In: Statistics and Probability
You may need to use the appropriate technology to answer this 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 | 9.0 | 11.0 | 9.8 |
7.9 | 7.6 | 10.2 | 12.5 |
5.4 | 9.6 | 9.4 | 11.9 |
7.6 | 10.4 | 10.0 | 10.6 |
8.6 | 9.6 | 8.9 | 11.2 |
7.8 | 10.2 | 8.7 | 11.2 |
(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.
Ha: μ1 =
μ2 = μ3 =
μ4H0:
μ1 = μ2 =
μ3 = μ4
Ha: μ1 ≠
μ2 ≠ μ3 ≠
μ4 H0:
μ1 = μ2 =
μ3 = μ4
Ha: Not all the population means are
equal.H0: At least two of the population means
are equal.
Ha: At least two of the population means are
different.H0: μ1 ≠
μ2 ≠ μ3 ≠
μ4
Ha: μ1 =
μ2 = μ3 =
μ4
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.
Do not 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 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.Reject H0. There is not 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.
Solution:
We have to test for any significant difference in the number of hours between breakdowns for four machines,
For testing this hypothesis, we perform one way ANOVA in spss.
a) The null and alternative hypotheses is
H0: μ1 =
μ2 = μ3 =
μ4
Ha: Not all the population means are equal.
From the ANOVA table,
The test statistic is 15.47
and the p-value is 0.000
# Conclusion:
Since p-value is 0.000 which is less than 0.05, so we reject Ho at 5% level of significance and conclude that There is sufficient evidence that the mean time between breakdowns is not the same for the four machines.
Hence, the correct option is:
Reject H0. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines
(b) Fisher's LSD Test
LSD = 0.005
the pairwise absolute difference between sample means for machines 2 and 4.
x2 − x4 = 1.800
Conclusion:
Since p-value is 0.005 which is less than 0.05, so
There is a significant difference between the means for machines 2 and 4