In the previous exercise, explain how the result of the 95% confidence interval in part b corresponds to the result of the decision using significance level 0.05 in part c.

Previous exercise

In 2014, the General Social Survey asked about the number of hours a week spent on the World Wide Web (variable denoted WWWHR). Some results are as follows:

Transcribed Image Text:

Group Mean StDev SE Mean 16.6 Male 621 12.1 0.027 13.7 Female 778 11.2 0.018 Difference = mu(Male) – mu (Female) Estimate for difference: 1.3 95% CI for difference: (-0.8, 2.5) T-Test of difference = 0 (vs ): T-value = 1.03 P-value 0.3044