1. Chapter Chi-Square Tests and the F-Distribution 1 of 91 10 © 2012 Pearson Education, Inc. All rights reserved.

2. Chapter Outline 10.1 Goodness of Fit 10.2 Independence 10.3 Comparing Two Variances 10.4 Analysis of Variance © 2012 Pearson Education, Inc. All rights reserved. 2 of 91

3. Section 10.1 Goodness of Fit © 2012 Pearson Education, Inc. All rights reserved. 3 of 91

4. Section 10.1 Objectives Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution © 2012 Pearson Education, Inc. All rights reserved. 4 of 91

5. Multinomial Experiments Multinomial experiment A probability experiment consisting of a fixed number of independent trials in which there are more than two possible outcomes for each trial. The probability for each outcome is fixed and each outcome is classified into categories. Recall that a binomial experiment had only two possible outcomes. © 2012 Pearson Education, Inc. All rights reserved. 5 of 91

6. Multinomial Experiments A tax preparation company wants to determine the proportions of people who used different methods to prepare their taxes. The company can perform a multinomial experiment. It wants to test a previous survey’s claim concerning the distribution of proportions of people who use different methods to prepare their taxes. It can compare the distribution of proportions obtained in the multinomial experiment with the previous survey’s specified distribution. It can perform a chi-square goodness-of-fit test. © 2012 Pearson Education, Inc. All rights reserved. 6 of 91

7. Chi-Square Goodness-of-Fit Test Used to test whether a frequency distribution fits an expected distribution. The null hypothesis states that the frequency distribution fits the specified distribution. The alternative hypothesis states that the frequency distribution does not fit the specified distribution. © 2012 Pearson Education, Inc. All rights reserved. 7 of 91

8. Multinomial Experiments Results of a survey of tax preparation methods. Distribution of tax preparation methods Accountant25% By hand20% Computer software35% Friend/family5% Tax preparation service15% Each outcome is classified into categories. The probability for each possible outcome is fixed. © 2012 Pearson Education, Inc. All rights reserved. 8 of 91

9. Chi-Square Goodness-of-Fit Test To test the previous survey’s claim, a company can perform a chi-square goodness-of-fit test using the following hypotheses. H 0 : The distribution of tax preparation methods is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend or family, and 15 % by tax preparation service. (claim) H a : The distribution of tax preparation methods differs from the claimed or expected distribution. © 2012 Pearson Education, Inc. All rights reserved. 9 of 91

10. Chi-Square Goodness-of-Fit Test To calculate the test statistic for the chi-square goodness-of-fit test, the observed frequencies and the expected frequencies are used. The observed frequency O of a category is the frequency for the category observed in the sample data. © 2012 Pearson Education, Inc. All rights reserved. 10 of 91

11. Chi-Square Goodness-of-Fit Test The expected frequency E of a category is the calculated frequency for the category.  Expected frequencies are obtained assuming the specified (or hypothesized) distribution. The expected frequency for the i th category is E i = np i where n is the number of trials (the sample size) and p i is the assumed probability of the i th category. © 2012 Pearson Education, Inc. All rights reserved. 11 of 91

12. Example: Finding Observed and Expected Frequencies A tax preparation company randomly selects 300 adults and asks them how they prepare their taxes. The results are shown at the right. Find the observed frequency and the expected frequency for each tax preparation method. (Adapted from National Retail Federation) Survey results (n = 300) Accountant71 By hand40 Computer software 101 Friend/family35 Tax preparation service 53 © 2012 Pearson Education, Inc. All rights reserved. 12 of 91

13. Solution: Finding Observed and Expected Frequencies Observed frequency: The number of adults in the survey naming a particular tax preparation method Survey results (n = 300) Accountant71 By hand40 Computer software 101 Friend/family35 Tax preparation service 53 observed frequency © 2012 Pearson Education, Inc. All rights reserved. 13 of 91

