Question

A television network commissioned a telephone poll of randomly sampled men. Of the 708 respondents who had​ children, 19​% said​ "yes" to the question​ "Are you a​ stay-at-home dad?" To help market commercial​ time, the network wants an accurate estimate of the true percentage of​ stay-at-home dads. Construct a 90​% confidence interval. left parenthesis nothing % comma nothing % right parenthesis

Answer 1

Answer: (16.6%, 21.4%)

Step-by-step explanation:

The confidence interval for proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Sample size : n= 708

The proportion of respondents who had​ children = $p=0.19$

Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Now, the 90​% confidence interval for proportion will be :-

$0.19\pm (1.645)\sqrt{\dfrac{0.19(1-0.19)}{708}}\approx0.19\pm 0.024\\\\=(0.19-0.024,0.19+0.024)=(0.166,\ 0.214)=(16.6\%,\ 21.4\%)$

Hence, the 90​% confidence interval for the proportion = (16.6%, 21.4%)