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	73	53
	53	89	60
	43	69	55
xj	52	78	59
sj2	80.67	78.67	44.67

(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: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃

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

H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃

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

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

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.

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.

Reject H₀. There is 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.

x₁ − x₂ = ____

x₁ − x3 = ____

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

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Source	SS	df	MS	F	P value
Between	1448.0000	2	724.000	10.65	0.0042
Within	612.0000	9	68.0000
Total	2060.00	11

a)

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

value of the test statistic F =10.65

p-value =0.0042

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

b)

critical value of t with 0.05 level and N-k=9 degree of freedom=				tN-k=	2.262
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =					13.19

Difference	Absolute Value	Conclusion
x1-x2	26.00	significant difference
x1-x3	7.00	not significant difference
x2-x3	19.00	significant difference

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.