Question

Suppose the Total Sum of Squares (SST) for a completely randomzied design with ?=4 treatments and n=12 total measurements is equal to 480. In each of the following cases, conduct an F-test of the null hypothesis that the mean responses for the 4 treatments are the same. Use α=0.025.

(a) The Treatment Sum of Squares (SSTR) is equal to 336 while the Total Sum of Squares (SST) is equal to 480.

The test statistic is F=

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.

(b) The Treatment Sum of Squares (SSTR) is equal to 48 while the Total Sum of Squares (SST) is equal to 480.

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 384 while the Total Sum of Squares (SST) is equal to 480.

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.

Answer 1

Ans:

df(treatments)=4-1=3

df(within)=12-4=8

MS=SS/df

F=MS(treatments)/MS(error)

When F>Fcritical,we reject null hypothesis.

1)

Source	df	SS	MS	F	Fcrit
treatments	3	336	112	6.222	5.416
within	8	144	18
Total	11	480

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.

2)

Source	df	SS	MS	F	Fcrit
treatments	3	48	16	0.296	5.416
within	8	432	54
Total	11	480

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

3)

Source	df	SS	MS	F	Fcrit
treatments	3	384	128	10.667	5.416
within	8	96	12
Total	11	480

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.