Question

A sample survey interviews SRSs of$500$female college students and $550$male college students. Each student is asked whether he or she worked for pay last summer. In all, $410$of the women and $484$of the men say “Yes.” The $95\%$confidence interval for the difference $p_M-p_F$in the proportions of college men and women who worked last summer is about

(a) $0.06\pm 0.00095$

(b) $0.06\pm 0.043.$

(c) $0.06\pm 0.036$

(d) $-0.06\pm 0.043$

(e)$-0.06\pm 0.036$

Short answer

The difference in the proportions of college men and women who worked last summer is option (b) $0.06\pm 0.043.$

Step-by-step solution

Given information

The number of women who worked last summer $=410$

The number of men who worked last summer $=484$

Find the difference in the proportions of college men and women who worked last summer.

Explanation

The given data is written as

$x_1=484$

$n_1=550$

$x_2=410$

$n_2=500$

By dividing the number of successes by the sample size, the sample proportion is calculated:

localid="1650360665085" $$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{484}{550} \\ =0.88 \end{gathered}$$

localid="1650360679467" $$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{410}{500} \\ =0.82 \end{gathered}$$

For confidence level $1-\alpha =0.95$,

find $z_{\alpha /2}=z_{0.025}$using table II

$z_{\alpha /2}=1.96$

The difference between the endpoints of the confidence $p_1-p_2$are:

localid="1650360715016" $$\begin{gathered} =\left({\hat{p}}_1-{\hat{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}} \\ =(0.88-0.82)\pm 1.96\sqrt{\frac{0.88(1-0.88)}{550}+\frac{0.82(1-0.82)}{500}} \end{gathered}$$

$\approx 0.06\pm 0.043.$