Question

The following data are from a completely randomized design. In the following calculations, use

α = 0.05.

	Treatment 1	Treatment 2	Treatment 3
	64	81	68
	48	71	55
	53	88	60
	43	64	49
xj	52	76	58
sj2	80.67	112.67	64.67

(a)Use analysis of variance to test for a significant difference among the means of the three treatments.

State the null and alternative hypotheses.

H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.

H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃   

H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃

H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal.

H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃

*Find the value of the test statistic. (Round your answer to two decimal places.)____

*Find the p-value. (Round your answer to three decimal places.)

p-value = ____

*State your conclusion.

Reject H₀. There is not sufficient evidence to conclude that the means of the three treatments are not equal.

Reject H₀. There is sufficient evidence to conclude that the means of the three treatments are not equal.    

Do not reject H₀. There is sufficient evidence to conclude that the means of the three treatments are not equal.

Do not reject H₀. There is not sufficient evidence to conclude that the means of the three treatments are not equal.

(b)Use Fisher's LSD procedure to determine which means are different.

Find the value of LSD. (Round your answer to two decimal places.)

LSD = ____

Find the pairwise absolute difference between sample means for each pair of treatments.

(abs val)x₁ − x₃=

(abs val)x₁ − x₂=

(abs val)x₂ − x₃=

*Which treatment means differ significantly? (Select all that apply.)

There is a significant difference between the means for treatments 1 and 2.

There is a significant difference between the means for treatments 1 and 3.

There is a significant difference between the means for treatments 2 and 3.

There are no significant differences.

Answer 1

a)

H₀: μ₁ = μ₂ = μ₃

H_a: Not all the population means are equal.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	std dev
1	4	208	52.000	81	9
2	4	304	76.000	113	11
3	4	232	58.000	65	8
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1248.0	2	624.0	7.26	0.0133	4.26
Within Groups	774.0	9	86.0
Total	2022.0	11

test stat = 7.26

p value = 0.013

Reject H₀. There is sufficient evidence to conclude that the means of the three treatments are not equal.    

b)

Level of significance=	0.0500
no. of treatments,k=	3
DF error =N-k=	9
MSE=	86.0000
t-critical value,t(α/2,df)=	2.2622

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj)) = 14.83

population mean difference
µ1-µ2	24
µ1-µ3	6
µ2-µ3	18

if absolute difference of means > critical value,means are significantly different ,otherwise not

There is a significant difference between the means for treatments 1 and 2.

There is a significant difference between the means for treatments 2 and 3.