Question

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 = 0.238. Another random sample of 29 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s2 = 0.112. 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.

Chi-square goodness-of-fit

Chi-square test of homogeneity

Chi-square for testing or estimating σ2 or σ

Two-way ANOVA F test for two variances 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.

Answer 1

F test for two variances

Part i)

H0: σ1² = σ2²; H1: σ1² ≠ σ2²

Part ii)

Test Statistic :-

f = 0.238 / 0.112
f = 2.12

Part iii)

P value = 2 * P ( f > 2.125 )= 0.0664 ( From f table )

0.050 < P-value < 0.100

Part iv)

Reject null hypothesis if P value < α = 0.05
Since P value = 0.0664 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.

Part v)

At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.