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.

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

Level of significance ∝= 0.90 or 0.10

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

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)

answer is more than one

step-by-step explanation:

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: