Suppose the Total Sum of Squares for a completely randomzied design with p = 6 t

Question

Suppose the Total Sum of Squares for a completely randomzied design with p = 6 treatments and n = 30 total measurements (SS(Total)) is equal to 420. In each of the following cases, conduct an F -test of the null hypothesis that the mean responses for the 6 treatments are the same. Use alpha = 0.05. F = (a) Sum of Squares for Treatment (SST) is 40% of SS(Total) F = Rejection region F > 3.12868 The final conclustion is A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ. B. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same. (b) Sum of Squares for Treatment (SST) is 20% of SS(Total) F = 1.25 Rejection region F > 2.60299 The final conclustion is A. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same. B. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ. (c) Sum of Squares for Treatment (SST) is 80% of SS(Total) F = Rejection region F > The final conclustion is A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ. B. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same.

Explanation / Answer

a)

F=23.2

F>2.2661

option A we can reject

b)

F=8.7

F>2.2661

option B)

c)

F=139.2

F>2.2661

optionA

SS df MS F Fcritical SSTR 168 5 33.6 23.2 2.2661 SSE 252 174 1.448276 SST 420 179

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