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

0.3876<p<0.4389

Explanation:

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

-We calculate the proportion of surveys returned:

`\hat p=(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)* \sqrt{(\hat p(1-\hat p))/(n)}\\\\=2.576* \sqrt{(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