Question

The 1996 gss asked, "if the husband in a family wants children, but the wife decides that she does not want any children, is it all right for the wife to refuse to have children?" of 720 respondents, 578 said yes. (a) find a 95% confidence interval for the population proportion who would say yes. round to two decimal places.

Answer 1

The confidence interval would be
$0.80\pm0.03$.

We use the formula
$p\pm z*\sigma_p$, where

$\sigma_p=\sqrt{\frac{p(1-p)}{N}}$, with p being the sample proportion and N being the sample size.

First we find the z-score associated with this level of confidence:
Convert 95% to a decimal:  95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

Since 578/720 said yes, this gives us p=0.80:
$\sigma_p=\sqrt{\frac{0.80(1-0.80)}{720}}=\sqrt{\frac{0.8(0.2)}{720}=0.015$

This gives us
$0.80\pm1.96(0.015)=0.80\pm 0.03$