Question

To test for any significant difference in the number of hours between breakdowns for four machines, the following data were obtained.

Machine 1	Machine 2	Machine 3	Machine 4
6.6	9.0	10.8	9.8
8.2	7.6	10.2	12.7
5.5	9.7	9.6	12.1
7.6	10.3	10.2	10.7
8.8	9.6	9.0	11.2
7.7	10.2	8.4	11.9

(a)

At the α = 0.05 level of significance, what is the difference, if any, in the population mean times among the four machines?

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 = μ₄
H_a: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄    

H₀: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄
H_a: μ₁ = μ₂ = μ₃ = μ₄

H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃ = μ₄

H₀: μ₁ = μ₂ = μ₃ = μ₄
H_a: Not all the population means are equal.

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

t stat =

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

p-value =

State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

Reject H₀. There is not sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.    

Reject H₀. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

Do not reject H₀. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

(b)

Use Fisher's LSD procedure to test for the equality of the means for machines 2 and 4. Use a 0.05 level of significance.

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

LSD =

Find the pairwise absolute difference between sample means for machines 2 and 4.

x₂ − x_{4 =}

What conclusion can you draw after carrying out this test? (Choose one)

There is a significant difference between the means for machines 2 and 4.

There is not a significant difference between the means for machines 2 and 4.    

Answer 1

a)

H₀: μ₁ = μ₂ = μ₃ = μ₄
H_a: Not all the population means are equal.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	std dev
1	6	44.4	7.400	1.396	1.182
2	6	56.4	9.400	0.996	0.998
3	6	58.2	9.700	0.78	0.883
4	6	68.4	11.40	1.10
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	48.4	3	16.1	15.09	0.0000	3.10
Within Groups	21.4	20	1.1
Total	69.8	23

test stat = 15.09

p value=0.0000

Reject H₀. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

b)

Level of significance=	0.0500
no. of treatments,k=	4
DF error =N-k=	20
MSE=	1.1
t-critical value,t(α/2,df)=	2.0860

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

LSD=1.25

pairwise absolute difference between sample means for machines 2 and 4.

x2 − x4 =2.00

(if absolute difference of means > critical value,means are significnantly different ,otherwise not )

  

There is a significant difference between the means for machines 2 and 4.