14. Solution: Finding Observed and Expected Frequencies Expected Frequency: E i = np i Tax preparation method % of people Observed frequency Expected frequency Accountant 25%71300(0.25) = 75 By hand20%40300(0.20) = 60 Computer Software35%101300(0.35) = 105 Friend/family 5%35300(0.05) = 15 Tax preparation service 15%53300(0.15) = 45 n = 300 © 2012 Pearson Education, Inc. All rights reserved. 14 of 91

15. Chi-Square Goodness-of-Fit Test For the chi-square goodness-of-fit test to be used, the following must be true. 1.The observed frequencies must be obtained by using a random sample. 2.Each expected frequency must be greater than or equal to 5. © 2012 Pearson Education, Inc. All rights reserved. 15 of 91

16. Chi-Square Goodness-of-Fit Test If these conditions are satisfied, then the sampling distribution for the goodness-of-fit test is approximated by a chi-square distribution with k – 1 degrees of freedom, where k is the number of categories. The test statistic for the chi-square goodness-of-fit test is where O represents the observed frequency of each category and E represents the expected frequency of each category. The test is always a right-tailed test. © 2012 Pearson Education, Inc. All rights reserved. 16 of 91

17. Chi - Square Goodness - of - Fit Test 1.Identify the claim. State the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom. 4.Determine the critical value. State H 0 and H a. Identify α. Use Table 6 in Appendix B. d.f. = k – 1 In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 17 of 91

18. Chi - Square Goodness - of - Fit Test If χ 2 is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region. 6.Calculate the test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 18 of 91

19. Example: Performing a Goodness of Fit Test Use the tax preparation method data to perform a chi- square goodness-of-fit test to test whether the distributions are different. Use α = 0.01. Survey results (n = 300) Accountant71 By hand40 Computer software 101 Friend/family35 Tax preparation service 53 Distribution of tax preparation methods Accountant25% By hand20% Computer software 35% Friend/family5% Tax preparation service 15% © 2012 Pearson Education, Inc. All rights reserved. 19 of 91

20. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: Conclusion: 0.01 5 – 1 = 4 The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/ family, and 15% by tax preparation service. (Claim) The distribution of tax preparation methods differs from the claimed or expected distribution. © 2012 Pearson Education, Inc. All rights reserved. 20 of 91

21. Solution: Performing a Goodness of Fit Test Tax preparation method Observed frequency Expected frequency Accountant7175 By hand4060 Computer hardware 101105 Friend/family3515 Tax preparation service 5345 © 2012 Pearson Education, Inc. All rights reserved. 21 of 91

22. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 5 – 1 = 4 The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/ family, and 15% by tax preparation service. (Claim) The distribution of tax preparation methods differs from the claimed or expected distribution. χ 2 ≈ 35.121 There is enough evidence at the 1% significance level to conclude that the distribution of tax preparation methods differs from the previous survey’s claimed or expected distribution. Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 22 of 91

23. Example: Performing a Goodness of Fit Test A researcher claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table on the next slide. Using α = 0.10, perform a chi-square goodness-of-fit test to test the claimed or expected distribution. What can you conclude? (Adapted from Mars Incorporated) © 2012 Pearson Education, Inc. All rights reserved. 23 of 91

24. Example: Performing a Goodness of Fit Test ColorFrequency Brown80 Yellow95 Red88 Blue83 Orange76 Green78 Solution: The claim is that the distribution is uniform, so the expected frequencies of the colors are equal. To find each expected frequency, divide the sample size by the number of colors. E = 500/6 ≈ 83.3 n = 500 © 2012 Pearson Education, Inc. All rights reserved. 24 of 91

25. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: Conclusion: 0.10 6 – 1 = 5 0.10 χ2χ2 0 9.236 Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim) Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform. © 2012 Pearson Education, Inc. All rights reserved. 25 of 91

26. Solution: Performing a Goodness of Fit Test Color Observed frequency Expected frequency Brown8083.33 Yellow9583.33 Red8883.33 Blue8383.33 Orange7683.33 Green7883.33 © 2012 Pearson Education, Inc. All rights reserved. 26 of 91

27. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 6 – 1 = 5 0.10 χ2χ2 0 9.236 χ 2 ≈ 3.016 3.016 There is not enough evidence at the 10% level of significance to reject the claim that the distribution is uniform. Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim) Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform. Fail to Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 27 of 91

28. Section 10.1 Summary Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution © 2012 Pearson Education, Inc. All rights reserved. 28 of 91

29. Section 10.2 Independence © 2012 Pearson Education, Inc. All rights reserved. 29 of 91

30. Section 10.2 Objectives Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent © 2012 Pearson Education, Inc. All rights reserved. 30 of 91

31. Contingency Tables r × c contingency table Shows the observed frequencies for two variables. The observed frequencies are arranged in r rows and c columns. The intersection of a row and a column is called a cell. © 2012 Pearson Education, Inc. All rights reserved. 31 of 91

32. Contingency Tables Example: The contingency table shows the results of a random sample of 2200 adults classified by their favorite way to eat ice cream and gender. (Adapted from Harris Interactive) © 2012 Pearson Education, Inc. All rights reserved. 32 of 91 Favorite way to eat ice cream Gender CupConeSundaeSandwichOther Male6002882042484 Female4103401802050

33. Finding the Expected Frequency Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell. The expected frequency for a cell E r,c in a contingency table is © 2012 Pearson Education, Inc. All rights reserved. 33 of 91

34. Example: Finding Expected Frequencies Find the expected frequency for each cell in the contingency table. Assume that the variables, favorite way to eat ice cream and gender, are independent. Favorite way to eat ice cream Gender CupConeSundaeSandwichOtherTotal Male60028820424841200 Female41034018020501000 Total1010628384441342200 marginal totals © 2012 Pearson Education, Inc. All rights reserved. 34 of 91

35. Solution: Finding Expected Frequencies Favorite way to eat ice cream Gender CupConeSundaeSandwichOtherTotal Male 60028820424841200 Female41034018020501000 Total1010628384441342200 © 2012 Pearson Education, Inc. All rights reserved. 35 of 91

36. Solution: Finding Expected Frequencies Favorite way to eat ice cream Gender CupConeSundaeSandwichOtherTotal Male 60028820424841200 Female41034018020501000 Total1010628384441342200 © 2012 Pearson Education, Inc. All rights reserved. 36 of 91

37. Solution: Finding Expected Frequencies Favorite way to eat ice cream Gender CupConeSundaeSandwichOtherTotal Male60028820424841200 Female41034018020501000 Total1010628384441342200 © 2012 Pearson Education, Inc. All rights reserved. 37 of 91

38. Chi-Square Independence Test Chi-square independence test Used to test the independence of two variables. Can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. © 2012 Pearson Education, Inc. All rights reserved. 38 of 91

39. Chi-Square Independence Test For the chi-square independence test to be used, the following must be true. 1.The observed frequencies must be obtained by using a random sample. 2.Each expected frequency must be greater than or equal to 5. © 2012 Pearson Education, Inc. All rights reserved. 39 of 91

40. Chi-Square Independence Test If these conditions are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chi-square distribution with (r – 1)(c – 1) degrees of freedom, where r and c are the number of rows and columns, respectively, of a contingency table. The test statistic for the chi-square independence test is where O represents the observed frequencies and E represents the expected frequencies. The test is always a right-tailed test. © 2012 Pearson Education, Inc. All rights reserved. 40 of 91

41. Chi - Square Independence Test 1.Identify the claim. State the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the degrees of freedom. 4.Determine the critical value. State H 0 and H a. Identify α. Use Table 6 in Appendix B. d.f. = (r – 1)(c – 1) In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 41 of 91

42. Chi - Square Independence Test If χ 2 is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region. 6.Calculate the test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 42 of 91

