Question 8 is from Gravetter and Wallnau\'s book, Chapter 13. For previous probl

Question

Question 8 is from Gravetter and Wallnau's book, Chapter 13.

For previous problem 7 the answers are:

a.

Source                          SS      df            MS                                    .

Between treatments     84        2            42          F(2,15)  = 6.00

Within treatments        105     15             7

Total                             189     17                                                      .

With alpha = .05, the critical value is F = 3.68.  Reject the null hypothesis and conclude that there are significant differences among the three treatments.

 b.  ita(squared) = 84/189 = 0.444.

 c.  Analysis of variance showed significant mean differences among the three treatments.

F(2,15) = 6.00, p < .05, ita(squared) = 0.444.

 

For problem 8, to create the following data, 3 points were added to each score in Treatment 1 of the previous problem. In the resulting data, the mean differences are much smaller than in the previous problem.

Treatment I Treatment II Treatment III .

n = 6 n = 6 n = 6
M = 4 M =5 M = 6 N = 18
T = 24 T = 30 T = 36 G = 90
SS = 30 SS = 35 SS = 40 SigmaX² = 567 .

a. Before beginning any calculation, predict how the change in the data should influence the outcome of the analysis. That is, how will the F-ratio and the value of  ita(squared) for these data compare with the values obtained in problem 7?

b. Use an ANOVA with alpha = .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in part a.?)

c. Calculate ita(squared) to measure the effect size for this study. (Does your answer agree with your prediction in part a.?)

Explanation / Answer

(b) xbar = G/n = 90/18 = 5 Within SS = 30 + 35 + 40 = 105 Total SS = 567 - G^2/n = 567 - 90^2/18 = 567 - 450 = 117 Between SS = Total SS - Within SS = 117 - 105 = 12 total DF = 18-1=17 Between DF = 3-1 =2 Within MS = 17-2 =15 Within MS = 105/15 = 7 Between MS = 12/2 = 6 F=6/7 = 0.8571 Fcritical = 3.68 ANOVA Source SS DF MS F Fcritical p-value Between 12 2 6 0.857143 3.68232 0.444146 Within 105 15 7 Total 17 F = 0.8571

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