PSY 520 Topic 6 Exercise Answers

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 The mean aggression scores for the population of sleep for 0, 24, & 48 hours H1:H0= false Decision: Reject H0 at the o.5 level of significance if F 5.14 Given df between= 2 ^ df within= 6 Hours of Sleep Deprivation 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 SS total =SS between +SS within 30 = 6+24 ANOVA Table Source SS df Ms F ration 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. ΣT2n Source SS Df Ms F ration Between 106.05 2 24.63 2.42 Within 34.87 9 10.18 Total 140.92 11 * Significant at the .05 level. (b) If appropriate, estimate the effect size with h2. n^2=106.05/140.92=.75 a large effect, according to the guidelines. If appropriate, use