Type I error?, f. What is the Type II error?, The next two questions refer to the following information:, A group of Statistics students have developed a technique that they feel will lower their anxiety level on, statistics exams. They measured their anxiety level at the start of the quarter and again at the end of the, quarter. Recorded is the paired data in that order: (1000, 900); (1200, 1050); (600, 700); (1300, 1100); (1000,, 900); (900, 900)., Exercise 11.14.25, This is a test of (pick the best answer):, , (Solution on p. 522.), , A. large samples, independent means, B. small samples, independent means, C. dependent means, Exercise 11.14.26, State the distribution to use for the test., , (Solution on p. 522.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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512, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , Figure 11.4, , 3. Calculate the following:, a. x =, b. s =, c. s2 =, , 11.15.3 Uniform Distribution, Test to see if grocery receipts follow the uniform distribution., 1. Using your lowest and highest values, X ∼ U (_______,_______), 2. Divide the distribution above into fifths., 3. Calculate the following:, a. Lowest value =, b. 20th percentile =, c. 40th percentile =, d. 60th percentile =, e. 80th percentile =, f. Highest value =, 4. For each fifth, count the observed number of receipts and record it. Then determine the expected, number of receipts and record that., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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513, , Fifth, , Observed, , Expected, , 1st, 2nd, 3rd, 4th, 5th, Table 11.36, , 5., 6., 7., 8., 9., 10., 11., , Ho :, Ha :, What distribution should you use for a hypothesis test?, Why did you choose this distribution?, Calculate the test statistic., Find the p-value., Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the pvalue., , Figure 11.5, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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514, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 12. State your decision., 13. State your conclusion in a complete sentence., , 11.15.4 Exponential Distribution, Test to see if grocery receipts follow the exponential distribution with decay parameter, , 1, x, , ., , 1. Using 1x as the decay parameter, X ∼ Exp (_______)., 2. Calculate the following:, a. Lowest value =, b. First quartile =, c. 37th percentile =, d. Median =, e. 63rd percentile =, f. 3rd quartile =, g. Highest value =, 3. For each cell, count the observed number of receipts and record it. Then determine the expected, number of receipts and record that., Cell, , Observed, , Expected, , 1st, 2nd, 3rd, 4th, 5th, 6th, Table 11.37, , 4., 5., 6., 7., 8., 9., 10., , Ho, Ha, What distribution should you use for a hypothesis test?, Why did you choose this distribution?, Calculate the test statistic., Find the p-value., Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the pvalue., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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515, , Figure 11.6, , 11. State your decision., 12. State your conclusion in a complete sentence., , 11.15.5 Discussion Questions, 1. Did your data fit either distribution? If so, which?, 2. In general, do you think it’s likely that data could fit more than one distribution? In complete sentences, explain why or why not., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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516, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , 11.16 Lab 2: Chi-Square Test for Independence16, Class Time:, Names:, , 11.16.1 Student Learning Outcome:, • The student will evaluate if there is a significant relationship between favorite type of snack and, gender., , 11.16.2 Collect the Data, 1. Using your class as a sample, complete the following chart., NOTE : You may need to combine two food categories so that each cell has an expected value, of at least 5, , Favorite type of snack, sweets (candy & baked goods), , ice cream, , chips & pretzels, , fruits & vegetables, , Total, , male, female, Total, Table 11.38, , 2. Looking at the above chart, does it appear to you that there is dependence between gender and favorite type of snack food? Why or why not?, , 11.16.3 Hypothesis Test, Conduct a hypothesis test to determine if the factors are independent, 1., 2., 3., 4., 5., 6., 7., , Ho :, Ha :, What distribution should you use for a hypothesis test?, Why did you choose this distribution?, Calculate the test statistic., Find the p-value., Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the pvalue., , 16 This, , content is available online at <http://cnx.org/content/m17050/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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517, , Figure 11.7, , 8. State your decision., 9. State your conclusion in a complete sentence., , 11.16.4 Discussion Questions, 1. Is the conclusion of your study the same as or different from your answer to (I2) above?, 2. Why do you think that occurred?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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518, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , Solutions to Exercises in Chapter 11, Solutions to Practice 1: Goodness-of-Fit Test, Solution to Exercise 11.10.4 (p. 493), degrees of freedom = 3, Solution to Exercise 11.10.5 (p. 493), 2016.14, Solution to Exercise 11.10.6 (p. 493), Rounded to 4 decimal places, the p-value is 0.0000., , Solutions to Practice 2: Contingency Tables, Solution to Exercise 11.11.1 (p., 12, Solution to Exercise 11.11.2 (p., 10301.8, Solution to Exercise 11.11.3 (p., 0, Solution to Exercise 11.11.4 (p., right, Solution to Exercise 11.11.6 (p., , 494), 494), 494), 494), 495), , a. Reject the null hypothesis, , Solutions to Practice 3: Test of a Single Variance, Solution to Exercise 11.12.2 (p., 225, Solution to Exercise 11.12.6 (p., 24, Solution to Exercise 11.12.7 (p., 36, Solution to Exercise 11.12.8 (p., 0.0549, , 496), 496), 496), 496), , Solutions to Homework, Solution to Exercise 11.13.3 (p. 498), a. The data fits the distribution, b. The data does not fit the distribution, c. 3, e. 19.27, f. 0.0002, h. Decision: Reject Null; Conclusion: Data does not fit the distribution., Solution to Exercise 11.13.5 (p. 499), c. 5, e. 13.4, f. 0.0199, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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519, g. Decision: Reject null when a = 0.05; Conclusion: Local data do not fit the AP Examinee Distribution., Decision: Do not reject null when a = 0.01; Conclusion: There is insufficient evidence to conclude, that Local data do not fit the AP Examinee Distribution., Solution to Exercise 11.13.7 (p. 500), c. 10, e. 11.48, f. 0.3214, h. Decision: Do not reject null when a = 0.05 and a = 0.01; Conclusion: There is insufficient evidence to, conclude that the distribution of majors by graduating females does not fit the distribution of expected, majors., Solution to Exercise 11.13.9 (p. 501), c. 4, e. 10.53, f. 0.0324, h. Decision: Reject null; Conclusion: Best ski area and level of skier are not independent., Solution to Exercise 11.13.11 (p. 501), c. 8, e. 33.55, f. 0, h. Decision: Reject null; Conclusion: Major and starting salary are not independent events., Solution to Exercise 11.13.13 (p. 502), c. 6, e. 25.21, f. 0.0003, h. Decision: Reject null, Solution to Exercise 11.13.15 (p. 503), c. 12, e. 125.74, f. 0, h. Decision: Reject null, Solution to Exercise 11.13.17 (p. 503), c: 4, d: Chi-Square with df = 4, e: 3.01, f: p-value = 0.5568, h: ii. Do not reject the null hypothesis., iv. There is insufficient evidence to conclude that the distribution of personality types is different for, business and social science majors., Solution to Exercise 11.13.18 (p. 504), c: 3, e: 4.01, f: p-value = 0.2601, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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520, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , h: ii. Do not reject the null hypothesis., iv. There is insufficient evidence to conclude that the distribution of breakfast ordered is different for, men and women., Solution to Exercise 11.13.19 (p. 504), c: 2, e: 7.05, f: p-value = 0.0294, h: ii. Reject the null hypothesis., iv. There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities., Solution to Exercise 11.13.20 (p. 504), c: 3, d: Chi-Square with df = 3, e: 11.75, f: p-value = 0.0083, h: ii. Reject the null hypothesis., iv. There is sufficient evidence to conclude that the distribution of fish in Green Valley Lake is not the, same as the distribution of fish in Echo Lake., Solution to Exercise 11.13.21 (p. 504), c. 83, d. Chi-Square with df = 83, e. 96.81, f. p-value = 0.1426; There is a 0.1426 probability that the sample standard deviation is 0.54 or more., h. Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the standard, deviation is more than 0.5 oz. It cannot be determined whether the equipment needs to be recalibrated, or not., Solution to Exercise 11.13.23 (p. 504), c. 4, d. Chi-Square with df = 4, e. 4.52, f. 0.3402, h. Decision: Do not reject null., Solution to Exercise 11.13.25 (p. 505), c. 49, d. Chi-Square with df = 49, e. 54.37, f. p-value = 0.2774; If the null hypothesis is true, there is a 0.2774 probability that the sample standard, deviation is 0.79 or more., h. Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the standard, deviation is more than 0.75. It cannot be determined if the standard deviation is greater than 0.75 or, not., Solution to Exercise 11.13.27 (p. 505), a. σ2 ≤ (1.5)2, c. 48, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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521, d. Chi-Square with df = 48, e. 85.33, f. 0.0007, h. Decision: Reject null., Solution to Exercise 11.13.28 (p., True, Solution to Exercise 11.13.29 (p., False, Solution to Exercise 11.13.30 (p., False, Solution to Exercise 11.13.31 (p., True, Solution to Exercise 11.13.32 (p., True, Solution to Exercise 11.13.33 (p., False, Solution to Exercise 11.13.34 (p., True, Solution to Exercise 11.13.35 (p., True, Solution to Exercise 11.13.36 (p., True, Solution to Exercise 11.13.37 (p., True, Solution to Exercise 11.13.38 (p., True, Solution to Exercise 11.13.39 (p., False, Solution to Exercise 11.13.40 (p., False, Solution to Exercise 11.13.41 (p., True, , 505), 505), 505), 505), 505), 505), 506), 506), 506), 506), 506), 506), 506), 506), , Solutions to Review, Solution to Exercise 11.14.1 (p., (0.0424, 0.0770), Solution to Exercise 11.14.2 (p., 2401, Solution to Exercise 11.14.4 (p., 7.5, Solution to Exercise 11.14.5 (p., 0.0122, Solution to Exercise 11.14.6 (p., N (7, 0.63), Solution to Exercise 11.14.7 (p., 0.9911, Solution to Exercise 11.14.8 (p., B, Solution to Exercise 11.14.9 (p., , 507), 507), 507), 507), 507), 507), 507), 507), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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522, , CHAPTER 11. THE CHI-SQUARE DISTRIBUTION, , a. True, b. False, c. False, Solution to Exercise 11.14.11 (p., student-t with df = 15, Solution to Exercise 11.14.12 (p., (560.07, 719.93), Solution to Exercise 11.14.13 (p., quantitative - continuous, Solution to Exercise 11.14.14 (p., quantitative - discrete, Solution to Exercise 11.14.15 (p., , 508), 508), 508), 508), 508), , b. P (4), c. 0.0183, Solution to Exercise 11.14.16 (p., greater than, Solution to Exercise 11.14.17 (p., No; P ( x = 8) = 0.0348, Solution to Exercise 11.14.18 (p., You will lose $5, Solution to Exercise 11.14.19 (p., Becca, Solution to Exercise 11.14.20 (p., 14, Solution to Exercise 11.14.21 (p., •., •., •., •., •., •., , 508), 508), 508), 508), 509), 509), , Sample mean = 3.2, Sample standard deviation = 1.85, Median = 3, Quartile 1 = 2, Quartile 3 = 5, IQR = 3, , Solution to Exercise 11.14.22 (p. 509), d. z = −1.19, e. 0.1171, f. Do not reject the null, Solution to Exercise 11.14.23 (p. 510), We conclude that the patient does have the HIV virus when, in fact, the patient does not., Solution to Exercise 11.14.24 (p. 510), c. z = 2.21 ; p = 0.0136, d. Reject the null, e. We conclude that the proportion of Californian professionals that wear jeans to work is greater than the, proportion of non-Californian professionals when, in fact, it is not greater., f. We cannot conclude that the proportion of Californian professionals that wear jeans to work is greater, than the proportion of non-Californian professionals when, in fact, it is greater., Solution to Exercise 11.14.25 (p. 510), C, Solution to Exercise 11.14.26 (p. 510), t5, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 12, , Linear Regression and Correlation, 12.1 Linear Regression and Correlation1, 12.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, , Discuss basic ideas of linear regression and correlation., Create and interpret a line of best fit., Calculate and interpret the correlation coefficient., Calculate and interpret outliers., , 12.1.2 Introduction, Professionals often want to know how two or more numeric variables are related. For example, is there a, relationship between the grade on the second math exam a student takes and the grade on the final exam?, If there is a relationship, what is it and how strong is the relationship?, In another example, your income may be determined by your education, your profession, your years of, experience, and your ability. The amount you pay a repair person for labor is often determined by an initial, amount plus an hourly fee. These are all examples in which regression can be used., The type of data described in the examples is bivariate data - "bi" for two variables. In reality, statisticians, use multivariate data, meaning many variables., In this chapter, you will be studying the simplest form of regression, "linear regression" with one independent variable (x). This involves data that fits a line in two dimensions. You will also study correlation which, measures how strong the relationship is., , 12.2 Linear Equations2, Linear regression for two variables is based on a linear equation with one independent variable. It has the, form:, y = a + bx, 1 This, 2 This, , content is available online at <http://cnx.org/content/m17089/1.6/>., content is available online at <http://cnx.org/content/m17086/1.4/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 523, , (12.0)
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524, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , where a and b are constant numbers., x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute, for the independent variable and then solve for the dependent variable., Example 12.1, The following examples are linear equations., y = 3 + 2x, , (12.0), , y = −0.01 + 1.2x, , (12.0), , The graph of a linear equation of the form y = a + bx is a straight line. Any line that is not vertical can be, described by this equation., Example 12.2, , Figure 12.1: Graph of the equation y = −1 + 2x., , Linear equations of this form occur in applications of life sciences, social sciences, psychology, business,, economics, physical sciences, mathematics, and other areas., Example 12.3, Aaron’s Word Processing Service (AWPS) does word processing. Its rate is $32 per hour plus a, $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to, do the word processing job., Problem, Find the equation that expresses the total cost in terms of the number of hours required to finish, the word processing job., Solution, Let x = the number of hours it takes to get the job done., Let y = the total cost to the customer., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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525, The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32) ( x ) is the cost of the, word processing only. The total cost is:, y = 31.50 + 32x, , 12.3 Slope and Y-Intercept of a Linear Equation3, For the linear equation y = a + bx, b = slope and a = y-intercept., From algebra recall that the slope is a number that describes the steepness of a line and the y-intercept is, the y coordinate of the point (0, a) where the line crosses the y-axis., , (a), , (b), , (c), , Figure 12.2: Three possible graphs of y = a + bx. (a) If b > 0, the line slopes upward to the right. (b) If, b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right., , Example 12.4, Svetlana tutors to make extra money for college. For each tutoring session, she charges a one, time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of, money Svetlana earns for each session she tutors is y = 25 + 15x., Problem, What are the independent and dependent variables? What is the y-intercept and what is the, slope? Interpret them using complete sentences., Solution, The independent variable (x) is the number of hours Svetlana tutors each session. The dependent, variable (y) is the amount, in dollars, Svetlana earns for each session., The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a one-time fee, of $25 (this is when x = 0). The slope is 15 (b = 15). For each session, Svetlana earns $15 for each, hour she tutors., , 3 This, , content is available online at <http://cnx.org/content/m17083/1.5/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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526, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.4 Scatter Plots4, Before we take up the discussion of linear regression and correlation, we need to examine a way to display, the relation between two variables x and y. The most common and easiest way is a scatter plot. The, following example illustrates a scatter plot., Example 12.5, From an article in the Wall Street Journal : In Europe and Asia, m-commerce is popular. Mcommerce users have special mobile phones that work like electronic wallets as well as provide, phone and Internet services. Users can do everything from paying for parking to buying a TV set, or soda from a machine to banking to checking sports scores on the Internet. For the years 2000, through 2004, was there a relationship between the year and the number of m-commerce users?, Construct a scatter plot. Let x = the year and let y = the number of m-commerce users, in millions., , x (year), , y (# of users), , 2000, , 0.5, , 2002, , 20.0, , 2003, , 33.0, , 2004, , (b), , 47.0, (a), , Figure 12.3: (a) Table showing the number of m-commerce users (in millions) by year. (b) Scatter plot, showing the number of m-commerce users (in millions) by year., , A scatter plot shows the direction and strength of a relationship between the variables. A clear direction, happens when there is either:, , • High values of one variable occurring with high values of the other variable or low values of one, variable occurring with low values of the other variable., • High values of one variable occurring with low values of the other variable., You can determine the strength of the relationship by looking at the scatter plot and seeing how close the, points are to a line, a power function, an exponential function, or to some other type of function., When you look at a scatterplot, you want to notice the overall pattern and any deviations from the pattern., The following scatterplot examples illustrate these concepts., 4 This, , content is available online at <http://cnx.org/content/m17082/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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527, , (a) Positive Linear Pattern (Strong), , (b) Linear Pattern w/ One Deviation, , Figure 12.4, , (a) Negative Linear Pattern (Strong), , (b) Negative Linear Pattern (Weak), , Figure 12.5, , (a) Exponential Growth Pattern, , (b) No Pattern, , Figure 12.6, , In this chapter, we are interested in scatter plots that show a linear pattern. Linear patterns are quite common. The linear relationship is strong if the points are close to a straight line. If we think that the points, show a linear relationship, we would like to draw a line on the scatter plot. This line can be calculated, through a process called linear regression. However, we only calculate a regression line if one of the variables helps to explain or predict the other variable. If x is the independent variable and y the dependent, variable, then we can use a regression line to predict y for a given value of x., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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528, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.5 The Regression Equation5, Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you, have a set of data whose scatter plot appears to "fit" a straight line. This is called a Line of Best Fit or Least, Squares Line., , 12.5.1 Optional Collaborative Classroom Activity, If you know a person’s pinky (smallest) finger length, do you think you could predict that person’s height?, Collect data from your class (pinky finger length, in inches). The independent variable, x, is pinky finger, length and the dependent variable, y, is height., For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then, "by eye" draw a line that appears to "fit" the data. For your line, pick two convenient points and use them, to find the slope of the line. Find the y-intercept of the line by extending your lines so they cross the y-axis., Using the slopes and the y-intercepts, write your equation of "best fit". Do you think everyone will have, the same equation? Why or why not?, Using your equation, what is the predicted height for a pinky length of 2.5 inches?, Example 12.6, A random sample of 11 statistics students produced the following data where x is the third exam, score, out of 80, and y is the final exam score, out of 200. Can you predict the final exam score of a, random student if you know the third exam score?, 5 This, , content is available online at <http://cnx.org/content/m17090/1.15/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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529, x (third exam score), , y (final exam score), , 65, , 175, , 67, , 133, , 71, , 185, , 71, , 163, , 66, , 126, , 75, , 198, , 67, , 153, , 70, , 163, , 71, , 159, , 69, , 151, , 69, , 159, (a), , (b), , Figure 12.7: (a) Table showing the scores on the final exam based on scores from the third exam. (b) Scatter, plot showing the scores on the final exam based on scores from the third exam., , The third exam score, x, is the independent variable and the final exam score, y, is the dependent variable., We will plot a regression line that best "fits" the data. If each of you were to fit a line "by eye", you would, draw different lines. We can use what is called a least-squares regression line to obtain the best fit line., Consider the following diagram. Each point of data is of the !, the form ( x, y)and each point of the line of, ^, best fit using least-squares linear regression has the form x, y ., ^, The y is read "y hat" and is the estimated value of y. It is the value of y obtained using the regression line., It is not generally equal to y from data., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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530, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Figure 12.8, , ^, The term y0 − y 0 = e0 is called the "error" or residual. It is not an error in the sense of a mistake. The, absolute value of a residual measures the vertical distance between the actual value of y and the estimated, value of y. In other words, it measures the vertical distance between the actual data point and the predicted, point on the line., If the observed data point lies above the line, the residual is positive, and the line underestimates the, actual data value for y. If the observed data point lies below the line, the residual is negative, and the line, overestimates that actual data value for y., ^, In the diagram above, y0 − y 0 = e0 is the residual for the point shown. Here the point lies above the line, and the residual is positive., e = the Greek letter epsilon, ^, For each data point, you can calculate the residuals or errors, yi − y i = ei for i = 1, 2, 3, ..., 11., Each |e| is a vertical distance., For the example about the third exam scores and the final exam scores for the 11 statistics students, there, are 11 data points. Therefore, there are 11 e values. If you square each e and add, you get, 11, , (e1 )2 + (e2 )2 + ... + (e11 )2 = Σ e2, i=1, , This is called the Sum of Squared Errors (SSE)., Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make, the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line, of best fit has the equation:, ^, y = a + bx, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (12.8)
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531, where a = y − b · x and b =, , Σ( x − x )·(y−y), ., Σ ( x − x )2, , x and y are the sample means of the x values and the y values, respectively. The best fit line always passes, through the point ( x, y)., , s, The slope b can be written as b = r · syx where sy = the standard deviation of the y values and s x = the, standard deviation of the x values. r is the correlation coefficient which is discussed in the next section., Least Squares Criteria for Best Fit, The process of fitting the best fit line is called linear regression. The idea behind finding the best fit line is, based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is, that the sum of the squared errors (SSE) is minimized, that is made as small as possible. Any other line you, might choose would have a higher SSE than the best fit line. This best fit line is called the least squares, regression line ., NOTE : Computer spreadsheets, statistical software, and many calculators can quickly calculate the, best fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions, to use the TI-83, TI-83+, and TI-84+ calculators to find the best fit line and create a scatterplot are, shown at the end of this section., , THIRD EXAM vs FINAL EXAM EXAMPLE:, The graph of the line of best fit for the third exam/final exam example is shown below:, , Figure 12.9, , The least squares regression line (best fit line) for the third exam/final exam example has the equation:, ^, y = −173.51 + 4.83x, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , (12.9)
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532, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, NOTE :, , Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that, there is a linear relationship between the variables, then it is reasonable to use a best fit line, to make predictions for y given x within the domain of x-values in the sample data, but not, necessarily for x-values outside that domain., You could use the line to predict the final exam score for a student who earned a grade of 73 on, the third exam., You should NOT use the line to predict the final exam score for a student who earned a grade of, 50 on the third exam, because 50 is not within the domain of the x-values in the sample data,, which are between 65 and 75., UNDERSTANDING SLOPE, The slope of the line, b, describes how changes in the variables are related. It is important to interpret, the slope of the line in the context of the situation represented by the data. You should be able to write a, sentence interpreting the slope in plain English., INTERPRETATION OF THE SLOPE: The slope of the best fit line tells us how the dependent variable (y), changes for every one unit increase in the independent (x) variable, on average., THIRD EXAM vs FINAL EXAM EXAMPLE, Slope: The slope of the line is b = 4.83., Interpretation: For a one point increase in the score on the third exam, the final exam score increases by, 4.83 points, on average., , 12.5.2 Using the TI-83+ and TI-84+ Calculators, Using the Linear Regression T Test: LinRegTTest, Step 1. In the STAT list editor, enter the X data in list L1 and the Y data in list L2, paired so that the corresponding (x,y) values are next to each other in the lists. (If a particular pair of values is repeated, enter, it as many times as it appears in the data.), Step 2. On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest. (Be careful to select, LinRegTTest as some calculators may also have a different item called LinRegTInt.), Step 3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1, Step 4. On the next line, at the prompt β or ρ, highlight "6= 0" and press ENTER, Step 5. Leave the line for "RegEq:" blank, Step 6. Highlight Calculate and press ENTER., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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533, , Figure 12.10, , The output screen contains a lot of information. For now we will focus on a few items from the output, and, will return later to the other items., The second line says y=a+bx. Scroll down to find the values a=-173.513, and b=4.8273 ; the equation of the, ^, best fit line is y = −173.51 + 4.83x, The two items at the bottom are r2 = .43969 and r=.663. For now, just note where to find these values; we, will discuss them in the next two sections., Graphing the Scatterplot and Regression Line, Step 1., Step 2., Step 3., Step 4., Step 5., Step 6., Step 7., , We are assuming your X data is already entered in list L1 and your Y data is in list L2, Press 2nd STATPLOT ENTER to use Plot 1, On the input screen for PLOT 1, highlight On and press ENTER, For TYPE: highlight the very first icon which is the scatterplot and press ENTER, Indicate Xlist: L1 and Ylist: L2, For Mark: it does not matter which symbol you highlight., Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the, window to the data, Step 8. To graph the best fit line, press the "Y=" key and type the equation -173.5+4.83X into equation Y1., (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it., Step 9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired, window using Xmin, Xmax, Ymin, Ymax, **With contributions from Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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534, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.6 Correlation Coefficient and Coefficient of Determination6, 12.6.1 The Correlation Coefficient r, Besides looking at the scatter plot and seeing that a line seems reasonable, how can you tell if the line is a, good predictor? Use the correlation coefficient as another indicator (besides the scatterplot) of the strength, of the relationship between x and y., The correlation coefficient, r, developed by Karl Pearson in the early 1900s, is a numerical measure of the, strength of association between the independent variable x and the dependent variable y., The correlation coefficient is calculated as, r = rh, , n · Σx · y − (Σx ) · (Σy), i h, i, n · Σx2 − (Σx )2 · n · Σy2 − (Σy)2, , (12.10), , where n = the number of data points., If you suspect a linear relationship between x and y, then r can measure how strong the linear relationship, is., What the VALUE of r tells us:, , • The value of r is always between -1 and +1: −1 ≤ r ≤ 1., • The size of the correlation r indicates the strength of the linear relationship between x and y. Values, of r close to -1 or to +1 indicate a stronger linear relationship between x and y., • If r = 0 there is absolutely no linear relationship between x and y (no linear correlation)., • If r = 1, there is perfect positive correlation. If r = −1, there is perfect negative correlation. In both, these cases, all of the original data points lie on a straight line. Of course, in the real world, this will, not generally happen., What the SIGN of r tells us, , • A positive value of r means that when x increases, y tends to increase and when x decreases, y tends, to decrease (positive correlation)., • A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends, to increase (negative correlation)., • The sign of r is the same as the sign of the slope, b, of the best fit line., NOTE : Strong correlation does not suggest that x causes y or y causes x. We say "correlation does, not imply causation." For example, every person who learned math in the 17th century is dead., However, learning math does not necessarily cause death!, 6 This, , content is available online at <http://cnx.org/content/m17092/1.12/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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535, , (a) Positive Correlation, , (b) Negative Correlation, , (c) Zero Correlation, , Figure 12.11: (a) A scatter plot showing data with a positive correlation. 0 < r < 1 (b) A scatter plot, showing data with a negative correlation. −1 < r < 0 (c) A scatter plot showing data with zero correlation., r=0, , The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can quickly calculate r. The correlation coefficient r is the bottom item in the output screens for the, LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions)., , 12.6.2 The Coefficient of Determination, r2 is called the coefficient of determination. r2 is the square of the correlation coefficient , but is usually, stated as a percent, rather than in decimal form. r2 has an interpretation in the context of the data:, , • r2 , when expressed as a percent, represents the percent of variation in the dependent variable y that, can be explained by variation in the independent variable x using the regression (best fit) line., • 1-r2 , when expressed as a percent, represents the percent of variation in y that is NOT explained by, variation in x using the regression line. This can be seen as the scattering of the observed data points, about the regression line., Consider the third exam/final exam example introduced in the previous section, ^, The line of best fit is: y = −173.51 + 4.83x, The correlation coefficient is r = 0.6631, The coefficient of determination is r2 = 0.66312 = 0.4397, Interpretation of r2 in the context of this example:, Approximately 44% of the variation (0.4397 is approximately 0.44) in the final exam grades can be explained by the variation in the grades on the third exam, using the best fit regression line., Therefore approximately 56% of the variation (1 - 0.44 = 0.56) in the final exam grades can NOT be explained by the variation in the grades on the third exam, using the best fit regression line. (This is, seen as the scattering of the points about the line.), **With contributions from Roberta Bloom., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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536, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.7 Testing the Significance of the Correlation Coefficient7, 12.7.1 Testing the Significance of the Correlation Coefficient, The correlation coefficient, r, tells us about the strength of the linear relationship between x and y. However,, the reliability of the linear model also depends on how many observed data points are in the sample. We, need to look at both the value of the correlation coefficient r and the sample size n, together., We perform a hypothesis test of the "significance of the correlation coefficient" to decide whether the, linear relationship in the sample data is strong enough to use to model the relationship in the population., The sample data is used to compute r, the correlation coefficient for the sample. If we had data for the, entire population, we could find the population correlation coefficient. But because we only have sample, data, we can not calculate the population correlation coefficient. The sample correlation coefficient, r, is our, estimate of the unknown population correlation coefficient., The symbol for the population correlation coefficient is ρ, the Greek letter "rho"., ρ = population correlation coefficient (unknown), r = sample correlation coefficient (known; calculated from sample data), The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is "close to, 0" or "significantly different from 0". We decide this based on the sample correlation coefficient r and the, sample size n., If the test concludes that the correlation coefficient is significantly different from 0, we say that the, correlation coefficient is "significant"., , • Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship, between x and y because the correlation coefficient is significantly different from 0.", • What the conclusion means: There is a significant linear relationship between x and y. We can use the, regression line to model the linear relationship between x and y in the population., If the test concludes that the correlation coefficient is not significantly different from 0 (it is close to 0),, we say that correlation coefficient is "not significant"., , • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship, between x and y because the correlation coefficient is not significantly different from 0.", • What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we can NOT use the regression line to model a linear relationship between x and y in the population., NOTE :, , • If r is significant and the scatter plot shows a linear trend, the line can be used to predict the, value of y for values of x that are within the domain of observed x values., • If r is not significant OR if the scatter plot does not show a linear trend, the line should not be, used for prediction., • If r is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate, or reliable for prediction OUTSIDE the domain of observed x values in the data., PERFORMING THE HYPOTHESIS TEST, SETTING UP THE HYPOTHESES:, , • Null Hypothesis: Ho : ρ = 0, 7 This, , content is available online at <http://cnx.org/content/m17077/1.15/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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537, , • Alternate Hypothesis: Ha : ρ 6= 0, What the hypotheses mean in words:, , •, , Null Hypothesis Ho : The population correlation coefficient IS NOT significantly different from 0., There IS NOT a significant linear relationship(correlation) between x and y in the population., • Alternate Hypothesis Ha : The population correlation coefficient IS significantly DIFFERENT FROM, 0. