Question

Given the following sample information, test the hypothesis that the treatment means are equal at the 0.01 significance level:

Treatment 1	Treatment 2	Treatment 3
3	9	6
2	6	3
5	5	5
1	6	5
3	8	5
1	5	4
	4	1
	7	5
	6
	4

a. State the null hypothesis and the alternative hypothesis.

H₀ : μ₁  (select one)  = / > / <

μ₂    (select one)   = / > / < μ₃

H₁ : Treatment means  (select one)  are not / are all the same.

b. What is the decision rule? (Round the final answer to 2 decimal places.)

Reject H₀ if F >:   

c. Compute SST, SSE, and SS total. (Round the final answers to 2 decimal places.)

SST =                

SSE =                

SS total =                  

d. Complete the ANOVA table. (Round the SS, MS, and F values to 2 decimal places.)

	Source	SS	DF	MS	F
	Factor
	Error
	Total

e. State your decision regarding the null hypothesis.

Decision:  (select one)  Reject / Do not reject  H_0.

f.Find the 95% confidence interval for the difference between treatment 2 and 3. (Round the final answers to 2 decimal places.)

95% confidence interval is: _ ± _  

We can conclude that the treatments 2 and 3 are  (select one)  different / the same

Answer 1

from the given data,

a)

b)

c)SST =46.96

  SSE = 53.00

SS total = 99.96

d)

Source	DF	Sum of Square	Mean Square	F Statistic	P-value
Groups (between groups)	2	46.96	23.48	9.30	0.001
Error (within groups)	21	53.00	2.52
Total	23	99.96	4.35

e)

Since p-value<α, H0 is rejected.

f)

Level of significance=   0.0500
no. of treatments,k=   3
DF error =N-k=   21
MSE=   2.5238
t-critical value,t(α/2,df)=   2.0796

critical value=tα/2,df √(MSE(1/ni+1/nj)) = 1.57

confidence interval = mean difference ± critical value

95% confidence interval = 1.75 ± 1.57

We can conclude that the treatments 2 and 3 are same.