Chapter 10: Q. 31 (page 626)

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

Expert verified

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" ${\hat{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{484}{550} = 0.88$

localid="1650360679467" ${\hat{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{410}{500} = 0.82$

For confidence level $1 - α = 0.95$ ,

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

$z_{α / 2} = 1.96$

The difference between the endpoints of the confidence $p_{1} - p_{2}$ are:

localid="1650360715016" $= ({\hat{p}}_{1} - {\hat{p}}_{2}) \pm z_{α / 2} \cdot \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}} = (0.88 - 0.82) \pm 1.96 \sqrt{\frac{0.88 (1 - 0.88)}{550} + \frac{0.82 (1 - 0.82)}{500}}$