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population., DRAWING A CONCLUSION:, There are two methods to make the decision. Both methods are equivalent and give the same result., Method 1: Using the p-value, Method 2: Using a table of critical values, In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05, Note: Using the p-value method, you could choose any appropriate significance level you want; you are, not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that, we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level, than 5% with the critical value method, we would need different tables of critical values that are not, provided in this textbook.), METHOD 1: Using a p-value to make a decision, The linear regression t-test LinRegTTEST on the TI-83+ or TI-84+ calculators calculates the p-value., On the LinRegTTEST input screen, on the line prompt for β or ρ, highlight "6= 0", The output screen shows the p-value on the line that reads "p ="., (Most computer statistical software can calculate the p-value.), If the p-value is less than the significance level (α = 0.05):, , • Decision: REJECT the null hypothesis., • Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship, between x and y because the correlation coefficient is significantly different from 0.", If the p-value is NOT less than the significance level (α = 0.05), , • Decision: DO NOT REJECT the null hypothesis., • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship, between x and y because the correlation coefficient is NOT significantly different from 0.", Calculation Notes:, You will use technology to calculate the p-value. The following describe the calculations to compute the, test statistics and the p-value:, The p-value is calculated using a t-distribution, with n − 2 degrees of freedom., √, , The formula for the test statistic is t = r√ n−22 . The value of the test statistic, t, is shown in the computer, 1−r, or calculator output along with the p-value. The test statistic t has the same sign as the correlation, coefficient r., The p-value is the combined area in both tails., An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99,, n-2) in 2nd DISTR., THIRD EXAM vs FINAL EXAM EXAMPLE: p value method, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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538, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , • Consider the third exam/final exam example., ^, • The line of best fit is: y = −173.51 + 4.83x with r = 0.6631 and there are n = 11 data points., • Can the regression line be used for prediction? Given a third exam score (x value), can we use the, line to predict the final exam score (predicted y value)?, Ho : ρ = 0, Ha : ρ 6 = 0, α = 0.05, The p-value is 0.026 (from LinRegTTest on your calculator or from computer software), The p-value, 0.026, is less than the significance level of α = 0.05, Decision: Reject the Null Hypothesis Ho, Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between, x and y because the correlation coefficient is significantly different from 0., Because r is significant and the scatter plot shows a linear trend, the regression line can be used to, predict final exam scores., METHOD 2: Using a table of Critical Values to make a decision, The 95% Critical Values of the Sample Correlation Coefficient Table (Section 12.10) at the end of this, chapter (before the Summary (Section 12.11)) may be used to give you a good idea of whether the computed value of r is significant or not. Compare r to the appropriate critical value in the table. If r is not, between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may want to use the line for prediction., Example 12.7, Suppose you computed r = 0.801 using n = 10 data points. df = n − 2 = 10 − 2 = 8. The, critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r >, positive critical value, then r is significant. Since r = 0.801 and 0.801 > 0.632, r is significant and, the line may be used for prediction. If you view this example on a number line, it will help you., , Figure 12.12: r is not significant between -0.632 and +0.632. r = 0.801 > + 0.632. Therefore, r is significant., , Example 12.8, Suppose you computed r = −0.624 with 14 data points. df = 14 − 2 = 12. The critical values are, -0.532 and 0.532. Since −0.624<−0.532, r is significant and the line may be used for prediction, , Figure 12.13: r = −0.624<−0.532. Therefore, r is significant., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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539, Example 12.9, Suppose you computed r = 0.776 and n = 6. df = 6 − 2 = 4. The critical values are -0.811, and 0.811. Since −0.811< 0.776 < 0.811, r is not significant and the line should not be used for, prediction., , Figure 12.14: −0.811<r = 0.776<0.811. Therefore, r is not significant., , THIRD EXAM vs FINAL EXAM EXAMPLE: critical value method, • Consider the third exam/final exam example., ^, • The line of best fit is: y = −173.51 + 4.83x with r = 0.6631 and there are n = 11 data points., • Can the regression line be used for prediction? Given a third exam score (x value), can we use the, line to predict the final exam score (predicted y value)?, Ho : ρ = 0, Ha : ρ 6 = 0, α = 0.05, Use the "95% Critical Value" table for r with df = n − 2 = 11 − 2 = 9, The critical values are -0.602 and +0.602, Since 0.6631 > 0.602, r is significant., Decision: Reject Ho :, Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between, x and y because the correlation coefficient is significantly different from 0., Because r is significant and the scatter plot shows a linear trend, the regression line can be used to, predict final exam scores., Example 12.10: Additional Practice Examples using Critical Values, Suppose you computed the following correlation coefficients. Using the table at the end of the, chapter, determine if r is significant and the line of best fit associated with each r can be used to, predict a y value. If it helps, draw a number line., 1. r = −0.567 and the sample size, n, is 19. The df = n − 2 = 17. The critical value is -0.456., −0.567<−0.456 so r is significant., 2. r = 0.708 and the sample size, n, is 9. The df = n − 2 = 7. The critical value is 0.666., 0.708 > 0.666 so r is significant., 3. r = 0.134 and the sample size, n, is 14. The df = 14 − 2 = 12. The critical value is 0.532., 0.134 is between -0.532 and 0.532 so r is not significant., 4. r = 0 and the sample size, n, is 5. No matter what the dfs are, r = 0 is between the two, critical values so r is not significant., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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540, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.7.2 Assumptions in Testing the Significance of the Correlation Coefficient, Testing the significance of the correlation coefficient requires that certain assumptions about the data are, satisfied. The premise of this test is that the data are a sample of observed points taken from a larger, population. We have not examined the entire population because it is not possible or feasible to do so. We, are examining the sample to draw a conclusion about whether the linear relationship that we see between, x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear, relationship between x and y in the population., The regression line equation that we calculate from the sample data gives the best fit line for our particular, sample. We want to use this best fit line for the sample as an estimate of the best fit line for the population., Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it, is appropriate to do this., The assumptions underlying the test of significance are:, , • There is a linear relationship in the population that models the average value of y for varying values, of x. In other words, the expected value of y for each particular value lies on a straight line in the, population. (We do not know the equation for the line for the population. Our regression line from, the sample is our best estimate of this line in the population.), • The y values for any particular x value are normally distributed about the line. This implies that, there are more y values scattered closer to the line than are scattered farther away. Assumption (1), above implies that these normal distributions are centered on the line: the means of these normal, distributions of y values lie on the line., • The standard deviations of the population y values about the line are equal for each value of x. In, other words, each of these normal distributions of y values has the same shape and spread about the, line., • The residual errors are mutually independent (no pattern)., , Figure 12.15: The y values for each x value are normally distributed about the line with the same standard, deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the, line than are scattered further away from the line., , **With contributions from Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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541, , 12.8 Prediction8, Recall the third exam/final exam example., We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best fit line for the final exam grade as a function of the grade on the third exam. We can now, use the least squares regression line for prediction., Suppose you want to estimate, or predict, the final exam score of statistics students who received 73 on the, third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75,, substitute x = 73 into the equation. Then:, ^, y = −173.51 + 4.83 (73) = 179.08, , (12.15), , We predict that statistic students who earn a grade of 73 on the third exam will earn a grade of 179.08 on, the final exam, on average., Example 12.11, Recall the third exam/final exam example., Problem 1, What would you predict the final exam score to be for a student who scored a 66 on the third, exam?, Solution, 145.27, , Problem 2, (Solution on p. 579.), What would you predict the final exam score to be for a student who scored a 90 on the third, exam?, **With contributions from Roberta Bloom, , 12.9 Outliers9, In some data sets, there are values (observed data points) called outliers. Outliers are observed data, points that are far from the least squares line. They have large "errors", where the "error" or residual is the, vertical distance from the line to the point., Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included, in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier, may hold valuable information about the population under study and should remain included in the data., The key is to carefully examine what causes a data point to be an outlier., Besides outliers, a sample may contain one or a few points that are called influential points. Influential, points are observed data points that are far from the other observed data points in the horizontal direction., These points may have a big effect on the slope of the regression line. To begin to identify an influential, point, you can remove it from the data set and see if the slope of the regression line is changed significantly., 8 This, 9 This, , content is available online at <http://cnx.org/content/m17095/1.8/>., content is available online at <http://cnx.org/content/m17094/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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542, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Computers and many calculators can be used to identify outliers from the data. Computer output for, regression analysis will often identify both outliers and influential points so that you can examine them., Identifying Outliers, We could guess at outliers by looking at a graph of the scatterplot and best fit line. However we would like, some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule, of thumb, we can flag any point that is located further than two standard deviations above or below the, best fit line as an outlier. The standard deviation used is the standard deviation of the residuals or errors., We can do this visually in the scatterplot by drawing an extra pair of lines that are two standard deviations, above and below the best fit line. Any data points that are outside this extra pair of lines are flagged as, potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the, standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure, is shown first, followed by the numerical calculations. You would generally only need to use one of these, methods., Example 12.12, In the third exam/final exam example, you can determine if there is an outlier or not. If there is, an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the, new line ought to fit the remaining data better. This means the SSE should be smaller and the, correlation coefficient ought to be closer to 1 or -1., Solution, Graphical Identification of Outliers, With the TI-83,83+,84+ graphing calculators, it is easy to identify the outlier graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point, on the line of best fit and that distance was equal to 2s or farther, then we would consider the data, point to be "too far" from the line of best fit. We need to find and graph the lines that are two, standard deviations below and above the regression line. Any points that are outside these two, lines are outliers. We will call these lines Y2 and Y3:, As we did with the equation of the regression line and the correlation coefficient, we will use, technology to calculate this standard deviation for us. Using the LinRegTTest with this data,, scroll down through the output screens to find s=16.412, Line Y2=-173.5+4.83x-2(16.4) and line Y3=-173.5+4.83x+2(16.4), ^, where y =-173.5+4.83x is the line of best fit. Y2 and Y3 have the same slope as the line of, best fit., Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and, Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not, between lines Y2 and Y3 is the point x=65, y=175. On the calculator screen it is just barely outside, these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final, exam; this point is further than 2 standard deviations away from the best fit line., Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell, if the point is between or outside the lines. On a computer, enlarging the graph may help; on a, small calculator screen, zooming in may make the graph clearer. Note that when the graph does, not give a clear enough picture, you can use the numerical comparisons to identify outliers., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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543, , Figure 12.16, , Numerical Identification of Outliers, In the table below, the first two columns are the third exam and final exam data. The third, ^, ^, column shows the predicted y values calculated from the line of best fit: y =-173.5+4.83x. The, residuals, or errors, have been calculated in the fourth column of the table: observed y value −, ^, predicted y value = y− y ., ^, s is the standard deviation of all the y− y = e values where n = the total number of data points. If, each residual is calculated and squared, and the results are added, we get the SSE. The standard, deviation of the residuals is calculated from the SSE as:, q, s = nSSE, −2, Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For, this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the, residuals 35; -17; 16; -6; -19; 9; 3; -1; -10; -9; -1., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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544, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , x, , y, , ^, y, , ^, y− y, , 65, , 175, , 140, , 175 − 140 = 35, , 67, , 133, , 150, , 133 − 150 = −17, , 71, , 185, , 169, , 185 − 169 = 16, , 71, , 163, , 169, , 163 − 169 = −6, , 66, , 126, , 145, , 126 − 145 = −19, , 75, , 198, , 189, , 198 − 189 = 9, , 67, , 153, , 150, , 153 − 150 = 3, , 70, , 163, , 164, , 163 − 164 = −1, , 71, , 159, , 169, , 159 − 169 = −10, , 69, , 151, , 160, , 151 − 160 = −9, , 69, , 159, , 160, , 159 − 160 = −1, , Table 12.1, , We are looking for all data points for which the residual is greater than 2s=2(16.4)=32.8 or less than, -32.8. Compare these values to the residuals in column 4 of the table. The only such data point is, the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for, this student is 35., How does the outlier affect the best fit line?, Numerically and graphically, we have identified the point (65,175) as an outlier. We should reexamine the data for this point to see if there are any problems with the data. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in, the data set. For this problem, we will suppose that we examined the data and found that this, outlier data was an error. Therefore we will continue on and delete the outlier, so that we can, explore how it affects the results, as a learning experience., Compute a new best-fit line and correlation coefficient using the 10 remaining points:, On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest,, the new line of best fit and the correlation coefficient are:, ^, y = −355.19 + 7.39x and r = 0.9121, The new line with r = 0.9121 is a stronger correlation than the original (r=0.6631) because r =, 0.9121 is closer to 1. This means that the new line is a better fit to the 10 remaining data values., The line can better predict the final exam score given the third exam score., , Numerical Identification of Outliers: Calculating s and Finding Outliers Manually, If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by, doing the following., ^, First, square each |y− y | (See the TABLE above):, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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545, The squares are 352 ; 172 ; 162 ; 62 ; 192 ; 92 ; 32 ; 12 ; 102 ; 92 ; 12, ^, Then, add (sum) all the |y− y | squared terms using the formula, 11, , Σ, , i=1, , ^, | yi − y i |, , !2, , 11, , = Σ ei 2, i=1, , ^, (Recall that yi − y i = ei .), , = 352 + 172 + 162 + 62 + 192 + 92 + 32 + 12 + 102 + 92 + 12, = 2440 = SSE. The result, SSE is the Sum of Squared Errors., ^, Next, calculate s, the standard deviation of all the y− y = e values where n = the total number of data, points., q, The calculation is s = nSSE, −2, For the third exam/final exam problem, s =, , q, , 2440, 11−2, , = 16.47, , Next, multiply s by 1.9:, (1.9) · (16.47) = 31.29, ^, 31.29 is almost 2 standard deviations away from the mean of the y− y values., If we were to measure the vertical distance from any data point to the corresponding point on the line of, best fit and that distance is at least 1.9s, then we would consider the data point to be "too far" from the line, of best fit. We call that point a potential outlier., ^, For the example, if any of the |y− y | values are at least 31.29, the corresponding ( x, y) data point is a, potential outlier., ^, For the third exam/final exam problem, all the |y− y |’s are less than 31.29 except for the first one which is, 35., 35 > 31.29, , ^, That is, |y− y | ≥ (1.9) · (s), , ^, The point which corresponds to |y− y | = 35 is (65, 175). Therefore, the data point (65, 175) is a potential, outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.), The next step is to compute a new best-fit line using the 10 remaining points. The new line of best, fit and the correlation coefficient are:, ^, y = −355.19 + 7.39x and r = 0.9121, Example 12.13, Using this new line of best fit (based on the remaining 10 data points), what would a student, who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the, prediction made using the original line?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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546, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, Solution, , ^, Using the new line of best fit, y = −355.19 + 7.39(73) = 184.28. A student who scored 73 points, on the third exam would expect to earn 184 points on the final exam., ^, The original line predicted y = −173.51 + 4.83(73) = 179.08 so the prediction using the, new line with the outlier eliminated differs from the original prediction., , Example 12.14, (From The Consumer Price Indexes Web site) The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The, CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as, a measure of inflation. By providing information about price changes in the Nation’s economy to, government, business, and labor, the CPI helps them to make economic decisions. The President,, Congress, and the Federal Reserve Board use the CPI’s trends to formulate monetary and fiscal, policies. In the following table, x is the year and y is the CPI., Data:, x, , y, , 1915, , 10.1, , 1926, , 17.7, , 1935, , 13.7, , 1940, , 14.7, , 1947, , 24.1, , 1952, , 26.5, , 1964, , 31.0, , 1969, , 36.7, , 1975, , 49.3, , 1979, , 72.6, , 1980, , 82.4, , 1986, , 109.6, , 1991, , 130.7, , 1999, , 166.6, , Table 12.2, , Problem, , • Make a scatterplot of the data., •, •, •, •, , ^, Calculate the least squares line. Write the equation in the form y = a + bx., Draw the line on the scatterplot., Find the correlation coefficient. Is it significant?, What is the average CPI for the year 1990?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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547, Solution, , • Scatter plot and line of best fit., ^, • y = −3204 + 1.662x is the equation of the line of best fit., • r = 0.8694, • The number of data points is n = 14. Use the 95% Critical Values of the Sample Correlation, Coefficient table at the end of Chapter 12. n − 2 = 12. The corresponding critical value is, 0.532. Since 0.8694 > 0.532, r is significant., ^, • y = −3204 + 1.662 (1990) = 103.4 CPI, • Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2=-3204+1.662X2(25.4) and Y3=-3204+1.662X+2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside, it.), , Figure 12.17, , NOTE : In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more, appropriate model to use than a line. In this example, a statistician should prefer to use other, methods to fit a curve to this data, rather than model the data with the line we found. In addition, to doing the calculations, it is always important to look at the scatterplot when deciding whether, a linear model is appropriate., , If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website, ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt ; our data is taken from the column entitled, "Annual Avg." (third column from the right). For example you could add more current years of, data. Try adding the more recent years 2004 : CPI=188.9, 2008 : CPI=215.3 and 2011: CPI=224.9., ^, See how it affects the model. (Check: y = −4436 + 2.295x. r = 0.9018. Is r significant? Is the fit, better with the addition of the new points?), **With contributions from Roberta Bloom, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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548, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.10 95% Critical Values of the Sample Correlation Coefficient Table10, , 10 This, , content is available online at <http://cnx.org/content/m17098/1.6/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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549, Degrees of Freedom: n − 2, , Critical Values: (+ and −), , 1, , 0.997, , 2, , 0.950, , 3, , 0.878, , 4, , 0.811, , 5, , 0.754, , 6, , 0.707, , 7, , 0.666, , 8, , 0.632, , 9, , 0.602, , 10, , 0.576, , 11, , 0.555, , 12, , 0.532, , 13, , 0.514, , 14, , 0.497, , 15, , 0.482, , 16, , 0.468, , 17, , 0.456, , 18, , 0.444, , 19, , 0.433, , 20, , 0.423, , 21, , 0.413, , 22, , 0.404, , 23, , 0.396, , 24, , 0.388, , 25, , 0.381, , 26, , 0.374, continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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550, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, 27, , 0.367, , 28, , 0.361, , 29, , 0.355, , 30, , 0.349, , 40, , 0.304, , 50, , 0.273, , 60, , 0.250, , 70, , 0.232, , 80, , 0.217, , 90, , 0.205, , 100, , 0.195, , Table 12.3, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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551, , 12.11 Summary11, Bivariate Data: Each data point has two values. The form is ( x, y)., ^, Line of Best Fit or Least Squares Line (LSL): y = a + bx, x = independent variable; y = dependent variable, ^, Residual: Actual y value − predicted y value = y− y, Correlation Coefficient r:, 1., 2., 3., 4., , Used to determine whether a line of best fit is good for prediction., Between -1 and 1 inclusive. The closer r is to 1 or -1, the closer the original points are to a straight line., If r is negative, the slope is negative. If r is positive, the slope is positive., If r = 0, then the line is horizontal., , Sum of Squared Errors (SSE): The smaller the SSE, the better the original set of points fits the line of best, fit., Outlier: A point that does not seem to fit the rest of the data., , 11 This, , content is available online at <http://cnx.org/content/m17081/1.4/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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552, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.12 Practice: Linear Regression12, 12.12.1 Student Learning Outcomes, • The student will evaluate bivariate data and determine if a line is an appropriate fit to the data., , 12.12.2 Given, Below are real data for the first two decades of AIDS reporting. (Source: Centers for Disease Control and, Prevention, National Center for HIV, STD, and TB Prevention), Adults and Adolescents only, United States, Year, , # AIDS cases diagnosed, , # AIDS deaths, , Pre-1981, , 91, , 29, , 1981, , 319, , 121, , 1982, , 1,170, , 453, , 1983, , 3,076, , 1,482, , 1984, , 6,240, , 3,466, , 1985, , 11,776, , 6,878, , 1986, , 19,032, , 11,987, , 1987, , 28,564, , 16,162, , 1988, , 35,447, , 20,868, , 1989, , 42,674, , 27,591, , 1990, , 48,634, , 31,335, , 1991, , 59,660, , 36,560, , 1992, , 78,530, , 41,055, , 1993, , 78,834, , 44,730, , 1994, , 71,874, , 49,095, , 1995, , 68,505, , 49,456, , 1996, , 59,347, , 38,510, , 1997, , 47,149, , 20,736, , 1998, , 38,393, , 19,005, , 1999, , 25,174, , 18,454, , 2000, , 25,522, , 17,347, , 2001, , 25,643, , 17,402, , 2002, , 26,464, , 16,371, , Total, , 802,118, , 489,093, Table 12.4, , 12 This, , content is available online at <http://cnx.org/content/m17088/1.12/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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553, NOTE : We will use the columns “year” and “# AIDS cases diagnosed” for all questions unless, otherwise stated., , 12.12.3 Graphing, Graph “year” vs. “# AIDS cases diagnosed.” Plot the points on the graph located below in the section, titled "Plot" . Do not include pre-1981. Label both axes with words. Scale both axes., , 12.12.4 Data, Exercise 12.12.1, Enter your data into your calculator or computer. The pre-1981 data should not be included. Why, is that so?, , 12.12.5 Linear Equation, Write the linear equation below, rounding to 4 decimal places:, NOTE : For any prediction questions, the answers are calculated using the least squares (best fit), line equation cited in the solution., , Exercise 12.12.2, Calculate the following:, a., b., c., d., , (Solution on p. 579.), , a=, b=, corr. =, n =(# of pairs), , Exercise 12.12.3, ^, equation: y =, , (Solution on p. 579.), , 12.12.6 Solve, Exercise 12.12.4, Solve., , (Solution on p. 579.), , ^, a. When x = 1985, y =, ^, b. When x = 1990, y =, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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554, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.12.7 Plot, Plot the 2 above points on the graph below. Then, connect the 2 points to form the regression line., , Obtain the graph on your calculator or computer., , 12.12.8 Discussion Questions, Look at the graph above., Exercise 12.12.5, Does the line seem to fit the data? Why or why not?, Exercise 12.12.6, Do you think a linear fit is best? Why or why not?, Exercise 12.12.7, Hand draw a smooth curve on the graph above that shows the flow of the data., Exercise 12.12.8, What does the correlation imply about the relationship between time (years) and the number of, diagnosed AIDS cases reported in the U.S.?, Exercise 12.12.9, Why is “year” the independent variable and “# AIDS cases diagnosed.” the dependent variable, (instead of the reverse)?, Exercise 12.12.10, Solve., , (Solution on p. 579.), , ^, a. When x = 1970, y =:, b. Why doesn’t this answer make sense?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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555, , 12.13 Homework13, Exercise 12.13.1, (Solution on p. 579.), For each situation below, state the independent variable and the dependent variable., a. A study is done to determine if elderly drivers are involved in more motor vehicle fatalities, than all other drivers. The number of fatalities per 100,000 drivers is compared to the age of, drivers., b. A study is done to determine if the weekly grocery bill changes based on the number of family, members., c. Insurance companies base life insurance premiums partially on the age of the applicant., d. Utility bills vary according to power consumption., e. A study is done to determine if a higher education reduces the crime rate in a population., , NOTE : For any prediction questions, the answers are calculated using the least squares (best fit), line equation cited in the solution., , Exercise 12.13.2, Recently,, the, annual, number, of, driver, deaths, per, 100,000, for, the, selected, age, groups, was, as, follows, (Source:, http://, http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html 14, ):, Age, , Number of Driver Deaths per 100,000, , 16-19, , 38, , 20-24, , 36, , 25-34, , 24, , 35-54, , 20, , 55-74, , 18, , 75+, , 28, Table 12.5, , a. For each age group, pick the midpoint of the interval for the x value. (For the 75+ group, use, 80.), b. Using “ages” as the independent variable and “Number of driver deaths per 100,000” as the, dependent variable, make a scatter plot of the data., ^, c. Calculate the least squares (best–fit) line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. Pick two ages and find the estimated fatality rates., f. Use the two points in (e) to plot the least squares line on your graph from (b)., g. Based on the above data, is there a linear relationship between age of a driver and driver fatality, rate?, h. What is the slope of the least squares (best-fit) line? Interpret the slope., 13 This, , content is available online at <http://cnx.org/content/m17085/1.14/>., , 14 http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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556, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, Exercise 12.13.3, (Solution on p. 579.), The average number of people in a family that received welfare for various years is given below., (Source: House Ways and Means Committee, Health and Human Services Department ), Year, , Welfare family size, , 1969, , 4.0, , 1973, , 3.6, , 1975, , 3.2, , 1979, , 3.0, , 1983, , 3.0, , 1988, , 3.0, , 1991, , 2.9, Table 12.6, , a. Using “year” as the independent variable and “welfare family size” as the dependent variable,, make a scatter plot of the data., ^, b. Calculate the least squares line. Put the equation in the form of: y = a + bx, c. Find the correlation coefficient. Is it significant?, d. Pick two years between 1969 and 1991 and find the estimated welfare family sizes., e. Use the two points in (d) to plot the least squares line on your graph from (b)., f. Based on the above data, is there a linear relationship between the year and the average number, of people in a welfare family?, g. Using the least squares line, estimate the welfare family sizes for 1960 and 1995. Does the least, squares line give an accurate estimate for those years? Explain why or why not., h. Are there any outliers in the above data?, i. What is the estimated average welfare family size for 1986? Does the least squares line give an, accurate estimate for that year? Explain why or why not., j. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.4, Use the AIDS data from the practice for this section (Section 12.12.2: Given), but this time use the, columns “year #” and “# new AIDS deaths in U.S.” Answer all of the questions from the practice, again, using the new columns., Exercise 12.13.5, (Solution on p. 580.), The height (sidewalk to roof) of notable tall buildings in America is compared to the number of, stories of the building (beginning at street level). (Source: Microsoft Bookshelf ), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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557, , Height (in feet), , Stories, , 1050, , 57, , 428, , 28, , 362, , 26, , 529, , 40, , 790, , 60, , 401, , 22, , 380, , 38, , 1454, , 110, , 1127, , 100, , 700, , 46, , Table 12.7, , a. Using “stories” as the independent variable and “height” as the dependent variable, make a, scatter plot of the data., b. Does it appear from inspection that there is a relationship between the variables?, ^, c. Calculate the least squares line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. Find the estimated heights for 32 stories and for 94 stories., f. Use the two points in (e) to plot the least squares line on your graph from (b)., g. Based on the above data, is there a linear relationship between the number of stories in tall, buildings and the height of the buildings?, h. Are there any outliers in the above data? If so, which point(s)?, i. What is the estimated height of a building with 6 stories? Does the least squares line give an, accurate estimate of height? Explain why or why not., j. Based on the least squares line, adding an extra story is predicted to add about how many feet, to a building?, k. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.6, Below is the life expectancy for an individual born in the United States in certain years. (Source:, National Center for Health Statistics), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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558, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Year of Birth, , Life Expectancy, , 1930, , 59.7, , 1940, , 62.9, , 1950, , 70.2, , 1965, , 69.7, , 1973, , 71.4, , 1982, , 74.5, , 1987, , 75, , 1992, , 75.7, , 2010, , 78.7, Table 12.8, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Draw a scatter plot of the ordered pairs., ^, c. Calculate the least squares line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. Find the estimated life expectancy for an individual born in 1950 and for one born in 1982., f. Why aren’t the answers to part (e) the values on the above chart that correspond to those years?, g. Use the two points in (e) to plot the least squares line on your graph from (b)., h. Based on the above data, is there a linear relationship between the year of birth and life expectancy?, i. Are there any outliers in the above data?, j. Using the least squares line, find the estimated life expectancy for an individual born in 1850., Does the least squares line give an accurate estimate for that year? Explain why or why not., k. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.7, (Solution on p. 580.), The percent of female wage and salary workers who are paid hourly rates is given below for the, years 1979 - 1992. (Source: Bureau of Labor Statistics, U.S. Dept. of Labor), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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559, , Year, , Percent of workers paid hourly rates, , 1979, , 61.2, , 1980, , 60.7, , 1981, , 61.3, , 1982, , 61.3, , 1983, , 61.8, , 1984, , 61.7, , 1985, , 61.8, , 1986, , 62.0, , 1987, , 62.7, , 1990, , 62.8, , 1992, , 62.9, Table 12.9, , a. Using “year” as the independent variable and “percent” as the dependent variable, make a, scatter plot of the data., b. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, c. Calculate the least squares line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. Find the estimated percents for 1991 and 1988., f. Use the two points in (e) to plot the least squares line on your graph from (b)., g. Based on the above data, is there a linear relationship between the year and the percent of, female wage and salary earners who are paid hourly rates?, h. Are there any outliers in the above data?, i. What is the estimated percent for the year 2050? Does the least squares line give an accurate, estimate for that year? Explain why or why not?, j. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.8, The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition, 10, for various pages is given below., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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560, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Page number, , Maximum value ($), , 4, , 16, , 14, , 19, , 25, , 15, , 32, , 17, , 43, , 19, , 57, , 15, , 72, , 16, , 85, , 15, , 90, , 17, Table 12.10, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Draw a scatter plot of the ordered pairs., ^, c. Calculate the least squares line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. Find the estimated maximum values for the restaurants on page 10 and on page 70., f. Use the two points in (e) to plot the least squares line on your graph from (b)., g. Does it appear that the restaurants giving the maximum value are placed in the beginning of, the “Fine Dining” section? How did you arrive at your answer?, h. Suppose that there were 200 pages of restaurants. What do you estimate to be the maximum, value for a restaurant listed on page 200?, i. Is the least squares line valid for page 200? Why or why not?, j. What is the slope of the least squares (best-fit) line? Interpret the slope., The next two questions refer to the following data: The cost of a leading liquid laundry detergent in, different sizes is given below., Size (ounces), , Cost ($), , 16, , 3.99, , 32, , 4.99, , 64, , 5.99, , 200, , 10.99, , Cost per ounce, , Table 12.11, , Exercise 12.13.9, , (Solution on p. 580.), , a. Using “size” as the independent variable and “cost” as the dependent variable, make a scatter, plot., b. