Question

Auto-car parts cannot tolerate much variance without negatively affecting driving performance. Production specifications recommend a variance of σ² = 0.0005 centimeters or less.

A sample of 23 parts is taken from the latest batch and it shows a sample variance s² = 0.0008 centimeters. Test the hypothesis that the sample variance of new batch would be meet the industry’s standards against the alternative that the new batch is outside the acceptable limit. US alpha = .05 for a test of variance. Which test would you use?

Set up the four steps properly and make the necessary calculations to reach the conclusion.

1. State the two Hypotheses

2.Determine the rejection criteria

3.Calculate the test statistic for variance

4.State your conclusion

(S.O.S)

Answer 1

Answer: Auto-car parts cannot tolerate much variance without negatively affecting driving performance. Production specifications recommend a variance of σ2 = 0.0005 centimeters or less.

A sample of 23 parts is taken from the latest batch and it shows a sample variance s2 = 0.0008 centimeters. Test the hypothesis that the sample variance of new batch would be meet the industry’s standards against the alternative that the new batch is outside the acceptable limit. Use alpha = .05 for a test of variance.

Solution:

The Chi-Square Statistic is used for test of variance.

1) The hypothesis test:

Null hypothesis, Ho: σ2 ≥ 0.0005

Alternative hypothesis, Ha: σ2 ≤ 0.0005

2) Rejection criteria:

At α = 0.05, and

df = n-1 = 23-1 = 22

Critical value of χ​​​​​​2 (0.05,22) = 33.924

Reject Ho, if χ​​​​​​2 > 33.924

3) The test statistic for variance:

χ​​​​​​2 = [ ( n - 1 ) * s2 ] / σ2

χ​​​​​​2 = [(23-1) * 0.0008] / 0.0005

χ​​​​​​2 = 35.20

Therefore, test statistic for variance, χ​​​​​​2 = 35.20

4) Conclusion:

Since, test statistic χ​​​​​​2 > critical value of χ2

Therefore, we reject the null hypothesis Ho.

There is sufficient evidence to concluded that the new batch is outside the acceptable limit.