Question

A sample of 500 nursing application included 60 men. Find the 90% confidence interval for the true proportion of men who applied for the nursing program

Answer 1

since we want 90% confidence interval so $\alpha =0.10$ . Since we may be wrong both high or low we need to divide  $\alpha$ by 2.

$n=500, x = 60, p = \frac{60}{500} = 0.12, q=1-p = 0.88$

Error term is, $E = Z_{\alpha /2} \sqrt{\frac{pq}{n}} = 1.645 \sqrt{\frac{0.12*0.88}{500}} = 0.02390632722$

so interval is

$0.12 -0.024 < p < 0.12+0.024$

$0.096 < p < 0.144$