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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561, ^, c. Calculate the least squares line. Put the equation in the form of: y = a + bx, d. Find the correlation coefficient. Is it significant?, e. If the laundry detergent were sold in a 40 ounce size, find the estimated cost., f. If the laundry detergent were sold in a 90 ounce size, find the estimated cost., g. Use the two points in (e) and (f) to plot the least squares line on your graph from (a)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Are there any outliers in the above data?, j. Is the least squares line valid for predicting what a 300 ounce size of the laundry detergent, would cost? Why or why not?, k. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.10, a. Complete the above table for the cost per ounce of the different sizes., b. Using “Size” as the independent variable and “Cost per ounce” as the dependent variable,, make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. Is it significant?, f. If the laundry detergent were sold in a 40 ounce size, find the estimated cost per ounce., g. If the laundry detergent were sold in a 90 ounce size, find the estimated cost per ounce., h. Use the two points in (f) and (g) to plot the least squares line on your graph from (b)., i. Does it appear that a line is the best way to fit the data? Why or why not?, j. Are there any outliers in the above data?, k. Is the least squares line valid for predicting what a 300 ounce size of the laundry detergent, would cost per ounce? Why or why not?, l. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.11, (Solution on p. 580.), According to flyer by a Prudential Insurance Company representative, the costs of approximate, probate fees and taxes for selected net taxable estates are as follows:, Net Taxable Estate ($), , Approximate Probate Fees and Taxes ($), , 600,000, , 30,000, , 750,000, , 92,500, , 1,000,000, , 203,000, , 1,500,000, , 438,000, , 2,000,000, , 688,000, , 2,500,000, , 1,037,000, , 3,000,000, , 1,350,000, Table 12.12, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Make a scatter plot of the data., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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562, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. Is it significant?, f. Find the estimated total cost for a net taxable estate of $1,000,000. Find the cost for $2,500,000., g. Use the two points in (f) to plot the least squares line on your graph from (b)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Are there any outliers in the above data?, j. Based on the above, what would be the probate fees and taxes for an estate that does not have, any assets?, k. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.12, The following are advertised sale prices of color televisions at Anderson’s., Size (inches), , Sale Price ($), , 9, , 147, , 20, , 197, , 27, , 297, , 31, , 447, , 35, , 1177, , 40, , 2177, , 60, , 2497, Table 12.13, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. Is it significant?, f. Find the estimated sale price for a 32 inch television. Find the cost for a 50 inch television., g. Use the two points in (f) to plot the least squares line on your graph from (b)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Are there any outliers in the above data?, j. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.13, (Solution on p. 580.), Below are the average heights for American boys. (Source: Physician’s Handbook, 1990 ), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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563, , Age (years), , Height (cm), , birth, , 50.8, , 2, , 83.8, , 3, , 91.4, , 5, , 106.6, , 7, , 119.3, , 10, , 137.1, , 14, , 157.5, Table 12.14, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. Is it significant?, f. Find the estimated average height for a one year–old. Find the estimated average height for an, eleven year–old., g. Use the two points in (f) to plot the least squares line on your graph from (b)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Are there any outliers in the above data?, j. Use the least squares line to estimate the average height for a sixty–two year–old man. Do you, think that your answer is reasonable? Why or why not?, k. What is the slope of the least squares (best-fit) line? Interpret the slope., Exercise 12.13.14, The following chart gives the gold medal times for every other Summer Olympics for the women’s, 100 meter freestyle (swimming)., Year, , Time (seconds), , 1912, , 82.2, , 1924, , 72.4, , 1932, , 66.8, , 1952, , 66.8, , 1960, , 61.2, , 1968, , 60.0, , 1976, , 55.65, , 1984, , 55.92, , 1992, , 54.64, , 2000, , 53.8, , 2008, , 53.1, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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564, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, Table 12.15, , a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. Is the decrease in times significant?, f. Find the estimated gold medal time for 1932. Find the estimated time for 1984., g. Why are the answers from (f) different from the chart values?, h. Use the two points in (f) to plot the least squares line on your graph from (b)., i. Does it appear that a line is the best way to fit the data? Why or why not?, j. Use the least squares line to estimate the gold medal time for the next Summer Olympics. Do, you think that your answer is reasonable? Why or why not?, The next three questions use the following state information., , State, , # letters in name, , Year entered the, Union, , Rank for entering, the Union, , Area, miles), , Alabama, , 7, , 1819, , 22, , 52,423, , Colorado, , 1876, , 38, , 104,100, , Hawaii, , 1959, , 50, , 10,932, , Iowa, , 1846, , 29, , 56,276, , Maryland, , 1788, , 7, , 12,407, , Missouri, , 1821, , 24, , 69,709, , New Jersey, , 1787, , 3, , 8,722, , Ohio, , 1803, , 17, , 44,828, , 1788, , 8, , 32,008, , Utah, , 1896, , 45, , 84,904, , Wisconsin, , 1848, , 30, , 65,499, , South Carolina, , 13, , (square, , Table 12.16, , Exercise 12.13.15, (Solution on p. 581.), We are interested in whether or not the number of letters in a state name depends upon the year, the state entered the Union., a. Decide which variable should be the independent variable and which should be the dependent, variable., b. Make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. What does it imply about the significance of the relationship?, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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565, f. Find the estimated number of letters (to the nearest integer) a state would have if it entered, the Union in 1900. Find the estimated number of letters a state would have if it entered the, Union in 1940., g. Use the two points in (f) to plot the least squares line on your graph from (b)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Use the least squares line to estimate the number of letters a new state that enters the Union this, year would have. Can the least squares line be used to predict it? Why or why not?, Exercise 12.13.16, We are interested in whether there is a relationship between the ranking of a state and the area of, the state., a. Let rank be the independent variable and area be the dependent variable., b. What do you think the scatter plot will look like? Make a scatter plot of the data., c. Does it appear from inspection that there is a relationship between the variables? Why or why, not?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. What does it imply about the significance of the relationship?, f. Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas?, g. Use the two points in (f) to plot the least squares line on your graph from (b)., h. Does it appear that a line is the best way to fit the data? Why or why not?, i. Are there any outliers?, j. Use the least squares line to estimate the area of a new state that enters the Union. Can the least, squares line be used to predict it? Why or why not?, k. Delete “Hawaii” and substitute “Alaska” for it. Alaska is the fortieth state with an area of, 656,424 square miles., l. Calculate the new least squares line., m. Find the estimated area for Alabama. Is it closer to the actual area with this new least squares, line or with the previous one that included Hawaii? Why do you think that’s the case?, n. Do you think that, in general, newer states are larger than the original states?, Exercise 12.13.17, (Solution on p. 581.), We are interested in whether there is a relationship between the rank of a state and the year it, entered the Union., a. Let year be the independent variable and rank be the dependent variable., b. What do you think the scatter plot will look like? Make a scatter plot of the data., c. Why must the relationship be positive between the variables?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. What does it imply about the significance of the relationship?, f. Let’s say a fifty-first state entered the union. Based upon the least squares line, when should, that have occurred?, g. Using the least squares line, how many states do we currently have?, h. Why isn’t the least squares line a good estimator for this year?, Exercise 12.13.18, Below are the percents of the U.S. labor force (excluding self-employed and unemployed ) that, are members of a union. We are interested in whether the decrease is significant. (Source: Bureau, of Labor Statistics, U.S. Dept. of Labor), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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566, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Year, , Percent, , 1945, , 35.5, , 1950, , 31.5, , 1960, , 31.4, , 1970, , 27.3, , 1980, , 21.9, , 1993, , 15.8, , 2011, , 11.8, , Table 12.17, , a. Let year be the independent variable and percent be the dependent variable., b. What do you think the scatter plot will look like? Make a scatter plot of the data., c. Why will the relationship between the variables be negative?, ^, d. Calculate the least squares line. Put the equation in the form of: y = a + bx, e. Find the correlation coefficient. What does it imply about the significance of the relationship?, f. Based on your answer to (e), do you think that the relationship can be said to be decreasing?, g. If the trend continues, when will there no longer be any union members? Do you think that, will happen?, The next two questions refer to the following information: The data below reflects the 1991-92 Reunion, Class Giving. (Source: SUNY Albany alumni magazine), Class Year, , Average Gift, , Total Giving, , 1922, , 41.67, , 125, , 1927, , 60.75, , 1,215, , 1932, , 83.82, , 3,772, , 1937, , 87.84, , 5,710, , 1947, , 88.27, , 6,003, , 1952, , 76.14, , 5,254, , 1957, , 52.29, , 4,393, , 1962, , 57.80, , 4,451, , 1972, , 42.68, , 18,093, , 1976, , 49.39, , 22,473, , 1981, , 46.87, , 20,997, , 1986, , 37.03, , 12,590, , Table 12.18, , Exercise 12.13.19, (Solution on p. 581.), We will use the columns “class year” and “total giving” for all questions, unless otherwise stated., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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567, a. What do you think the scatter plot will look like? Make a scatter plot of the data., ^, b. Calculate the least squares line. Put the equation in the form of: y = a + bx, c. Find the correlation coefficient. What does it imply about the significance of the relationship?, d. For the class of 1930, predict the total class gift., e. For the class of 1964, predict the total class gift., f. For the class of 1850, predict the total class gift. Why doesn’t this value make any sense?, Exercise 12.13.20, We will use the columns “class year” and “average gift” for all questions, unless otherwise stated., a. What do you think the scatter plot will look like? Make a scatter plot of the data., ^, b. Calculate the least squares line. Put the equation in the form of: y = a + bx, c. Find the correlation coefficient. What does it imply about the significance of the relationship?, d. For the class of 1930, predict the average class gift., e. For the class of 1964, predict the average class gift., f. For the class of 2010, predict the average class gift. Why doesn’t this value make any sense?, Exercise 12.13.21, (Solution on p. 581.), We are interested in exploring the relationship between the weight of a vehicle and its fuel efficiency (gasoline mileage). The data in the table show the weights, in pounds, and fuel efficiency,, measured in miles per gallon, for a sample of 12 vehicles., Weight, , Fuel Efficiency, , 2715, , 24, , 2570, , 28, , 2610, , 29, , 2750, , 38, , 3000, , 25, , 3410, , 22, , 3640, , 20, , 3700, , 26, , 3880, , 21, , 3900, , 18, , 4060, , 18, , 4710, , 15, Table 12.19, , a., b., c., d., e., , Graph a scatterplot of the data., Find the correlation coefficient and determine if it is significant., Find the equation of the best fit line., Write the sentence that interprets the meaning of the slope of the line in the context of the data., What percent of the variation in fuel efficiency is explained by the variation in the weight of the, vehicles, using the regression line? (State your answer in a complete sentence in the context, of the data.), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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568, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, f. Accurately graph the best fit line on your scatterplot., g. For the vehicle that weights 3000 pounds, find the residual (y-yhat). Does the value predicted, by the line underestimate or overestimate the observed data value?, h. Identify any outliers, using either the graphical or numerical procedure demonstrated in the, textbook., i. The outlier is a hybrid car that runs on gasoline and electric technology, but all other vehicles, in the sample have engines that use gasoline only. Explain why it would be appropriate to, remove the outlier from the data in this situation. Remove the outlier from the sample data., Find the new correlation coefficient, coefficient of determination, and best fit line., j. Compare the correlation coefficients and coefficients of determination before and after removing, the outlier, and explain in complete sentences what these numbers indicate about how the, model has changed., Exercise 12.13.22, (Solution on p. 581.), The four data sets below were created by statistician Francis Anscomb. They show why it is important to examine the scatterplots for your data, in addition to finding the correlation coefficient,, in order to evaluate the appropriateness of fitting a linear model., Set 1, , Set 2, , Set 3, , Set 4, , x, , y, , x, , y, , x, , y, , x, , y, , 10, , 8.04, , 10, , 9.14, , 10, , 7.46, , 8, , 6.58, , 8, , 6.95, , 8, , 8.14, , 8, , 6.77, , 8, , 5.76, , 13, , 7.58, , 13, , 8.74, , 13, , 12.74, , 8, , 7.71, , 9, , 8.81, , 9, , 8.77, , 9, , 7.11, , 8, , 8.84, , 11, , 8.33, , 11, , 9.26, , 11, , 7.81, , 8, , 8.47, , 14, , 9.96, , 14, , 8.10, , 14, , 8.84, , 8, , 7.04, , 6, , 7.24, , 6, , 6.13, , 6, , 6.08, , 8, , 5.25, , 4, , 4.26, , 4, , 3.10, , 4, , 5.39, , 19, , 12.50, , 12, , 10.84, , 12, , 9.13, , 12, , 8.15, , 8, , 5.56, , 7, , 4.82, , 7, , 7.26, , 7, , 6.42, , 8, , 7.91, , 5, , 5.68, , 5, , 4.74, , 5, , 5.73, , 8, , 6.89, , Table 12.20, , a. For each data set, find the least squares regression line and the correlation coefficient. What did, you discover about the lines and values of r?, For each data set, create a scatter plot and graph the least squares regression line. Use the graphs, to answer the following questions:, b. For which data set does it appear that a curve would be a more appropriate model than a line?, c. Which data set has an influential point (point close to or on the line that greatly influences the, best fit line)?, d. Which data set has an outlier (obviously visible on the scatter plot with best fit line graphed)?, e. Which data set appears to be the most appropriate to model using the least squares regression, line?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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569, , 12.13.1 Try these multiple choice questions, Exercise 12.13.23, (Solution on p. 582.), A correlation coefficient of -0.95 means there is a ____________ between the two variables., A., B., C., D., , Strong positive correlation, Weak negative correlation, Strong negative correlation, No Correlation, , Exercise 12.13.24, (Solution on p. 582.), According to the data reported by the New York State Department of Health regarding West Nile, Virus (http://www.health.state.ny.us/nysdoh/westnile/update/update.htm) for the years 20002008, the least squares line equation for the number of reported dead birds (x) versus the number, ^, of human West Nile virus cases (y) is y = −10.2638 + 0.0491x. If the number of dead birds reported, in a year is 732, how many human cases of West Nile virus can be expected? r = 0.5490, A., B., C., D., , No prediction can be made., 19.6, 15, 38.1, , The next three questions refer to the following data: (showing the number of hurricanes by category to, directly strike the mainland U.S. each decade) obtained from www.nhc.noaa.gov/gifs/table6.gif 15 A major, hurricane is one with a strength rating of 3, 4 or 5., Decade, , Total Number of Hurricanes, , Number of Major Hurricanes, , 1941-1950, , 24, , 10, , 1951-1960, , 17, , 8, , 1961-1970, , 14, , 6, , 1971-1980, , 12, , 4, , 1981-1990, , 15, , 5, , 1991-2000, , 14, , 5, , 2001 – 2004, , 9, , 3, Table 12.21, , Exercise 12.13.25, (Solution on p. 582.), Using only completed decades (1941 – 2000), calculate the least squares line for the number of, major hurricanes expected based upon the total number of hurricanes., ^, A. y = −1.67x + 0.5, ^, B. y = 0.5x − 1.67, ^, C. y = 0.94x − 1.67, ^, D. y = −2x + 1, 15 http://www.nhc.noaa.gov/gifs/table6.gif, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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570, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, Exercise 12.13.26, (Solution on p. 582.), The correlation coefficient is 0.942. Is this considered significant? Why or why not?, A., B., C., D., , No, because 0.942 is greater than the critical value of 0.707, Yes, because 0.942 is greater than the critical value of 0.707, No, because 0942 is greater than the critical value of 0.811, Yes, because 0.942 is greater than the critical value of 0.811, , Exercise 12.13.27, (Solution on p. 582.), The data for 2001-2004 show 9 hurricanes have hit the mainland United States. The line of best fit, predicts 2.83 major hurricanes to hit mainland U.S. Can the least squares line be used to make this, prediction?, A., B., C., D., , No, because 9 lies outside the independent variable values, Yes, because, in fact, there have been 3 major hurricanes this decade, No, because 2.83 lies outside the dependent variable values, Yes, because how else could we predict what is going to happen this decade., , **Exercises 21 and 22 contributed by Roberta Bloom, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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571, , 12.14 Lab 1: Regression (Distance from School)16, Class Time:, Names:, , 12.14.1 Student Learning Outcomes:, • The student will calculate and construct the line of best fit between two variables., • The student will evaluate the relationship between two variables to determine if that relationship is, significant., , 12.14.2 Collect the Data, Use 8 members of your class for the sample. Collect bivariate data (distance an individual lives from school,, the cost of supplies for the current term)., 1. Complete the table., Distance from school, , Cost of supplies this term, , Table 12.22, , 2. Which variable should be the dependent variable and which should be the independent variable?, Why?, 3. Graph “distance” vs. “cost.” Plot the points on the graph. Label both axes with words. Scale both, axes., 16 This, , content is available online at <http://cnx.org/content/m17080/1.11/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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572, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Figure 12.18, , 12.14.3 Analyze the Data, Enter your data into your calculator or computer. Write the linear equation below, rounding to 4 decimal, places., 1. Calculate the following:, a=, b=, correlation =, n=, ^, e. equation: y =, f. Is the correlation significant? Why or why not? (Answer in 1-3 complete sentences.), a., b., c., d., , 2. Supply an answer for the following senarios:, a. For a person who lives 8 miles from campus, predict the total cost of supplies this term:, b. For a person who lives 80 miles from campus, predict the total cost of supplies this term:, 3. Obtain the graph on your calculator or computer. Sketch the regression line below., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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573, , Figure 12.19, , 12.14.4 Discussion Questions, 1. Answer each with 1-3 complete sentences., a. Does the line seem to fit the data? Why?, b. What does the correlation imply about the relationship between the distance and the cost?, 2. Are there any outliers? If so, which point is an outlier?, 3. Should the outlier, if it exists, be removed? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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574, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.15 Lab 2: Regression (Textbook Cost)17, Class Time:, Names:, , 12.15.1 Student Learning Outcomes:, • The student will calculate and construct the line of best fit between two variables., • The student will evaluate the relationship between two variables to determine if that relationship is, significant., , 12.15.2 Collect the Data, Survey 10 textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook)., 1. Complete the table., Number of pages, , Cost of textbook, , Table 12.23, , 2. Which variable should be the dependent variable and which should be the independent variable?, Why?, 3. Graph “distance” vs. “cost.” Plot the points on the graph in "Analyze the Data". Label both axes with, words. Scale both axes., , 12.15.3 Analyze the Data, Enter your data into your calculator or computer. Write the linear equation below, rounding to 4 decimal, places., 1. Calculate the following:, a., b., c., d., 17 This, , a=, b=, correlation =, n=, , content is available online at <http://cnx.org/content/m17087/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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575, e. equation: y =, f. Is the correlation significant? Why or why not? (Answer in 1-3 complete sentences.), 2. Supply an answer for the following senarios:, a. For a textbook with 400 pages, predict the cost:, b. For a textbook with 600 pages, predict the cost:, 3. Obtain the graph on your calculator or computer. Sketch the regression line below., , Figure 12.20, , 12.15.4 Discussion Questions, 1. Answer each with 1-3 complete sentences., a. Does the line seem to fit the data? Why?, b. What does the correlation imply about the relationship between the number of pages and the cost?, 2. Are there any outliers? If so, which point(s) is an outlier?, 3. Should the outlier, if it exists, be removed? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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576, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 12.16 Lab 3: Regression (Fuel Efficiency)18, Class Time:, Names:, , 12.16.1 Student Learning Outcomes:, • The student will calculate and construct the line of best fit between two variables., • The student will evaluate the relationship between two variables to determine if that relationship is, significant., , 12.16.2 Collect the Data, Use the most recent April issue of Consumer Reports. It will give the total fuel efficiency (in miles per, gallon) and weight (in pounds) of new model cars with automatic transmissions. We will use this data to, determine the relationship, if any, between the fuel efficiency of a car and its weight., 1. Which variable should be the independent variable and which should be the dependent variable?, Explain your answer in one or two complete sentences., 2. Using your random number generator, randomly select 20 cars from the list and record their weights, and fuel efficiency into the table below., Weight, , 18 This, , Fuel Efficiency, , content is available online at <http://cnx.org/content/m17079/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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577, Table 12.24, , 3. Which variable should be the dependent variable and which should be the independent variable?, Why?, 4. By hand, do a scatterplot of “weight” vs. “fuel efficiency”. Plot the points on graph paper. Label both, axes with words. Scale both axes accurately., , Figure 12.21, , 12.16.3 Analyze the Data, Enter your data into your calculator or computer. Write the linear equation below, rounding to 4 decimal, places., 1. Calculate the following:, a=, b=, correlation =, n=, ^, e. equation: y =, , a., b., c., d., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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578, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , 2. Obtain the graph of the regression line on your calculator. Sketch the regression line on the same axes, as your scatterplot., , 12.16.4 Discussion Questions, 1. Is the correlation significant? Explain how you determined this in complete sentences., 2. Is the relationship a positive one or a negative one? Explain how you can tell and what this means in, terms of weight and fuel efficiency., 3. In one or two complete sentences, what is the practical interpretation of the slope of the least squares, line in terms of fuel efficiency and weight?, 4. For a car that weighs 4000 pounds, predict its fuel efficiency. Include units., 5. Can we predict the fuel efficiency of a car that weighs 10000 pounds using the least squares line?, Explain why or why not., 6. Questions. Answer each in 1 to 3 complete sentences., a. Does the line seem to fit the data? Why or why not?, b. What does the correlation imply about the relationship between fuel efficiency and weight of a, car? Is this what you expected?, 7. Are there any outliers? If so, which point is an outlier?, ** This lab was designed and contributed by Diane Mathios., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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579, , Solutions to Exercises in Chapter 12, Solution to Example 12.11, Problem 2 (p. 541), The x values in the data are between 65 and 75. 90 is outside of the domain of the observed x values in, the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even, though it is possible to enter x into the equation and calculate a y value, you should not do so!), To really understand how unreliable the prediction can be outside of the observed x values in the, data, make the substitution x = 90 into the equation., ^, y = −173.51 + 4.83 (90) = 261.19, The final exam score is predicted to be 261.19. The largest the final exam score can be is 200., NOTE : The process of predicting inside of the observed x values in the data is called interpolation., The process of predicting outside of the observed x values in the data is called extrapolation., , Solutions to Practice: Linear Regression, Solution to Exercise 12.12.2 (p. 553), a., b., c., d., , a = -3,448,225, b = 1750, corr. = 0.4526, n = 22, , Solution to Exercise 12.12.3 (p. 553), ^, y = -3,448,225 +1750x, Solution to Exercise 12.12.4 (p. 553), a. 25,525, b. 34,275, Solution to Exercise 12.12.10 (p. 554), a. -725, , Solutions to Homework, Solution to Exercise 12.13.1 (p. 555), a. Independent: Age; Dependent: Fatalities, d. Independent: Power Consumption; Dependent: Utility, Solution to Exercise 12.13.3 (p. 556), ^, b. y = 88.7206 − 0.0432x, c. -0.8533, Yes, g. No, h. No., i. 2.93, Yes, j. slope = -0.0432. As the year increases by one, the welfare family size tends to decrease by 0.0432 people., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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580, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Solution to Exercise 12.13.5 (p. 556), b. Yes, ^, c. y = 102.4287 + 11.7585x, d. 0.9436; yes, e. 478.70 feet; 1207.73 feet, g. Yes, h. Yes; (57, 1050), i. 172.98; No, j. 11.7585 feet, k. slope = 11.7585. As the number of stories increases by one, the height of the building tends to increase, by 11.7585 feet., Solution to Exercise 12.13.7 (p. 558), b. Yes, ^, c. y = −266.8863 + 0.1656x, d. 0.9448; Yes, e. 62.8233; 62.3265, h. yes; (1987, 62.7), i. 72.5937; No, j. slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase, by 0.1656., Solution to Exercise 12.13.9 (p. 560), b. Yes, ^, c. y = 3.5984 + 0.0371x, d. 0.9986; Yes, e. $5.08, f. $6.93, i. No, j. Not valid, k. slope = 0.0371. As the number of ounces increases by one, the cost of liquid detergent tends to increase, by $0.0371 or is predicted to increase by $0.0371 (about 4 cents)., Solution to Exercise 12.13.11 (p. 561), c. Yes, ^, d. y = −337, 424.6478 + 0.5463x, e. 0.9964; Yes, f. $208,875.35; $1,028,325.35, h. Yes, i. No, k. slope = 0.5463. As the net taxable estate increases by one dollar, the approximate probate fees and taxes, tend to increase by 0.5463 dollars (about 55 cents)., Solution to Exercise 12.13.13 (p. 562), c. Yes, ^, d. y = 65.0876 + 7.0948x, e. 0.9761; yes, f. 72.2 cm; 143.13 cm, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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581, h. Yes, i. No, j. 505.0 cm; No, k. slope = 7.0948. As the age of an American boy increases by one year, the average height tends to increase, by 7.0948 cm., Solution to Exercise 12.13.15 (p. 564), c. No, ^, d. y = 47.03 − 0.0216x, e. -0.4280, f. 6; 5, Solution to Exercise 12.13.17 (p. 565), ^, d. y = −480.5845 + 0.2748x, e. 0.9553, f. 1934, Solution to Exercise 12.13.19 (p. 566), ^, b. y = −569, 770.2796 + 296.0351x, c. 0.8302, d. $1577.46, e. $11,642.66, f. -$22,105.34, Solution to Exercise 12.13.21 (p. 567), b. r = -0.8, significant, c. yhat = 48.4-0.00725x, d. For every one pound increase in weight, the fuel efficiency tends to decrease (or is predicted to decrease), by 0.00725 miles per gallon. (For every one thousand pounds increase in weight, the fuel efficiency, tends to decrease by 7.25 miles per gallon.), e. 64% of the variation in fuel efficiency is explained by the variation in weight using the regression line., g. yhat=48.4-0.00725(3000)=26.65 mpg. y-yhat=25-26.65=-1.65. Because yhat=26.5 is greater than y=25, the, line overestimates the observed fuel efficiency., h. (2750,38) is the outlier. Be sure you know how to justify it using the requested graphical or numerical, methods, not just by guessing., i. yhat = 42.4-0.00578x, j. Without outlier, r=-0.885, rsquare=0.76; with outlier, r=-0.8, rsquare=0.64. The new linear model is a, better fit, after the outlier is removed from the data, because the new correlation coefficient is farther, from 0 and the new coefficient of determination is larger., Solution to Exercise 12.13.22 (p. 568), a. All four data sets have the same correlation coefficient r=0.816 and the same least squares regression line, yhat=3+0.5x, b. Set 2 ; c. Set 4 ; d. Set 3 ; e. Set 1, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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582, , CHAPTER 12. LINEAR REGRESSION AND CORRELATION, , Figure 12.22, , Solution to Exercise 12.13.23 (p., C, Solution to Exercise 12.13.24 (p., A, Solution to Exercise 12.13.25 (p., B, Solution to Exercise 12.13.26 (p., D, Solution to Exercise 12.13.27 (p., A, , 569), 569), 569), 570), 570), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Chapter 13, , F Distribution and One-Way ANOVA, 13.1 F Distribution and One-Way ANOVA1, 13.1.1 Student Learning Outcomes, By the end of this chapter, the student should be able to:, , •, •, •, •, , Interpret the F probability distribution as the number of groups and the sample size change., Discuss two uses for the F distribution: One-Way ANOVA and the test of two variances., Conduct and interpret One-Way ANOVA., Conduct and interpret hypothesis tests of two variances., , 13.1.2 Introduction, Many statistical applications in psychology, social science, business administration, and the natural sciences, involve several groups. For example, an environmentalist is interested in knowing if the average amount of, pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a, person earns varies according to his or her upbringing. A consumer looking for a new car might compare, the average gas mileage of several models., For hypothesis tests involving more than two averages, statisticians have developed a method called Analysis of Variance" (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called, single factor or One-Way ANOVA. You will also study the F distribution, used for One-Way ANOVA, and, the test of two variances. This is just a very brief overview of One-Way ANOVA. You will study this topic, in much greater detail in future statistics courses., , • One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer., • For further information about One-Way ANOVA, use the online link ANOVA2 . Use the back button, to return here. (The url is http://en.wikipedia.org/wiki/Analysis_of_variance.), 1 This, , content is available online at <http://cnx.org/content/m17065/1.11/>., , 2 http://en.wikipedia.org/wiki/Analysis_of_variance, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 583
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584, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , 13.2 One-Way ANOVA3, 13.2.1 F Distribution and One-Way ANOVA: Purpose and Basic Assumptions of OneWay ANOVA, The purpose of a One-Way ANOVA test is to determine the existence of a statistically significant difference, among several group means. The test actually uses variances to help determine if the means are equal or, not., In order to perform a One-Way ANOVA test, there are five basic assumptions to be fulfilled:, , •, •, •, •, •, , Each population from which a sample is taken is assumed to be normal., Each sample is randomly selected and independent., The populations are assumed to have equal standard deviations (or variances)., The factor is the categorical variable., The response is the numerical variable., , 13.2.2 The Null and Alternate Hypotheses, The null hypothesis is simply that all the group population means are the same. The alternate hypothesis, is that at least one pair of means is different. For example, if there are k groups:, Ho : µ1 = µ2 = µ3 = ... = µk, Ha : At least two of the group means µ1 , µ2 , µ3 , ..., µk are not equal., The graphs help in the understanding of the hypothesis test. In the first graph (red box plots),, Ho : µ1 = µ2 = µ3 and the three populations have the same distribution if the null hypothesis is true. The, variance of the combined data is approximately the same as the variance of each of the populations., If the null hypothesis is false, then the variance of the combined data is larger which is caused by, the different means as shown in the second graph (green box plots)., 3 This, , content is available online at <http://cnx.org/content/m17068/1.10/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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585, , 13.3 The F Distribution and the F Ratio4, The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir, Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of, freedom; one for the numerator and one for the denominator., 4 This, , content is available online at <http://cnx.org/content/m17076/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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586, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , For example, if F follows an F distribution and the degrees of freedom for the numerator are 4 and the, degrees of freedom for the denominator are 10, then F ∼ F4,10 ., NOTE : The F distribution is derived from the Student’s-t distribution. One-Way ANOVA expands, the t-test for comparing more than two groups. The scope of that derivation is beyond the level of, this course., , To calculate the F ratio, two estimates of the variance are made., 1. Variance between samples: An estimate of σ2 that is the variance of the sample means multiplied by, n (when there is equal n). If the samples are different sizes, the variance between samples is weighted, to account for the different sample sizes. The variance is also called variation due to treatment or, explained variation., 2. Variance within samples: An estimate of σ2 that is the average of the sample variances (also known, as a pooled variance). When the sample sizes are different, the variance within samples is weighted., The variance is also called the variation due to error or unexplained variation., , • SSbetween = the sum of squares that represents the variation among the different samples., • SSwithin = the sum of squares that represents the variation within samples that is due to chance., To find a "sum of squares" means to add together squared quantities which, in some cases, may be weighted., We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive, Statistics., MS means "mean square." MSbetween is the variance between groups and MSwithin is the variance within, groups., Calculation of Sum of Squares and Mean Square, , •, •, •, •, •, •, , k = the number of different groups, n j = the size of the jth group, s j = the sum of the values in the jth group, n = total number of all the values combined. (total sample size: ∑ n j ), x = one value: ∑ x = ∑ s j, Sum of squares of all values from every group combined: ∑ x2, , • Between group variability: SStotal = ∑ x2 −, 2, , ( ∑ x )2, n, , • Total sum of squares: ∑ x2 − (∑nx), • Explained, variation- sum of squares representing variation among the different samples SSbetween =, , , 2, 2, (∑ s j ), sj, ( ), ∑ nj − n, • Unexplained variation- sum of squares representing variation within samples due to chance:, SSwithin = SStotal − SSbetween, • df’s for different groups (df’s for the numerator): dfbetween = k − 1, • Equation for errors within samples (df’s for the denominator): dfwithin = n − k, between, • Mean square (variance estimate) explained by the different groups: MSbetween = SS, df, between, , • Mean square (variance estimate) that is due to chance (unexplained): MSwithin =, MSbetween and MSwithin can be written as follows:, SSbetween, SSbetween, d f between =, k −1, SSwithin, SSwithin, =, d f within, n−k, , • MSbetween =, • MSwithin =, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , SSwithin, dfwithin
