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%)

Explanation:

The confidence interval for proportion is given by :-

`p\pm z_(\alpha/2)\sqrt{(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{(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%)