Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in centimeters) are being examined. A random sample of 21 roller bearings from the old manufacturing process showed the sample variance of diameters to be s2 = 0.217. Another random sample of 26 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s2 = 0.132. 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 ANOVA Chi-square for testing or estimating σ2 or σ Chi-square test of independence Chi-square test of homogeneity One-way ANOVA Chi-square goodness-of-fit F test for two variances (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 statistic software output for this problem is :
F test for two variances
(i)
Level of significance = 0.05
H0: σ12 = σ22; H1: σ12 ≠ σ22
(ii)
Test statistics = 1.64
(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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