Problem 10 is from Gravetter and Wallnau, Chapter 13. Previous problem 9 answers

Question

Problem 10 is from Gravetter and Wallnau, Chapter 13.

Previous problem 9 answers are:

a. The sample variances are 4, 5, and 6.

b.

Source                              SS          df             MS                              

Between treatments          90           2              45       F(2,12) = 9.00

Within treatments             60          12              5

Totals                               150         14                                             

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.

Problem 10 summary table

Treatment I Treatment II Treatment III .

n = 5 n = 5 n = 5
M = 2 M =5 M = 8 N = 15
T = 10 T = 25 T = 40 G = 75
SS = 64 SS = 80 SS = 96 SigmaX² = 705 .

a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 9.

b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained in problem 9.

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

Explanation / Answer

(a) Var1 = SS/(n - 1) = 64/(5 - 1) = 16, Var2 = SS/(n - 1) = 80/(5 - 1) = 20, Var3 = SS/(n - 1) = 96/(5 - 1) = 24 These sample variances are higher than those of problem (9). Each sample variance is 4 times that in problem (9) -------------------------------------------------------------------------------------------------- (b) Since sample variances have increased, the MS within increases, and therefore, F decreases. -------------------------------------------------------------------------------------------------- (c) SS Total = S X^2 - (G^2 /N) = 705 - (75^2 /15) = 330 SS between = (t1^2 /n1) + (t2^2 /n2) + (t3^2 /n3) - (G^2 /N) = (10^2 /5) + (25^2 /5) + (40^2 /5) - (75^2 /15) = 90 SS within samples = 330 - 90 = 240 [Note: SS within can also be calculated as the sum of all the SS values = 64 + 80 + 96 = 240] MS between samples = 90/2 = 45 MS within samples = 240/12 = 20 F = 45/20 = 2.25 Critical F- score for a = 0.05 and Df (2, 12) is 3.8853 Since 2.25 < 3.8853, we fail to reject Ho There is no sufficient evidence of a significant difference between the treatment means.

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