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587, The One-Way ANOVA test depends on the fact that MSbetween can be influenced by population differences, among means of the several groups. Since MSwithin compares values of each group to its own group mean,, the fact that group means might be different does not affect MSwithin ., The null hypothesis says that all groups are samples from populations having the same normal distribution., The alternate hypothesis says that at least two of the sample groups come from populations with different, normal distributions. If the null hypothesis is true, MSbetween and MSwithin should both estimate the same, value., NOTE : The null hypothesis says that all the group population means are equal. The hypothesis of, equal means implies that the populations have the same normal distribution because it is assumed, that the populations are normal and that they have equal variances., , F-Ratio or F Statistic, F=, , MSbetween, MSwithin, , (13.0), , If MSbetween and MSwithin estimate the same value (following the belief that Ho is true), then the F-ratio, should be approximately equal to 1. Mostly just sampling errors would contribute to variations away, from 1. As it turns out, MSbetween consists of the population variance plus a variance produced from the, differences between the samples. MSwithin is an estimate of the population variance. Since variances are, always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin . Then the, F-ratio will be larger than 1. However, if the population effect size is small it is not unlikely that MSwithin, will be larger in a give sample., The above calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F ratio can be written as:, F-Ratio Formula when the groups are the same size, F=, , n · s x_ 2, s2 pooled, , (13.0), , where ..., , •, •, •, •, •, , n =the sample size, dfnumerator = k − 1, dfdenominator = n − k, s2 pooled = the mean of the sample variances (pooled variance), s x 2 = the variance of the sample means, , The data is typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this, manner by computer software., Source of, Variation, , Sum of Squares, (SS), , Degrees of, Freedom (df), , Mean Square, (MS), continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , F
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588, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, Factor, (Between), , SS(Factor), , k-1, , MS(Factor) =, SS(Factor)/(k-1), , Error, (Within), , SS(Error), , n-k, , MS(Error) =, SS(Error)/(n-k), , Total, , SS(Total), , n-1, , F=, MS(Factor)/MS(Error), , Table 13.1, , Example 13.1, Three different diet plans are to be tested for mean weight loss. The entries in the table are the, weight losses for the different plans. The One-Way ANOVA table is shown below., Plan 1, , Plan 2, , Plan 3, , 5, , 3.5, , 8, , 4.5, , 7, , 4, , 4, 3, , 3.5, 4.5, Table 13.2, , One-Way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between) and SS(Error) =, SS(Within) are shown above. This same information is provided by the TI calculator hypothesis, test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the, data from Plan 1, Plan 2, Plan 3 respectively)., Source of, Variation, , Sum of Squares, (SS), , Factor, (Between), , SS(Factor), = SS(Between), =2.2458, , Error, (Within), , SS(Error), = SS(Within), = 20.8542, , Total, , SS(Total), = 2.9792 + 20.8542, =23.1, , Degrees of, Freedom (df), k-1, = 3 groups - 1, =2, n-k, = 10 total data - 3, groups, =7, , Mean Square, (MS), MS(Factor), = SS(Factor)/(k-1), = 2.2458/2, = 1.1229, , F, F=, MS(Factor)/MS(Error), = 1.1229/2.9792, = 0.3769, , MS(Error), = SS(Error)/(n-k), = 20.8542/7, = 2.9792, , n-1, = 10 total data - 1, =9, Table 13.3, , The One-Way ANOVA hypothesis test is always right-tailed because larger F-values are way out in the, right tail of the F-distribution curve and tend to make us reject Ho ., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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589, , 13.3.1 Notation, The notation for the F distribution is F ∼ Fdf(num),df(denom), where df(num) = d f between and df(denom) = d f within, The mean for the F distribution is µ =, , d f (num), d f (denom)−1, , 13.4 Facts About the F Distribution5, 1., 2., 3., 4., , The curve is not symmetrical but skewed to the right., There is a different curve for each set of dfs., The F statistic is greater than or equal to zero., As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal., 5. Other uses for the F distribution include comparing two variances and Two-Way Analysis of Variance., Comparing two variances is discussed at the end of the chapter. Two-Way Analysis is mentioned for, your information only., , (a), , (b), , Figure 13.1, , Example 13.2, One-Way ANOVA: Four sororities took a random sample of sisters regarding their grade means, for the past term. The results are shown below:, 5 This, , content is available online at <http://cnx.org/content/m17062/1.14/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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590, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , MEAN GRADES FOR FOUR SORORITIES, Sorority 1, , Sorority 2, , Sorority 3, , Sorority 4, , 2.17, , 2.63, , 2.63, , 3.79, , 1.85, , 1.77, , 3.78, , 3.45, , 2.83, , 3.25, , 4.00, , 3.08, , 1.69, , 1.86, , 2.55, , 2.26, , 3.33, , 2.21, , 2.45, , 3.18, , Table 13.4, , Problem, Using a significance level of 1%, is there a difference in mean grades among the sororities?, Solution, Let µ1 , µ2 , µ3 , µ4 be the population means of the sororities. Remember that the null hypothesis, claims that the sorority groups are from the same normal distribution. The alternate hypothesis, says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each size 5., NOTE : This is an example of a balanced design, since each factor (i.e. Sorority) has the same, number of observations., , Ho : µ1 = µ2 = µ3 = µ4, Ha : Not all of the means µ1 , µ2 , µ3 , µ4 are equal., Distribution for the test: F3,16, where k = 4 groups and n = 20 samples in total, d f (num) = k − 1 = 4 − 1 = 3, d f (denom) = n − k = 20 − 4 = 16, Calculate the test statistic: F = 2.23, Graph:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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591, , Figure 13.2, , Probability statement: p-value = P ( F > 2.23) = 0.1241, Compare α and the p − value: α = 0.01, , p-value = 0.1241, , α < p-value, , Make a decision: Since α < p-value, you cannot reject Ho ., Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean, grades for the sororities., TI-83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS., Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4). The F statistic is 2.2303 and the, p-value is 0.1241. df(numerator) = 3 (under "Factor") and df(denominator) = 16 (under Error)., , Example 13.3, A fourth grade class is studying the environment. One of the assignments is to grow bean plants, in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed, with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery., Nick chose to grow his bean plants in soil from his mother’s garden. No chemicals were used, on the plants, only water. They were grown inside the classroom next to a large window. Each, child grew 5 plants. At the end of the growing period, each plant was measured, producing the, following data (in inches):, Tommy’s Plants, , Tara’s Plants, , Nick’s Plants, , 24, , 25, , 23, , 21, , 31, , 27, , 23, , 23, , 22, , 30, , 20, , 30, , 23, , 28, , 20, , Table 13.5, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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592, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, Problem 1, Does it appear that the three media in which the bean plants were grown produce the same mean, height? Test at a 3% level of significance., Solution, This time, we will perform the calculations that lead to the F’ statistic. Notice that each group has, the same number of plants so we will use the formula F’ =, , n·s x_ 2, ., s2 pooled, , First, calculate the sample mean and sample variance of each group., Tommy’s Plants, , Tara’s Plants, , Nick’s Plants, , Sample Mean, , 24.2, , 25.4, , 24.4, , Sample Variance, , 11.7, , 18.3, , 16.3, , Table 13.6, , Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and, 24.4). Variance of the group means = 0.413 = s x 2, Then MSbetween = ns x 2 = (5) (0.413) where n = 5 is the sample size (number of plants each child, grew)., Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean, of the sample variances = 15.433 = s2 pooled, Then MSwithin = s2 pooled = 15.433., The F statistic (or F ratio) is F =, , MSbetween, MSwithin, , =, , n·s x_ 2, s2 pooled, , =, , (5)·(0.413), 15.433, , = 0.134, , The dfs for the numerator = the number of groups − 1 = 3 − 1 = 2, The dfs for the denominator = the total number of samples − the number of groups = 15 − 3 = 12, The distribution for the test is F2,12 and the F statistic is F = 0.134, The p-value is P ( F > 0.134) = 0.8759., Decision: Since α = 0.03 and the p-value = 0.8759, do not reject Ho . (Why?), Conclusion: With a 3% the level of significance, from the sample data, the evidence is not sufficient, to conclude that the mean heights of the bean plants are different., (This experiment was actually done by three classmates of the son of one of the authors.), Another fourth grader also grew bean plants but this time in a jelly-like mass. The heights were, (in inches) 24, 28, 25, 30, and 32., Problem 2, (Solution on p. 610.), Do a One-Way ANOVA test on the 4 groups. You may use your calculator or computer to, perform the test. Are the heights of the bean plants different? Use a solution sheet (Section 14.5.4)., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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593, , 13.4.1 Optional Classroom Activity, From the class, create four groups of the same size as follows: men under 22, men at least 22, women under, 22, women at least 22. Have each member of each group record the number of states in the United States, he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four, groups are the same. Test at a 1% level of significance. Use one of the solution sheets (Section 14.5.4) at the, end of the chapter (after the homework)., , 13.5 Test of Two Variances6, Another of the uses of the F distribution is testing two variances. It is often desirable to compare two, variances rather than two averages. For instance, college administrators would like two college professors, grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation, in the lid and the container should be the same. A supermarket might be interested in the variability of, check-out times for two checkers., In order to perform a F test of two variances, it is important that the following are true:, 1. The populations from which the two samples are drawn are normally distributed., 2. The two populations are independent of each other., Suppose we sample randomly from two independent normal populations. Let σ12 and σ22 be the population, variances and s21 and s22 be the sample variances. Let the sample sizes be n1 and n2 . Since we are interested, in comparing the two sample variances, we use the F ratio, , F=, , ", , ( s1 )2, (σ1 )2, , #, , ", , ( s2 )2, (σ2 )2, , #, , F has the distribution F ∼ F (n1 − 1, n2 − 1), where n1 − 1 are the degrees of freedom for the numerator and n2 − 1 are the degrees of freedom for the, denominator., If the null hypothesis is σ12 = σ22 , then the F-Ratio becomes F =, , NOTE :, , The F ratio could also be, , ( s2 )2, ., ( s1 )2, , ", , ( s1 )2, (σ1 )2, , #, , ", , 2, , #, , ( s2 ), (σ2 )2, , =, , ( s1 )2, ., ( s2 )2, , It depends on Ha and on which sample variance is larger., , If the two populations have equal variances, then s21 and s22 are close in value and F =, , ( s1 )2, ( s2 )2, , is close to 1. But, , if the two population variances are very different, s21 and s22 tend to be very different, too. Choosing s21 as, the larger sample variance causes the ratio, is a large number., , ( s1 )2, ( s2 )2, , to be greater than 1. If s21 and s22 are far apart, then F =, , ( s1 )2, ( s2 )2, , Therefore, if F is close to 1, the evidence favors the null hypothesis (the two population variances are equal)., But if F is much larger than 1, then the evidence is against the null hypothesis., 6 This, , content is available online at <http://cnx.org/content/m17075/1.8/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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594, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , A test of two variances may be left, right, or two-tailed., Example 13.4, Two college instructors are interested in whether or not there is any variation in the way they, grade math exams. They each grade the same set of 30 exams. The first instructor’s grades have a, variance of 52.3. The second instructor’s grades have a variance of 89.9., Problem, Test the claim that the first instructor’s variance is smaller. (In most colleges, it is desirable for, the variances of exam grades to be nearly the same among instructors.) The level of significance is, 10%., Solution, Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively., n1 = n2 = 30., Ho : σ12 = σ22 and Ha : σ12 <σ22, , Calculate the test statistic: By the null hypothesis σ12 = σ22 , the F statistic is, , F=, , ", , ( s1 )2, (σ1 )2, , #, , ", , 2, , #, , ( s2 ), (σ2 )2, , =, , ( s1 )2, ( s2 )2, , =, , 52.3, 89.9, , = 0.5818, , Distribution for the test: F29,29, Graph:, , where n1 − 1 = 29 and n2 − 1 = 29., , This test is left tailed., , Draw the graph labeling and shading appropriately., , Figure 13.3, , Probability statement: p-value = P (F < 0.5818) = 0.0753, Compare α and the p-value: α = 0.10, , α > p-value., , Make a decision: Since α > p-value, reject Ho ., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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595, Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude, that the variance in grades for the first instructor is smaller., TI-83+ and TI-84: Press STAT and arrow over to TESTS. Arrow down p, to D:2-SampFTest, . Press, p, ENTER. Arrow to Stats and press ENTER. For Sx1, n1, Sx2, and n2, enter (52.3), 30, (89.9), and, 30. Press ENTER after each. Arrow to σ1: and <σ2. Press ENTER. Arrow down to Calculate and, press ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try Draw instead of, Calculate., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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596, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , 13.6 Summary7, • A One-Way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with 2 different degrees of freedom., Assumptions:, a. Each population from which a sample is taken is assumed to be normal., b. Each sample is randomly selected and independent., c. The populations are assumed to have equal standard deviations (or variances), , • A Test of Two Variances hypothesis test determines if two variances are the same. The distribution, for the hypothesis test is the F distribution with 2 different degrees of freedom., Assumptions:, a. The populations from which the two samples are drawn are normally distributed., b. The two populations are independent of each other., , 7 This, , content is available online at <http://cnx.org/content/m17072/1.4/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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597, , 13.7 Practice: One-Way ANOVA8, 13.7.1 Student Learning Outcome, • The student will conduct a One-Way ANOVA hypothesis test., , 13.7.2 Given, Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected, from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses., Northeast, , South, , West, , Central, , East, , 16.3, , 16.9, , 16.4, , 16.2, , 17.1, , 16.1, , 16.5, , 16.5, , 16.6, , 17.2, , 16.4, , 16.4, , 16.6, , 16.5, , 16.6, , 16.5, , 16.2, , 16.1, , 16.4, , 16.8, , x=, , ________, , ________, , ________, , ________, , ________, , s2 =, , ________, , ________, , ________, , ________, , ________, , Table 13.7, , 13.7.3 Hypothesis, Exercise 13.7.1, State the hypotheses., Ho :, Ha :, , 13.7.4 Data Entry, Enter the data into your calculator or computer., Exercise 13.7.2, degrees of freedom - numerator: df (n) =, , (Solution on p. 610.), , Exercise 13.7.3, degrees of freedom - denominator: df (d) =, , (Solution on p. 610.), , Exercise 13.7.4, F test statistic =, , (Solution on p. 610.), , Exercise 13.7.5, p-value =, , (Solution on p. 610.), , 8 This, , content is available online at <http://cnx.org/content/m17067/1.10/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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598, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , 13.7.5 Decisions and Conclusions, State the decisions and conclusions (in complete sentences) for the following preconceived levels of α ., Exercise 13.7.6, α = 0.05, Decision:, Conclusion:, Exercise 13.7.7, α = 0.01, Decision:, Conclusion:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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599, , 13.8 Homework9, D IRECTIONS : Use a solution sheet to conduct the following hypothesis tests. The solution sheet, can be found in the Table of Contents 14. Appendix., Exercise 13.8.1, (Solution on p. 610.), Three students, Linda, Tuan, and Javier, are given 5 laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats, Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is, weighed again and the net gain in grams is recorded. Using a significance level of 10%, test the, hypothesis that the three formulas produce the same mean weight gain., Weights of Student Lab Rats, Linda’s rats, , Tuan’s rats, , Javier’s rats, , 43.5, , 47.0, , 51.2, , 39.4, , 40.5, , 40.9, , 41.3, , 38.9, , 37.9, , 46.0, , 46.3, , 45.0, , 38.2, , 44.2, , 48.6, , Table 13.8, , Exercise 13.8.2, A grassroots group opposed to a proposed increase in the gas tax claimed that the increase, would hurt working-class people the most, since they commute the farthest to work. Suppose, that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are below. Using a 5% significance level, test the hypothesis that the 3, mean commuting mileages are the same., working-class, , professional (middle incomes), , professional (wealthy), , 17.8, , 16.5, , 8.5, , 26.7, , 17.4, , 6.3, , 49.4, , 22.0, , 4.6, , 9.4, , 7.4, , 12.6, , 65.4, , 9.4, , 11.0, , 47.1, , 2.1, , 28.6, , 19.5, , 6.4, , 15.4, , 51.2, , 13.9, , 9.3, , Table 13.9, , Exercise 13.8.3, (Solution on p. 610.), Refer to Exercise 13.8.1. Determine whether or not the variance in weight gain is statistically the, same among Javier’s and Linda’s rats., 9 This, , content is available online at <http://cnx.org/content/m17063/1.15/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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600, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, Exercise 13.8.4, Refer to Exercise 13.8.2 above (Exercise 13.8.2). Determine whether or not the variance in mileage, driven is statistically the same among the working class and professional (middle income) groups., , For the next two problems, refer to the data from Terri Vogel’s Log Book., http://cnx.org/content/m17132/latest/?collection=col10522/latest/ 10, Exercise 13.8.5, (Solution on p. 610.), Examine the 7 practice laps. Determine whether the mean lap time is statistically the same for the, 7 practice laps, or if there is at least one lap that has a different mean time from the others., Exercise 13.8.6, Examine practice laps 3 and 4. Determine whether or not the variance in lap time is statistically, the same for those practice laps., For the next four problems, refer to the following data., The following table lists the number of pages in four different types of magazines., home decorating, , news, , health, , computer, , 172, , 87, , 82, , 104, , 286, , 94, , 153, , 136, , 163, , 123, , 87, , 98, , 205, , 106, , 103, , 207, , 197, , 101, , 96, , 146, , Table 13.10, , Exercise 13.8.7, (Solution on p. 610.), Using a significance level of 5%, test the hypothesis that the four magazine types have the same, mean length., Exercise 13.8.8, Eliminate one magazine type that you now feel has a mean length different than the others. Redo, the hypothesis test, testing that the remaining three means are statistically the same. Use a new, solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same?, Exercise 13.8.9, Which two magazine types do you think have the same variance in length?, Exercise 13.8.10, Which two magazine types do you think have different variances in length?, Exercise 13.8.11, (Solution on p. 611.), A researcher wants to know if the mean time (in minutes) that people watch their favorite news, station are the same. Suppose that the table below shows the results of a study., 10 http://cnx.org/content/m17132/latest/?collection=col10522/latest/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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601, , CNN, , FOX, , Local, , 45, , 15, , 72, , 12, , 43, , 37, , 18, , 68, , 56, , 38, , 50, , 60, , 23, , 31, , 51, , 35, , 22, Table 13.11, , Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05., Exercise 13.8.12, Are the means for the final exams the same for all statistics class delivery types? The table below, shows the scores on final exams from several randomly selected classes that used the different, delivery types., Online, , Hybrid, , Face-to-Face, , 72, , 83, , 80, , 84, , 73, , 78, , 77, , 84, , 84, , 80, , 81, , 81, , 81, , 86, 79, 82, Table 13.12, , Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05., Exercise 13.8.13, (Solution on p. 611.), Are the mean number of times a month a person eats out same for whites, blacks, Hispanics and, Asians? Suppose that the table below shows the results of a study., White, , Black, , Hispanic, , Asian, , 6, , 4, , 7, , 8, , 8, , 1, , 3, , 3, , 2, , 5, , 5, , 5, , 4, , 2, , 4, , 1, , 6, , 7, , 6, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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602, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, Table 13.13, , Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05., Exercise 13.8.14, Are the mean number of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that the table below shows the results of a study., Powder, , Machine Made, , Hard Packed, , 1210, , 2107, , 2846, , 1080, , 1149, , 1638, , 1537, , 862, , 2019, , 941, , 1870, , 1178, , 1528, , 2233, , 1382, Table 13.14, , Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05., Exercise 13.8.15, (Solution on p. 611.), Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall, the same as the variance for the amount of money that shoppers spend on Sundays at the mall?, Suppose that the table below shows the results of a study., Saturday, , Sunday, , 75, , 44, , 62, , 137, , 18, , 58, , 0, , 82, , 150, , 61, , 124, , 39, , 94, , 19, , 50, , 127, , 62, , 99, , 31, , 141, , 73, , 60, , 118, , 73, 89, , Table 13.15, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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603, Assume that both distributions are normal. Use a level of significance of 0.05., Exercise 13.8.16, Are the variances for incomes on the East Coast and the West Coast the same? Suppose that the, table below shows the results of a study. Income is shown in thousands of dollars., East, , West, , 38, , 71, , 47, , 126, , 30, , 42, , 82, , 51, , 75, , 44, , 52, , 90, , 115, , 88, , 67, Table 13.16, , Assume that both distributions are normal. Use a level of significance of 0.05., **Exercises 11 - 16 were contributed by Dr. Larry Green, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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604, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , 13.9 Review11, The next two questions refer to the following situation:, Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a, drought occurs, the probability of water rationing is 10%. However, in any year, the probability of water, rationing is 5%., Exercise 13.9.1, What is the probability of both a drought and water rationing occurring?, , (Solution on p. 611.), , Exercise 13.9.2, (Solution on p. 611.), Out of the years with water rationing, find the probability that there is a drought., The next three questions refer to the following survey:, Favorite Type of Pie by Gender, apple, , pumpkin, , pecan, , female, , 40, , 10, , 30, , male, , 20, , 30, , 10, , Table 13.17, , Exercise 13.9.3, (Solution on p. 611.), Suppose that one individual is randomly chosen. Find the probability that the person’s favorite, pie is apple or the person is male., Exercise 13.9.4, (Solution on p. 611.), Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan., Exercise 13.9.5, (Solution on p. 611.), Conduct a hypothesis test to determine if favorite pie type and gender are independent., The next two questions refer to the following situation:, Let’s say that the probability that an adult watches the news at least once per week is 0.60., Exercise 13.9.6, (Solution on p. 611.), We randomly survey 14 people. On average, how many people do we expect to watch the news, at least once per week?, Exercise 13.9.7, (Solution on p. 611.), We randomly survey 14 people. Of interest is the number that watch the news at least once per, week. State the distribution of X. X ∼, Exercise 13.9.8, (Solution on p. 611.), The following histogram is most likely to be a result of sampling from which distribution?, 11 This, , content is available online at <http://cnx.org/content/m17070/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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605, , Figure 13.4, , A., B., C., D., , Chi-Square, Geometric, Uniform, Binomial, , Exercise 13.9.9, (Solution on p. 611.), The ages of De Anza evening students is known to be normally distributed with a population, mean of 40 and a population standard deviation of 6. A sample of 6 De Anza evening students, reported their ages (in years) as: 28; 35; 47; 45; 30; 50. Find the probability that the mean of 6 ages, of randomly chosen students is less than 35 years. Hint: Find the sample mean., The next three questions refer to the following situation:, The amount of money a customer spends in one trip to the supermarket is known to have an exponential, distribution. Suppose the mean amount of money a customer spends in one trip to the supermarket is $72., Exercise 13.9.10, (Solution on p. 612.), Find the probability that one customer spends less than $72 in one trip to the supermarket?, Exercise 13.9.11, (Solution on p. 612.), Suppose 5 customers pool their money. (They are poor college students.) How much money, altogether would you expect the 5 customers to spend in one trip to the supermarket (in dollars)?, Exercise 13.9.12, (Solution on p. 612.), State the distribution to use if you want to find the probability that the mean amount spent by 5, customers in one trip to the supermarket is less than $60., Exercise 13.9.13, (Solution on p. 612.), A math exam was given to all the fifth grade children attending Country School. Two random, samples of scores were taken. The null hypothesis is that the mean math scores for boys and girls, in fifth grade are the same. Conduct a hypothesis test., n, , x, , s2, , Boys, , 55, , 82, , 29, , Girls, , 60, , 86, , 46, , Table 13.18, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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606, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, Exercise 13.9.14, (Solution on p. 612.), In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed,, 25 had played an organized sport growing up. We are interested in whether the proportion for, males is higher than the proportion for females. Conduct a hypothesis test., Exercise 13.9.15, Which of the following is preferable when designing a hypothesis test?, A., B., C., D., , (Solution on p. 612.), , Maximize α and minimize β, Minimize α and maximize β, Maximize α and β, Minimize α and β, , The next three questions refer to the following situation:, 120 people were surveyed as to their favorite beverage (non-alcoholic). The results are below., Preferred Beverage by Age, 0–9, , 10 – 19, , 20 – 29, , 30 +, , Totals, , Milk, , 14, , 10, , 6, , 0, , 30, , Soda, , 3, , 8, , 26, , 15, , 52, , Juice, , 7, , 12, , 12, , 7, , 38, , Totals, , 24, , 30, , 44, , 22, , 120, , Table 13.19, , Exercise 13.9.16, Are the events of milk and 30+:, , (Solution on p. 612.), , a. Independent events? Justify your answer., b. Mutually exclusive events? Justify your answer., Exercise 13.9.17, (Solution on p. 612.), Suppose that one person is randomly chosen. Find the probability that person is 10 – 19 given, that he/she prefers juice., Exercise 13.9.18, (Solution on p. 612.), Are Preferred Beverage and Age independent events? Conduct a hypothesis test., Exercise 13.9.19, (Solution on p. 612.), Given the following histogram, which distribution is the data most likely to come from?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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607, , Figure 13.5, , A., B., C., D., , uniform, exponential, normal, chi-square, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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608, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , 13.10 Lab: One-Way ANOVA12, Class Time:, Names:, , 13.10.1 Student Learning Outcome:, • The student will conduct a simple One-Way ANOVA test involving three variables., , 13.10.2 Collect the Data, 1. Record the price per pound of 8 fruits, 8 vegetables, and 8 breads in your local supermarket., Fruits, , Vegetables, , Breads, , Table 13.20, , 2. Explain how you could try to collect the data randomly., , 13.10.3 Analyze the Data and Conduct a Hypothesis Test, 1. Compute the following:, a. Fruit:, i. x =, ii. s x =, iii. n =, a. Vegetables:, i. x =, ii. s x =, iii. n=, a. Bread:, i. x =, ii. s x =, iii. n =, 12 This, , content is available online at <http://cnx.org/content/m17061/1.9/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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609, 2. Find the following:, a. d f (num) =, b. d f (denom) =, 3. State the approximate distribution for the test., 4. Test statistic: F =, 5. Sketch a graph of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s), corresponding to the p-value., 6. p-value =, 7. Test at α = 0.05. State your decision and conclusion., 8. a. Decision: Why did you make this decision?, b. Conclusion (write a complete sentence)., c. Based on the results of your study, is there a need to further investigate any of the food groups’, prices? Why or why not?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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610, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , Solutions to Exercises in Chapter 13, Solution to Example 13.3, Problem 2 (p. 592), , • F = 0.9496, • p − value = 0.4402, From the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants, are different., , Solutions to Practice: One-Way ANOVA, Solution to Exercise 13.7.2 (p., d f (1) = 4, Solution to Exercise 13.7.3 (p., d f (2) = 15, Solution to Exercise 13.7.4 (p., Test statistic = F = 4.22, Solution to Exercise 13.7.5 (p., 0.017, , 597), 597), 597), 597), , Solutions to Homework, Solution to Exercise 13.8.1 (p. 599), a. Ho : µ L = µ T = µ J, c. df (n) = 2; df (d) = 12, e. 0.67, f. 0.5305, h. Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the means are, different., Solution to Exercise 13.8.3 (p. 599), c. df (n) = 4; df (d) = 4, e. 3.00, f. 2 (0.1563) = 0.3126. Using the TI-83+/84+ function 2-SampFtest, you get the the test statistic as 2.9986, and p-value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335, but the p-value is the same because this is a two-tailed test., h. Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the variances are, different., Solution to Exercise 13.8.5 (p. 600), c. df (n) = 6; df (d) = 98, e. 1.69, f. 0.1319, h. Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the mean lap, times are different., Solution to Exercise 13.8.7 (p. 600), a. Ho : µd = µn = µh = µc, b. Alternate Hypothesis: At least one pair of means is different, c. df (n) = 3; df (d) = 16, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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611, e. 8.69, f. 0.0012, h. Decision: Reject null; Conclusion: There is sufficient evidence to conclude that the mean lengths are, different., Solution to Exercise 13.8.11 (p. 600), c: df(n) = 2; df(d) = 14, d: F2,14, e: 4.08, f: 0.0401, h:, ii: Reject the null hypothesis, iv: There is sufficient evidence to conclude that the mean times are different., Solution to Exercise 13.8.13 (p. 601), c: df(n) = 3; df(d) = 15, d: F3,15, e: 0.8853, f: 0.4711, h:, ii: Do not reject the null hypothesis, iv: There is insufficient evidence to conclude that the mean number of times are different., Solution to Exercise 13.8.15 (p. 602), c: df(n) = 11; df(d) = 12, d: F11,12, e: 1.35, f: 0.6090, h:, ii: Do not reject the null hypothesis, iv: There is insufficient evidence to conclude that the variances are different., , Solutions to Review, Solution to Exercise 13.9.1 (p. 604), 0.02, Solution to Exercise 13.9.2 (p. 604), 0.40, Solution to Exercise 13.9.3 (p. 604), 100, 140, , Solution to Exercise 13.9.4 (p. 604), 10, 60, , Solution to Exercise 13.9.5 (p. 604), p-value = 0; Reject null; Conclude dependent events, Solution to Exercise 13.9.6 (p. 604), 8.4, Solution to Exercise 13.9.7 (p. 604), B (14, 0.60), Solution to Exercise 13.9.8 (p. 604), D, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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612, , CHAPTER 13. F DISTRIBUTION AND ONE-WAY ANOVA, , Solution to Exercise 13.9.9 (p. 605), 0.3669, Solution to Exercise 13.9.10 (p. 605), 0.6321, Solution to Exercise 13.9.11 (p. 605), $360, Solution, to Exercise 13.9.12 (p. 605), , , 72, N 72, √, 5, Solution to Exercise 13.9.13 (p. 605), p-value = 0.0006; Reject null; Conclude averages are not equal, Solution to Exercise 13.9.14 (p. 606), p-value = 0; Reject null; Conclude proportion of males is higher, Solution to Exercise 13.9.15 (p. 606), D, Solution to Exercise 13.9.16 (p. 606), , a. No, b. Yes, P (M and 30+) = 0, Solution to Exercise 13.9.17 (p. 606), 12, 38, , Solution to Exercise 13.9.18 (p. 606), No; p-value = 0, Solution to Exercise 13.9.19 (p. 606), A, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Appendix, 14.1 Practice Final Exam 11, Questions 1-2 refer to the following:, An experiment consists of tossing two 12-sided dice (the numbers 1-12 are printed on the sides of each, dice)., , • Let Event A = both dice show an even number, • Let Event B = both dice show a number more than 8, Exercise 14.1.1, Events A and B are:, A., B., C., D., , (Solution on p. 668.), , Mutually exclusive., Independent., Mutually exclusive and independent., Neither mutually exclusive nor independent., , Exercise 14.1.2, Find P ( A| B), A., B., C., D., , (Solution on p. 668.), , 2, 4, 16, 144, 4, 16, 2, 144, , Exercise 14.1.3, (Solution on p. 668.), Which of the following are TRUE when we perform a hypothesis test on matched or paired samples?, A., B., C., D., , Sample sizes are almost never small., Two measurements are drawn from the same pair of individuals or objects., Two sample means are compared to each other., Answer choices B and C are both true., , Questions 4 - 5 refer to the following:, 118 students were asked what type of color their bedrooms were painted: light colors, dark colors or vibrant, colors. The results were tabulated according to gender., 1 This, , content is available online at <http://cnx.org/content/m16304/1.17/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , 613
