Exam 12: Inferences for Population Proportions

A poll investigating the level of public support for proposed gun control legislation reported that 68% of the respondents were in favor. The pollsters reported a sampling error of $\pm 5 \%$ %. When the responses were broken down by gender, support was 3% higher among women than men. The pollsters said the margin of error for this difference was $\pm$ 6%. Why is the margin of error larger for the difference in support between the genders than for the overall level of support?

Use the two-proportion plus-four z-interval procedure to find the required confidence interval. -A survey of randomly chosen adults found that 27 of the 86 women and 37 of the 87 men follow regular exercise programs. Construct a 95% confidence interval for the difference in the Proportions of women and men who have regular exercise programs.

Use the two-proportions z-interval procedure to obtain the required confidence interval for the difference between twopopulation proportions. Assume that independent simple random samples have been selected from the two populations. -A state university found that it retained 25 students out of 352 in 2003 and 26 students out of 334 in 2004. Construct a 90% confidence interval for the difference in the proportions of students retained In 2003 and 2004.

Use the one-proportion z-test to perform the specified hypothesis test. Use the critical-value approach. - $\mathrm { x } = 18 , \mathrm { n } = 100 , \mathrm { H } _ { 0 } : \mathrm { p } = 0.2 , \mathrm { H } _ { \mathrm { a } } : \mathrm { p } \neq 0.2 , \alpha = 0.01$