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

Topic 6 Exercises Chapter 16 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. Statistical hypothesis = H0:M0= M24= M48 This represents the mean aggression scores for the population of sleep. H1:H0 = false Given this decision, we need to reject the H0 at the 0.5 level of significance if F> 5.14 Given df between = 2 ^ df within = 6 Sum of squares= SS between = ∑ T2/n - G2/N [(15)2/3 + (18)2/3 + (12)2/3] - (45)2/9 [225/3 + 324/3 + 144/3] – 2025/9 [75 + 108 + 48] – 225 = 231 – 225 = 6 SS within = ∑ X^2 - ∑T2/n (3)^2 + (5)^2 + (7)^2+ (4)^2+ (8)^2 + (6)^2+ (2)^2+ (4)^2 + (6)^2– [(15)2/3 + (18)2/3 + (12)2/3] 9 + 25 + 49 + 16 + 64 + 36 + 4 + 16 + 36 – [75 + 108 + 48] 255-231=24 SS total =∑ X^2 - G2/N 255-225=30 Check= SS total = SS between + SS within 30 = 6+24 ANOVA Table Source SS df Ms F ratio Treatment 6 K-1=2 3 0.75 Error 24 n-k=6 4 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.