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614, , APPENDIX, , Light colors, , Dark colors, , Vibrant colors, , Female, , 20, , 22, , 28, , Male, , 10, , 30, , 8, , Table 14.1, , Exercise 14.1.4, (Solution on p. 668.), Find the probability that a randomly chosen student is male or has a bedroom painted with light, colors., A., B., C., D., , 10, 118, 68, 118, 48, 118, 10, 48, , Exercise 14.1.5, (Solution on p. 668.), Find the probability that a randomly chosen student is male given the student’s bedroom is, painted with dark colors., A., B., C., D., , 30, 118, 30, 48, 22, 118, 30, 52, , Questions 6 – 7 refer to the following:, We are interested in the number of times a teenager must be reminded to do his/her chores each week. A, survey of 40 mothers was conducted. The table below shows the results of the survey., x, , P (x), , 0, , 2, 40, 5, 40, , 1, 2, 3, 4, 5, , 14, 40, 7, 40, 4, 40, , Table 14.2, , Exercise 14.1.6, Find the probability that a teenager is reminded 2 times., , (Solution on p. 668.), , A. 8, 8, B. 40, 6, C. 40, D. 2, Exercise 14.1.7, (Solution on p. 668.), Find the expected number of times a teenager is reminded to do his/her chores., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, A., B., C., D., , 615, , 15, 2.78, 1.0, 3.13, , Questions 8 – 9 refer to the following:, On any given day, approximately 37.5% of the cars parked in the De Anza parking structure are parked, crookedly. (Survey done by Kathy Plum.) We randomly survey 22 cars. We are interested in the number of, cars that are parked crookedly., Exercise 14.1.8, (Solution on p. 668.), For every 22 cars, how many would you expect to be parked crookedly, on average?, A., B., C., D., , 8.25, 11, 18, 7.5, , Exercise 14.1.9, What is the probability that at least 10 of the 22 cars are parked crookedly., A., B., C., D., , (Solution on p. 668.), , 0.1263, 0.1607, 0.2870, 0.8393, , Exercise 14.1.10, (Solution on p. 668.), Using a sample of 15 Stanford-Binet IQ scores, we wish to conduct a hypothesis test. Our claim, is that the mean IQ score on the Stanford-Binet IQ test is more than 100. It is known that the, standard deviation of all Stanford-Binet IQ scores is 15 points. The correct distribution to use for, the hypothesis test is:, A., B., C., D., , Binomial, Student’s-t, Normal, Uniform, , Questions 11 – 13 refer to the following:, De Anza College keeps statistics on the pass rate of students who enroll in math classes. In a sample of 1795, students enrolled in Math 1A (1st quarter calculus), 1428 passed the course. In a sample of 856 students, enrolled in Math 1B (2nd quarter calculus), 662 passed. In general, are the pass rates of Math 1A and Math, 1B statistically the same? Let A = the subscript for Math 1A and B = the subscript for Math 1B., Exercise 14.1.11, (Solution on p. 668.), If you were to conduct an appropriate hypothesis test, the alternate hypothesis would be:, A., B., C., D., , Ha : p A = p B, Ha : p A > p B, Ho : p A = p B, Ha : p A 6 = p B, , Exercise 14.1.12, The Type I error is to:, , (Solution on p. 668.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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616, , APPENDIX, A. conclude that the pass rate for Math 1A is the same as the pass rate for Math 1B when, in fact,, the pass rates are different., B. conclude that the pass rate for Math 1A is different than the pass rate for Math 1B when, in fact,, the pass rates are the same., C. conclude that the pass rate for Math 1A is greater than the pass rate for Math 1B when, in fact,, the pass rate for Math 1A is less than the pass rate for Math 1B., D. conclude that the pass rate for Math 1A is the same as the pass rate for Math 1B when, in fact,, they are the same., Exercise 14.1.13, The correct decision is to:, , (Solution on p. 668.), , A. reject Ho, B. not reject Ho, C. There is not enough information given to conduct the hypothesis test, Kia, Alejandra, and Iris are runners on the track teams at three different schools. Their running times, in, minutes, and the statistics for the track teams at their respective schools, for a one mile run, are given in the, table below:, Running Time, , School Average Running Time, , School Standard Deviation, , Kia, , 4.9, , 5.2, , .15, , Alejandra, , 4.2, , 4.6, , .25, , Iris, , 4.5, , 4.9, , .12, Table 14.3, , Exercise 14.1.14, (Solution on p. 668.), Which student is the BEST when compared to the other runners at her school?, A., B., C., D., , Kia, Alejandra, Iris, Impossible to determine, , Questions 15 – 16 refer to the following:, The following adult ski sweater prices are from the Gorsuch Ltd. Winter catalog:, , {$212, $292, $278, $199$280, $236}, Assume the underlying sweater price population is approximately normal. The null hypothesis is that the, mean price of adult ski sweaters from Gorsuch Ltd. is at least $275., Exercise 14.1.15, The correct distribution to use for the hypothesis test is:, A., B., C., D., , (Solution on p. 668.), , Normal, Binomial, Student’s-t, Exponential, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 617, , Exercise 14.1.16, The hypothesis test:, A., B., C., D., , (Solution on p. 668.), , is two-tailed, is left-tailed, is right-tailed, has no tails, , Exercise 14.1.17, (Solution on p. 668.), Sara, a statistics student, wanted to determine the mean number of books that college professors, have in their office. She randomly selected 2 buildings on campus and asked each professor in the, selected buildings how many books are in his/her office. Sara surveyed 25 professors. The type, of sampling selected is a:, A., B., C., D., , simple random sampling, systematic sampling, cluster sampling, stratified sampling, , Exercise 14.1.18, (Solution on p. 668.), A clothing store would use which measure of the center of data when placing orders for the, typical "middle" customer?, A., B., C., D., , Mean, Median, Mode, IQR, , Exercise 14.1.19, In a hypothesis test, the p-value is, , (Solution on p. 668.), , A. the probability that an outcome of the data will happen purely by chance when the null hypothesis is true., B. called the preconceived alpha., C. compared to beta to decide whether to reject or not reject the null hypothesis., D. Answer choices A and B are both true., Questions 20 - 22 refer to the following:, A community college offers classes 6 days a week: Monday through Saturday. Maria conducted a study of, the students in her classes to determine how many days per week the students who are in her classes come, to campus for classes. In each of her 5 classes she randomly selected 10 students and asked them how many, days they come to campus for classes. Each of her classes are the same size. The results of her survey are, summarized in the table below., Number of Days on Campus, , Frequency, , Relative Frequency, , 1, , 2, , 2, , 12, , .24, , 3, , 10, , .20, , 4, , Cumulative Relative Frequency, , .98, , 5, , 0, , 6, , 1, , .02, , 1.00, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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618, , APPENDIX, Table 14.4, , Exercise 14.1.20, (Solution on p. 668.), Combined with convenience sampling, what other sampling technique did Maria use?, A., B., C., D., , simple random, systematic, cluster, stratified, , Exercise 14.1.21, How many students come to campus for classes 4 days a week?, A., B., C., D., , 49, 25, 30, 13, , Exercise 14.1.22, What is the 60th percentile for the this data?, A., B., C., D., , (Solution on p. 668.), , (Solution on p. 668.), , 2, 3, 4, 5, , The next two questions refer to the following:, The following data are the results of a random survey of 110 Reservists called to active duty to increase, security at California airports., Number of Dependents, , Frequency, , 0, , 11, , 1, , 27, , 2, , 33, , 3, , 20, , 4, , 19, Table 14.5, , Exercise 14.1.23, (Solution on p. 668.), Construct a 95% Confidence Interval for the true population mean number of dependents of, Reservists called to active duty to increase security at California airports., A., B., C., D., , (1.85, 2.32), (1.80, 2.36), (1.97, 2.46), (1.92, 2.50), , Exercise 14.1.24, The 95% confidence Interval above means:, , (Solution on p. 668.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 619, , A. 5% of Confidence Intervals constructed this way will not contain the true population aveage, number of dependents., B. We are 95% confident the true population mean number of dependents falls in the interval., C. Both of the above answer choices are correct., D. None of the above., Exercise 14.1.25, X ∼U (4, 10). Find the 30th percentile., A., B., C., D., , 0.3000, 3, 5.8, 6.1, , Exercise 14.1.26, If X ∼ Exp (0.8), then P ( x < µ) =, A., B., C., D., , (Solution on p. 669.), , (Solution on p. 669.), , 0.3679, 0.4727, 0.6321, cannot be determined, , Exercise 14.1.27, (Solution on p. 669.), The lifetime of a computer circuit board is normally distributed with a mean of 2500 hours and a, standard deviation of 60 hours. What is the probability that a randomly chosen board will last at, most 2560 hours?, A., B., C., D., , 0.8413, 0.1587, 0.3461, 0.6539, , Exercise 14.1.28, (Solution on p. 669.), A survey of 123 Reservists called to active duty as a result of the September 11, 2001, attacks, was conducted to determine the proportion that were married. Eighty-six reported being married., Construct a 98% confidence interval for the true population proportion of reservists called to active, duty that are married., A., B., C., D., , (0.6030, 0.7954), (0.6181, 0.7802), (0.5927, 0.8057), (0.6312, 0.7672), , Exercise 14.1.29, (Solution on p. 669.), Winning times in 26 mile marathons run by world class runners average 145 minutes with a standard deviation of 14 minutes. A sample of the last 10 marathon winning times is collected., Let x = mean winning times for 10 marathons., The distribution for x is:, , , A. N 145, √14, 10, , B. N (145, 14), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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620, , APPENDIX, C. t9, D. t10, Exercise 14.1.30, (Solution on p. 669.), Suppose that Phi Beta Kappa honors the top 1% of college and university seniors. Assume that, grade point means (G.P.A.) at a certain college are normally distributed with a 2.5 mean and a, standard deviation of 0.5. What would be the minimum G.P.A. needed to become a member of, Phi Beta Kappa at that college?, A., B., C., D., , 3.99, 1.34, 3.00, 3.66, , The number of people living on American farms has declined steadily during this century. Here are data, on the farm population (in millions of persons) from 1935 to 1980., Year, , 1935, , 1940, , 1945, , 1950, , 1955, , 1960, , 1965, , 1970, , 1975, , 1980, , Population, , 32.1, , 30.5, , 24.4, , 23.0, , 19.1, , 15.6, , 12.4, , 9.7, , 8.9, , 7.2, , Table 14.6, , The linear regression equation is y-hat = 1166.93 – 0.5868x, Exercise 14.1.31, What was the expected farm population (in millions of persons) for 1980?, A., B., C., D., , 7.2, 5.1, 6.0, 8.0, , Exercise 14.1.32, In linear regression, which is the best possible SSE?, A., B., C., D., , (Solution on p. 669.), , (Solution on p. 669.), , 13.46, 18.22, 24.05, 16.33, , Exercise 14.1.33, (Solution on p. 669.), In regression analysis, if the correlation coefficient is close to 1 what can be said about the best fit, line?, A., B., C., D., , It is a horizontal line. Therefore, we can not use it., There is a strong linear pattern. Therefore, it is most likely a good model to be used., The coefficient correlation is close to the limit. Therefore, it is hard to make a decision., We do not have the equation. Therefore, we can not say anything about it., , Question 34-36 refer to the following:, A study of the career plans of young women and men sent questionnaires to all 722 members of the senior, class in the College of Business Administration at the University of Illinois. One question asked which, major within the business program the student had chosen. Here are the data from the students who, responded., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 621, , Female, , Male, , Accounting, , 68, , 56, , Administration, , 91, , 40, , Ecomonics, , 5, , 6, , Finance, , 61, , 59, , Table 14.7: Does the data suggest that there is a relationship between the gender of students and their, , choice of major?, Exercise 14.1.34, The distribution for the test is:, A., B., C., D., , Chi2 8, Chi2 3, t721, N (0, 1), , Exercise 14.1.35, The expected number of female who choose Finance is :, A., B., C., D., , (Solution on p. 669.), , (Solution on p. 669.), , 37, 61, 60, 70, , Exercise 14.1.36, (Solution on p. 669.), The p-value is 0.0127 and the level of significance is 0.05. The conclusion to the test is:, A. There is insufficient evidence to conclude that the choice of major and the gender of the student, are not independent of each other., B. There is sufficient evidence to conclude that the choice of major and the gender of the student, are not independent of each other., C. There is sufficient evidence to conclude that students find Economics very hard., D. There is in sufficient evidence to conclude that more females prefer Administration than males., Exercise 14.1.37, (Solution on p. 669.), An agency reported that the work force nationwide is composed of 10% professional, 10% clerical,, 30% skilled, 15% service, and 35% semiskilled laborers. A random sample of 100 San Jose residents, indicated 15 professional, 15 clerical, 40 skilled, 10 service, and 20 semiskilled laborers. At α = .10, does the work force in San Jose appear to be consistent with the agency report for the nation?, Which kind of test is it?, A., B., C., D., , Chi2 goodness of fit, Chi2 test of independence, Independent groups proportions, Unable to determine, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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622, , APPENDIX, , 14.2 Practice Final Exam 22, Exercise 14.2.1, (Solution on p. 669.), A study was done to determine the proportion of teenagers that own a car. The population, proportion of teenagers that own a car is the, A., B., C., D., , statistic, parameter, population, variable, , The next two questions refer to the following data:, value, , frequency, , 0, , 1, , 1, , 4, , 2, , 7, , 3, , 9, , 6, , 4, Table 14.8, , Exercise 14.2.2, The box plot for the data is:, , (Solution on p. 669.), , A., , B., , C., 2 This, , content is available online at <http://cnx.org/content/m16303/1.16/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 623, , D., Exercise 14.2.3, (Solution on p. 669.), If 6 were added to each value of the data in the table, the 15th percentile of the new list of values, is:, A., B., C., D., , 6, 1, 7, 8, , The next two questions refer to the following situation:, Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a, drought occurs, the probability of water rationing is 10%. However, in any year, the probability of water, rationing is 5%., Exercise 14.2.4, What is the probability of both a drought and water rationing occurring?, A., B., C., D., , (Solution on p. 669.), , 0.05, 0.01, 0.02, 0.30, , Exercise 14.2.5, Which of the following is true?, , (Solution on p. 669.), , A. drought and water rationing are independent events, B. drought and water rationing are mutually exclusive events, C. none of the above, The next two questions refer to the following situation:, Suppose that a survey yielded the following data:, Favorite Pie Type, gender, , apple, , pumpkin, , pecan, , female, , 40, , 10, , 30, , male, , 20, , 30, , 10, , Table 14.9, , Exercise 14.2.6, (Solution on p. 669.), Suppose that one individual is randomly chosen. The probability that the person’s favorite pie is, apple or the person is male is:, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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624, , APPENDIX, A., B., C., D., , 40, 60, 60, 140, 120, 140, 100, 140, , Exercise 14.2.7, Suppose Ho is: Favorite pie type and gender are independent., , (Solution on p. 669.), , The p-value is:, A., B., C., D., , ≈0, 1, 0.05, cannot be determined, , The next two questions refer to the following situation:, Let’s say that the probability that an adult watches the news at least once per week is 0.60. We randomly, survey 14 people. Of interest is the number that watch the news at least once per week., Exercise 14.2.8, Which of the following statements is FALSE?, A., B., C., D., , (Solution on p. 669.), , X ∼ B (14, 0.60), The values for x are: {1, 2, 3, ..., 14}, µ = 8.4, P ( X = 5) = 0.0408, , Exercise 14.2.9, Find the probability that at least 6 adults watch the news., , (Solution on p. 669.), , 6, A. 14, B. 0.8499, C. 0.9417, D. 0.6429, , Exercise 14.2.10, (Solution on p. 669.), The following histogram is most likely to be a result of sampling from which distribution?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , A., B., C., D., , 625, , Chi-Square with df = 6, Exponential, Uniform, Binomial, , The ages of campus day and evening students is known to be normally distributed. A sample of 6 campus, day and evening students reported their ages (in years) as: {18, 35, 27, 45, 20, 20}, Exercise 14.2.11, (Solution on p. 670.), What is the error bound for the 90% confidence interval of the true average age?, A., B., C., D., , 11.2, 22.3, 17.5, 8.7, , Exercise 14.2.12, (Solution on p. 670.), If a normally distributed random variable has µ = 0 and σ = 1 , then 97.5% of the population, values lie above:, A., B., C., D., , -1.96, 1.96, 1, -1, , The next three questions refer to the following situation:, The amount of money a customer spends in one trip to the supermarket is known to have an exponential, distribution. Suppose the average amount of money a customer spends in one trip to the supermarket is, $72., Exercise 14.2.13, (Solution on p. 670.), What is the probability that one customer spends less than $72 in one trip to the supermarket?, A. 0.6321, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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626, , APPENDIX, B. 0.5000, C. 0.3714, D. 1, Exercise 14.2.14, (Solution on p. 670.), How much money altogether would you expect next 5 customers to spend in one trip to the, supermarket (in dollars)?, A. 72, 2, B. 725, C. 5184, D. 360, Exercise 14.2.15, (Solution on p. 670.), If you want to find the probability that the mean of 50 customers is less than $60, the distribution, to use is:, A. N (72, 72) , B. N 72, √72, 50, , C. Exp (72), 1, D. Exp 72, The next three questions refer to the following situation:, The amount of time it takes a fourth grader to carry out the trash is uniformly distributed in the interval, from 1 to 10 minutes., Exercise 14.2.16, (Solution on p. 670.), What is the probability that a randomly chosen fourth grader takes more than 7 minutes to take, out the trash?, A., B., C., D., , 3, 9, 7, 9, 3, 10, 7, 10, , Exercise 14.2.17, (Solution on p. 670.), Which graph best shows the probability that a randomly chosen fourth grader takes more than 6, minutes to take out the trash given that he/she has already taken more than 3 minutes?, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 627, , Exercise 14.2.18, (Solution on p. 670.), We should expect a fourth grader to take how many minutes to take out the trash?, A., B., C., D., , 4.5, 5.5, 5, 10, , The next three questions refer to the following situation:, At the beginning of the quarter, the amount of time a student waits in line at the campus cafeteria is normally distributed with a mean of 5 minutes and a standard deviation of 1.5 minutes., Exercise 14.2.19, What is the 90th percentile of waiting times (in minutes)?, A., B., C., D., , 1.28, 90, 7.47, 6.92, , Exercise 14.2.20, The median waiting time (in minutes) for one student is:, A., B., C., D., , (Solution on p. 670.), , (Solution on p. 670.), , 5, 50, 2.5, 1.5, , Exercise 14.2.21, (Solution on p. 670.), Find the probability that the average wait time of 10 students is at most 5.5 minutes., A. 0.6301, B. 0.8541, C. 0.3694, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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628, , APPENDIX, D. 0.1459, Exercise 14.2.22, (Solution on p. 670.), A sample of 80 software engineers in Silicon Valley is taken and it is found that 20% of them earn, approximately $50,000 per year. A point estimate for the true proportion of engineers in Silicon, Valley who earn $50,000 per year is:, A., B., C., D., , 16, 0.2, 1, 0.95, , Exercise 14.2.23, If P ( Z < zα ) = 0. 1587 where Z ∼ N (0, 1) , then α is equal to:, A., B., C., D., , (Solution on p. 670.), , -1, 0.1587, 0.8413, 1, , Exercise 14.2.24, (Solution on p. 670.), A professor tested 35 students to determine their entering skills. At the end of the term, after, completing the course, the same test was administered to the same 35 students to study their, improvement. This would be a test of:, A., B., C., D., , independent groups, 2 proportions, matched pairs, dependent groups, exclusive groups, , Exercise 14.2.25, (Solution on p. 670.), A math exam was given to all the third grade children attending ABC School. Two random, samples of scores were taken., n, , x, , s, , Boys, , 55, , 82, , 5, , Girls, , 60, , 86, , 7, , Table 14.10, , Which of the following correctly describes the results of a hypothesis test of the claim, “There is, a difference between the mean scores obtained by third grade girls and boys at the 5 % level of, significance”?, A. Do not reject Ho . There is insufficient evidence to conclude that there is a difference in the, mean scores., B. Do not reject Ho . There is sufficient evidence to conclude that there is a difference in the mean, scores., C. Reject Ho . There is insufficient evidence to conclude that there is no difference in the mean, scores., D. Reject Ho . There is sufficient evidence to conclude that there is a difference in the mean scores., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 629, , Exercise 14.2.26, (Solution on p. 670.), In a survey of 80 males, 45 had played an organized sport growing up. Of the 70 females surveyed,, 25 had played an organized sport growing up. We are interested in whether the proportion for, males is higher than the proportion for females. The correct conclusion is:, A. There is insufficient information to conclude that the proportion for males is the same as the, proportion for females., B. There is insufficient information to conclude that the proportion for males is not the same as, the proportion for females., C. There is sufficient evidence to conclude that the proportion for males is higher than the proportion for females., D. Not enough information to determine., Exercise 14.2.27, (Solution on p. 670.), Note: Chi-Square Test of a Single Variance; Not all classes cover this topic. From past experience,, a statistics teacher has found that the average score on a midterm is 81 with a standard deviation, of 5.2. This term, a class of 49 students had a standard deviation of 5 on the midterm. Do the data, indicate that we should reject the teacher’s claim that the standard deviation is 5.2? Use α = 0.05., A. Yes, B. No, C. Not enough information given to solve the problem, Exercise 14.2.28, (Solution on p. 670.), Note: F Distribution Test of ANOVA; Not all classes cover this topic. Three loading machines, are being compared. Ten samples were taken for each machine. Machine I took an average of 31, minutes to load packages with a standard deviation of 2 minutes. Machine II took an average of 28, minutes to load packages with a standard deviation of 1.5 minutes. Machine III took an average of, 29 minutes to load packages with a standard deviation of 1 minute. Find the p-value when testing, that the average loading times are the same., A. the p–value is close to 0, B. p–value is close to 1, C. Not enough information given to solve the problem, The next three questions refer to the following situation:, A corporation has offices in different parts of the country. It has gathered the following information concerning the number of bathrooms and the number of employees at seven sites:, Number of employees x, , 650, , 730, , 810, , 900, , 102, , 107, , 1150, , Number of bathrooms y, , 40, , 50, , 54, , 61, , 82, , 110, , 121, , Table 14.11, , Exercise 14.2.29, (Solution on p. 670.), Is the correlation between the number of employees and the number of bathrooms significant?, A. Yes, B. No, C. Not enough information to answer question, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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630, , APPENDIX, Exercise 14.2.30, The linear regression equation is:, A., B., C., D., , (Solution on p. 670.), , ŷ = 0.0094 − 79.96x, ŷ = 79.96 + 0.0094x, ŷ = 79.96 − 0.0094x, ŷ = −0.0094 + 79.96x, , Exercise 14.2.31, (Solution on p. 670.), If a site has 1150 employees, approximately how many bathrooms should it have?, A., B., C., D., , 69, 91, 91,954, We should not be estimating here., , Exercise 14.2.32, (Solution on p. 670.), Note: Chi-Square Test of a Single Variance; Not all classes cover this topic. Suppose that a sample, of size 10 was collected, with x = 4.4 and s = 1.4 ., Ho : σ2 = 1.6 vs. Ha : σ2 6= 1.6. Which graph best describes the results of the test?, , Exercise 14.2.33, (Solution on p. 670.), 64 backpackers were asked the number of days their latest backpacking trip was. The number of, days is given in the table below:, # of days, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , Frequency, , 5, , 9, , 6, , 12, , 7, , 10, , 5, , 10, , Table 14.12, , Conduct an appropriate test to determine if the distribution is uniform., A. The p–value is > 0.10. There is insufficient information to conclude that the distribution is not, uniform., B. The p–value is < 0.01. There is sufficient information to conclude the distribution is not uniform., C. The p–value is between 0.01 and 0.10, but without alpha (α) there is not enough information, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 631, , D. There is no such test that can be conducted., Exercise 14.2.34, (Solution on p. 670.), Note: F Distribution test of One-Way ANOVA; Not all classes cover this topic. Which of the, following statements is true when using one-way ANOVA?, A., B., C., D., , The populations from which the samples are selected have different distributions., The sample sizes are large., The test is to determine if the different groups have the same means., There is a correlation between the factors of the experiment., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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632, , APPENDIX, , 14.3 Data Sets3, 14.3.1 Lap Times, The following tables provide lap times from Terri Vogel’s Log Book. Times are recorded in seconds for, 2.5-mile laps completed in a series of races and practice runs., Race Lap Times (in Seconds), Lap 1, , Lap 2, , Lap 3, , Lap 4, , Lap 5, , Lap 6, , Lap 7, , Race 1, , 135, , 130, , 131, , 132, , 130, , 131, , 133, , Race 2, , 134, , 131, , 131, , 129, , 128, , 128, , 129, , Race 3, , 129, , 128, , 127, , 127, , 130, , 127, , 129, , Race 4, , 125, , 125, , 126, , 125, , 124, , 125, , 125, , Race 5, , 133, , 132, , 132, , 132, , 131, , 130, , 132, , Race 6, , 130, , 130, , 130, , 129, , 129, , 130, , 129, , Race 7, , 132, , 131, , 133, , 131, , 134, , 134, , 131, , Race 8, , 127, , 128, , 127, , 130, , 128, , 126, , 128, , Race 9, , 132, , 130, , 127, , 128, , 126, , 127, , 124, , Race 10, , 135, , 131, , 131, , 132, , 130, , 131, , 130, , Race 11, , 132, , 131, , 132, , 131, , 130, , 129, , 129, , Race 12, , 134, , 130, , 130, , 130, , 131, , 130, , 130, , Race 13, , 128, , 127, , 128, , 128, , 128, , 129, , 128, , Race 14, , 132, , 131, , 131, , 131, , 132, , 130, , 130, , Race 15, , 136, , 129, , 129, , 129, , 129, , 129, , 129, , Race 16, , 129, , 129, , 129, , 128, , 128, , 129, , 129, , Race 17, , 134, , 131, , 132, , 131, , 132, , 132, , 132, , Race 18, , 129, , 129, , 130, , 130, , 133, , 133, , 127, , Race 19, , 130, , 129, , 129, , 129, , 129, , 129, , 128, , Race 20, , 131, , 128, , 130, , 128, , 129, , 130, , 130, , Table 14.13, 3 This, , content is available online at <http://cnx.org/content/m17132/1.5/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 633, Practice Lap Times (in Seconds), Lap 1, , Lap 2, , Lap 3, , Lap 4, , Lap 5, , Lap 6, , Lap 7, , Practice 1, , 142, , 143, , 180, , 137, , 134, , 134, , 172, , Practice 2, , 140, , 135, , 134, , 133, , 128, , 128, , 131, , Practice 3, , 130, , 133, , 130, , 128, , 135, , 133, , 133, , Practice 4, , 141, , 136, , 137, , 136, , 136, , 136, , 145, , Practice 5, , 140, , 138, , 136, , 137, , 135, , 134, , 134, , Practice 6, , 142, , 142, , 139, , 138, , 129, , 129, , 127, , Practice 7, , 139, , 137, , 135, , 135, , 137, , 134, , 135, , Practice 8, , 143, , 136, , 134, , 133, , 134, , 133, , 132, , Practice 9, , 135, , 134, , 133, , 133, , 132, , 132, , 133, , Practice 10, , 131, , 130, , 128, , 129, , 127, , 128, , 127, , Practice 11, , 143, , 139, , 139, , 138, , 138, , 137, , 138, , Practice 12, , 132, , 133, , 131, , 129, , 128, , 127, , 126, , Practice 13, , 149, , 144, , 144, , 139, , 138, , 138, , 137, , Practice 14, , 133, , 132, , 137, , 133, , 134, , 130, , 131, , Practice 15, , 138, , 136, , 133, , 133, , 132, , 131, , 131, , Table 14.14, , 14.3.2 Stock Prices, The following table lists initial public offering (IPO) stock prices for all 1999 stocks that at least doubled in, value during the first day of trading. This is historical data., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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634, , APPENDIX, IPO Offer Prices, $17.00, , $23.00, , $14.00, , $16.00, , $12.00, , $26.00, , $20.00, , $22.00, , $14.00, , $15.00, , $22.00, , $18.00, , $18.00, , $21.00, , $21.00, , $19.00, , $15.00, , $21.00, , $18.00, , $17.00, , $15.00, , $25.00, , $14.00, , $30.00, , $16.00, , $10.00, , $20.00, , $12.00, , $16.00, , $17.44, , $16.00, , $14.00, , $15.00, , $20.00, , $20.00, , $16.00, , $17.00, , $16.00, , $15.00, , $15.00, , $19.00, , $48.00, , $16.00, , $18.00, , $9.00, , $18.00, , $18.00, , $20.00, , $8.00, , $20.00, , $17.00, , $14.00, , $11.00, , $16.00, , $19.00, , $15.00, , $21.00, , $12.00, , $8.00, , $16.00, , $13.00, , $14.00, , $15.00, , $14.00, , $13.41, , $28.00, , $21.00, , $17.00, , $28.00, , $17.00, , $19.00, , $16.00, , $17.00, , $19.00, , $18.00, , $17.00, , $15.00, , $14.00, , $21.00, , $12.00, , $18.00, , $24.00, , $15.00, , $23.00, , $14.00, , $16.00, , $12.00, , $24.00, , $20.00, , $14.00, , $14.00, , $15.00, , $14.00, , $19.00, , $16.00, , $38.00, , $20.00, , $24.00, , $16.00, , $8.00, , $18.00, , $17.00, , $16.00, , $15.00, , $7.00, , $19.00, , $12.00, , $8.00, , $23.00, , $12.00, , $18.00, , $20.00, , $21.00, , $34.00, , $16.00, , $26.00, , $14.00, , Table 14.15, NOTE : Data compiled by Jay R. Ritter of Univ. of Florida using data from Securities Data Co. and, Bloomberg., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 635, , 14.4 Group Projects, 14.4.1 Group Project: Univariate Data4, 14.4.1.1 Student Learning Objectives, , • The student will design and carry out a survey., • The student will analyze and graphically display the results of the survey., , 14.4.1.2 Instructions, As you complete each task below, check it off. Answer all questions in your summary., ____ Decide what data you are going to study., E XAMPLES : Here are two examples, but you may NOT use them: number of M&M’s per, small bag, number of pencils students have in their backpacks., ____ Are your data discrete or continuous? How do you know?, ____ Decide how you are going to collect the data (for instance, buy 30 bags of M&M’s; collect data from, the World Wide Web)., ____ Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random, (using a random number generator) sampling. Do not use convenience sampling. What method did, you use? Why did you pick that method?, ____ Conduct your survey. Your data size must be at least 30., ____ Summarize your data in a chart with columns showing data value, frequency, relative frequency, and cumulative relative frequency., ____ Answer the following (rounded to 2 decimal places):, 1., 2., 3., 4., 5., , x=, s=, First quartile =, Median =, 70th percentile =, , What value is 2 standard deviations above the mean?, What value is 1.5 standard deviations below the mean?, Construct a histogram displaying your data., In complete sentences, describe the shape of your graph., Do you notice any potential outliers? If so, what values are they? Show your work in how you used, the potential outlier formula in Chapter 2 (since you have univariate data) to determine whether or, not the values might be outliers., ____ Construct a box plot displaying your data., ____ Does the middle 50% of the data appear to be concentrated together or spread apart? Explain how, you determined this., ____ Looking at both the histogram and the box plot, discuss the distribution of your data., ____, ____, ____, ____, ____, , 14.4.1.3 Assignment Checklist, You need to turn in the following typed and stapled packet, with pages in the following order:, ____ Cover sheet: name, class time, and name of your study, 4 This, , content is available online at <http://cnx.org/content/m17142/1.8/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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636, , APPENDIX, , ____ Summary page: This should contain paragraphs written with complete sentences. It should include, answers to all the questions above. It should also include statements describing the population under, study, the sample, a parameter or parameters being studied, and the statistic or statistics produced., ____ URL for data, if your data are from the World Wide Web., ____ Chart of data, frequency, relative frequency and cumulative relative frequency., ____ Page(s) of graphs: histogram and box plot., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 637, , 14.4.2 Group Project: Continuous Distributions and Central Limit Theorem5, 14.4.2.1 Student Learning Objectives, , • The student will collect a sample of continuous data., • The student will attempt to fit the data sample to various distribution models., • The student will validate the Central Limit Theorem., , 14.4.2.2 Instructions, As you complete each task below, check it off. Answer all questions in your summary., 14.4.2.3 Part I: Sampling, ____ Decide what continuous data you are going to study. (Here are two examples, but you may NOT use, them: the amount of money a student spends on college supplies this term or the length of a long, distance telephone call.), ____ Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random, (using a random number generator) sampling. Do not use convenience sampling. What method did, you use? Why did you pick that method?, ____ Conduct your survey. Gather at least 150 pieces of continuous quantitative data., ____ Define (in words) the random variable for your data. X = _______, ____ Create 2 lists of your data: (1) unordered data, (2) in order of smallest to largest., ____ Find the sample mean and the sample standard deviation (rounded to 2 decimal places)., −, , 1. x =, 2. s =, ____ Construct a histogram of your data containing 5 - 10 intervals of equal width. The histogram should, be a representative display of your data. Label and scale it., , 14.4.2.4 Part II: Possible Distributions, ____ Suppose that X followed the theoretical distributions below. Set up each distribution using the appropriate information from your data., ____ Uniform: X ~ U ____________ Use the lowest and highest values as a and b., ____ Exponential: X ~ Exp ____________Use x to estimate µ ., ____ Normal: X ~ N ____________ Use x to estimate for µ and s to estimate for σ., ____ Must your data fit one of the above distributions? Explain why or why not., ____ Could the data fit 2 or 3 of the above distributions (at the same time)? Explain., ____ Calculate the value k(an X value) that is 1.75 standard deviations above the sample mean. k =, _________ (rounded to 2 decimal places) Note: k = x + (1.75) ∗ s, ____ Determine the relative frequencies (RF) rounded to 4 decimal places., 1., 2., 3., 4., , frequency, , RF = total number surveyed, RF ( X < k) =, RF ( X > k) =, RF ( X = k) =, , Use a separate piece of paper for EACH distribution (uniform, exponential, normal) to respond to the, following questions., 5 This, , content is available online at <http://cnx.org/content/m17141/1.9/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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638, , APPENDIX, NOTE : You should have one page for the uniform, one page for the exponential, and one page for, the normal, , ____ State the distribution: X ∼ _________, ____ Draw a graph for each of the three theoretical distributions. Label the axes and mark them appropriately., ____ Find the following theoretical probabilities (rounded to 4 decimal places)., 1. P(X < k ) =, 2. P(X > k ) =, 3. P(X = k ) =, ____ Compare the relative frequencies to the corresponding probabilities. Are the values close?, ____ Does it appear that the data fit the distribution well? Justify your answer by comparing the probabilities to the relative frequencies, and the histograms to the theoretical graphs., 14.4.2.5 Part III: CLT Experiments, ______ From your original data (before ordering), use a random number generator to pick 40 samples of, size 5. For each sample, calculate the average., ______ On a separate page, attached to the summary, include the 40 samples of size 5, along with the 40, sample averages., ______ List the 40 averages in order from smallest to largest., ______ Define the random variable, X , in words. X =, ______ State the approximate theoretical distribution of X. X ∼, ______ Base this on the mean and standard deviation from your original data., ______ Construct a histogram displaying your data. Use 5 to 6 intervals of equal width. Label and scale it., Calculate the value k (an X value) that is 1.75 standard deviations above the sample mean. k= _____, (rounded to 2 decimal places), Determine the relative frequencies (RF) rounded to 4 decimal places., 1. RF( X < k ) =, 2. RF(X > k ) =, 3. RF(X = k ) =, Find the following theoretical probabilities (rounded to 4 decimal places)., •. P(X < k ) =, •. P(X > k ) =, •. P(X = k ) =, ______ Draw the graph of the theoretical distribution of X., ______ Answer the questions below., ______ Compare the relative frequencies to the probabilities. Are the values close?, −, , ______ Does it appear that the data of averages fit the distribution of X well? Justify your answer by, comparing the probabilities to the relative frequencies, and the histogram to the theoretical graph., ______ In 3 - 5 complete sentences for each, answer the following questions. Give thoughtful explanations., ______ In summary, do your original data seem to fit the uniform, exponential, or normal distributions?, Answer why or why not for each distribution. If the data do not fit any of those distributions, explain, why., ______ What happened to the shape and distribution when you averaged your data? In theory, what, should have happened? In theory, would “it” always happen? Why or why not?, ______ Were the relative frequencies compared to the theoretical probabilities closer when comparing the, −, , X or X distributions? Explain your answer., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 639, , 14.4.2.6 Assignment Checklist, You need to turn in the following typed and stapled packet, with pages in the following order:, ____ Cover sheet: name, class time, and name of your study, ____ Summary pages: These should contain several paragraphs written with complete sentences that describe the experiment, including what you studied and your sampling technique, as well as answers, to all of the questions above., ____ URL for data, if your data are from the World Wide Web., ____ Pages, one for each theoretical distribution, with the distribution stated, the graph, and the probability questions answered, ____ Pages of the data requested, ____ All graphs required, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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640, , APPENDIX, , 