43. Example: Performing a χ 2 Independence Test Using the gender/favorite way to eat ice cream contingency table, can you conclude that the adults favorite ways to eat ice cream are related to gender? Use α = 0.01. Expected frequencies are shown in parentheses. Favorite way to eat ice cream Gender CupConeSundaeSandwichOtherTotal Male 600 (550.91) 228 (342.55) 204 (209.45) 24 (24) 84 (73.09) 1200 Female 410 (459.09) 340 (285.45) 180 (174.55) 20 (20) 50 (60.91) 1000 Total 1010628384441342200 © 2012 Pearson Education, Inc. All rights reserved. 43 of 91

44. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 (2 – 1)(5 – 1) = 4 The adults’ favorite ways to eat ice cream are independent of gender. The adults’ favorite ways to eat ice cream are dependent on gender. (Claim) © 2012 Pearson Education, Inc. All rights reserved. 44 of 91

45. Solution: Performing a Goodness of Fit Test © 2012 Pearson Education, Inc. All rights reserved. 45 of 91

46. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 (2 – 1)(5 – 1) = 4 The adults’ favorite ways to eat ice cream are independent of gender. The adults’ favorite ways to eat ice cream are dependent on gender. (Claim) χ 2 ≈ 32.630 There is enough evidence at the 1% level of significance to conclude that the adults’ favorite ways to eat ice cream and gender are dependent. Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 46 of 91

47. Section 10.2 Summary Used a contingency table to find expected frequencies Used a chi-square distribution to test whether two variables are independent © 2012 Pearson Education, Inc. All rights reserved. 47 of 91

48. Section 10.3 Comparing Two Variances © 2012 Pearson Education, Inc. All rights reserved. 48 of 91

49. Section 10.3 Objectives Interpret the F-distribution and use an F-table to find critical values Perform a two-sample F-test to compare two variances © 2012 Pearson Education, Inc. All rights reserved. 49 of 91

50. F-Distribution Let represent the sample variances of two different populations. If both populations are normal and the population variances are equal, then the sampling distribution of is called an F-distribution. © 2012 Pearson Education, Inc. All rights reserved. 50 of 91

51. Properties of the F-Distribution 1.The F-distribution is a family of curves each of which is determined by two types of degrees of freedom:  The degrees of freedom corresponding to the variance in the numerator, denoted d.f. N  The degrees of freedom corresponding to the variance in the denominator, denoted d.f. D 2. F-distributions are positively skewed. 3.The total area under each curve of an F-distribution is equal to 1. © 2012 Pearson Education, Inc. All rights reserved. 51 of 91

52. Properties of the F-Distribution 4. F-values are always greater than or equal to 0. 5.For all F-distributions, the mean value of F is approximately equal to 1. d.f. N = 1 and d.f. D = 8 d.f. N = 8 and d.f. D = 26 d.f. N = 16 and d.f. D = 7 d.f. N = 3 and d.f. D = 11 F 1234 © 2012 Pearson Education, Inc. All rights reserved. 52 of 91 F-Distributions

53. Finding Critical Values for the F-Distribution 1.Specify the level of significance α. 2.Determine the degrees of freedom for the numerator, d.f. N. 3.Determine the degrees of freedom for the denominator, d.f. D. 4.Use Table 7 in Appendix B to find the critical value. If the hypothesis test is a.one-tailed, use the α F-table. b.two-tailed, use the ½α F-table. © 2012 Pearson Education, Inc. All rights reserved. 53 of 91

54. Example: Finding Critical F-Values Find the critical F-value for a right-tailed test when α = 0.10, d.f. N = 5 and d.f. D = 28. The critical value is F 0 = 2.06. Solution: © 2012 Pearson Education, Inc. All rights reserved. 54 of 91

55. Example: Finding Critical F-Values Find the critical F-value for a two-tailed test when α = 0.05, d.f. N = 4 and d.f. D = 8. Solution: When performing a two-tailed hypothesis test using the F-distribution, you need only to find the right- tailed critical value. You must remember to use the ½α table. © 2012 Pearson Education, Inc. All rights reserved. 55 of 91

56. Solution: Finding Critical F-Values ½α = 0.025, d.f. N = 4 and d.f. D = 8 The critical value is F 0 = 5.05. © 2012 Pearson Education, Inc. All rights reserved. 56 of 91

57. Two-Sample F-Test for Variances To use the two-sample F-test for comparing two population variances, the following must be true. 1.The samples must be randomly selected. 2.The samples must be independent. 3.Each population must have a normal distribution. © 2012 Pearson Education, Inc. All rights reserved. 57 of 91

58. Two-Sample F-Test for Variances Test Statistic where represent the sample variances with The degrees of freedom for the numerator is d.f. N = n 1 – 1 where n 1 is the size of the sample having variance The degrees of freedom for the denominator is d.f. D = n 2 – 1, and n 2 is the size of the sample having variance © 2012 Pearson Education, Inc. All rights reserved. 58 of 91

59. Two - Sample F - Test for Variances 1.Identify the claim. State the null and alternative hypotheses. 2.Specify the level of significance. 3.Determine the degrees of freedom. 4.Determine the critical value. State H 0 and H a. Identify α. Use Table 7 in Appendix B. d.f. N = n 1 – 1 d.f. D = n 2 – 1 In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 59 of 91

60. Two - Sample F - Test for Variances If F is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region. 6.Calculate the test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 60 of 91

61. Example: Performing a Two-Sample F-Test A restaurant manager is designing a system that is intended to decrease the variance of the time customers wait before their meals are served. Under the old system, a random sample of 10 customers had a variance of 400. Under the new system, a random sample of 21 customers had a variance of 256. At α = 0.10, is there enough evidence to convince the manager to switch to the new system? Assume both populations are normally distributed. © 2012 Pearson Education, Inc. All rights reserved. 61 of 91

62. Solution: Performing a Two-Sample F-Test H 0 : H a : α = d.f. N = d.f. D = Rejection Region: Test Statistic: Decision: σ12 ≤ σ22σ12 ≤ σ22 σ 1 2 > σ 2 2 (Claim) 0.10 920 0F1.96 0.10 Because 400 > 256, There is not enough evidence at the 10% level of significance to convince the manager to switch to the new system. 1.96 1.56 Fail to Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 62 of 91

63. Example: Performing a Two-Sample F-Test You want to purchase stock in a company and are deciding between two different stocks. Because a stock’s risk can be associated with the standard deviation of its daily closing prices, you randomly select samples of the daily closing prices for each stock to obtain the results. At α = 0.05, can you conclude that one of the two stocks is a riskier investment? Assume the stock closing prices are normally distributed. Stock AStock B n 2 = 30n 1 = 31 s 2 = 3.5s 1 = 5.7 © 2012 Pearson Education, Inc. All rights reserved. 63 of 91

64. Solution: Performing a Two-Sample F-Test H 0 : H a : ½α = d.f. N = d.f. D = Rejection Region: Test Statistic: Decision: σ12 = σ22σ12 = σ22 σ 1 2 ≠ σ 2 2 (Claim) 0. 025 3029 0F2.09 0.025 Because 5.7 2 > 3.5 2, There is enough evidence at the 5% level of significance to support the claim that one of the two stocks is a riskier investment. 2.09 2.652 Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 64 of 91

65. Section 10.3 Summary Interpreted the F-distribution and used an F-table to find critical values Performed a two-sample F-test to compare two variances © 2012 Pearson Education, Inc. All rights reserved. 65 of 91

66. Section 10.4 Analysis of Variance © 2012 Pearson Education, Inc. All rights reserved. 66 of 91

67. Section 10.4 Objectives Use one-way analysis of variance to test claims involving three or more means Introduce two-way analysis of variance © 2012 Pearson Education, Inc. All rights reserved. 67 of 91

