Chapter 13 1. In an independent-measures ANOVA, individual differences contribute to the variance in the numerator and in the denominator of the F-ratio. For a repeated-measures ANOVA, what happens to the individual differences in the numerator of the F-ratio. In a repeated measures ANOVA test, individual differences in a numerator don't exist because the same people take part in all of the offered treatments. 2. What is the correct denominator for the repeated-measures F-ratio? In repeated measures ANOVA tests, the denominator is the error variance that measures the variance expected (when no systematic effects/individual diff are affecting the variance) 3. The results from a repeated-measures ANOVA are reported as, F(8, 30) = 4.75, p < .05. For this study, how many treatment conditions were compared? 8+1=9 4. A repeated-measures analysis of variance produces SStotal = 70 and SSwithin treatments = 45. For this analysis, what is SSbetween treatments? 70-45= 25 5. For a repeated-measures study comparing four treatment conditions with a sample of n = 4 participants, what are the degrees of freedom for the repeated-measures F-ratio? 4-1= 3 between treatments 4-1/\\4= 12 within treatments 4-1 = 3 between subjects Df error = 12-3 = 9 Df total = 12 + 3 = 15 6. A repeated-measures study uses a sample of n = 14 participants to evaluate the mean differences among four treatment conditions. In the analysis of variance for this study, what is the value for dfbetween subjects? N-1 = 13 - 1 = 12 7. A repeated-measures ANOVA has SSwithin treatments = 46 and SSbetween subjects = 12. For this analysis, what is the value for SSerror? 46-12= 36 is SS error 8. A repeated-measures analysis of variance for a study comparing three treatment conditions with a sample of n = 10 participants, produces an F-ratio of F = 3.60. For this result, which of the following is the correct statistical decision for alpha = .05 and alpha = .01 (3 points) a) alpha of .05 reject the null hypothesis because F=3.60 > CV=3.55 b) alpha of .01 fail to reject null because F=3.60