14.4.3 Partner Project: Hypothesis Testing - Article6, 14.4.3.1 Student Learning Objectives, , • The student will identify a hypothesis testing problem in print., • The student will conduct a survey to verify or dispute the results of the hypothesis test., • The student will summarize the article, analysis, and conclusions in a report., , 14.4.3.2 Instructions, As you complete each task below, check it off. Answer all questions in your summary., ____ Find an article in a newspaper, magazine or on the internet which makes a claim about ONE population mean or ONE population proportion. The claim may be based upon a survey that the article was, reporting on. Decide whether this claim is the null or alternate hypothesis., ____ Copy or print out the article and include a copy in your project, along with the source., ____ State how you will collect your data. (Convenience sampling is not acceptable.), ____ Conduct your survey. You must have more than 50 responses in your sample. When you hand in, your final project, attach the tally sheet or the packet of questionnaires that you used to collect data., Your data must be real., ____ State the statistics that are a result of your data collection: sample size, sample mean, and sample, standard deviation, OR sample size and number of successes., ____ Make 2 copies of the appropriate solution sheet., ____ Record the hypothesis test on the solution sheet, based on your experiment. Do a DRAFT solution, first on one of the solution sheets and check it over carefully. Have a classmate check your solution, to see if it is done correctly. Make your decision using a 5% level of significance. Include the 95%, confidence interval on the solution sheet., ____ Create a graph that illustrates your data. This may be a pie or bar chart or may be a histogram or box, plot, depending on the nature of your data. Produce a graph that makes sense for your data and gives, useful visual information about your data. You may need to look at several types of graphs before, you decide which is the most appropriate for the type of data in your project., ____ Write your summary (in complete sentences and paragraphs, with proper grammar and correct, spelling) that describes the project. The summary MUST include:, 1. Brief discussion of the article, including the source., 2. Statement of the claim made in the article (one of the hypotheses)., 3. Detailed description of how, where, and when you collected the data, including the sampling technique. Did you use cluster, stratified, systematic, or simple random sampling (using a random, number generator)? As stated above, convenience sampling is not acceptable., 4. Conclusion about the article claim in light of your hypothesis test. This is the conclusion of your, hypothesis test, stated in words, in the context of the situation in your project in sentence form,, as if you were writing this conclusion for a non-statistician., 5. Sentence interpreting your confidence interval in the context of the situation in your project., , 14.4.3.3 Assignment Checklist, Turn in the following typed (12 point) and stapled packet for your final project:, ____ Cover sheet containing your name(s), class time, and the name of your study., ____ Summary, which includes all items listed on summary checklist., ____ Solution sheet neatly and completely filled out. The solution sheet does not need to be typed., 6 This, , content is available online at <http://cnx.org/content/m17140/1.8/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 641, , ____ Graphic representation of your data, created following the guidelines discussed above. Include only, graphs which are appropriate and useful., ____ Raw data collected AND a table summarizing the sample data (n, xbar and s; or x, n, and p’, as, appropriate for your hypotheses). The raw data does not need to be typed, but the summary does., Hand in the data as you collected it. (Either attach your tally sheet or an envelope containing your, questionnaires.), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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642, , APPENDIX, , 14.4.4 Partner Project: Hypothesis Testing - Word Problem7, 14.4.4.1 Student Learning Objectives, , • The student will write, edit, and solve a hypothesis testing word problem., , 14.4.4.2 Instructions, Write an original hypothesis testing problem for either ONE population mean or ONE population proportion. As you complete each task, check it off. Answer all questions in your summary. Look at the homework, for the Hypothesis Testing: Single Mean and Single Proportion chapter for examples (poems, two acts of a, play, a work related problem). The problems with names attached to them are problems written by students, in past quarters. Some other examples that are not in the homework include: a soccer hypothesis testing, poster, a cartoon, a news reports, a children’s story, a song., ____ Your problem must be original and creative. It also must be in proper English. If English is difficult, for you, have someone edit your problem., ____ Your problem must be at least ½ page, typed and singled spaced. This DOES NOT include the data., Data will make the problem longer and that is fine. For this problem, the data and story may be real, or fictional., ____ In the narrative of the problem, make it very clear what the null and alternative hypotheses are., ____ Your sample size must be LARGER THAN 50 (even if it is fictional)., ____ State in your problem how you will collect your data., ____ Include your data with your word problem., ____ State the statistics that are a result of your data collection: sample size, sample mean, and sample, standard deviation, OR sample size and number of successes., ____ Create a graph that illustrates your problem. This may be a pie or bar chart or may be a histogram, or box plot, depending on the nature of your data. Produce a graph that makes sense for your data, and gives useful visual information about your data. You may need to look at several types of graphs, before you decide which is the most appropriate for your problem., ____ Make 2 copies of the appropriate solution sheet., ____ Record the hypothesis test on the solution sheet, based on your problem. Do a DRAFT solution first, on one of the solution sheets and check it over carefully. Make your decision using a 5% level of, significance. Include the 95% confidence interval on the solution, , 14.4.4.3 Assignment Checklist, You need to turn in the following typed (12 point) and stapled packet for your final project:, Cover sheet containing your name, the name of your problem, and the date, The problem, Data for the problem, Solution sheet neatly and completely filled out. The solution sheet does not need to be typed., Graphic representation of the data, created following the guidelines discussed above. Include only, graphs that are appropriate and useful., ____ Sentences interpreting the results of the hypothesis test and the confidence interval in the context, of the situation in the project., , ____, ____, ____, ____, ____, , 7 This, , content is available online at <http://cnx.org/content/m17144/1.7/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 643, , 14.4.5 Group Project: Bivariate Data, Linear Regression, and Univariate Data8, 14.4.5.1 Student Learning Objectives, , •, •, •, •, , The students will collect a bivariate data sample through the use of appropriate sampling techniques., The student will attempt to fit the data to a linear model., The student will determine the appropriateness of linear fit of the model., The student will analyze and graph univariate data., , 14.4.5.2 Instructions, 1. As you complete each task below, check it off. Answer all questions in your introduction or summary., 2. Check your course calendar for intermediate and final due dates., 3. Graphs may be constructed by hand or by computer, unless your instructor informs you otherwise., All graphs must be neat and accurate., 4. All other responses must be done on the computer., 5. Neatness and quality of explanations are used to determine your final grade., , 14.4.5.3 Part I: Bivariate Data, Introduction, ____ State the bivariate data your group is going to study., E XAMPLES : Here are two examples, but you may NOT use them: height vs. weight and age, vs. running distance., ____ Describe how your group is going to collect the data (for instance, collect data from the web, survey, students on campus)., ____ Describe your sampling technique in detail. Use cluster, stratified, systematic, or simple random, sampling (using a random number generator) sampling. Convenience sampling is NOT acceptable., ____ Conduct your survey. Your number of pairs must be at least 30., ____ Print out a copy of your data., Analysis, On a separate sheet of paper construct a scatter plot of the data. Label and scale both axes., State the least squares line and the correlation coefficient., On your scatter plot, in a different color, construct the least squares line., Is the correlation coefficient significant? Explain and show how you determined this., Interpret the slope of the linear regression line in the context of the data in your project. Relate the, explanation to your data, and quantify what the slope tells you., ____ Does the regression line seem to fit the data? Why or why not? If the data does not seem to be linear,, explain if any other model seems to fit the data better., ____ Are there any outliers? If so, what are they? Show your work in how you used the potential outlier, formula in the Linear Regression and Correlation chapter (since you have bivariate data) to determine, whether or not any pairs might be outliers., ____, ____, ____, ____, ____, , 8 This, , content is available online at <http://cnx.org/content/m17143/1.6/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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644, , APPENDIX, , 14.4.5.4 Part II: Univariate Data, In this section, you will use the data for ONE variable only. Pick the variable that is more interesting to, analyze. For example: if your independent variable is sequential data such as year with 30 years and one, piece of data per year, your x-values might be 1971, 1972, 1973, 1974, . . ., 2000. This would not be interesting, to analyze. In that case, choose to use the dependent variable to analyze for this part of the project., _____ Summarize your data in a chart with columns showing data value, frequency, relative frequency,, and cumulative relative frequency., _____ Answer the following, rounded to 2 decimal places:, 1., 2., 3., 4., 5., 6., 7., 8., , Sample mean =, Sample standard deviation =, First quartile =, Third quartile =, Median =, 70th percentile =, Value that is 2 standard deviations above the mean =, Value that is 1.5 standard deviations below the mean =, , _____ Construct a histogram displaying your data. Group your data into 6 – 10 intervals of equal width., Pick regularly spaced intervals that make sense in relation to your data. For example, do NOT group, data by age as 20-26,27-33,34-40,41-47,48-54,55-61 . . . Instead, maybe use age groups 19.5-24.5, 24.529.5, . . . or 19.5-29.5, 29.5-39.5, 39.5-49.5, . . ., _____ In complete sentences, describe the shape of your histogram., _____ Are there any potential outliers? Which values are they? Show your work and calculations as to, how you used the potential outlier formula in chapter 2 (since you are now using univariate data) to, determine which values might be outliers., _____ Construct a box plot of your data., _____ Does the middle 50% of your data appear to be concentrated together or spread out? Explain how, you determined this., _____ Looking at both the histogram AND the box plot, discuss the distribution of your data. For example:, how does the spread of the middle 50% of your data compare to the spread of the rest of the data represented in the box plot; how does this correspond to your description of the shape of the histogram;, how does the graphical display show any outliers you may have found; does the histogram show any, gaps in the data that are not visible in the box plot; are there any interesting features of your data that, you should point out., , 14.4.5.5 Due Dates, • Part I, Intro: __________ (keep a copy for your records), • Part I, Analysis: __________ (keep a copy for your records), • Entire Project, typed and stapled: __________, Cover sheet: names, class time, and name of your study., Part I: label the sections “Intro” and “Analysis.”, Part II:, Summary page containing several paragraphs written in complete sentences describing the experiment, including what you studied and how you collected your data. The summary page, should also include answers to ALL the questions asked above., ____ All graphs requested in the project., ____ All calculations requested to support questions in data., ____ Description: what you learned by doing this project, what challenges you had, how you overcame the challenges., ____, ____, ____, ____, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 645, , NOTE : Include answers to ALL questions asked, even if not explicitly repeated in the items, above., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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646, , APPENDIX, , 14.5 Solution Sheets, 14.5.1 Solution Sheet: Hypothesis Testing for Single Mean and Single Proportion9, Class Time:, Name:, a. Ho :, b. Ha :, c. In words, CLEARLY state what your random variable X or P’ represents., d. State the distribution to use for the test., e. What is the test statistic?, f. What is the p-value? In 1 – 2 complete sentences, explain what the p-value means for this problem., g. Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value., , Figure 14.1, , h. Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write, an appropriate conclusion, using complete sentences., i. Alpha:, ii. Decision:, iii. Reason for decision:, iv. Conclusion:, i. Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the graph of, the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval., , Figure 14.2, , 9 This, , content is available online at <http://cnx.org/content/m17134/1.6/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 647, , 14.5.2 Solution Sheet: Hypothesis Testing for Two Means, Paired Data, and Two, Proportions10, Class Time:, Name:, a. Ho : _______, b. Ha : _______, c. In words, clearly state what your random variable X 1 − X 2 , P1 ’ − P2 ’- or X d represents., d. State the distribution to use for the test., e. What is the test statistic?, f. What is the p-value? In 1 – 2 complete sentences, explain what the p-value means for this problem., g. Use the previous information to sketch a picture of this situation. CLEARLY label and scale the horizontal axis and shade the region(s) corresponding to the p-value., , Figure 14.3, , h. Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write, an appropriate conclusion, using complete sentences., i. Alpha:, ii. Decision:, iii. Reason for decision:, iv. Conclusion:, i. In complete sentences, explain how you determined which distribution to use., , 10 This, , content is available online at <http://cnx.org/content/m17133/1.6/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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648, , APPENDIX, , 14.5.3 Solution Sheet: The Chi-Square Distribution11, Class Time:, Name:, a. Ho : _______, b. Ha : _______, c. What are the degrees of freedom?, d. State the distribution to use for the test., e. What is the test statistic?, f. What is the p-value? In 1 – 2 complete sentences, explain what the p-value means for this problem., g. Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal, axis and shade the region(s) corresponding to the p-value., , Figure 14.4, , h. Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and write appropriate conclusions, using complete sentences., i. Alpha:, ii. Decision:, iii. Reason for decision:, iv. Conclusion:, , 11 This, , content is available online at <http://cnx.org/content/m17136/1.6/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 649, , 14.5.4 Solution Sheet: F Distribution and One-Way ANOVA12, Class Time:, Name:, a. Ho :, b. Ha :, c. df (n) = ______ df (d) = _______, d. State the distribution to use for the test., e. What is the test statistic?, f. What is the p-value?, g. Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal, axis and shade the region(s) corresponding to the p-value., , Figure 14.5, , h. Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and write appropriate conclusions, using complete sentences., i. Alpha:, ii. Decision:, iii. Reason for decision:, iv. Conclusion:, , 12 This, , content is available online at <http://cnx.org/content/m17135/1.7/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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650, , APPENDIX, , 14.6 English Phrases Written Mathematically13, 14.6.1 English Phrases Written Mathematically, When the English says:, , Interpret this as:, , Xis at least 4., , X≥4, , The minimum of X is 4., , X≥4, , X is no less than 4., , X≥4, , X is greater than or equal to 4., , X≥4, , X is at most 4., , X≤4, , The maximum of X is 4., , X≤4, , Xis no more than 4., , X≤4, , X is less than or equal to 4., , X≤4, , Xdoes not exceed 4., , X≤4, , Xis greater than 4., , X>4, , X is more than 4., , X>4, , Xexceeds 4., , X>4, , Xis less than 4., , X<4, , There are fewer X than 4., , X<4, , Xis 4., , X=4, , Xis equal to 4., , X=4, , Xis the same as 4., , X=4, , Xis not 4., , X 6= 4, , Xis not equal to 4., , X 6= 4, , Xis not the same as 4., , X 6= 4, , Xis different than 4., , X 6= 4, , Table 14.16, , 13 This, , content is available online at <http://cnx.org/content/m16307/1.6/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 651, , 14.7 Symbols and their Meanings14, Symbols and their Meanings, Chapter (1st used), Sampling and Data, , Symbol, , √, , Spoken, , Meaning, , The square root of, , same, , Sampling and Data, , π, , Pi, , 3.14159. . . (a specific, number), , Descriptive Statistics, , Q1, , Quartile one, , the first quartile, , Descriptive Statistics, , Q2, , Quartile two, , the second quartile, , Descriptive Statistics, , Q3, , Quartile three, , the third quartile, , Descriptive Statistics, , IQR, , inter-quartile range, , Q3-Q1=IQR, , Descriptive Statistics, , x, , x-bar, , sample mean, , Descriptive Statistics, , µ, , mu, , population mean, , Descriptive Statistics, , s s x sx, , s, , sample standard deviation, , Descriptive Statistics, , s2 s2x, , s-squared, , sample variance, , Descriptive Statistics, , σ σx σx, , sigma, , population, deviation, , Descriptive Statistics, , σ2 σx2, , sigma-squared, , population variance, , Descriptive Statistics, , Σ, , capital sigma, , sum, , Probability Topics, , {}, , brackets, , set notation, , Probability Topics, , S, , S, , sample space, , Probability Topics, , A, , Event A, , event A, , Probability Topics, , P ( A), , probability of A, , probability of A occurring, , Probability Topics, , P ( A | B), , probability of A given B, , prob. of A occurring, given B has occurred, , Probability Topics, , P ( AorB), , prob. of A or B, , prob. of A or B or both, occurring, , standard, , continued on next page, , 14 This, , content is available online at <http://cnx.org/content/m16302/1.9/>., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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652, , APPENDIX, , Probability Topics, , P ( AandB), , prob. of A and B, , prob. of both A and B, occurring (same time), , Probability Topics, , A’, , A-prime, complement, of A, , complement of A, not A, , Probability Topics, , P ( A’), , prob. of complement of, A, , same, , Probability Topics, , G1, , green on first pick, , same, , Probability Topics, , P ( G1 ), , prob. of green on first, pick, , same, , Discrete Random Variables, , PDF, , prob. distribution function, , same, , Discrete Random Variables, , X, , X, , the random variable X, , Discrete Random Variables, , X∼, , the distribution of X, , same, , Discrete Random Variables, , B, , binomial distribution, , same, , Discrete Random Variables, , G, , geometric distribution, , same, , Discrete Random Variables, , H, , hypergeometric dist., , same, , Discrete Random Variables, , P, , Poisson dist., , same, , Discrete Random Variables, , λ, , Lambda, , average of Poisson distribution, , Discrete Random Variables, , ≥, , greater than or equal to, , same, , Discrete Random Variables, , ≤, , less than or equal to, , same, , Discrete Random Variables, , =, , equal to, , same, , Discrete Random Variables, , 6=, , not equal to, , same, continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 653, , Continuous, Variables, , Random, , f (x), , f of x, , function of x, , Continuous, Variables, , Random, , pd f, , prob. density function, , same, , Continuous, Variables, , Random, , U, , uniform distribution, , same, , Continuous, Variables, , Random, , Exp, , exponential, tion, , same, , Continuous, Variables, , Random, , k, , k, , critical value, , Continuous, Variables, , Random, , f (x) =, , f of x equals, , same, , Continuous, Variables, , Random, , m, , m, , decay rate (for exp., dist.), , The Normal Distribution, , N, , normal distribution, , same, , The Normal Distribution, , z, , z-score, , same, , The Normal Distribution, , Z, , standard normal dist., , same, , The Central Limit Theorem, , CLT, , Central Limit Theorem, , same, , The Central Limit Theorem, , X, , X-bar, , the random variable Xbar, , The Central Limit Theorem, , µx, , mean of X, , the average of X, , The Central Limit Theorem, , µx, , mean of X-bar, , the average of X-bar, , The Central Limit Theorem, , σx, , standard deviation of X, , same, , The Central Limit Theorem, , σx, , standard deviation of, X-bar, , same, , The Central Limit Theorem, , ΣX, , sum of X, , same, , distribu-, , continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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654, , APPENDIX, , The Central Limit Theorem, , Σx, , sum of x, , same, , Confidence Intervals, , CL, , confidence level, , same, , Confidence Intervals, , CI, , confidence interval, , same, , Confidence Intervals, , EBM, , error bound for a mean, , same, , Confidence Intervals, , EBP, , error bound for a proportion, , same, , Confidence Intervals, , t, , student-t distribution, , same, , Confidence Intervals, , df, , degrees of freedom, , same, , Confidence Intervals, , tα, , student-t with a/2 area, in right tail, , same, , Confidence Intervals, , ^, p’ p, , p-prime; p-hat, , sample proportion of, success, , Confidence Intervals, , ^, q’ q, , q-prime; q-hat, , sample proportion of, failure, , Hypothesis Testing, , H0, , H-naught, H-sub 0, , null hypothesis, , Hypothesis Testing, , Ha, , H-a, H-sub a, , alternate hypothesis, , Hypothesis Testing, , H1, , H-1, H-sub 1, , alternate hypothesis, , Hypothesis Testing, , α, , alpha, , probability of Type I error, , Hypothesis Testing, , β, , beta, , probability of Type II, error, , Hypothesis Testing, , X1 − X2, , X1-bar minus X2-bar, , difference, means, , µ1 − µ2, , mu-1 minus mu-2, , difference in population means, , P’1 − P’2, , P1-prime, prime, , p1 − p2, , p1 minus p2, , difference in population proportions, , X2, , Ky-square, , Chi-square, , Chi-Square, tion, , 2, , Distribu-, , minus, , P2-, , in, , difference in, proportions, , continued on next page, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , sample, , sample
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APPENDIX, , 655, , Linear Regression and, Correlation, , F-Distribution, ANOVA, , and, , O, , Observed, , Observed frequency, , E, , Expected, , Expected frequency, , y = a + bx, , y equals a plus b-x, , equation of a line, , ^, y, , y-hat, , estimated value of y, , r, , correlation coefficient, , same, , e, , error, , same, , SSE, , Sum of Squared Errors, , same, , 1.9s, , 1.9 times s, , cut-off value for outliers, , F, , F-ratio, , F ratio, , Table 14.17, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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656, , APPENDIX, , 14.8 Formulas15, Formula 14.1: Factorial, n! = n (n − 1) (n − 2) ... (1), 0! = 1, Formula, 14.2: Combinations, n, n!, =, r, (n−r )!r!, Formula 14.3: Binomial Distribution, X ∼ B (n, p), , P ( X = x ) = nx p x qn− x , for x = 0, 1, 2, ..., n, Formula 14.4: Geometric Distribution, X ∼ G ( p), P ( X = x ) = q x−1 p , for x = 1, 2, 3, ..., Formula 14.5: Hypergeometric Distribution, X ∼ H (r, b, n), , !, ( xr ) n−b x, , , P (X = x) =, r +b, n, , Formula 14.6: Poisson Distribution, X ∼ P (µ), µ x e−µ, , P ( X = x ) = x!, Formula 14.7: Uniform Distribution, X ∼ U ( a, b), f ( X ) = b−1 a , a < x < b, Formula 14.8: Exponential Distribution, X ∼ Exp (m), f ( x ) = me−mx , m > 0, x ≥ 0, Formula 14.9:Normal Distribution, X ∼ N µ, σ2, f (x) =, , √1 e, σ 2π, , −( x −µ)2, 2σ2, , ,, , −∞ < x < ∞, , FormulaR14.10: Gamma Function, ∞, Γ (z) = 0 x z−1 e− x dx z > 0, √, Γ 12 = π, Γ (m + 1) = m! for m, a nonnegative integer, otherwise: Γ ( a + 1) = aΓ ( a), Formula 14.11: Student-t Distribution, X ∼ td f, 15 This, , content is available online at <http://cnx.org/content/m16301/1.7/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 657, , , f (x) =, X=, , 2, , 1+ xn, , −(n+1), 2, , √, , nπΓ(, , n, 2, , 1, Γ ( n+, 2 ), , ), , qZ, , Y, n, , Z ∼ N (0, 1) , Y ∼ X2d f ,n = degrees of freedom, Formula 14.12: Chi-Square Distribution, X ∼ X2d f, n −2, , f (x) =, , −x, , x 2 e 2, n, 2 2 Γ( n2 ), , , x > 0 , n = positive integer and degrees of freedom, , Formula 14.13: F Distribution, X ∼ Fd f (n),d f (d), d f (n) =degrees of freedom for the numerator, d f (d) =degrees of freedom for the denominator, f (x) =, X=, , Yu, Wv, , v, Γ ( u+, 2 ), , Γ(, , u, 2, , v, 2, , )Γ( ), , u, v, , u ( u −1 ) h, 2x 2, 1+, , u, v, , , , x −.5(u+v), , i, , , Y, W are chi-square, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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658, , APPENDIX, , 14.9 Notes for the TI-83, 83+, 84 Calculator16, 14.9.1 Quick Tips, Legend, represents a button press, •, • [ ] represents yellow command or green letter behind a key, • < > represents items on the screen, To adjust the contrast, Press, , , then hold, , to increase the contrast or, , to decrease the contrast., , To capitalize letters and words, Press, , to get one capital letter, or press, , , then, , letters. You can return to the top-level button values by pressing, , to set all button presses to capital, again., , To correct a mistake, If you hit a wrong button, just hit, , and start again., , To write in scientific notation, Numbers in scientific notation are expressed on the TI-83, 83+, and 84 using E notation, such that..., , • 4.321 E 4 = 4.321 × 104, • 4.321 E -4 = 4.321 × 10−4, To transfer programs or equations from one calculator to another:, Both calculators: Insert your respective end of the link cable cable and press, , , then [LINK]., , Calculator receiving information:, Step 1. Use the arrows to navigate to and select <RECEIVE>, Step 2. Press, Calculator sending information:, Step 1. Press appropriate number or letter., Step 2. Use up and down arrows to access the appropriate item., Step 3. Press, to select item to transfer., Step 4. Press right arrow to navigate to and select <TRANSMIT>., Step 5. Press, NOTE :, , ERROR 35 LINK generally means that the cables have not been inserted far enough., , Both calculators: Insert your respective end of the link cable cable Both calculators: press, [QUIT] To exit when done., 16 This, , content is available online at <http://cnx.org/content/m19710/1.7/>., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>, , , then
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APPENDIX, , 659, , 14.9.2 Manipulating One-Variable Statistics, NOTE :, , These directions are for entering data with the built-in statistical program., Sample Data, Data, , Frequency, , -2, , 10, , -1, , 3, , 0, , 4, , 1, , 5, , 3, , 8, , Table 14.18: We are manipulating 1-variable statistics., , To begin:, Step 1. Turn on the calculator., Step 2. Access statistics mode., Step 3. Select <4:ClrList> to clear data from lists, if desired., ,, Step 4. Enter list [L1] to be cleared., , [L1] ,, Step 5. Display last instruction., , [ENTRY], Step 6. Continue clearing remaining lists in the same fashion, if desired., ,, , [L2] ,, Step 7. Access statistics mode., Step 8. Select <1:Edit ., , ., , .>, , Step 9. Enter data. Data values go into [L1]. (You may need to arrow over to [L1]), , • Type in a data value and enter it. (For negative numbers, use the negate (-) key at the bottom of, the keypad), ,, ,, • Continue in the same manner until all data values are entered., Step 10. In [L2], enter the frequencies for each data value in [L1]., , • Type in a frequency and enter it. (If a data value appears only once, the frequency is "1"), ,, • Continue in the same manner until all data values are entered., Step 11. Access statistics mode., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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660, , APPENDIX, , Step 12. Navigate to <CALC>, Step 13. Access <1:1-var Stats>, Step 14. Indicate that the data is in [L1]..., , [L1] ,, Step 15. ...and indicate that the frequencies are in [L2]., , [L2] ,, Step 16. The statistics should be displayed. You may arrow down to get remaining statistics. Repeat as necessary., , 14.9.3 Drawing Histograms, NOTE :, , We will assume that the data is already entered, , We will construct 2 histograms with the built-in STATPLOT application. The first way will use the default, ZOOM. The second way will involve customizing a new graph., Step 1. Access graphing mode., , [STAT PLOT], Step 2. Select <1:plot 1> To access plotting - first graph., Step 3. Use the arrows navigate go to <ON> to turn on Plot 1., , <ON> ,, Step 4. Use the arrows to go to the histogram picture and select the histogram., Step 5. Use the arrows to navigate to <Xlist>, Step 6. If "L1" is not selected, select it., , [L1] ,, Step 7. Use the arrows to navigate to <Freq>., Step 8. Assign the frequencies to [L2]., , [L2] ,, Step 9. Go back to access other graphs., , [STAT PLOT], Step 10. Use the arrows to turn off the remaining plots., Step 11. Be sure to deselect or clear all equations before graphing., To deselect equations:, Step 1. Access the list of equations., Step 2. Select each equal sign (=)., Step 3. Continue, until all equations are deselected., To clear equations:, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 661, , Step 1. Access the list of equations., Step 2. Use the arrow keys to navigate to the right of each equal sign (=) and clear them., Step 3. Repeat until all equations are deleted., To draw default histogram:, Step 1. Access the ZOOM menu., Step 2. Select <9:ZoomStat>, Step 3. The histogram will show with a window automatically set., To draw custom histogram:, Step 1. Access, to set the graph parameters., Step 2. • Xmin = −2.5, • Xmax = 3.5, • Xscl = 1 (width of bars), • Ymin = 0, • Ymax = 10, • Yscl = 1 (spacing of tick marks on y-axis), • Xres = 1, to see the histogram., , Step 3. Access, To draw box plots:, , Step 1. Access graphing mode., , [STAT PLOT], Step 2. Select <1:Plot 1> to access the first graph., Step 3. Use the arrows to select <ON> and turn on Plot 1., Step 4. Use the arrows to select the box plot picture and enable it., Step 5. Use the arrows to navigate to <Xlist>, Step 6. If "L1" is not selected, select it., , [L1] ,, Step 7. Use the arrows to navigate to <Freq>., Step 8. Indicate that the frequencies are in [L2]., , [L2] ,, Step 9. Go back to access other graphs., , [STAT PLOT], Step 10. Be sure to deselect or clear all equations before graphing using the method mentioned above., Step 11. View the box plot., , [STAT PLOT], Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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662, , APPENDIX, , 14.9.4 Linear Regression, 14.9.4.1 Sample Data, The following data is real. The percent of declared ethnic minority students at De Anza College for selected, years from 1970 - 1995 was:, Year, , Student Ethnic Minority Percentage, , 1970, , 14.13, , 1973, , 12.27, , 1976, , 14.08, , 1979, , 18.16, , 1982, , 27.64, , 1983, , 28.72, , 1986, , 31.86, , 1989, , 33.14, , 1992, , 45.37, , 1995, , 53.1, , Table 14.19: The independent variable is "Year," while the independent variable is "Student Ethnic Minority, , Percent.", , Student Ethnic Minority Percentage, , Figure 14.6: By hand, verify the scatterplot above., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 663, , NOTE : The TI-83 has a built-in linear regression feature, which allows the data to be edited.The, x-values will be in [L1]; the y-values in [L2]., , To enter data and do linear regression:, Step 1. ON Turns calculator on, Step 2. Before accessing this program, be sure to turn off all plots., , • Access graphing mode., , [STAT PLOT], • Turn off all plots., ,, Step 3. Round to 3 decimal places. To do so:, , • Access the mode menu., , [STAT PLOT], • Navigate to <Float> and then to the right to <3>., • All numbers will be rounded to 3 decimal places until changed., Step 4. Enter statistics mode and clear lists [L1] and [L2], as describe above., ,, Step 5. Enter editing mode to insert values for x and y., ,, Step 6. Enter each value. Press, , to continue., , To display the correlation coefficient:, Step 1. Access the catalog., , [CATALOG], Step 2. Arrow down and select <DiagnosticOn>, ... ,, ,, Step 3. r and r2 will be displayed during regression calculations., Step 4. Access linear regression., Step 5. Select the form of y = a + bx, ,, The display will show:, LinReg, , •, •, •, •, •, , y = a + bx, a = −3176.909, b = 1.617, r2 = 0.924, r = 0.961, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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664, , APPENDIX, , This means the Line of Best Fit (Least Squares Line) is:, , • y = −3176.909 + 1.617x, • Percent = −3176.909 + 1.617(year #), The correlation coefficient r = 0.961, To see the scatter plot:, Step 1. Access graphing mode., , [STAT PLOT], Step 2. Select <1:plot 1> To access plotting - first graph., Step 3. Navigate and select <ON> to turn on Plot 1., , <ON>, Step 4. Navigate to the first picture., Step 5. Select the scatter plot., Step 6. Navigate to <Xlist>, , [L1] to select it., Step 7. If [L1] is not selected, press, Step 8. Confirm that the data values are in [L1]., , <ON>, Step 9. Navigate to <Ylist>, Step 10. Select that the frequencies are in [L2]., , [L2] ,, Step 11. Go back to access other graphs., , [STAT PLOT], Step 12. Use the arrows to turn off the remaining plots., Step 13. Access, , to set the graph parameters., , • Xmin = 1970, • Xmax = 2000, • Xscl = 10 (spacing of tick marks on x-axis), • Ymin = −0.05, • Ymax = 60, • Yscl = 10 (spacing of tick marks on y-axis), • Xres = 1, Step 14. Be sure to deselect or clear all equations before graphing, using the instructions above., Step 15. Press, , to see the scatter plot., , To see the regression graph:, Step 1. Access the equation menu. The regression equation will be put into Y1., Step 2. Access the vars menu and navigate to <5:, , Statistics>, , ,, Step 3. Navigate to <EQ>., Step 4. <1: RegEQ> contains the regression equation which will be entered in Y1., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 665, , Step 5. Press, , . The regression line will be superimposed over scatter plot., , To see the residuals and use them to calculate the critical point for an outlier:, Step 1. Access the list. RESID will be an item on the menu. Navigate to it., , [LIST], <RESID>, Step 2. Confirm twice to view the list of residuals. Use the arrows to select them., ,, Step 3. The critical point for an outlier is: 1.9V nSSE, −2 where:, , • n = number of pairs of data, • SSE = sum of the squared errors, • ∑ residual 2, Step 4. Store the residuals in [L3]., ,, Step 5. Calculate the, , , [L3] ,, , (residual )2, ., n −2, , Note that n − 2 = 8, , , [L3] ,, ,, Step 6. Store this value in [L4]., , ,, , ,, , [L4] ,, Step 7. Calculate the critical value using the equation above., ,, , ,, , ,, , , [V] ,, , ,, , , [LIST], , ,, , ,, , ,, , , [L4] ,, ,, ,, Step 8. Verify that the calculator displays: 7.642669563. This is the critical value., Step 9. Compare the absolute value of each residual value in [L3] to 7.64 . If the absolute value is greater, than 7.64, then the (x, y) corresponding point is an outlier. In this case, none of the points is an outlier., To obtain estimates of y for various x-values:, There are various ways to determine estimates for "y". One way is to substitute values for "x" in the, equation. Another way is to use the, , on the graph of the regression line., , 14.9.5 TI-83, 83+, 84 instructions for distributions and tests, 14.9.5.1 Distributions, Access DISTR (for "Distributions")., For technical assistance, visit the Texas Instruments website at http://www.ti.com17 and enter your calculator model into the "search" box., Binomial Distribution, , • binompdf(n,p,x) corresponds to P(X = x), • binomcdf(n,p,x) corresponds to P(X ≤ x), • To see a list of all probabilities for x: 0, 1, . . . , n, leave off the "x" parameter., Poisson Distribution, , • poissonpdf(λ,x) corresponds to P(X = x), • poissoncdf(λ,x) corresponds to P(X ≤ x), 17 http://www.ti.com, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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666, , APPENDIX, , Continuous Distributions (general), , • −∞ uses the value -1EE99 for left bound, • ∞ uses the value 1EE99 for right bound, Normal Distribution, , • normalpdf(x,µ,σ) yields a probability density function value (only useful to plot the normal curve,, in which case "x" is the variable), • normalcdf(left bound, right bound, µ,σ) corresponds to P(left bound < X < right bound), • normalcdf(left bound, right bound) corresponds to P(left bound < Z < right bound) - standard, normal, • invNorm(p,µ,σ) yields the critical value, k: P(X < k) = p, • invNorm(p) yields the critical value, k: P(Z < k) = p for the standard normal, Student-t Distribution, , • tpdf(x,df) yields the probability density function value (only useful to