Question

6. Concerned about campus safety, college officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. How many randomly selected students must be surveyed in order to be 97% confident that the sample percentage is within 2 percentage points of the true population percentage. A previous study resulted in this confidence interval: 0.048 <p< 0.102

Please show clear and organized work so I can follow along.

Answer 1

Given : Margin of error =2%=0.02

Significance level=$\alpha=0.03$

Now , $Z_{\alpha/2}=Z_{0.03/2}=2.17$ ; From standard notmal distribution table

From the confidence interval ,

The previous estimate of the proportion is ,

$\hat{p}$=(upper confidence limit+lower confidence limit)/2=(0.102+0.048)/2=0.075

Therefore , the required sample size is ,

$n=\frac{Z_{\alpha/2}^2*\hat{p}*(1-\hat{p})}{E^2}=\frac{2.17^2*0.075*(1-0.075)}{0.02^2}=816.6998\approx 817$