1. Chapter 10 Chi-Square Tests and the F- Distribution 1 Larson/Farber 4th ed

2. Chapter Outline 10.1 Goodness of Fit 10.2 Independence 10.3 Comparing Two Variances 10.4 Analysis of Variance 2 Larson/Farber 4th ed

3. Section 10.1 Goodness of Fit 3 Larson/Farber 4th ed

4. Section 10.1 Objectives Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution 4 Larson/Farber 4th ed

5. Multinomial Experiments Multinomial experiment A probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial. A binomial experiment had only two possible outcomes. The probability for each outcome is fixed and each outcome is classified into categories. 5 Larson/Farber 4th ed

6. Multinomial Experiments Example: A radio station claims that the distribution of music preferences for listeners in the broadcast region is as shown below. Distribution of music Preferences Classical4%Oldies2% Country36%Pop18% Gospel11%Rock29% Each outcome is classified into categories. The probability for each possible outcome is fixed. 6 Larson/Farber 4th ed

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. 7 Larson/Farber 4th ed

8. Chi-Square Goodness-of-Fit Test Example: To test the radio station’s claim, the executive can perform a chi-square goodness-of-fit test using the following hypotheses. H 0 : The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock. (claim) H a : The distribution of music preferences differs from the claimed or expected distribution. 8 Larson/Farber 4th ed

9. 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. 9 Larson/Farber 4th ed

10. 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. 10 Larson/Farber 4th ed

11. Example: Finding Observed and Expected Frequencies A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown at the right. Find the observed frequencies and the expected frequencies for each type of music. Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125 11 Larson/Farber 4th ed

12. Solution: Finding Observed and Expected Frequencies Observed frequency: The number of radio music listeners naming a particular type of music Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125 observed frequency 12 Larson/Farber 4th ed

13. Solution: Finding Observed and Expected Frequencies Expected Frequency: E i = np i Type of music % of listeners Observed frequency Expected frequency Classical 4%8 Country36%210 Gospel11%72 Oldies 2%10 Pop18%75 Rock29%125 n = 500 500(0.04) = 20 500(0.36) = 180 500(0.11) = 55 500(0.02) = 10 500(0.18) = 90 500(0.29) = 145 13 Larson/Farber 4th ed

14. 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. 14 Larson/Farber 4th ed

15. 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. 15 Larson/Farber 4th ed

16. 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 . Use Table 6 in Appendix B. d.f. = k – 1 In WordsIn Symbols 16 Larson/Farber 4th ed

17. 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 17 Larson/Farber 4th ed

18. Example: Performing a Goodness of Fit Test Use the music preference data to perform a chi-square goodness-of-fit test to test whether the distributions are different. Use α = 0.01. Survey results (n = 500) Classical8 Country210 Gospel72 Oldies10 Pop75 Rock125 Distribution of music preferences Classical4% Country36% Gospel11% Oldies2% Pop18% Rock29% 18 Larson/Farber 4th ed

19. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: Conclusion: 0.01 6 – 1 = 5 0.01 χ2χ2 0 15.086 music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution 19 Larson/Farber 4th ed

20. Solution: Performing a Goodness of Fit Test Type of music Observed frequency Expected frequency Classical820 Country210180 Gospel7255 Oldies10 Pop7590 Rock125145 20 Larson/Farber 4th ed

21. Solution: Performing a Goodness of Fit Test H 0 : H a : α = d.f. = Rejection Region Test Statistic: Decision: 0.01 6 – 1 = 5 0.01 χ2χ2 0 15.086 music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution χ 2 = 22.713 22.713 There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution. Reject H 0 21 Larson/Farber 4th ed

22. Example: Performing a Goodness of Fit Test The manufacturer of M&M’s candies 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) 22 Larson/Farber 4th ed

23. 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 23 Larson/Farber 4th ed n = 500

24. 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&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform 24 Larson/Farber 4th ed