plot the student-t curve, in, which case "x" is the variable), • tcdf(left bound, right bound, df) corresponds to P(left bound < t < right bound), Chi-square Distribution, , • X2 pdf(x,df) yields the probability density function value (only useful to plot the chi2 curve, in which, case "x" is the variable), • X2 cdf(left bound, right bound, df) corresponds to P(left bound < X2 < right bound), F Distribution, , • Fpdf(x,dfnum,dfdenom) yields the probability density function value (only useful to plot the F curve,, in which case "x" is the variable), • Fcdf(left bound,right bound,dfnum,dfdenom) corresponds to P(left bound < F < right bound), 14.9.5.2 Tests and Confidence Intervals, Access STAT and TESTS., For the Confidence Intervals and Hypothesis Tests, you may enter the data into the appropriate lists and, press DATA to have the calculator find the sample means and standard deviations. Or, you may enter the, sample means and sample standard deviations directly by pressing STAT once in the appropriate tests., Confidence Intervals, , • ZInterval is the confidence interval for mean when σ is known, • TInterval is the confidence interval for mean when σ is unknown; s estimates σ., • 1-PropZInt is the confidence interval for proportion, NOTE : The confidence levels should be given as percents (ex. enter "95" or ".95" for a 95% confidence level)., , Hypothesis Tests, , • Z-Test is the hypothesis test for single mean when σ is known, • T-Test is the hypothesis test for single mean when σ is unknown; s estimates σ., • 2-SampZTest is the hypothesis test for 2 independent means when both σ’s are known, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , •, •, •, •, •, •, , 667, , 2-SampTTest is the hypothesis test for 2 independent means when both σ’s are unknown, 1-PropZTest is the hypothesis test for single proportion., 2-PropZTest is the hypothesis test for 2 proportions., X2 -Test is the hypothesis test for independence., X2 GOF-Test is the hypothesis test for goodness-of-fit (TI-84+ only)., LinRegTTEST is the hypothesis test for Linear Regression (TI-84+ only)., NOTE : Input the null hypothesis value in the row below "Inpt." For a test of a single mean, "µ∅", represents the null hypothesis. For a test of a single proportion, "p∅" represents the null hypothesis. Enter the alternate hypothesis on the bottom row., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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668, , APPENDIX, , Solutions to Exercises in Chapter 14, Solutions to Practice Final Exam 1, Solution to Exercise 14.1.1 (p. 613), B: Independent., Solution to Exercise 14.1.2 (p. 613), 4, C: 16, Solution to Exercise 14.1.3 (p. 613), B: Two measurements are drawn from the same pair of individuals or objects., Solution to Exercise 14.1.4 (p. 614), 68, B: 118, Solution to Exercise 14.1.5 (p. 614), 30, D: 52, Solution to Exercise 14.1.6 (p. 614), 8, B: 40, Solution to Exercise 14.1.7 (p. 614), B: 2.78, Solution to Exercise 14.1.8 (p. 615), A: 8.25, Solution to Exercise 14.1.9 (p. 615), C: 0.2870, Solution to Exercise 14.1.10 (p. 615), C: Normal, Solution to Exercise 14.1.11 (p. 615), D: Ha : p A 6= p B, Solution to Exercise 14.1.12 (p. 615), B: conclude that the pass rate for Math 1A is different than the pass rate for Math 1B when, in fact, the pass, rates are the same., Solution to Exercise 14.1.13 (p. 616), B: not reject Ho, Solution to Exercise 14.1.14 (p. 616), C: Iris, Solution to Exercise 14.1.15 (p. 616), C: Student’s-t, Solution to Exercise 14.1.16 (p. 617), B: is left-tailed, Solution to Exercise 14.1.17 (p. 617), C: cluster sampling, Solution to Exercise 14.1.18 (p. 617), B: Median, Solution to Exercise 14.1.19 (p. 617), A: the probability that an outcome of the data will happen purely by chance when the null hypothesis is, true., Solution to Exercise 14.1.20 (p. 618), D: stratified, Solution to Exercise 14.1.21 (p. 618), B: 25, Solution to Exercise 14.1.22 (p. 618), C: 4, Solution to Exercise 14.1.23 (p. 618), A: (1.85, 2.32), Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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APPENDIX, , 669, , Solution to Exercise 14.1.24 (p., C: Both above are correct., Solution to Exercise 14.1.25 (p., C: 5.8, Solution to Exercise 14.1.26 (p., C: 0.6321, Solution to Exercise 14.1.27 (p., A: 0.8413, Solution to Exercise 14.1.28 (p., A: (0.6030, 0.7954), Solution, 14.1.29 (p., , to Exercise, , 618), 619), 619), 619), 619), 619), , √14, 10, , A: N 145,, Solution to Exercise 14.1.30 (p. 620), D: 3.66, Solution to Exercise 14.1.31 (p. 620), B: 5.1, Solution to Exercise 14.1.32 (p. 620), A: 13.46, Solution to Exercise 14.1.33 (p. 620), B: There is a strong linear pattern. Therefore, it is most likely a good model to be used., Solution to Exercise 14.1.34 (p. 621), B: Chi2 3, Solution to Exercise 14.1.35 (p. 621), D: 70, Solution to Exercise 14.1.36 (p. 621), B: There is sufficient evidence to conclude that the choice of major and the gender of the student are not, independent of each other., Solution to Exercise 14.1.37 (p. 621), A: Chi2 goodness of fit, , Solutions to Practice Final Exam 2, Solution to Exercise 14.2.1 (p. 622), B: parameter, Solution to Exercise 14.2.2 (p. 622), A, Solution to Exercise 14.2.3 (p. 623), C: 7, Solution to Exercise 14.2.4 (p. 623), C: 0.02, Solution to Exercise 14.2.5 (p. 623), C: none of the above, Solution to Exercise 14.2.6 (p. 623), 100, D: 140, Solution to Exercise 14.2.7 (p. 624), A: ≈ 0, Solution to Exercise 14.2.8 (p. 624), B: The values for x are: {1, 2, 3, ..., 14}, Solution to Exercise 14.2.9 (p. 624), C: 0.9417, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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670, , APPENDIX, , Solution to Exercise 14.2.10 (p., D: Binomial, Solution to Exercise 14.2.11 (p., D: 8.7, Solution to Exercise 14.2.12 (p., A: -1.96, Solution to Exercise 14.2.13 (p., A: 0.6321, Solution to Exercise 14.2.14 (p., D: 360, Solution, 14.2.15 (p., , to Exercise, , 624), 625), 625), 625), 626), 626), , √72, 50, , B: N 72,, Solution to Exercise 14.2.16 (p. 626), A: 39, Solution to Exercise 14.2.17 (p. 626), D, Solution to Exercise 14.2.18 (p. 627), B: 5.5, Solution to Exercise 14.2.19 (p. 627), D: 6.92, Solution to Exercise 14.2.20 (p. 627), A: 5, Solution to Exercise 14.2.21 (p. 627), B: 0.8541, Solution to Exercise 14.2.22 (p. 628), B: 0.2, Solution to Exercise 14.2.23 (p. 628), A: -1, Solution to Exercise 14.2.24 (p. 628), C: matched pairs, dependent groups, Solution to Exercise 14.2.25 (p. 628), D: Reject Ho . There is sufficient evidence to conclude that there is a difference in the mean scores., Solution to Exercise 14.2.26 (p. 629), C: There is sufficient evidence to conclude that the proportion for males is higher than the proportion for, females., Solution to Exercise 14.2.27 (p. 629), B: No, Solution to Exercise 14.2.28 (p. 629), B: p-value is close to 1., Solution to Exercise 14.2.29 (p. 629), B: No, Solution to Exercise 14.2.30 (p. 630), C: ŷ = 79.96x − 0.0094, Solution to Exercise 14.2.31 (p. 630), D: We should not be estimating here., Solution to Exercise 14.2.32 (p. 630), A, Solution to Exercise 14.2.33 (p. 630), A: The p–value is > 0.10. There is insufficient information to conclude that the distribution is not uniform., Solution to Exercise 14.2.34 (p. 631), C: The test is to determine if the different groups have the same means., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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672, , GLOSSARY, , Glossary, , A Analysis of Variance, Also referred to as ANOVA. A method of testing whether or not the means of three or more, populations are equal. The method is applicable if:, , • All populations of interest are normally distributed., • The populations have equal standard deviations., • Samples (not necessarily of the same size) are randomly and independently selected from, each population., The test statistic for analysis of variance is the F-ratio., Average, A number that describes the central tendency of the data. There are a number of specialized, averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean., , B Bernoulli Trials, An experiment with the following characteristics:, , • There are only 2 possible outcomes called “success” and “failure” for each trial., • The probability p of a success is the same for any trial (so the probability q = 1 − p of a, failure is the same for any trial)., Binomial Distribution, A discrete random variable (RV) which arises from Bernoulli trials. There are a fixed number, n,, of independent trials. “Independent” means that the result of any trial (for example, trial 1), does not affect the results of the following trials, and all trials are conducted under the same, conditions. Under these circumstances the binomial RV X is defined as the number of successes, in n trials. The notation is: X ∼ B (n, p). The mean is µ = np and the standard deviation, is, , √, σ = npq. The probability of exactly x successes in n trials is P ( X = x ) = nx p x qn− x ., , C Central Limit Theorem, Given a random variable (RV) with known mean µ and known standard deviation σ. We are, sampling with size n and we are interested in two new RVs - the sample mean,, X, and, the, , sample sum, ΣX. If the size n of the sample is sufficiently large, then X ∼ N µ, √σn and ΣX ∼, √ , N nµ, nσ . If the size n of the sample is sufficiently large, then the distribution of the sample, means and the distribution of the sample sums will approximate a normal distribution, regardless of the shape of the population. The mean of the sample means will equal the, population mean and the mean of the sample sums will equal n times the population mean., The standard deviation of the distribution of the sample means, √σn , is called the standard error, of the mean., , Coefficient of Correlation, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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GLOSSARY, , 673, , A measure developed by Karl Pearson (early 1900s) that gives the strength of association, between the independent variable and the dependent variable. The formula is:, r = rh, , n ∑ xy − (∑ x ) (∑ y), ih, i,, n ∑ x 2 − ( ∑ x )2 n ∑ y2 − ( ∑ y )2, , (12.11), , where n is the number of data points. The coefficient cannot be more then 1 and less then -1., The closer the coefficient is to ±1, the stronger the evidence of a significant linear relationship, between x and y., Conditional Probability, The likelihood that an event will occur given that another event has already occurred., Confidence Interval (CI), An interval estimate for an unknown population parameter. This depends on:, , • The desired confidence level., • Information that is known about the distribution (for example, known standard deviation)., • The sample and its size., Confidence Level (CL), The percent expression for the probability that the confidence interval contains the true, population parameter. For example, if the CL = 90%, then in 90 out of 100 samples the interval, estimate will enclose the true population parameter., Contingency Table, The method of displaying a frequency distribution as a table with rows and columns to show, how two variables may be dependent (contingent) upon each other. The table provides an easy, way to calculate conditional probabilities., Continuous Random Variable, A random variable (RV) whose outcomes are measured., Example: The height of trees in the forest is a continuous RV., Cumulative Relative Frequency, The term applies to an ordered set of observations from smallest to largest. The Cumulative, Relative Frequency is the sum of the relative frequencies for all values that are less than or equal, to the given value., , D Data, A set of observations (a set of possible outcomes). Most data can be put into two groups:, qualitative (hair color, ethnic groups and other attributes of the population) and quantitative, (distance traveled to college, number of children in a family, etc.). Quantitative data can be, separated into two subgroups: discrete and continuous. Data is discrete if it is the result of, counting (the number of students of a given ethnic group in a class, the number of books on a, shelf, etc.). Data is continuous if it is the result of measuring (distance traveled, weight of, luggage, etc.), Degrees of Freedom (df), The number of objects in a sample that are free to vary., Discrete Random Variable, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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674, , GLOSSARY, A random variable (RV) whose outcomes are counted., , E Equally Likely, Each outcome of an experiment has the same probability., Error Bound for a Population Mean (EBM), The margin of error. Depends on the confidence level, sample size, and known or estimated, population standard deviation., Error Bound for a Population Proportion(EBP), The margin of error. Depends on the confidence level, sample size, and the estimated (from the, sample) proportion of successes., Event, A subset in the set of all outcomes of an experiment. The set of all outcomes of an experiment is, called a sample space and denoted usually by S. An event is any arbitrary subset in S. It can, contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, etc., Standard notations for events are capital letters such as A, B, C, etc., Expected Value, Expected arithmetic average when an experiment is repeated many times. (Also called the, mean). Notations: E ( x ) , µ. For a discrete random variable (RV) with probability distribution, function P ( x ) ,the definition can also be written in the form E ( x ) = µ = ∑ xP ( x ) ., Experiment, A planned activity carried out under controlled conditions., Exponential Distribution, A continuous random variable (RV) that appears when we are interested in the intervals of time, between some random events, for example, the length of time between emergency arrivals at a, hospital. Notation: X~Exp (m). The mean is µ = m1 and the standard deviation is σ = m1 . The, probability density function is f ( x ) = me−mx , x ≥ 0 and the cumulative distribution function, is P ( X ≤ x ) = 1 − e−mx ., , F Frequency, The number of times a value of the data occurs., , G Geometric Distribution, A discrete random variable (RV) which arises from the Bernoulli trials. The trials are repeated, until the first success. The geometric variable X is defined as the number of trials until the first, success. Notation: X ∼ G ( p). The mean is µ = 1p and the standard deviation is, r , , σ = 1p · 1p − 1 The probability of exactly x failures before the first success is given by the, formula: P ( X = x ) = p (1 − p) x−1 ., , H Hypergeometric Distribution, A discrete random variable (RV) that is characterized by, , • A fixed number of trials., • The probability of success is not the same from trial to trial., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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GLOSSARY, , 675, , We sample from two groups of items when we are interested in only one group. X is defined as, the number of successes out of the total number of items chosen. Notation: X~H (r, b, n) .,, where r = the number of items in the group of interest, b = the number of items in the group not, of interest, and n = the number of items chosen., Hypothesis, A statement about the value of a population parameter. In case of two hypotheses, the statement, assumed to be true is called the null hypothesis (notation H0 ) and the contradictory statement is, called the alternate hypothesis (notation Ha )., Hypothesis Testing, Based on sample evidence, a procedure to determine whether the hypothesis stated is a, reasonable statement and cannot be rejected, or is unreasonable and should be rejected., , I, , Independent Events, The occurrence of one event has no effect on the probability of the occurrence of any other event., Events A and B are independent if one of the following is true: (1). P ( A| B) = P ( A) ; (2), P ( B| A) = P ( B) ; (3) P ( AandB) = P ( A) P ( B)., Inferential Statistics, Also called statistical inference or inductive statistics. This facet of statistics deals with, estimating a population parameter based on a sample statistic. For example, if 4 out of the 100, calculators sampled are defective we might infer that 4 percent of the production is defective., Interquartile Range (IRQ), The distance between the third quartile (Q3) and the first quartile (Q1). IQR = Q3 - Q1., , L Level of Significance of the Test, Probability of a Type I error (reject the null hypothesis when it is true). Notation: α. In, hypothesis testing, the Level of Significance is called the preconceived α or the preset α., , M Mean, A number that measures the central tendency. A common name for mean is ’average.’ The term, ’mean’ is a shortened form of ’arithmetic mean.’ By definition, the mean for a sample (denoted, Sum of all values in the sample, by x) is x = Number of values in the sample , and the mean for a population (denoted by µ) is, µ=, , Sum of all values in the population, Number of values in the population ., , Median, A number that separates ordered data into halves. Half the values are the same number or, smaller than the median and half the values are the same number or larger than the median., The median may or may not be part of the data., Mode, The value that appears most frequently in a set of data., Mutually Exclusive, An observation cannot fall into more than one class (category). Being in more than one category, prevents being in a mutually exclusive category., , N Normal Distribution, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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676, , GLOSSARY, A continuous random variable (RV) with pdf f(x) =, , √1 e−( x −µ) 2 /2σ2 ,, σ 2π, , where µ is the mean of, , the distribution and σ is the standard deviation. Notation: X ∼ N (µ, σ ). If µ = 0 and σ = 1, the, RV is called the standard normal distribution., , O Outcome (observation), A particular result of an experiment., Outlier, An observation that does not fit the rest of the data., , P p-value, The probability that an event will happen purely by chance assuming the null hypothesis is true., The smaller the p-value, the stronger the evidence is against the null hypothesis., Parameter, A numerical characteristic of the population., Percentile, A number that divides ordered data into hundredths., Example: Let a data set contain 200 ordered observations starting with {2.3, 2.7, 2.8, 2.9, 2.9, 3.0...}., (2.7+2.8), = 2.75, because 1% of the data is to the left of this point on, Then the first percentile is, 2, (2.9+2.9), , the number line and 99% of the data is on its right. The second percentile is, = 2.9., 2, Percentiles may or may not be part of the data. In this example, the first percentile is not in the, data, but the second percentile is. The median of the data is the second quartile and the 50th, percentile. The first and third quartiles are the 25th and the 75th percentiles, respectively., Point Estimate, A single number computed from a sample and used to estimate a population parameter., Poisson Distribution, A discrete random variable (RV) that counts the number of times a certain event will occur in a, specific interval. Characteristics of the variable:, , • The probability that the event occurs in a given interval is the same for all intervals., • The events occur with a known mean and independently of the time since the last event., The distribution is defined by the mean µ of the event in the interval. Notation: X~P (µ). The, √, mean is µ = np. The standard deviation is σ = µ. The probability of having exactly x, µx, , successes in r trials is P ( X = x ) = e−µ x! . The Poisson distribution is often used to approximate, the binomial distribution when n is “large” and p is “small” (a general rule is that n should be, greater than or equal to 20 and p should be less than or equal to .05)., Population, The collection, or set, of all individuals, objects, or measurements whose properties are being, studied., Probability, A number between 0 and 1, inclusive, that gives the likelihood that a specific event will occur., The foundation of statistics is given by the following 3 axioms (by A. N. Kolmogorov, 1930’s):, Let S denote the sample space and A and B are two events in S . Then:, , • 0 ≤ P ( A) ≤ 1;., Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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GLOSSARY, , 677, , • If A and B are any two mutually exclusive events, then P ( AorB) = P ( A) + P ( B)., • P (S) = 1., Probability Distribution Function (PDF), A mathematical description of a discrete random variable (RV), given either in the form of an, equation (formula) , or in the form of a table listing all the possible outcomes of an experiment, and the probability associated with each outcome., Example: A biased coin with probability 0.7 for a head (in one toss of the coin) is tossed 5 times., We are interested in the number of, heads, (the RV X = the number of heads). X is Binomial, so, X ∼ B (5, 0.7) and P ( X = x ) = , , 5, , x, , .7x .35− x or in the form of the table:, x, , P (X = x), , 0, , 0.0024, , 1, , 0.0284, , 2, , 0.1323, , 3, , 0.3087, , 4, , 0.3602, , 5, , 0.1681, Table 4.3, , Proportion, , • As a number: A proportion is the number of successes divided by the total number in the, sample., • As a probability distribution: Given a binomial random variable (RV), X ∼ B (n, p), consider, the ratio of the number X of successes in n Bernouli trials to the number n of trials. P’ = Xn ., This new RV, is called a proportion, and if the number of trials, n, is large enough, P’, ∼ N p, pq, n ., , Q Qualitative Data, See Data., Quantitative Data, Quartiles, The numbers that separate the data into quarters. Quartiles may or may not be part of the data., The second quartile is the median of the data., , R Random Variable (RV), see Variable, Relative Frequency, The ratio of the number of times a value of the data occurs in the set of all outcomes to the, number of all outcomes., , S Sample, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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678, , GLOSSARY, A portion of the population understudy. A sample is representative if it characterizes the, population being studied., Sample Space, The set of all possible outcomes of an experiment., Standard Deviation, A number that is equal to the square root of the variance and measures how far data values are, from their mean. Notation: s for sample standard deviation and σ for population standard, deviation., Standard Error of the Mean, The standard deviation of the distribution of the sample means,, , √σ ., n, , Standard Normal Distribution, A continuous random variable (RV) X~N (0, 1) .. When X follows the standard normal, distribution, it is often noted as Z~N (0, 1)., Statistic, A numerical characteristic of the sample. A statistic estimates the corresponding population, parameter. For example, the average number of full-time students in a 7:30 a.m. class for this, term (statistic) is an estimate for the average number of full-time students in any class this term, (parameter)., Student’s-t Distribution, Investigated and reported by William S. Gossett in 1908 and published under the pseudonym, Student. The major characteristics of the random variable (RV) are:, • It is continuous and assumes any real values., • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at, the apex than the normal distribution., • It approaches the standard normal distribution as n gets larger., • There is a "family" of t distributions: every representative of the family is completely, defined by the number of degrees of freedom which is one less than the number of data., Student-t Distribution, , T Tree Diagram, The useful visual representation of a sample space and events in the form of a “tree” with, branches marked by possible outcomes simultaneously with associated probabilities, (frequencies, relative frequencies)., Type 1 Error, The decision is to reject the Null hypothesis when, in fact, the Null hypothesis is true., Type 2 Error, The decision is to not reject the Null hypothesis when, in fact, the Null hypothesis is false., , U Uniform Distribution, A continuous random variable (RV) that has equally likely outcomes over the domain,, a < x < b. Often referred as the Rectangular distribution because the graph of the pdf has the, form of a rectangle. Notation: X~U ( a, b). The mean is µ = a+2 b and the standard deviation is, q, ( b − a )2, σ=, The probability density function is f ( X ) = b−1 a for a < x < b or a ≤ x ≤ b. The, 12, −a ., cumulative distribution is P ( X ≤ x ) = xb−, a, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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GLOSSARY, , 679, , V Variable (Random Variable), A characteristic of interest in a population being studied. Common notation for variables are, upper case Latin letters X, Y, Z,...; common notation for a specific value from the domain (set of, all possible values of a variable) are lower case Latin letters x, y, z,.... For example, if X is the, number of children in a family, then x represents a specific integer 0, 1, 2, 3, .... Variables in, statistics differ from variables in intermediate algebra in two following ways., , • The domain of the random variable (RV) is not necessarily a numerical set; the domain may, be expressed in words; for example, if X = hair color then the domain is {black, blond, gray,, green, orange}., • We can tell what specific value x of the Random Variable X takes only after performing the, experiment., Variance, Mean of the squared deviations from the mean. Square of the standard deviation. For a set of, data, a deviation can be represented as x − x where x is a value of the data and x is the sample, mean. The sample variance is equal to the sum of the squares of the deviations divided by the, difference of the sample size and 1., Venn Diagram, The visual representation of a sample space and events in the form of circles or ovals showing, their intersections., , Z z-score, x −µ, , The linear transformation of the form z = σ . If this transformation is applied to any normal, distribution X~N (µ, σ ) , the result is the standard normal distribution Z~N (0, 1). If this, transformation is applied to any specific value x of the RV with mean µ and standard deviation, σ , the result is called the z-score of x. Z-scores allow us to compare data that are normally, distributed but scaled differently., , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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680, , INDEX, , Index of Keywords and Terms, Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords, do not necessarily appear in the text of the page. They are merely associated with that section. Ex., apples, § 1.1 (1) Terms are referenced by the page they appear on. Ex. apples, 1, , ", , "hypothesis testing.", 377, , A A AND B, § 3.2(122), A OR B, § 3.2(122), accessibility, § (5), addition, § 3.4(127), additional, § (5), adoption, § (5), alternate hypothesis, § 13.2(584), § 13.3(585), ANOVA, § 13.1(583), § 13.2(584), § 13.3(585),, § 13.4(589), § 13.5(593), § 14.5.4(649), answer, § 1.8(33), appendix, § 14.3(632), area, § 5.2(223), article, § 14.4.3(640), average, § 1.4(16), § 4.3(169), , B bar, § 2.4(63), Bernoulli, § 4.5(172), § 4.15(192), Bernoulli Trial, 173, binomial, § 4.4(172), § 4.5(172), § 4.9(183),, § 4.15(192), binomial distribution, 381, binomial probability distribution, 173, bivariate, § 14.4.5(643), box, § 2.5(68), § 2.11(90), § 2.13(95), § 5.10(254), boxes, § 2.4(63), , C cards, § 4.17(205), categorical, § 1.4(16), center, § 2.6(71), central, § 14.4.2(637), Central Limit Theorem, § 7.2(290), § 7.3(293),, § 7.10(318), 387, chance, § 3.2(122), § 3.3(124), chi, § 11.4(474), § 11.5(481), Chi-Square, § 11.7(488), § 14.5.3(648), CLT, 294, cluster, § 1.10(38), § 1.14(52), cluster sample, § 1.6(24), collaborative, § (1), § (5), collection, 1, , condition, § 3.3(124), conditional, § 3.2(122), § 3.11(144), 230, conditional probability, 123, confidence interval, 328, 336, confidence intervals, 339, 377, confidence level, 329, 339, contingency, § 3.5(131), § 3.9(141), contingency table, 131, 481, continuity correction factor, 299, Continuous, § 1.5(18), 18, § 1.10(38), § 1.12(42),, § 5.1(221), § 5.2(223), § 5.3(226), § 5.4(233),, § 5.5(239), § 5.6(240), § 5.7(243), § 5.8(245),, § 5.9(251), § 5.10(254), § 14.4.2(637), continuous random variable, 233, convenience, § 1.10(38), Convenience sampling, § 1.6(24), Counting, § 1.5(18), critical value, 267, cumulative, § 1.9(33), § 1.10(38), § 1.12(42),, § 1.13(50), cumulative distribution function (CDF), 234, Cumulative relative frequency, 34, curve, § 13.4(589), , D Data, § 1.1(13), § 1.2(13), 13, § 1.4(16), 17,, § 1.5(18), § 1.7(31), § 1.10(38), § 1.11(39),, § 1.12(42), § 1.13(50), § 2.1(59), § 2.2(59),, § 2.4(63), § 14.3(632), § 14.4.1(635),, § 14.4.5(643), § 14.5.2(647), degrees of freedom, 336, § 13.3(585),, § 13.4(589), § 13.5(593), degrees of freedom (df), 431, descriptive, § 1.2(13), § 2.2(59), § 2.3(60),, § 2.6(71), § 2.11(90), § 2.13(95), deviation, § 2.11(90), § 2.13(95), diagram, § 3.6(134), § 3.7(135), dice, § 4.18(209), Discrete, § 1.5(18), 18, § 1.10(38), § 1.12(42),, § 4.1(167), § 4.2(168), § 4.3(169), § 4.4(172),, § 4.5(172), § 4.6(175), § 4.7(178), § 4.8(180),, § 4.9(183), § 4.10(185), § 4.15(192), § 4.16(202),, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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INDEX, , 681, , § 4.17(205), § 4.18(209), display, § 2.2(59), distribution, § 4.1(167), § 4.2(168), § 4.5(172),, § 4.6(175), § 4.7(178), § 4.8(180), § 4.10(185),, § 4.16(202), § 4.17(205), § 4.18(209), § 5.1(221),, § 5.2(223), § 5.3(226), § 5.4(233), § 5.5(239),, § 5.6(240), § 5.7(243), § 5.8(245), § 5.9(251),, § 5.10(254), § 11.5(481), § 14.4.2(637),, § 14.5.3(648), distribution is binomial, 339, dot plot, § 1.2(13), , E elementary, § (1), § (5), § (9), § 2.1(59),, § 2.2(59), § 2.3(60), § 2.6(71), § 2.7(76),, § 2.8(79), § 2.9(81), § 2.10(89), § 2.11(90),, § 2.12(93), § 2.14(112), § 3.4(127), § 3.5(131),, § 3.6(134), § 3.7(135), § 3.8(140), § 3.9(141),, § 3.10(143), § 3.11(144), § 3.12(155), § 3.13(157),, § 4.1(167), § 4.2(168), § 4.3(169), § 4.4(172),, § 4.5(172), § 4.6(175), § 4.7(178), § 4.8(180),, § 4.9(183), § 4.10(185), § 4.12(188), § 4.13(189),, § 4.14(191), § 4.15(192), § 4.16(202), § 4.17(205),, § 4.18(209), § 5.1(221), § 5.2(223), § 5.3(226),, § 5.4(233), § 5.5(239), § 5.6(240), § 5.7(243),, § 5.8(245), § 5.9(251), § 5.10(254), § 6.1(261),, § 6.2(262), § 6.3(263), § 6.5(265), § 6.6(269),, § 6.7(270), § 6.8(272), § 6.9(278), § 6.10(280),, § 6.11(283), § 7.1(289), § 7.4(294), § 7.5(301),, § 7.6(302), § 7.7(305), § 7.8(312), § 7.9(314),, § 8.1(327), § 8.2(329), § 8.3(336), § 8.4(339),, § 8.5(344), § 8.6(345), § 8.7(347), § 8.8(349),, § 8.9(351), § 8.10(361), § 8.11(364), § 8.12(367),, § 8.13(369), § 9.1(377), § 9.2(378), § 9.3(379),, § 9.4(380), § 9.5(381), § 9.6(381), § 9.7(382),, § 9.8(383), § 9.9(383), § 9.10(385), § 9.11(385),, § 9.12(396), § 9.13(397), § 9.14(399), § 9.15(401),, § 9.16(403), § 9.17(416), § 9.18(419), § 10.1(429),, § 10.2(430), § 10.3(433), § 10.4(435), § 10.5(437),, § 10.6(442), § 10.7(443), § 10.8(445), § 10.9(447),, § 10.10(459), § 10.11(461), § 11.1(471),, § 11.2(472), § 11.3(472), § 11.4(474), § 11.5(481),, § 11.8(488), § 11.9(491), § 11.10(492),, § 11.11(494), § 11.12(496), § 11.13(498),, § 11.14(507), § 11.15(511), § 11.16(516),, § 12.1(523), § 12.4(526), § 12.5(528), § 12.6(534),, § 12.7(536), § 12.8(541), § 12.9(541),, § 12.10(548), § 12.12(552), § 12.13(555),, § 12.14(571), § 12.15(574), § 12.16(576),, § 13.6(596), § 13.7(597), § 13.8(599), § 13.9(604),, § 13.10(608), § 14.1(613), § 14.2(622),, § 14.3(632), § 14.4.1(635), § 14.4.2(637),, § 14.4.3(640), § 14.4.4(642), § 14.4.5(643),, , § 14.5.1(646), § 14.5.2(647), § 14.5.3(648),, § 14.5.4(649), § 14.6(650), § 14.7(651),, § 14.8(656), elementary statistics, § (11), § 2.13(95), empirical, § 5.10(254), equally likely, § 3.2(122), 122, error bound for a population mean, 329, 337, error bound„ 339, event, § 3.2(122), 122, § 3.11(144), exclusive, § 3.3(124), § 3.10(143), § 3.11(144), exercise, § 1.14(52), § 2.13(95), § 3.9(141),, § 3.13(157), § 4.15(192), § 4.16(202), § 4.17(205),, § 4.18(209), § 5.6(240), § 5.7(243), § 5.9(251),, § 5.10(254), exercises, § 3.12(155), expected, § 4.3(169), expected value, 169, experiment, § 3.2(122), 122, § 4.1(167),, § 4.5(172), § 4.7(178), § 4.8(180), § 4.17(205),, § 4.18(209), exponential, § 5.1(221), § 5.4(233), 233,, § 5.5(239), § 5.7(243), exponential distribution, 297, , F F Distribution, § 13.1(583), § 13.2(584),, § 13.3(585), § 13.4(589), § 13.5(593),, § 14.5.4(649), F Ratio, § 13.3(585), fit, § 11.4(474), formula, § 3.8(140), § 4.9(183), § 5.5(239), frequency, § 1.9(33), 34, § 1.10(38), § 1.11(39),, § 1.12(42), § 1.13(50), § 1.14(52), 63, § 2.11(90),, § 2.13(95), § 3.2(122), function, § 4.2(168), § 4.4(172), § 4.5(172),, § 4.6(175), § 4.7(178), § 4.9(183), § 4.16(202),, § 5.1(221), § 5.3(226), § 5.4(233), § 5.6(240),, § 5.7(243), § 5.8(245), § 5.9(251), functions, § 5.2(223), , G geometric, § 4.4(172), § 4.6(175), § 4.9(183),, § 4.15(192), good, § 11.4(474), Goodness-of-Fit, § 11.7(488), graph, § 2.2(59), § 2.3(60), § 5.1(221), § 5.6(240),, § 5.7(243), § 5.8(245), § 14.4.1(635), guide, § (5), , H histogram, § 2.4(63), § 2.11(90), § 2.13(95),, § 5.10(254), Homework, § 1.12(42), § 2.11(90), § 2.13(95),, § 3.9(141), § 3.12(155), § 3.13(157), § 4.15(192),, § 4.16(202), § 4.17(205), § 4.18(209), § 5.6(240),, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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682, , INDEX, § 5.7(243), § 5.8(245), § 5.9(251), § 5.10(254), homogeneity, § 11.6(485), § 11.7(488), hypergeometric, § 4.4(172), § 4.7(178),, § 4.15(192), hypergeometric probability, 178, hypergeometrical, § 4.9(183), hypotheses, 378, hypothesis, § 14.4.3(640), § 14.4.4(642),, § 14.5.1(646), § 14.5.2(647), hypothesis test, 381, 383, 385, § 13.2(584),, § 13.3(585), , I, , independence, § 11.5(481), § 11.7(488), independent, § 3.3(124), 124, 128, § 3.10(143),, § 3.11(144), inferential, § 1.2(13), inferential statistics, 327, interquartile range, 71, Introduction, § 1.1(13), § 3.1(121), § 4.1(167), IQR, § 2.6(71), , K key terms, § 3.2(122), L lab, § 1.14(52), § 3.13(157), § 4.17(205),, § 4.18(209), § 5.10(254), § 7.10(318),, § 14.4.1(635), § 14.4.2(637), § 14.4.5(643), large, § 4.3(169), law, § 4.3(169), Law of Large Numbers, 294, leaf, § 2.3(60), likelihood, § 1.3(16), limit, § 14.4.2(637), location, § 2.6(71), long term, § 3.2(122), long-term, § 3.13(157), § 4.3(169), , M mean, 17, § 2.6(71), 76, § 2.11(90), § 2.13(95),, § 4.3(169), 169, 290, 292, 295, § 14.5.1(646), means, § 14.5.2(647), means square, § 13.3(585), measurement, § 1.7(31), Measuring, § 1.5(18), median, § 2.1(59), § 2.5(68), 68, § 2.6(71), 76,, § 2.11(90), § 2.13(95), mode, § 2.6(71), 77, § 2.11(90), § 2.13(95), modules, 1, multiplication, § 3.4(127), mutually, § 3.3(124), § 3.10(143), § 3.11(144), mutually exclusive, 125, 128, , N nonsampling errors, § 1.6(24), Normal Approximation to the Binomial, 299, normal distribution, 336, 339, 380, § 13.5(593), , normally distributed, 290, 293, 381, null hypothesis, 381, 382, 383, § 13.2(584),, § 13.3(585), numbers, § 4.3(169), numerical, § 1.4(16), , O One-Way Analysis of Variance, § 13.1(583),, § 13.2(584), § 13.3(585), One-Way ANOVA, 584, § 14.5.4(649), outcome, § 3.2(122), 122, § 3.3(124), outlier, 60, 72, outliers, 71, 541, , P p-value, 382, 383, 383, 383, 385, pair, § 14.5.2(647), parameter, § 1.4(16), 17, § 1.10(38), 327, PDF, § 4.2(168), percentile, § 2.6(71), § 2.11(90), § 2.13(95), percentiles, 71, plot, § 2.11(90), § 2.13(95), § 5.10(254), point estimate, 327, Poisson, § 4.4(172), § 4.8(180), § 4.9(183),, § 4.15(192), Poisson probability distribution, 180, population, § 1.4(16), 16, 31, § 1.10(38),, § 2.13(95), § 5.9(251), § 13.2(584), § 13.3(585), populations, § 13.5(593), practice, § 1.11(39), § 2.11(90), § 3.9(141),, § 3.10(143), § 3.12(155), § 4.16(202), § 5.6(240),, § 5.7(243), § 5.9(251), probability, § 1.3(16), 16, § 1.10(38), § 3.1(121),, § 3.2(122), 122, § 3.3(124), § 3.4(127), § 3.5(131),, § 3.6(134), § 3.7(135), § 3.8(140), § 3.9(141),, § 3.10(143), § 3.11(144), § 3.12(155), § 3.13(157),, § 4.2(168), § 4.3(169), § 4.4(172), § 4.5(172),, § 4.6(175), § 4.7(178), § 4.8(180), § 4.9(183),, § 4.10(185), § 4.16(202), § 4.17(205), § 4.18(209),, § 5.1(221), § 5.2(223), § 5.3(226), § 5.5(239),, § 5.6(240), § 5.7(243), § 5.9(251), probability distribution function, 168, problem, § 2.13(95), § 4.15(192), § 14.4.4(642), project, § 14.4.1(635), § 14.4.2(637),, § 14.4.5(643), proportion, § 1.4(16), 17, § 14.5.1(646),, § 14.5.2(647), , Q Qualitative, § 1.5(18), § 1.10(38), § 1.12(42), Qualitative data, 18, Quantitative, § 1.5(18), § 1.10(38), § 1.12(42), Quantitative data, 18, quartile, § 2.6(71), § 2.11(90), § 2.13(95), quartiles, § 2.5(68), 68, 71, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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INDEX, , 683, , R random, § 1.3(16), § 1.10(38), § 1.12(42),, § 1.14(52), § 4.1(167), § 4.2(168), § 4.3(169),, § 4.4(172), § 4.5(172), § 4.6(175), § 4.7(178),, § 4.8(180), § 4.9(183), § 4.10(185), § 4.15(192),, § 4.16(202), § 4.17(205), § 4.18(209), § 5.1(221),, § 5.2(223), § 5.3(226), § 5.4(233), § 5.5(239),, § 5.6(240), § 5.7(243), § 5.8(245), § 5.9(251), random sampling, § 1.6(24), random variable, 167, 431, 434, randomness, § 1.3(16), relative, § 1.9(33), § 1.10(38), § 1.12(42),, § 1.13(50), § 2.13(95), § 3.2(122), relative frequency, 34, 63, replacement, § 1.10(38), § 3.13(157), representative, § 1.4(16), resources, § (5), review, § 3.12(155), § 4.16(202), § 5.9(251), round, § 1.8(33), rounding, § 1.8(33), rule, § 3.4(127), , S sample, § 1.4(16), 16, § 1.6(24), § 1.7(31),, § 1.10(38), § 1.12(42), § 1.14(52), § 2.13(95),, § 13.2(584), § 13.3(585), § 13.5(593),, § 14.4.5(643), Sample Means, § 7.2(290), sample space, § 3.2(122), 122, 127, 136, samples, 31, Sampling, § 1.1(13), § 1.4(16), 16, § 1.6(24),, § 