7.24 One question in the 2002 General Social Survey asked participants whom they voted for in the 2000 election. Of the 980 women who voted, 459 voted for Bush. Of the 759 men who voted, 426 voted for Bush. a) Find a \( 95 \% \) confidence interval for the proportion of women who voted for Bush. b) Find a \( 95 \% \) confidence interval for the proportion of men who voted for Bush. c) Find a 95\% confidence interval for the difference in proportions and interpret your interval.
Solution
Step 1 ：1. Calculate \(p_w\) and \(p_m\). \(p_w = \frac{459}{980} \approx 0.4684,\ p_m=\frac{426}{759} \approx 0.5618.\)
Step 2 ：2. Compute \(95\%\ CI\) for women and men. CI_w: \(p_w \pm z \times \sqrt{\frac{p_w(1-p_w)}{n_w}} \), CI_m: \(p_m \pm z \times \sqrt{\frac{p_m(1-p_m)}{n_m}} \), where z=1.96.
Step 3 ：3. Calculate the \(95\%\ CI\) for each group: CI_w: \(0.4684 \pm 1.96 \times \sqrt{\frac{0.4684(1-0.4684)}{980}} \approx (0.4455, 0.4913)\). CI_m: \(0.5618 \pm 1.96 \times \sqrt{\frac{0.5618(1-0.5618)}{759}} \approx (0.5331, 0.5905)\).
Step 4 ：4. Find a \(95\% CI\) for the difference in proportions: \(CI_{diff} = (p_w - p_m) \pm z \times \sqrt{\frac{variance_w}{n_w} + \frac{variance_m}{n_m}}\), where variance_w=\(p_w(1-p_w)\), variance_m=\(p_m(1-p_m)\), and z=1.96.
Step 5 ：5. Calculate the \(95\% CI\) for the difference in proportions: \((-0.0934) \pm 1.96 \times \sqrt{\frac{0.4684(1-0.4684)}{980} + \frac{0.5618(1-0.5618)}{759}} \approx (-0.1353, -0.0515)\).
