Question

A company is experimenting with synthetic fibers as a substitute for natural fibers.

The quality characteristic of interest is the breaking strength. A random sample of 8 natural fibers yields an average breaking strength of 540 kg with a standard deviation of 55 kg. A random sample of 10 synthetic fibers gives a mean breaking strength of 610 kg with a standard deviation of 22 kg.

(a) Can you conclude that the variances of the breaking strengths of natural and synthetic fibers are different? Use a level of significance α of 0.05.

(b) Find a two-sided 95% confidence interval for the ratio of the variances of the breaking strengths of natural and synthetic fibers.

Answer 1

(a) For this we conduct an F test for 2 variances (Since we want to test variation)

Given

s₁ = 55, s₁² = 3025, n₁ = 8, df₁ = 8-1 = 7

s₂ = 22, s₂² = 484, n₂ = 10, df₂ = 10-1 = 9

The Hypothesis:

H0: : There is no difference in variances of the breaking strengths of natural and synthetic fibers.

Ha:: There is a difference in variances of the breaking strengths of natural and synthetic fibers.

The Test Statistic:

F = s₁²/s₂² = 3025/484 = 6.25

The Critical Values: at = 0.05, df1 = 7, df2 = 9

Lower Critical (F_0.975,6,8) = 0.207

Upper critical (F_0.025,6,8) = 4.197

The p value: At F = 6.25, df = 6,8; p value = 0.014

The Decision Rule:

Critical Value Method: If Fobserved is < 0.207 or if Fobserved is > 4.197, Then reject H0.

The p value method: If p value is < , then reject H0.

The Decision:

Critical Value Method: Since Fobserved is > 4.197, We Reject H0.

The p value method: Since p value is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that there is a difference in variances of the breaking strengths of natural and synthetic fibers.

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(b) 95% CI for Variances

s₁²/s₂² = 3025/484 = 6.25

(F_0.975,6,8) = 0.207

(F_0.025,6,8) = 4.197

The Confidence Interval is given by

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