Question

b 1. For each of the following studies, decide whether you can reject the null hy- pothesis that the groups come from identical populations. Use the .05 level. Note that study (b) provides S, not S. (a) n Group 1 10 7.4 .82 Group 2 10 6.8 .90 Group 3 10 6.8 .80 M S2 (b) n Group 1 25 94 24 M Group 2 25 101 28 Group 3 25 124 31 Group 4 25 105 25 S 2. For each of the followin..

Answer 1

The Hypothesis: HO: The groups come from identical populations. Ha: At least 1 group is different from the others. The Test Statistic: The ANOVA table is as below. The critical value is calculated for a = 0.05 for df1 = 2 and df2 = 27; Critical value = 3.3541 The p value is calculated for F = 1.43 for df1 = 2 and df2 = 27; p value = 0.2571 Source SS DF Mean Square F Fcv p Between 2.40 2 1.20 1.43 3.3541 0.2571 Within/Error 22.68 27 0.84 Total 25.08 29 Fobserved = 1.43 The Decision Rule: If Fobserved is > Fcritical, Then reject HO. Also if p-value is <a, Then reject HO. The Decision: Since Fobserved (1.43) is < Fcritical (3.3541), We Fail to Reject Ho. Also since p-value (0.2571) is > a (0.05), We Fail to Reject HO. The Conclusion: There isn't sufficient evidence at the 95% level of significance to warrant rejection of the claim that the groups come from different populations. Calculations For the ANOVA Table: Overall Mean = [(10 * 7.4) + (10 * 6.8) + (10 * 6.8)]/30 = 7 SS treatment = SUM n* ( īj - overall mean)2 = 10 * (7.4 - 7)2 + 10 *(6.8 - 7)2 + 10 * (6.8 - 712 = 24 Calculations For the ANOVA Table: Overall Mean = [(10 * 7.4) + (10 * 6.8) + (10 * 6.8)]/30 = 7 SS treatment = SUM n* (ū; - overall mean)2 = 10 * (7.4 - 7)2 + 10 * (6.8 - 7)2 + 10 * (6.8 - 7)2 = 2.4 df1 - k-1 = 3 - 1 = 2 MSTR = SS treatment / df1 = 2.4 / 2 = 1.2 SSerror = SUM ((n - 1) * Variance] = 9*0.82 +9*0.9 +9* 0.8 = 22.68 df2 = N-k= 30 - 3 = 27 Therefore MS error = $Serror / df2 = 22.68 / 27 = 0.84 F = MSTR/MSE = 1.2 / 0.84 = 1.43 Scanned with CamScanner

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