25. Solution: Performing a Goodness of Fit Test Color Observed frequency Expected frequency Brown8083.3 Yellow9583.3 Red8883.3 Blue8383.3 Orange7683.3 Green7883.3 25 Larson/Farber 4th ed

26. 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 to dispute the claim that the distribution is uniform. Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform Fail to Reject H 0 26 Larson/Farber 4th ed

27. Section 10.1 Summary Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution 27 Larson/Farber 4th ed

28. Section 10.2 Independence 28 Larson/Farber 4th ed

29. Section 10.2 Objectives Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent 29 Larson/Farber 4th ed

30. 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. 30 Larson/Farber 4th ed

31. Contingency Tables Example: The contingency table shows the results of a random sample of 550 company CEOs classified by age and size of company.(Adapted from Grant Thornton LLP, The Segal Company) Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Small / Midsize 42691086021 Large5188512022 31 Larson/Farber 4th ed

32. 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 32 Larson/Farber 4th ed

33. Example: Finding Expected Frequencies Find the expected frequency for each cell in the contingency table. Assume that the variables, age and company size, are independent. Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Total Small / Midsize 42691086021300 Large5188512022250 Total478719318043550 marginal totals 33 Larson/Farber 4th ed

34. Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Total Small / Midsize 42691086021300 Large5188512022250 Total478719318043550 34 Larson/Farber 4th ed

35. Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Total Small / Midsize 42691086021300 Large5188512022250 Total478719318043550 35 Larson/Farber 4th ed

36. Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Total Small / Midsize 42691086021300 Large5188512022250 Total478719318043550 36 Larson/Farber 4th ed

37. 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. 37 Larson/Farber 4th ed

38. 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. 38 Larson/Farber 4th ed

39. 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. 39 Larson/Farber 4th ed

40. Chi - Square Independence 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 . Use Table 6 in Appendix B. d.f. = (r – 1)(c – 1) In WordsIn Symbols 40 Larson/Farber 4th ed

41. 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 41 Larson/Farber 4th ed

42. Example: Performing a χ 2 Independence Test Using the age/company size contingency table, can you conclude that the CEOs ages are related to company size? Use α = 0.01. Expected frequencies are shown in parentheses. Age Company size 39 and under 40 - 4950 - 5960 - 69 70 and over Total Small / Midsize 42 (25.64) 69 (47.45) 108 (105.27) 60 (98.18) 21 (23.45) 300 Large 5 (21.36) 18 (39.55) 85 (87.73) 120 (81.82) 22 (19.55) 250 Total 478719318043550 42 Larson/Farber 4th ed

43. 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 0.01 χ2χ2 013.277 CEOs’ ages are independent of company size CEOs’ ages are dependent on company size 43 Larson/Farber 4th ed

44. Solution: Performing a Goodness of Fit Test 44 Larson/Farber 4th ed

45. 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 0.01 χ2χ2 013.277 CEOs’ ages are independent of company size CEOs’ ages are dependent on company size χ 2 = 77.9 There is enough evidence to conclude CEOs’ ages are dependent on company size. 77.9 Reject H 0 45 Larson/Farber 4th ed

46. Section 10.2 Summary Used a contingency table to find expected frequencies Used a chi-square distribution to test whether two variables are independent 46 Larson/Farber 4th ed

47. Section 10.3 Comparing Two Variances 47 Larson/Farber 4th ed

48. 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 48 Larson/Farber 4th ed

49. 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. 49 Larson/Farber 4th ed

50. 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. 50 Larson/Farber 4th ed

51. 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 51 Larson/Farber 4th ed

52. Critical Values for the F-Distribution 1.Specify the level of significance . 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  F-table. b.two-tailed, use the ½  F-table. 52 Larson/Farber 4th ed

53. Example: Finding Critical F-Values Find the critical F-value for a right-tailed test when α = 0.05, d.f. N = 6 and d.f. D = 29. The critical value is F 0 = 2.43. 53 Larson/Farber 4th ed Solution:

54. 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. 54 Larson/Farber 4th ed

