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
6.5	8.9	10.7	9.9
7.8	7.7	10.0	12.9
5.4	9.6	9.4	12.0
7.5	10.2	10.0	10.7
8.4	9.5	8.9	11.2
7.6	9.9	8.6	11.7

(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: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃ = μ₄H₀: 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 = μ₄H₀: μ₁ = μ₂ = μ₃ = μ₄
H_a: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄H₀: 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.)

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 mean time between breakdowns is not the same for the four machines.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.    Do not reject H₀. There is 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.

(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 − x₄

=  

What conclusion can you draw after carrying out this test?

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

We input this data set in MS Excel and use the "Anova: Single Factor" option under Data > Data Analysis to carry out the one way Anova problem and answer the given questions. The output is given below.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Machine 1	6	43.2	7.2	1.156
Machine 2	6	55.8	9.3	0.804
Machine 3	6	57.6	9.6	0.612
Machine 4	6	68.4	11.4	1.096
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	53.325	3.000	17.775	19.384	0.000	3.098
Within Groups	18.340	20.000	0.917
Total	71.665	23.000

(a) H₀: μ₁ = μ₂ = μ₃ = μ₄, H_a: Not all the population means are equal.
Value of test statistic = F = 19.38.
p-value = 0.000.
Reject H₀. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

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