Question

n a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2472 subjects randomly selected from an online group involved with ears. 1022 surveys were returned. Construct a 99​% confidence interval for the proportion of returned surveys.

Answer 1

Answer:

0.3876<p<0.4389

Step-by-step explanation:

-Given $n=2472, \ x=1022 , \ CI=0.99$

-We calculate the proportion of surveys returned:

$\hat p=\frac{1022}{2472}\\\\=0.4134$

For a 99% confidence interval:

$z_{\alpha/2}=2.576$

#The margin of error is calculated as;

$ME=z_{0.005}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\=2.576\times \sqrt{\frac{0.4134(1-0.4134)}{2472}}\\\\=0.0255$

The confidence interval are then:

$CI=\hat p\pm ME\\\\=0.4134\pm 0.0255\\\\=[0.3876,0.4389]$

Hence, the confidence interval is 0.3876<p<0.4389