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 = 0.237. Another random sample of 28 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s² = 0.128. 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 test of homogeneity

Chi-square test of independence     

Chi-square for testing or estimating σ² or σ

One-way ANOVA

F test for two variances

Chi-square goodness-of-fit

Two-way ANOVA

(i) Give the value of the level of significance.
0.05

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

..................

Null and alternative hypothesis:  
Hₒ : σ₁² = σ₂²  
H₁ : σ₁² ≠ σ₂²  
Test statistic:  
F = s₁² / s₂² = 0.237 / 21 =    1.85
Degree of freedom:  
df₁ = n₁-1 =    20
df₂ = n₂-1 =    27

P-value :  
P-value = 2*F.DIST.RT(1.8516, 20, 27) =    0.1356

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