68. One-Way ANOVA One-way analysis of variance A hypothesis-testing technique that is used to compare means from three or more populations. Analysis of variance is usually abbreviated ANOVA. Hypotheses:  H 0 : μ 1 = μ 2 = μ 3 =…= μ k (all population means are equal)  H a : At least one of the means is different from the others. © 2012 Pearson Education, Inc. All rights reserved. 68 of 91

69. One-Way ANOVA In a one-way ANOVA test, the following must be true. 1.Each sample must be randomly selected from a normal, or approximately normal, population. 2.The samples must be independent of each other. 3.Each population must have the same variance. © 2012 Pearson Education, Inc. All rights reserved. 69 of 91

70. One-Way ANOVA 1.The variance between samples MS B measures the differences related to the treatment given to each sample and is sometimes called the mean square between. 2.The variance within samples MS W measures the differences related to entries within the same sample. This variance, sometimes called the mean square within, is usually due to sampling error. © 2012 Pearson Education, Inc. All rights reserved. 70 of 91

71. One-Way Analysis of Variance Test If the conditions for a one-way analysis of variance are satisfied, then the sampling distribution for the test is approximated by the F-distribution. The test statistic is The degrees of freedom for the F-test are d.f. N = k – 1 and d.f. D = N – k where k is the number of samples and N is the sum of the sample sizes. © 2012 Pearson Education, Inc. All rights reserved. 71 of 91

72. Test Statistic for a One-Way ANOVA 1.Find the mean and variance of each sample. 2.Find the mean of all entries in all samples (the grand mean). 3.Find the sum of squares between the samples. 4.Find the sum of squares within the samples. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 72 of 91

73. Test Statistic for a One-Way ANOVA 5.Find the variance between the samples. 6.Find the variance within the samples 7.Find the test statistic. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 73 of 91

74. Performing a One-Way ANOVA Test 1.Identify the claim. State the null and alternative hypotheses. 2.Specify the level of significance. 3.Identify the degrees of freedom. 4.Determine the critical value. State H 0 and H a. Identify α. Use Table 7 in Appendix B. d.f. N = k – 1 d.f. D = N – k In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 74 of 91

75. Performing a One-Way ANOVA Test If F is in the rejection region, reject H 0. Otherwise, fail to reject H 0. 5.Determine the rejection region. 6.Calculate the test statistic. 7.Make a decision to reject or fail to reject the null hypothesis. 8.Interpret the decision in the context of the original claim. In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 75 of 91

76. ANOVA Summary Table A table is a convenient way to summarize the results in a one-way ANOVA test. d.f. D SS W Within d.f. N SS B Between F Mean squares Degrees of freedom Sum of squares Variation © 2012 Pearson Education, Inc. All rights reserved. 76 of 91

77. Example: Performing a One-Way ANOVA A medical researcher wants to determine whether there is a difference in the mean length of time it takes three types of pain relievers to provide relief from headache pain. Several headache sufferers are randomly selected and given one of the three medications. Each headache sufferer records the time (in minutes) it takes the medication to begin working. The results are shown on the next slide. At α = 0.01, can you conclude that the mean times are different? Assume that each population of relief times is normally distributed and that the population variances are equal. © 2012 Pearson Education, Inc. All rights reserved. 77 of 91

78. Example: Performing a One-Way ANOVA Medication 1Medication 2Medication 3 121614 151417 2120 1215 19 Solution: k = 3 (3 samples) N = n 1 + n 2 + n 3 = 4 + 5 + 4 = 13 (sum of sample sizes) © 2012 Pearson Education, Inc. All rights reserved. 78 of 91

79. Solution: Performing a One-Way ANOVA H 0 : H a : α = d.f. N = d.f. D = Rejection Region: Test Statistic: Decision: μ 1 = μ 2 = μ 3 At least one mean is different. (Claim) 0. 01 3 – 1 = 2 13 – 3 = 10 © 2012 Pearson Education, Inc. All rights reserved. 79 of 91

80. Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. © 2012 Pearson Education, Inc. All rights reserved. 80 of 91

81. Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. © 2012 Pearson Education, Inc. All rights reserved. 81 of 91

82. Solution: Performing a One-Way ANOVA H 0 : H a : α = d.f. N = d.f. D = Rejection Region: Test Statistic: Decision: μ 1 = μ 2 = μ 3 At least one mean is different. (Claim) 0. 01 3 – 1 = 2 13 – 3 = 10 There is not enough evidence at the 1% level of significance to conclude that there is a difference in the mean length of time it takes the three pain relievers to provide relief from headache pain. 1.50 Fail to Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 82 of 91 7.56

83. Example: Using the TI-83/84 to Perform a One-Way ANOVA A researcher believes that the mean earnings of top-paid actors, athletes, and musicians are the same. The earnings (in millions of dollars) for several randomly selected from each category are shown in the table in next slide. Assume that the populations are normally distributed, the samples are independent, and the population variances are equal. At α = 0.01, can you conclude that the mean earnings are the same for the three categories? Use a technology tool. (Source: Forbes.com LLC) © 2012 Pearson Education, Inc. All rights reserved. 83 of 91

84. Example: Using the TI-83/84 to Perform a One-Way ANOVA © 2012 Pearson Education, Inc. All rights reserved. 84 of 91

85. Solution: Using the TI-83/84 to Perform a One-Way ANOVA H 0 : H a : Store data into lists L 1, L 2, and L 3 μ 1 = μ 2 = μ 3 (Claim) At least one mean is different. Decision: There is enough evidence at the 10% level of significance to reject the claim that the mean earnings are the same. P ≈ 0.06 so P < α Reject H 0 © 2012 Pearson Education, Inc. All rights reserved. 85 of 91

86. Two-Way ANOVA Two-way analysis of variance A hypothesis-testing technique that is used to test the effect of two independent variables, or factors, on one dependent variable. © 2012 Pearson Education, Inc. All rights reserved. 86 of 91

87. Two-Way ANOVA Example: Suppose a medical researcher wants to test the effect of gender and type of medication on the mean length of time it takes pain relievers to provide relief. Males taking type IFemales taking type I Males taking type IIFemales taking type II Males taking type IIIFemales taking type III Gender MaleFemale I II III Type of medication © 2012 Pearson Education, Inc. All rights reserved. 87 of 91

88. Two-Way ANOVA Hypotheses Main effect The effect of one independent variable on the dependent variable. Interaction effect The effect of both independent variables on the dependent variable. © 2012 Pearson Education, Inc. All rights reserved. 88 of 91

89. Two-Way ANOVA Hypotheses Hypotheses for main effects: H 0 : Gender has no effect on the mean length of time it takes a pain reliever to provide relief. H a : Gender has an effect on the mean length of time it takes a pain reliever to provide relief. H 0 : Type of medication has no effect on the mean length of time it takes a pain reliever to provide relief. H a : Type of medication has an effect on the mean length of time it takes a pain reliever to provide relief. © 2012 Pearson Education, Inc. All rights reserved. 89 of 91

90. Two-Way ANOVA Hypotheses Hypotheses for interaction effects: H 0 : There is no interaction effect between gender and type of medication on the mean length of time it takes a pain reliever to provide relief. H a : There is an interaction effect between gender and type of medication on the mean length of time it takes a pain reliever to provide relief. © 2012 Pearson Education, Inc. All rights reserved. 90 of 91 Perform a two-way ANOVA test, calculating the F-statistic for each hypothesis. It is possible to reject none, one, two, or all of the null hypotheses. The statistics involved with a two-way ANOVA test is beyond the scope of this course. You can use a technology tool such as MINITAB to perform the test.

91. Section 10.4 Summary Used one-way analysis of variance to test claims involving three or more means Introduced two-way analysis of variance © 2012 Pearson Education, Inc. All rights reserved. 91 of 91