Question

The accompanying table gives results from a study of words spoken in a day by men and women. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. Men Women 186 210 15,668.6 16,215.3 8,632.9 7,301.2 What are the null and alternative hypotheses? O A. Ho: 67 = 63 - H₂:0 < 63 OC. He: 0770 Hy: 0; = 0 OB. Ho:03 = 02 Hy: 0; > o OD. Hp: 0; -o Hy: 02 Identify the test statistic. F= (Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Reject Ho. There is sufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. O C. Reject Ho. There is insufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by O B. Fail to reject Ho. There is sufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. D . Fail to reject Ho. There is insufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day O

Answer 1

The statistical software output for this problem is:

Two Sample variance summary hypothesis test: G : Variance of population 1 G+ : Variance of population 2 0," Ratio of two variances H:0,20, 21 Hypothesis test results: Ratio Num. DF Den. DF Sample Ratio F-Stat P-value og 185 209 23980572 1.3990572 0.0094

From above output:

Hypotheses: Option B is correct.

F = 1.40

P - value = 0.009

Conclusion: Option A is correct.