Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in centimeters) are being examined.

Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in centimeters) are being examined. A random sample of 21roller bearings from the old manufacturing process showed the sample variance of diameters to bes2 = .

Another random sample of27roller bearings from the new manufacturing process showed the sample variance of their diameters to bes2 = .

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 estimatingσ2orσ,Ftest for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following.

One-way ANOVA

Two-way ANOVA

Chi-square test of independence

F test for two variances

Chi-square test of homogeneity

Chi-square goodness-of-fit

Chi-square for testing or estimating σ2 or σ

(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.