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 30 softballs from the production line and finds that s = 0.059 inch. The Anderson-Darling test of normality gives a P-value of 0.0632. 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 = 0.05 level of significance? On a separate sheet of paper, write down the hypotheses (H0 and Ha) to be tested.

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

Rejection Region: To test the given hypotheses, we will use a (left / right / two) [b]-tailed test. The appropriate critical value(s) for this test is/are [c]. (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.) On a separate sheet of paper, sketch the rejection region(s) for this test. You will need this sketch in Question 6.

Refer to the data given in Question 3.

The test statistic for this test is χ²₀=[a].  (Calculate this value in a single step in your calculator, and report your answer rounded to 3 decimal places.)

Label the test statistic in your sketch from Question 3. Use this sketch to conclude the hypothesis test.

We [b] (reject / fail to reject) H₀.
The given data [c] (does / does not) provide significant evidence that the standard deviation of circumferences of softballs from his manufacturer exceeds 0.05 inch.

Answer 1

Result:

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 30 softballs from the production line and finds that s = 0.059 inch. The Anderson-Darling test of normality gives a P-value of 0.0632. 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 = 0.05 level of significance? On a separate sheet of paper, write down the hypotheses (H0 and Ha) to be tested.

Conditions: The conditions for the χ2 ("chi-square") test for standard deviations [a] (are ) satisfied for this data.

Rejection Region: To test the given hypotheses, we will use a ( right ) [b]-tailed test.

The appropriate critical value(s) for this test is/are 42.557 [c]. (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.) On a separate sheet of paper, sketch the rejection region(s) for this test. You will need this sketch in Question 6.

Refer to the data given in Question 3.

The test statistic for this test is χ²₀= 40.380 [a].  (Calculate this value in a single step in your calculator, and report your answer rounded to 3 decimal places.)

Label the test statistic in your sketch from Question 3. Use this sketch to conclude the hypothesis test.

We [b] (reject / fail to reject) H₀.
The given data [c] (does not) provide significant evidence that the standard deviation of circumferences of softballs from his manufacturer exceeds 0.05 inch.

Calculations:

Ho: σ =0.05    Ha: σ >0.05

= (29*0.059²) / 0.05² = 40.3796

DF=30-1=29

Chi-Square Test of Variance
Data
Null Hypothesis                        s^2=	0.0025
Level of Significance	0.05
Sample Size	30
Sample Standard Deviation	0.059
Intermediate Calculations
Degrees of Freedom	29
Half Area	0.025
Chi-Square Statistic	40.3796
Upper-Tail Test
Upper Critical Value	42.5570
p-Value	0.0779
Do not reject the null hypothesis