Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in centimeters) are being examined. A random sample of 26 roller bearings from the old manufacturing process showed the sample variance of diameters to be s2 = 0.231. Another random sample of 28 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s2 = 0.146. Use a 5% level of significance to test the claim that there is a difference (either way) in the population variances between the old and new manufacturing processes.
Classify the problem as being a Chi-square test of independence or homogeneity, Chi-square goodness-of-fit, Chi-square for testing or estimating σ2 or σ, F test for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following.
Two-way ANOVAF test for two variances
Chi-square test of homogeneity
Chi-square for testing or estimating σ2 or σ
Chi-square goodness-of-fit
One-way ANOVA
Chi-square test of independence
(i) Give the value of the level of
significance.
State the null and alternate hypotheses.
H0: σ12 = σ22; H1: σ12 > σ22
H0: σ12 = σ22; H1: σ12 < σ22
H0: σ12 = σ22; H1: σ12 ≠ σ22
H0: σ12 < σ22; H1: σ12 = σ22
(ii) Find the sample test statistic. (Round your answer to
two decimal places.)
(iii) Find the P-value of the sample test
statistic.
P-value > 0.200
0.100 < P-value < 0.200
0.050 < P-value < 0.100
0.020 < P-value < 0.050
0.002 < P-value < 0.020
P-value < 0.002
(iv) Conclude the test.
Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.
Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.
Since the P-value is less than the level of significance α = 0.05, we fail to reject the null hypothesis.
Since the P-value is greater than or equal to the level of significance α = 0.05, we reject the null hypothesis.
(v) Interpret the conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.
At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is not different.
At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is not different.
At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is different.
The statistical software output for this problem is:
Hence,
Two-way ANOVA F test for two variances
i) Level of significance = 0.05
Hypotheses: H0: σ12 = σ22; H1: σ12 ≠ σ22
ii) Test statistic = 1.58
iii) P-value > 0.200
iv) Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.
v) At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.
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