1.7(31), § 1.10(38), § 1.11(39), § 1.12(42),, § 1.13(50), § 1.14(52), sampling distribution, 78, sampling errors, § 1.6(24), sampling variability of a statistic, 83, set, § 14.3(632), sheet, § 14.5.1(646), § 14.5.2(647), § 14.5.3(648), simple, § 1.10(38), simple random sampling, § 1.6(24), single, § 14.5.1(646), Sir Ronald Fisher, § 13.3(585), size, § 1.7(31), skew, § 2.6(71), § 13.4(589), solution, § 14.5.1(646), § 14.5.2(647),, § 14.5.3(648), § 14.5.4(649), spread, § 2.6(71), square, § 11.4(474), § 11.5(481), standard, § 2.11(90), § 2.13(95), standard deviation, 81, 336, 380, 381, 382, 385, standard error, 430, standard error of the mean., 291, standard normal distribution, 262, statistic, § 1.4(16), 17, § 1.10(38), 79, , statistics, § (1), § (5), § (9), § 1.1(13), § 1.2(13),, 13, § 1.3(16), § 1.5(18), § 1.6(24), § 1.7(31),, § 1.8(33), § 1.9(33), § 1.10(38), § 1.11(39),, § 1.12(42), § 1.13(50), § 1.14(52), § 2.1(59),, § 2.2(59), § 2.3(60), § 2.6(71), § 2.7(76),, § 2.8(79), § 2.9(81), § 2.10(89), § 2.11(90),, § 2.12(93), § 2.13(95), § 2.14(112), § 3.1(121),, § 3.2(122), § 3.3(124), § 3.4(127), § 3.5(131),, § 3.6(134), § 3.7(135), § 3.8(140), § 3.9(141),, § 3.10(143), § 3.11(144), § 3.12(155), § 3.13(157),, § 4.1(167), § 4.2(168), § 4.3(169), § 4.4(172),, § 4.5(172), § 4.6(175), § 4.7(178), § 4.8(180),, § 4.9(183), § 4.10(185), § 4.12(188), § 4.13(189),, § 4.14(191), § 4.15(192), § 4.16(202), § 4.17(205),, § 4.18(209), § 5.1(221), § 5.2(223), § 5.3(226),, § 5.4(233), § 5.5(239), § 5.6(240), § 5.7(243),, § 5.8(245), § 5.9(251), § 5.10(254), § 6.1(261),, § 6.2(262), § 6.3(263), § 6.5(265), § 6.6(269),, § 6.7(270), § 6.8(272), § 6.9(278), § 6.10(280),, § 6.11(283), § 7.1(289), § 7.2(290), § 7.3(293),, § 7.4(294), § 7.5(301), § 7.6(302), § 7.7(305),, § 7.8(312), § 7.9(314), § 7.10(318), § 8.1(327),, § 8.2(329), § 8.3(336), § 8.4(339), § 8.5(344),, § 8.6(345), § 8.7(347), § 8.8(349), § 8.9(351),, § 8.10(361), § 8.11(364), § 8.12(367), § 8.13(369),, § 9.1(377), § 9.2(378), § 9.3(379), § 9.4(380),, § 9.5(381), § 9.6(381), § 9.7(382), § 9.8(383),, § 9.9(383), § 9.10(385), § 9.11(385), § 9.12(396),, § 9.13(397), § 9.14(399), § 9.15(401), § 9.16(403),, § 9.17(416), § 9.18(419), § 10.1(429), § 10.2(430),, § 10.3(433), § 10.4(435), § 10.5(437), § 10.6(442),, § 10.7(443), § 10.8(445), § 10.9(447),, § 10.10(459), § 10.11(461), § 11.1(471),, § 11.2(472), § 11.3(472), § 11.4(474), § 11.5(481),, § 11.8(488), § 11.9(491), § 11.10(492),, § 11.11(494), § 11.12(496), § 11.13(498),, § 11.14(507), § 11.15(511), § 11.16(516),, § 12.1(523), § 12.4(526), § 12.5(528), § 12.6(534),, § 12.7(536), § 12.8(541), § 12.9(541),, § 12.10(548), § 12.12(552), § 12.13(555),, § 12.14(571), § 12.15(574), § 12.16(576),, § 13.1(583), § 13.2(584), § 13.3(585), § 13.4(589),, § 13.5(593), § 13.6(596), § 13.7(597), § 13.8(599),, § 13.9(604), § 13.10(608), § 14.1(613),, § 14.2(622), § 14.3(632), § 14.4.1(635),, § 14.4.2(637), § 14.4.3(640), § 14.4.4(642),, § 14.4.5(643), § 14.5.1(646), § 14.5.2(647),, § 14.5.3(648), § 14.5.4(649), § 14.6(650),, § 14.7(651), § 14.8(656), stem, § 2.3(60), stemplot, § 2.3(60), , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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684, , INDEX, stratified, § 1.10(38), § 1.14(52), stratified sample, § 1.6(24), Student’s-t distribution, 336, 381, student’s-t distribution., 380, sum of squares, § 13.3(585), summary, § 5.5(239), Sums, § 7.3(293), supplemental, § (5), survey, § 1.12(42), § 14.4.1(635), § 14.4.3(640), systematic, § 1.10(38), § 1.13(50), § 1.14(52), systematic sample, § 1.6(24), , T table, § 3.5(131), § 3.9(141), teacher, § (5), Terminology, § 3.2(122), test, § 11.4(474), § 11.5(481), § 14.4.3(640),, § 14.4.4(642), § 14.5.1(646), § 14.5.2(647), Test for Homogeneity, § 11.6(485), Test of Two Variances, § 13.1(583), § 13.5(593), The Central Limit Theorem, 289, 290, theorem, § 14.4.2(637), tree, § 3.7(135), tree diagram, 135, trial, § 4.5(172), two, § 14.5.2(647), Two-Way Analysis of Variance, § 13.4(589), Type I error, 379, 383, Type II error, 379, , U uniform, § 5.1(221), § 5.3(226), § 5.5(239),, , § 5.6(240), § 5.10(254), uniform distribution, 226, univariate, § 14.4.1(635), § 14.4.5(643), usage, § (5), , V variability, § 1.7(31), variable, § 1.4(16), 17, § 1.10(38), § 4.1(167),, § 4.2(168), § 4.3(169), § 4.4(172), § 4.5(172),, § 4.6(175), § 4.7(178), § 4.8(180), § 4.9(183),, § 4.10(185), § 4.15(192), § 4.16(202), § 4.17(205),, § 4.18(209), § 5.1(221), § 5.3(226), § 5.4(233),, § 5.5(239), § 5.6(240), § 5.7(243), § 5.9(251), variables, § 5.8(245), variance, 82, § 13.2(584), § 13.3(585),, § 13.5(593), variances, 584, variation, § 1.7(31), Venn, § 3.6(134), Venn diagram, 134, version, § (5), video, § (5), , W with replacement, § 1.6(24), without replacement, § 1.6(24), word, § 14.4.4(642), , Z z-score, 336, z-scores, 262, , α α, 383, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 687, , Module: "Sampling and Data: Answers and Rounding Off", Used here as: "Answers and Rounding Off", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16006/1.8/, Page: 33, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Sampling and Data: Frequency, Relative Frequency, and Cumulative Frequency", Used here as: "Frequency", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16012/1.20/, Pages: 33-37, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Sampling and Data: Summary", Used here as: "Summary", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16023/1.10/, Page: 38, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Sampling and Data: Practice 1", Used here as: "Practice: Sampling and Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16016/1.16/, Pages: 39-41, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Sampling and Data: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16010/1.19/, Pages: 42-49, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Sampling and Data: Data Collection Lab I", Used here as: "Lab 1: Data Collection", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16004/1.11/, Pages: 50-51, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 689, , Module: "Descriptive Statistics: Measuring the Location of the Data", Used here as: "Measures of the Location of the Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16314/1.18/, Pages: 71-76, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Descriptive Statistics: Measuring the Center of the Data", Used here as: "Measures of the Center of the Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17102/1.13/, Pages: 76-79, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Descriptive Statistics: Skewness and the Mean, Median, and Mode", Used here as: "Skewness and the Mean, Median, and Mode", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17104/1.9/, Pages: 79-80, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Descriptive Statistics: Measuring the Spread of the Data", Used here as: "Measures of the Spread of the Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17103/1.15/, Pages: 81-88, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Descriptive Statistics: Summary of Formulas", Used here as: "Summary of Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16310/1.9/, Page: 89, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Descriptive Statistics: Practice 1", Used here as: "Practice 1: Center of the Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16312/1.12/, Pages: 90-92, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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692, , ATTRIBUTIONS, , Module: "Probability Topics: Practice II", Used here as: "Practice 2: Calculating Probabilities", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16840/1.12/, Page: 143, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Probability Topics: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16836/1.21/, Pages: 144-154, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Probability Topics: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16842/1.9/, Pages: 155-156, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Probability Topics: Probability Lab", Used here as: "Lab: Probability Topics", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16841/1.15/, Pages: 157-159, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Discrete Random Variables: Introduction", Used here as: "Discrete Random Variables", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16825/1.14/, Pages: 167-168, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Probability Distribution Function (PDF) for a Discrete Random Variable", Used here as: "Probability Distribution Function (PDF) for a Discrete Random Variable", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16831/1.14/, Pages: 168-169, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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694, , ATTRIBUTIONS, , Module: "Discrete Random Variables: Summary of the Discrete Probability Functions", Used here as: "Summary of Functions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16833/1.11/, Pages: 183-184, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Practice 1: Discrete Distributions", Used here as: "Practice 1: Discrete Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16830/1.14/, Page: 185, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Practice 2: Binomial Distribution", Used here as: "Practice 2: Binomial Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17107/1.18/, Pages: 186-187, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Practice 3: Poisson Distribution", Used here as: "Practice 3: Poisson Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17109/1.15/, Page: 188, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Practice 4: Geometric Distribution", Used here as: "Practice 4: Geometric Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17108/1.17/, Pages: 189-190, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Practice 5: Hypergeometric Distribution", Used here as: "Practice 5: Hypergeometric Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17106/1.13/, Page: 191, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 695, , Module: "Discrete Random Variables: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16823/1.21/, Pages: 192-201, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16832/1.11/, Pages: 202-204, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Lab I", Used here as: "Lab 1: Discrete Distribution (Playing Card Experiment)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16827/1.12/, Pages: 205-208, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Discrete Random Variables: Lab II", Used here as: "Lab 2: Discrete Distribution (Lucky Dice Experiment)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16826/1.12/, Pages: 209-212, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Introduction", Used here as: "Continuous Random Variables", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16808/1.12/, Pages: 221-223, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Continuous Probability Functions", Used here as: "Continuous Probability Functions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16805/1.9/, Pages: 223-225, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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696, , ATTRIBUTIONS, , Module: "Continuous Random Variables: The Uniform Distribution", Used here as: "The Uniform Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16819/1.17/, Pages: 226-233, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: The Exponential Distribution", Used here as: "The Exponential Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16816/1.15/, Pages: 233-238, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Summary of The Uniform and Exponential Probability Distributions", Used here as: "Summary of the Uniform and Exponential Probability Distributions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16813/1.10/, Page: 239, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Continuous Random Variables: Practice 1", Used here as: "Practice 1: Uniform Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16812/1.14/, Pages: 240-242, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Practice 2", Used here as: "Practice 2: Exponential Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16811/1.11/, Pages: 243-244, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16807/1.14/, Pages: 245-250, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 697, , Module: "Continuous Random Variables: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16810/1.11/, Pages: 251-253, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Continuous Random Variables: Lab I", Used here as: "Lab: Continuous Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16803/1.13/, Pages: 254-256, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Normal Distribution: Introduction", Used here as: "The Normal Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16979/1.12/, Pages: 261-262, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Normal Distribution: Standard Normal Distribution", Used here as: "The Standard Normal Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16986/1.7/, Page: 262, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Normal Distribution: Z-scores", Used here as: "Z-scores", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16991/1.10/, Pages: 263-264, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Normal Distribution: Areas to the Left and Right of x", Used here as: "Areas to the Left and Right of x", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16976/1.5/, Page: 265, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 699, , Module: "Normal Distribution: Normal Distribution Lab II", Used here as: "Lab 2: Normal Distribution (Pinkie Length)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16980/1.16/, Pages: 283-284, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Introduction", Used here as: "The Central Limit Theorem", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16953/1.17/, Pages: 289-290, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Central Limit Theorem for Sample Means", Used here as: "The Central Limit Theorem for Sample Means (Averages)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16947/1.23/, Pages: 290-292, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Central Limit Theorem for Sums", Used here as: "The Central Limit Theorem for Sums", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16948/1.16/, Pages: 293-294, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Using the Central Limit Theorem", Used here as: "Using the Central Limit Theorem", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16958/1.21/, Pages: 294-300, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Summary of Formulas", Used here as: "Summary of Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16956/1.8/, Page: 301, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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700, , ATTRIBUTIONS, , Module: "Central Limit Theorem: Practice", Used here as: "Practice: The Central Limit Theorem", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16954/1.12/, Pages: 302-304, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16952/1.24/, Pages: 305-311, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16955/1.12/, Pages: 312-313, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Central Limit Theorem Lab I", Used here as: "Lab 1: Central Limit Theorem (Pocket Change)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16950/1.10/, Pages: 314-317, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Central Limit Theorem: Central Limit Theorem Lab II", Used here as: "Lab 2: Central Limit Theorem (Cookie Recipes)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16945/1.11/, Pages: 318-322, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Introduction", Used here as: "Confidence Intervals", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16967/1.16/, Pages: 327-329, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 701, , Module: "Confidence Intervals: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal", Used here as: "Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16962/1.23/, Pages: 329-336, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s-t", Used here as: "Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s-t", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16959/1.24/, Pages: 336-339, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Confidence Interval for a Population Proportion", Used here as: "Confidence Interval for a Population Proportion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16963/1.20/, Pages: 339-343, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Summary of Formulas", Used here as: "Summary of Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16973/1.8/, Page: 344, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Practice 1", Used here as: "Practice 1: Confidence Intervals for Means, Known Population Standard Deviation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16970/1.13/, Pages: 345-346, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Practice 2", Used here as: "Practice 2: Confidence Intervals for Means, Unknown Population Standard Deviation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16971/1.14/, Pages: 347-348, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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702, , ATTRIBUTIONS, , Module: "Confidence Intervals: Practice 3", Used here as: "Practice 3: Confidence Intervals for Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16968/1.13/, Pages: 349-350, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16966/1.16/, Pages: 351-360, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16972/1.10/, Pages: 361-363, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Confidence Interval Lab I", Used here as: "Lab 1: Confidence Interval (Home Costs)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16960/1.11/, Pages: 364-366, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Confidence Interval Lab II", Used here as: "Lab 2: Confidence Interval (Place of Birth)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16961/1.11/, Pages: 367-368, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Confidence Intervals: Confidence Interval Lab III", Used here as: "Lab 3: Confidence Interval (Womens’ Heights)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16964/1.12/, Pages: 369-370, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 703, , Module: "Hypothesis Testing of Single Mean and Single Proportion: Introduction", Used here as: "Hypothesis Testing: Single Mean and Single Proportion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16997/1.11/, Pages: 377-378, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Null and Alternate Hypotheses", Used here as: "Null and Alternate Hypotheses", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16998/1.14/, Pages: 378-379, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Outcomes and the Type I and Type II, Errors", Used here as: "Outcomes and the Type I and Type II Errors", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17006/1.8/, Pages: 379-380, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Distribution Needed for Hypothesis, Testing", Used here as: "Distribution Needed for Hypothesis Testing", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17017/1.13/, Pages: 380-381, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Assumptions", Used here as: "Assumption", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17002/1.16/, Page: 381, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Rare Events", Used here as: "Rare Events", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16994/1.8/, Page: 381, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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704, , ATTRIBUTIONS, , Module: "Hypothesis Testing of Single Mean and Single Proportion: Using the Sample to Test the Null, Hypothesis", Used here as: "Using the Sample to Test the Null Hypothesis", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16995/1.17/, Pages: 382-383, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Decision and Conclusion", Used here as: "Decision and Conclusion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16992/1.11/, Page: 383, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Additional Information", Used here as: "Additional Information", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16999/1.13/, Pages: 383-384, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Summary of the Hypothesis Test", Used here as: "Summary of the Hypothesis Test", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16993/1.6/, Page: 385, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Examples", Used here as: "Examples", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17005/1.25/, Pages: 385-395, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Summary of Formulas", Used here as: "Summary of Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16996/1.9/, Page: 396, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 705, , Module: "Hypothesis Testing of Single Mean and Single Proportion: Practice 1", Used here as: "Practice 1: Single Mean, Known Population Standard Deviation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17004/1.11/, Pages: 397-398, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Practice 2", Used here as: "Practice 2: Single Mean, Unknown Population Standard Deviation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17016/1.13/, Pages: 399-400, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Practice 3", Used here as: "Practice 3: Single Proportion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17003/1.15/, Pages: 401-402, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17001/1.14/, Pages: 403-415, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17013/1.12/, Pages: 416-418, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Single Mean and Single Proportion: Lab", Used here as: "Lab: Hypothesis Testing of a Single Mean and Single Proportion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17007/1.12/, Pages: 419-422, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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706, , ATTRIBUTIONS, , Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Introduction", Used here as: "Hypothesis Testing: Two Population Means and Two Population Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17029/1.9/, Pages: 429-430, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Comparing Two, Independent Population Means with Unknown Population Standard Deviations", Used here as: "Comparing Two Independent Population Means with Unknown Population Standard Deviations", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17025/1.18/, Pages: 430-433, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Comparing Two, Independent Population Means with Known Population Standard Deviations", Used here as: "Comparing Two Independent Population Means with Known Population Standard Deviations", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17042/1.10/, Pages: 433-435, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Comparing Two, Independent Population Proportions", Used here as: "Comparing Two Independent Population Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17043/1.12/, Pages: 435-437, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Matched or Paired, Samples", Used here as: "Matched or Paired Samples", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17033/1.15/, Pages: 437-441, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 707, , Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Summary of Types, of Hypothesis Tests", Used here as: "Summary of Types of Hypothesis Tests", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17044/1.5/, Page: 442, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Practice 1", Used here as: "Practice 1: Hypothesis Testing for Two Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17027/1.13/, Pages: 443-444, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing: Two Population Means and Two Population Proportions: Practice 2", Used here as: "Practice 2: Hypothesis Testing for Two Averages", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17039/1.12/, Pages: 445-446, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Two Means and Two Proportions: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17023/1.21/, Pages: 447-458, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Two Means and Two Proportions: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17021/1.9/, Pages: 459-460, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Hypothesis Testing of Two Means and Two Proportions: Lab I", Used here as: "Lab: Hypothesis Testing for Two Means and Two Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17022/1.13/, Pages: 461-465, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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708, , ATTRIBUTIONS, , Module: "The Chi-Square Distribution: Introduction", Used here as: "The Chi-Square Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17048/1.9/, Pages: 471-472, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Notation", Used here as: "Notation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17052/1.6/, Page: 472, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Facts About The Chi-Square Distribution", Used here as: "Facts About the Chi-Square Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17045/1.6/, Pages: 472-473, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Goodness-of-Fit Test", Used here as: "Goodness-of-Fit Test", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17192/1.8/, Pages: 474-481, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Test of Independence", Used here as: "Test of Independence", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17191/1.12/, Pages: 481-485, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Test for Homogeneity", By: Susan Dean, URL: http://cnx.org/content/m43655/1.2/, Pages: 485-487, Copyright: Susan Dean, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Comparison Summary of the Chi-Square Tests Goodness-of-Fit,, Independence and Homogeneity", By: Susan Dean, URL: http://cnx.org/content/m43654/1.2/, Page: 488, Copyright: Susan Dean, License: http://creativecommons.org/licenses/by/3.0/, Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 709, , Module: "The Chi-Square Distribution: Test of a Single Variance", Used here as: "Test of a Single Variance", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17059/1.7/, Pages: 488-490, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Summary of Formulas", Used here as: "Summary of Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17058/1.8/, Page: 491, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Practice 1", Used here as: "Practice 1: Goodness-of-Fit Test", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17054/1.12/, Pages: 492-493, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Practice 2", Used here as: "Practice 2: Contingency Tables", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17056/1.13/, Pages: 494-495, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Practice 3", Used here as: "Practice 3: Test of a Single Variance", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17053/1.8/, Pages: 496-497, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17028/1.20/, Pages: 498-506, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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710, , ATTRIBUTIONS, , Module: "The Chi-Square Distribution: Review", Used here as: "Review", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17057/1.11/, Pages: 507-510, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Lab I", Used here as: "Lab 1: Chi-Square Goodness-of-Fit", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17049/1.9/, Pages: 511-515, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "The Chi-Square Distribution: Lab II", Used here as: "Lab 2: Chi-Square Test for Independence", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17050/1.11/, Pages: 516-517, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Introduction", Used here as: "Linear Regression and Correlation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17089/1.6/, Page: 523, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Linear Equations", Used here as: "Linear Equations", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17086/1.4/, Pages: 523-525, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Linear Regression and Correlation: Slope and Y-Intercept of a Linear Equation", Used here as: "Slope and Y-Intercept of a Linear Equation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17083/1.5/, Page: 525, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 711, , Module: "Linear Regression and Correlation: Scatter Plots", Used here as: "Scatter Plots", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17082/1.8/, Pages: 526-527, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: The Regression Equation", Used here as: "The Regression Equation", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17090/1.15/, Pages: 528-533, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Correlation Coefficient and Coefficient of Determination", Used here as: "Correlation Coefficient and Coefficient of Determination", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17092/1.12/, Pages: 534-535, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Testing the Significance of the Correlation Coefficient", Used here as: "Testing the Significance of the Correlation Coefficient", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17077/1.15/, Pages: 536-540, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Prediction", Used here as: "Prediction", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17095/1.8/, Page: 541, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Outliers", Used here as: "Outliers", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17094/1.14/, Pages: 541-547, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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712, , ATTRIBUTIONS, , Module: "Linear Regression and Correlation: 95% Critical Values of the Sample Correlation Coefficient, Table", Used here as: "95% Critical Values of the Sample Correlation Coefficient Table", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17098/1.6/, Pages: 548-550, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Summary", Used here as: "Summary", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17081/1.4/, Page: 551, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Linear Regression and Correlation: Practice", Used here as: "Practice: Linear Regression", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17088/1.12/, Pages: 552-554, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Homework", Used here as: "Homework", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17085/1.14/, Pages: 555-570, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Regression Lab I", Used here as: "Lab 1: Regression (Distance from School)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17080/1.11/, Pages: 571-573, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Linear Regression and Correlation: Regression Lab II", Used here as: "Lab 2: Regression (Textbook Cost)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17087/1.9/, Pages: 574-575, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 713, , Module: "Linear Regression and Correlation: Regression Lab III", Used here as: "Lab 3: Regression (Fuel Efficiency)", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17079/1.8/, Pages: 576-578, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "F Distribution and One-Way ANOVA: Introduction", Used here as: "F Distribution and One-Way ANOVA", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17065/1.11/, Page: 583, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "F Distribution and One-Way ANOVA: Purpose and Basic Assumptions of One-Way ANOVA", Used here as: "One-Way ANOVA", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17068/1.10/, Pages: 584-585, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "F Distribution and One-Way ANOVA: The F Distribution And The F Ratio", Used here as: "The F Distribution and the F Ratio", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17076/1.14/, Pages: 585-589, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "F Distribution and One-Way ANOVA: Facts About the F Distribution", Used here as: "Facts About the F Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17062/1.14/, Pages: 589-593, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "F Distribution and One-Way ANOVA: Test of Two Variances", Used here as: "Test of Two Variances", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17075/1.8/, Pages: 593-595, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 715, , Module: "Collaborative Statistics: Practice Final Exam 2", Used here as: "Practice Final Exam 2", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16303/1.16/, Pages: 622-631, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: Data Sets", Used here as: "Data Sets", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17132/1.5/, Pages: 632-634, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: Projects: Univariate Data", Used here as: "Group Project: Univariate Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17142/1.8/, Pages: 635-636, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Projects: Continuous Distributions & Central Limit Theorem", Used here as: "Group Project: Continuous Distributions and Central Limit Theorem", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17141/1.9/, Pages: 637-639, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Projects: Hypothesis Testing Article", Used here as: "Partner Project: Hypothesis Testing - Article", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17140/1.8/, Pages: 640-641, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: Projects: Hypothesis Testing Word Problem", Used here as: "Partner Project: Hypothesis Testing - Word Problem", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17144/1.7/, Page: 642, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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716, , ATTRIBUTIONS, , Module: "Collaborative Statistics: Projects: Bivariate Data, Linear Regression and Univariate Data", Used here as: "Group Project: Bivariate Data, Linear Regression, and Univariate Data", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17143/1.6/, Pages: 643-645, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Solution Sheets: Hypothesis Testing: Single Mean and Single Proportion", Used here as: "Solution Sheet: Hypothesis Testing for Single Mean and Single Proportion", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17134/1.6/, Page: 646, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Solution Sheets: Hypothesis Testing: Two Means, Paired Data, Two, Proportions", Used here as: "Solution Sheet: Hypothesis Testing for Two Means, Paired Data, and Two Proportions", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17133/1.6/, Page: 647, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Solution Sheets: The Chi-Square Distribution", Used here as: "Solution Sheet: The Chi-Square Distribution", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17136/1.6/, Page: 648, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: Solution Sheets: F Distribution and One-Way ANOVA", Used here as: "Solution Sheet: F Distribution and One-Way ANOVA", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m17135/1.7/, Page: 649, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: English Phrases Written Mathematically", Used here as: "English Phrases Written Mathematically", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16307/1.6/, Page: 650, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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ATTRIBUTIONS, , 717, , Module: "Collaborative Statistics: Symbols and their Meanings", Used here as: "Symbols and their Meanings", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16302/1.9/, Pages: 651-655, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/2.0/, Module: "Collaborative Statistics: Formulas", Used here as: "Formulas", By: Susan Dean, Barbara Illowsky, Ph.D., URL: http://cnx.org/content/m16301/1.7/, Pages: 656-657, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Collaborative Statistics: Notes for the TI-83, 83+, 84 Calculator", Used here as: "Notes for the TI-83, 83+, 84 Calculator", By: Barbara Illowsky, Ph.D., Susan Dean, URL: http://cnx.org/content/m19710/1.7/, Pages: 658-667, Copyright: Maxfield Foundation, License: http://creativecommons.org/licenses/by/3.0/, Module: "Tables", By: Susan Dean, URL: http://cnx.org/content/m19138/1.3/, Page: 671, Copyright: Susan Dean, License: http://creativecommons.org/licenses/by/2.0/, , Available for free at Connexions <http://cnx.org/content/col10522/1.40>
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Collaborative Statistics, Collaborative Statistics was written by Barbara Illowsky and Susan Dean, faculty members at De Anza, College in Cupertino, California. The textbook was developed over several years and has been used in, regular and honors-level classroom settings and in distance learning classes. This textbook is intended for, introductory statistics courses being taken by students at two– and four–year colleges who are majoring in, fields other than math or engineering. Intermediate algebra is the only prerequisite. The book focuses on, applications of statistical knowledge rather than the theory behind it., , About OpenStax-CNX, Rhaptos is a web-based collaborative publishing system for educational material.
