Question

A sample of 500 nursing applications included 60
from men. Find the 90% confielence interval
for the
true proportion of men who applied to the nursing
program.​

Answer 1

Answer:

90% confidence interval for the true proportion of men who applied to the nursing   program.​

(0.09674 ,0.14326)

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 500

sample proportion

                $p = \frac{x}{n} = \frac{60}{500} = 0.12$

Level of significance ∝= 0.90 or 0.10

90% confidence interval for the true proportion of men who applied to the nursing   program.​

$(p - Z_{\frac{0.10}{2} } \sqrt{\frac{p(1-p)}{n} } , p + Z_{\frac{0.10}{2} } \sqrt{\frac{p(1-p)}{n} })$

$(p - Z_{0.05 } \sqrt{\frac{p(1-p)}{n} } , p + Z_{0.05 } \sqrt{\frac{p(1-p)}{n} })$

$(0.12 - 1.645 \sqrt{\frac{0.12(1-0.12)}{500} } , 0.12 + 1.645 \sqrt{\frac{0.12(1-0.12)}{500} })$

On calculation , we get

( 0.12 - 0.02326 , 0.12 + 0.02326)

(0.09674 ,0.14326)

Final answer:-

90% confidence interval for the true proportion of men who applied to the nursing   program.​

(0.09674 ,0.14326)