Grand Canyon University_PSY 520 Topic 6 Exercise:Chapter 16, 17, 18 COMPLETE SOLUTION

Chapter 16, numbers 16.9, 16.10, 16.12 and 16.14 16.9 Given the aggression scores below for Outcome A of the sleep deprivation experiment, verify that, as suggested earlier, these mean differences shouldn’t be taken seriously by testing the null hypothesis at the .05 level of significance. Use the computation formulas for the various sums of squares and summarize results with an ANOVA table. ANOVA: Single Factor SUMMARY Groups Coun t Su m Averag e Varianc e zero 3 15 5 4 twenty-four 3 18 6 4 forty-eight 3 12 4 4 ANOVA Source of Variation SS df MS F Pvalue F crit Between Groups 6 2 3 0.75 0.512 5. 5 Within Groups 24 6 4 Total 30 8 H0: μ 0 = μ 24 = μ 48 H1: H0 is false Based on this ANOVA Table we would fail to reject the null hypothesis because F=0.75 and the Fcrit = 5.. 16.10 Another psychologist conducts a sleep deprivation experiment. For reasons beyond his control, unequal numbers of subjects occupy the different groups. (Therefore, when calculating in SS between and SS within , you must adjust the denominator term, n , to reflect the unequal numbers of subjects in the group totals.) (a) Summarize the results with an ANOVA table. You need not do a step-by step hypothesis test procedure. zero twenty-four forty-eight 1 4 7 3 7 12 6 5 10 2 9 1 Anova: Single Factor SUMMARY Groups Count Sum Average Variance zero 5 13 2.6 4.3 twenty-four 3 16 5. 3 2. 3 forty-eight 4 38 9.5 4. 3 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 106.05 2 53.025 13. 0. 0 4. Within Groups 34. 7 9 3. 4 Total 140. 7 11 H0: μ 0 = μ 24 = μ 48 H1: H0 is false Based on this ANOVA Table we would reject the null hypothesis because F=13. and the Fcrit = 4.. (b) If appropriate, estimate the effect size with eta-squared.  2  2= 0. This indicates a large effect size. (c) If appropriate, use Tukey’s HSD test (with n´ = 4 for the sample size, n)to identify pairs of means that contribute to the significant F, given that X´ 0 = 2.60, X´ 24 = 5.33, and X´ 48 = 9.50. HSD= 3.95*(√3.8740/4) = 3. x0 x24 x48 x0 0 2.73 6.9 x24 0 4.17 x48 0 Only the pairs with 48 hours of sleep deprivation contribute to the significant F. (d) If appropriate, estimate effect sizes with Cohen’s d. d(x48, x0)= 6.9/√3.87 = 3.5 d(x48, x24)= 4.17/√3.87 = 2.12 These estimates indicate a large effect size because they are much bigger than .5. (e) Indicate how all of the above results would be reported in the literature, given sample standard deviations of s 0 = 2.07, s 24 = 1.53, and s 48 = 2.08. Aggression scores for subjects deprived of sleep for 0 hours ( X´ 0 = 2.60, s 0 = 2.07), for 24 hours ( X´ 24 = 5.33, s 24 = 1.53), for 48 hours ( X´ 48 = 9.50, s 48 = 2.08) differ significantly [ F (2,9) = 13.70, MSE = 3.87, p < .05, η 2 = .75]. According to Tukey’s HSD test, the mean differences between the 24- and 48-hour groups (4.17) and between the 0- and 48-hour groups (6.90) are significant ( HSD = 3.87, p < .05, d = 2.12, 3.50). *16.12 For some experiment, imagine four possible outcomes, as described in the following ANOVA table. (a) How many groups are in Outcome D? 4 (b) Assuming groups of equal size, what’s the size of each group in Outcome C? dftotal= # of scores minus one, so there are 84 scores, dfbetween= # of groups minus one, so there are 4 groups, therefore 84/4=21 (c) Which outcome(s) would cause the null hypothesis to be rejected at the .05 level of significance? a,b,c) fcrit=2.75, d) fcrit=6.59 Outcomes A and B would be rejected because the F values exceed the fcrit. (d) Which outcome provides the least information about a possible treatment effect? Outcome D because the samples are so small that a treatment effect is difficult to detect. (e) Which outcome would be the least likely to stimulate additional research? Outcome C because it did not detect an effect and has a large sample size. (f) Specify the approximate p -values for each of these outcomes. Approximate p-values are: a) p< .05, b) p< .01, c) p>.05, d) p> .05 16.14 The F test describes the ratio of two sources of variability: that for subjects treated differently and that for subjects treated similarly. Is there any sense in which the t test for two independent groups can be viewed likewise? Yes, a t-test could be run multiple times to find the paired comparisons, however this also increases the level of significance and increase the chance of type-I errors. 2. Chapter 17, numbers 17.6, 1