Question

In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2455 subjects randomly selected from an online group involved with ears. 931 surveys were returned. Construct a 99​% confidence interval for the proportion of returned surveys.

Answer 1

Answer: 0.355,0.405)

Step-by-step explanation:

Given : Significance level : $\alpha=1-0.99=0.01$

Number of subjects (n) = 2455

Number of surveys returned = 931

The probability of surveys get return will be :-

$p=\dfrac{931}{2455}=0.379226069246\approx0.38$

The confidence interval for proportion is given by :-

$p\pm z_{\alpha/2}\times\sqrt{\dfrac{p(1-p)}{n}}$

$0.38\pm z_{0.005}\times\sqrt{\dfrac{0.38(1-0.38)}{2455}}\\\\=0.38\pm(2.576)0.0098\\\approx0.38\pm0.025=(0.355,0.405)$

Hence, the 99​% confidence interval for the proportion of returned surveys : (0.355,0.405)