In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2617 subjects randomly selected from an online group involved with ears. surveys were returned. Construct a 99% confidence interval for the proportion of returned surveys.

Answer:

Given Data :

Favorable cases $(x)=932$

Sample size $(n)=2,617$

Confidence interval level $=99%$

Solution:

The sample proportion $(\hat{p} )= \frac{x}{n} = \frac{932}{2{,}617} = 0.35613$

The significance level $(\alpha ) = 1 – 0.99 = 0.01$

The confidence interval :

$$CI = \hat{p} \pm z_c \times \sqrt{\frac{\hat{p}(1 – \hat{p})}{n}} $$ $$ = 0.3561 \pm 2.576 \times \sqrt{\frac{0.3561(1 – 0.3561)}{2{,}617}} $$ $$= (0.332, 0.380)$$