Statistics > QUESTIONS & ANSWERS > Grand Canyon University_PSY 520 Topic 6 Exercise:Chapter 16, 17, 18 COMPLETE SOLUTION (All)
Grand Canyon University_PSY 520 Topic 6 Exercise:Chapter 16, 17, 18 COMPLETE SOLUTION
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1. 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 differenc... es 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 5.1432528 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 = 5.14325285. 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 5.33333333 3 2.33333333 3 forty-eight 4 38 9.5 4.33333333 3 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 106.05 2 53.025 13.6871414 0.0018643 0 4.256494729 Within Groups 34.8666666 7 9 3.87407407 4 Total 140.916666 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=13.6871414 and the Fcrit = 4.256494729. (b) If appropriate, estimate the effect size with eta-squared. 2 2= 0.752572 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) = 3.887327 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 Outc [Show More]
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