In a 2012 survey, Gallup asked a sample of U.S. adults if they would prefer to have a job outside the home, or if they would prefer to stay home to care for the family and home. Partial results for the individuals who expressed a preference, broken down by gender, are displayed in the two-way table.

Job Outside of the Home Stay at Home Total
Male 391 ??? 504
Female 254 219 473
Total 645 332 977

8- Find the number of males who would prefer to stay at home.

9- What proportion of respondents would prefer a job outside of the home? Round your answer to two decimal places.

10- Compute the difference in the proportion of men who would prefer a job outside of the home and the proportion of females who would prefer a job outside of the home. Use two decimal places in your answer.

11- Use Statekey to calculate a 90% bootstrap percentile confidence interval for the difference in the proportion of men who would prefer a job outside of the home and the proportion of women who would prefer a job outside of the home.

12- Calculate a 90 % confidence interval for the difference in the proportion of men who would prefer a job outside of the home and proportion of women who would prefer a job outside of the home using the standard error (formula from ch 6). Show your work. How does it compare to your previous answer?

13- Are you justified in using the interval calculated in the previous problem? Explain briefly.

14- If you wanted to test to test the null hypothesis of no difference between men and women and you generated a randomization distribution to do that, where would that randomization distribution be centered?