Question

NCAA rules require the circumference of a softball to be 12 ± 0.125 inches. A softball manufacturer bidding on an NCAA contract is shown to meet the requirements for mean circumference. Suppose that the NCAA also requires that the standard deviation of the softball circumferences not exceed 0.05 inch. A representative from the NCAA believes the manufacturer does not meet this requirement. She collects a random sample of 25 softballs from the production line and finds that s = 0.076 inch. The Anderson-Darling test of normality gives a P-value of 0.632. Is there enough evidence to support the representative’s belief (i.e., that the standard deviation of circumferences exceeds 0.05 inch) at the α = 0.01 level of significance?

On a separate sheet of paper, write down the hypotheses (H₀ and H_a) to be tested.

Conditions:
The conditions for the χ² ("chi-square") test for standard deviations  (are / are not) satisfied for this data.

Rejection Region:
To test the given hypotheses, we will use a (left / right / two) -tailed test.  
The appropriate critical value(s) for this test is/are _____.  (Report your answer exactly as it appears in Table VII. For two-tailed tests, report both critical values in the answer blank separated by only a single space.)

Answer 1

The conditions for the χ2 ("chi-square") test for standard deviations are satisfied for this data.

The provided sample variance is s^2 =0.005776 and the sample size is given by n=25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:σ^2 =0.0025
Ha:σ^2>0.0025 (RIGHT TAILED TEST)
This corresponds to a right-tailed test, for which a Chi-Square test for one population variance will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the the rejection region for this right-tailed test is R={χ^2:χ^2>42.98}.

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

χ^2 =(n−1)s^2/ σ^2 =(25−1)⋅0.005776/0.0025 =55.45
(4) Decision about the null hypothesis

Critical value= 42.98

Since it is observed that χ^2=55.45>χ^2=42.98,

it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population variance σ^2 is greater than 0.0025, at the 0.01 significance level.

NOTE: I DON'T KNOW IN WHAT WAY CRITICAL VALUE APPEARS IN TABLE VII. SO I ROUNDED IT TWO DECIMAL PLACES.