Chapter 13: Introduction to Analysis of Variance

Chapter 13: Introduction to Analysis of Variance 1. When there is no treatment effect, the numerator and the denominator of the F-ratio are both measuring the same sources of variability (random, unsystematic differences from sampling error). In this case, the F-ratio is balanced and should have a value near 1.00. 2. Both the F-ratio and the t statistic are comparing the actual mean differences between samples (numerator) with the differences that would be expected if there is no treatment effect (H0 is true). If the numerator is sufficiently bigger than the denominator, we conclude that there is a significant difference between treatments. 3. a. As the differences between sample means increase, MSbetween also increases, and the Fratio increases. b. Increases in sample variability cause MSwithin to increase and, thereby, decrease the Fratio. 4. With 3 or more treatment conditions, you need three or more t tests to evaluate all the mean differences. Each test involves a risk of a Type I error. The more tests you do, the more risk there is of a Type I error. The ANOVA performs all of the tests simultaneously with a single, fixed alpha level. 5. a. Post tests are used to determine exactly which treatment conditions are significantly different. b. If there are only two treatments, then there is no questions as to which two treatments are different. c. If the decision is to fail to reject H0, then there are no significant differences. 6. a. The three means produce s2 = 4. b. The three means produce s2 = 49. 7. a. The three means produce SS = 14. b. n(SSmeans) = 140 c. SSbetween = 140 8. Source SS df MS Between Treatments 70 2 35 F(2, 12) = 17.50 Within Treatments 24 12 2 Total 94 14 With α = .05, the critical value is F = 3.88. Reject the null hypothesis and conclude that there are significant differences among the three treatments. Solutions - Chapter 13 - page 454 9. a. Reducing the differences between treatments should reduce the value of MSbetween and should lower the value of the F-ratio. b. Source SS df MS Between Treatments 10 2 5 F(2, 12) = 2.50 Within Treatments 24 12 2 Total 34 11 With α = .05, the critical value is F = 3.88. Fail to reject the null hypothesis and conclude that there are no significant differences among the three treatments. Notice that the F-ratio is substantially smaller than it was in problem 8. 10. a. Source SS df MS Between Treatments 8 2 4 F(2, 9) = 1.00 Within Treatments 36 9 4 Total 44 11 With α = .05, the critical value is F = 4.26. Fail to reject the null hypothesis and conclude that there are no significant differences among the three treatments. b. For these data, eta squared is 8/44 = 0.18 or 18%. 11. a. Source SS df MS Between Treatments 72 2 36 F(2, 9) = 9.00 Within Treatments 36 9 4 Total 108 11 With α = .05, the critical value is F = 4.26. Reject the null hypothesis and conclude that there are significant differences among the three treatments. Increasing the mean differences between treatments produced a larger F-ratio. b. For these data, eta squared is 72/108 = 0.667 or 66.7%. A larger difference between treatments produces a larger effect size. 12. a. Source SS df MS Between Treatments 32 2 16 F(2,9) = 10.26 Within Treatments 14 9 1.56 Total 46 11 With α = .05, the critical value is F = 4.26. Reject the null hypothesis and conclude that there are significant differences among the three treatments. b. For these data, eta squared is 32/46 = 0.70. Solutions - Chapter 13 - page 455 13. a. Increasing the variance within each treatment will increase MSwithin treatments and will lower the value of the F-ratio. b. Source SS df MS Between Treatments 32 2 16 F(2,9) = 4.00 Within Treatments 36 9 4 Total 68 11 With α = .05, the critical value is F = 4.26. Fail to reject the null hypothesis and conclude that there are not significant differences among the three treatments. As predicted, the F-ratio is smaller than in Problem 12. 14. a. With dfbetween = 2, there must be 3 treatment conditions. b. Adding the two df values (2 and 24) gives dftotal = 26. Therefore, the total number of subjects must be N = 27. 15. a. With dfbetween = 3, there must be 4 treatments. b. Adding the two df values (3 and 36) gives dftotal = 39. Therefore, the total number of subjects must be N = 40. 16. a. Source SS df MS Between Treatments 40 3 13.33 F(3, 12) = 7.98 Within Treatments 20 12 1.67 Total 60 15 With α = .05, the critical value is F = 3.49. Reject the null hypothesis and conclude that there are significant differences among the four levels. b. η2 = 40/60 = 0.67 17. SStotal = 40 (same as SSbetween from problem 16) 18. a. If there are no differences between treatments, then SSbetween treatments should be zero. b. SStotal = MSwithin treatments = 42. c. Source SS df MS Between Treatments 0 2 0 F=0 Within Treatments 42 9 4.67 Total 42 11 There are no significant differences between treatments. 19. Source SS df MS Between Treatments 72 2 36 F(2, 21) = 7.20 Within Treatments 105 21 5 Total 177 23 With α = .05, the critical value is F = 3.47. Reject the null hypothesis and conclude that there are significant differences among the three types of teachers. Solutions - Chapter 13 - page 456 20. 21. 22. Source Between Treatments Within Treatments Total SS df MS 18 2 9 99 33 3 117 35 Source Between Treatments Within Treatments Total SS df MS 30 2 15 54 27 2 84 29 Source SS df MS Between Treatments 81 3 27 Within Treatments 168 56 3 Total 249 59 F=3 F = 7.50 F=9 23. Source SS df MS Between Treatments 23 2 11.5 F(2, 11) = 3.83 Within Treatments 33 11 3 Total 56 13 With df = 2, 11, the critical value for α = .05 is 3.98. Fail to reject the null hypothesis. 24. a. The pooled variance is 6, the estimated standard error is 1.50 and t(10) = 4.00. With df = 10, the critical value is 2.228. Reject the null hypothesis. b. Source SS df MS Between Treatments 96 1 96 F(1, 10) = 16 Within Treatments 60 10 6 Total 156 11 With df = 1, 10, the critical value is 4.96. Reject the null hypothesis. Note that F = t2. 25. a. The pooled variance is 2, the estimated standard error is 0.89 and t(8) = 4.49. With df = 8, the critical value is 2.306. Reject the null hypothesis. b. Source SS df MS Between Treatments 40 1 40 F(1, 8) = 20 Within Treatments 16 8 2 Total 56 9 With df = 1, 8, the critical value is 5.32. Reject the null hypothesis. Within rounding error, note that F = t2. Solutions - Chapter 13 - page 457 26. a. Source SS df MS Between Treatments 252 2 126 F(2, 15) = 19.30 Within Treatments 98 15 6.53 Total 350 17 With df = 2, 15 the critical value is 3.68. Reject the null hypothesis. b. The percentage of variance explained by the mean differences is η2 = 0.72 or 72%. c. With k = 3 groups and df = 15, q = 3.67. The HSD = 3.83. The non-migrating birds are significantly different from either other group, but there is no significant difference between the short- and long-distance migrants. 27. a. The means and SS values are Single Twin Triplet M=8 M=6 M=5 SS = 10 SS = 18 SS = 14 Source SS df MS Between Treatments 28 2 14 F(2, 15) = 5.00 Within Treatments 42 15 2.8 Total 70 17 With df = 2, 15 the critical value is 3.68. Reject the null hypothesis. b. For single versus twin, the Scheffé F = 6/2.8 = 2.14 (not significant). For single versus triplet, the Scheffé F = 13.5/2.8 = 4.82 (significant). For twin versus triplet, the Scheffé F = 1.5/2.8 = 0.54 (not significant). Solutions - Chapter 13 - page 458