Question

Suppose the Total Sum of Squares (SST) for a completely randomzied design with k=6 treatments and n=24 total measurements is equal to 400. In each of the following cases, conduct an FF-test of the null hypothesis that the mean responses for the 66 treatments are the same. Use α=0.01.

(a) The Treatment Sum of Squares (SSTR) is equal to 200 while the Total Sum of Squares (SST) is equal to 400.

The test statistic is F=

The critical value is F=

The final conclusion 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.

(b) The Treatment Sum of Squares (SSTR) is equal to 280 while the Total Sum of Squares (SST) is equal to 400.

The test statistic is F=

The critical value is F=

The final conclusion 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) The Treatment Sum of Squares (SSTR) is equal to 240 while the Total Sum of Squares (SST) is equal to 400.

The test statistic isF=

The critical value is F=

The final conclusion 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.

Answer 1

a)

Source	SS	df	MS	F	p vlaue	Fcrit
treatment	200.000	5	40.000	3.600	0.0197	4.248
error	200.000	18	11.111
total	400.000	23

The test statistic is F=3.600

The critical value is F=4.248

A. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same.

b)

The test statistic is F=8.400

The critical value is F=4.248

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)

The test statistic is F=5.400

The critical value is F=4.248

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