Question

Suppose the Total Sum of Squares (SST) for a completely randomzied design with ?=6k=6 treatments and ?=18n=18 total measurements is equal to 500500. In each of the following cases, conduct an ?F-test of the null hypothesis that the mean responses for the 66 treatments are the same. Use ?=0.025α=0.025.

(a) The Treatment Sum of Squares (SSTR) is equal to 350350 while the Total Sum of Squares (SST) is equal to 500500.

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 100100 while the Total Sum of Squares (SST) is equal to 500500.

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 5050 while the Total Sum of Squares (SST) is equal to 500500.

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

Given there are k = 6 treatments

the total number of observations n = 18 which means there are 3 observations in each treatment level.

The format of ANOVA table is as follows

ANOVA
Source of variation	SS	df	MS	F
Treatments	SSTR	k-1 (6-1=5)	SSTR/(k-1) = MSTR	MSTR/MSE
Error	SSE	(n-1) - (k-1)= 12	SSE / (n-k) = MSE
Total	SST	n-1 (18-1 =17)

SSE = SST - SSTR

Given alpha = 0.025

critical value = F_{0.025,k-1,n-k} = F_0.025,5,12 = 3.89

Rejection criteria:- If the test statistic is more than critical value, Reject null Hypothesis.

a)

(a) The Treatment Sum of Squares (SSTR) is equal to 350 while the Total Sum of Squares (SST) is equal to 500.

ANOVA
Source of variation	SS	df	MS	F
Treatments	350	5	70	5.6
Error	150	12	12.5
Total	500	17

The test statistic is ?=5.60

The critical value is ?=3.89

Conclusion:- 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 100 while the Total Sum of Squares (SST) is equal to 500.

ANOVA
Source of variation	SS	df	MS	F
Treatments	100	5	20	0.60
Error	400	12	33.33
Total	500	17

The test statistic is ?=0.60

The critical value is ?=3.89

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

c)  The Treatment Sum of Squares (SSTR) is equal to 50 while the Total Sum of Squares (SST) is equal to 500.

ANOVA
Source of variation	SS	df	MS	F
Treatments	50	5	10	0.267
Error	450	12	37.5
Total	500	17

The test statistic is ?=0.27

The critical value is ?=3.89

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