55. 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. 55 Larson/Farber 4th ed

56. 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. 56 Larson/Farber 4th ed

57. 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 57 Larson/Farber 4th ed

58. Two - Sample F - Test for Variances 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 . Use Table 7 in Appendix B. d.f. N = n 1 – 1 d.f. D = n 2 – 1 In WordsIn Symbols 58 Larson/Farber 4th ed

59. 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 59 Larson/Farber 4th ed

60. 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. 60 Larson/Farber 4th ed

61. 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 σ12 > σ22σ12 > σ22 0.10 920 0F1.96 0.10 Because 400 > 256, There is not enough evidence to convince the manager to switch to the new system. 1.96 1.56 Fail to Reject H 0 61 Larson/Farber 4th ed

62. 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 62 Larson/Farber 4th ed

63. 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 σ12 ≠ σ22σ12 ≠ σ22 0. 025 3029 0F2.09 0.025 Because 5.7 2 > 3.5 2, There is enough evidence to support the claim that one of the two stocks is a riskier investment. 2.09 2.65 Reject H 0 63 Larson/Farber 4th ed

64. 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 64 Larson/Farber 4th ed

65. Section 10.4 Analysis of Variance 65 Larson/Farber 4th ed

66. Section 10.4 Objectives Use one-way analysis of variance to test claims involving three or more means Introduce two-way analysis of variance 66 Larson/Farber 4th ed

67. 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. 67 Larson/Farber 4th ed

68. 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. 68 Larson/Farber 4th ed

69. 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. 69 Larson/Farber 4th ed

70. 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. 70 Larson/Farber 4th ed

71. 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 71 Larson/Farber 4th ed

72. 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 72 Larson/Farber 4th ed

73. 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 . Use Table 7 in Appendix B. d.f. N = k – 1 d.f. D = N – k In WordsIn Symbols 73 Larson/Farber 4th ed

74. 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 74 Larson/Farber 4th ed

75. 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 75 Larson/Farber 4th ed

76. 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. 76 Larson/Farber 4th ed

77. 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) 77 Larson/Farber 4th ed

78. 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 0. 01 3 – 1 = 2 13 – 3 = 10 0F 7.56 0.01 78 Larson/Farber 4th ed

79. Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. 79 Larson/Farber 4th ed

80. Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. 80 Larson/Farber 4th ed

81. 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 0. 01 3 – 1 = 2 13 – 3 = 10 0F 7.56 0.01 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 81 Larson/Farber 4th ed

82. Example: Using the TI-83/84 to Perform a One-Way ANOVA Three airline companies offer flights between Corydon and Lincolnville. Several randomly selected flight times (in minutes) between the towns for each airline are shown on the next slide. Assume that the populations of flight times are normally distributed, the samples are independent, and the population variances are equal. At α = 0.01, can you conclude that there is a difference in the means of the flight times? Use a TI-83/84. 82 Larson/Farber 4th ed

83. Example: Using the TI-83/84 to Perform a One-Way ANOVA Airline 1Airline 2Airline 3 122119120 135133158 126143155 131149126 125114147 116124164 120126134 108131151 142140131 113136141 83 Larson/Farber 4th ed

84. 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 At least one mean is different Decision: There is enough evidence to support the claim. You can conclude that there is a difference in the means of the flight times. P-value < α Reject H 0 84 Larson/Farber 4th ed

85. 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. 85 Larson/Farber 4th ed

86. 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 86 Larson/Farber 4th ed

87. 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. 87 Larson/Farber 4th ed

88. 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. 88 Larson/Farber 4th ed

89. 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. 89 Larson/Farber 4th ed

90. Two-Way ANOVA A two-way ANOVA test calculates an F-test statistic for each hypothesis. It is possible to reject none, one, two, or all of the null hypotheses. Can use a technology tool such as MINITAB to perform a two-way ANOVA test. 90 Larson/Farber 4th ed

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 91 Larson/Farber 4th ed