Question

Given the following sample information, test the hypothesis that the treatment means are equal at the 0.10 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  (Click to select)  =  >  <  μ₂    (Click to select)  =  >  <  μ₃

H₁ : Treatment means  (Click to select)  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:  (Click to select)  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  (Click to select)  different  the same   .

Answer 1

a)

Ho: μ₁ = μ₂   = μ₃

b)

df treatments =	number of treatments-1=		2
df error =	N-k=	24-3=	21

for (2,21) df and 0.1 level of significance: critical value F =2.57

Reject H₀ if F >   2.57

c)

i	ni	x̅i	ni*(Xi-Xgrand)2	SS=(ni-1)*s2
treatment 1	6	2.5000	25.010	11.500
treatment 2	10	6.0000	21.267	24.000
treatment 3	8	4.2500	0.681	17.500
Total	24		46.9583	53.0000
			SSTr	SSE

SST =46.96

SSE =53.000

SStotal =99.96

d)

Source	SS	df	MS	F
Between	46.96	2	23.48	9.30
Within	53.00	21	2.52
Total	99.96	23

e)

Reject Ho (since test statistic >critical value)

f)

critical value of t with 0.05 level and N-k=21 degree of freedom=				tN-k=	2.080
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =					1.57

95% confidence interval is =1.75 +/- 1.57 or (0.18 to 3.32)

We can conclude that the treatments 2 and 3 are different