Question

In a Gallup poll, 1025 randomly selected adults were surveyed and 29% of them said that they used the Internet for shopping at least a few times a year. Find a 99% confidence interval estimate of the proportion of adults who use the Internet for shopping.

Answer 1

Answer:   (0.25,0.33)

Step-by-step explanation:

A 99% confidence interval for population proportion is given by:-

$\hat{p}\pm 2.576\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ = sample proportion, $n$ = sample size.

Given: $n=1025, \hat{p}=0.29$

A 99% confidence interval estimate of the proportion of adults who use the Internet for shopping:

$0.29\pm 2.576\sqrt{\dfrac{0.29(1-0.29)}{1025}}\\\\=0.29\pm 2.576\sqrt{0.00020087804878}\\\\=0.29\pm2.576(0.01417)\\\\=0.29\pm0.03650192\\\\=(0.29-0.03650192,\ 0.29+0.03650192)\\\\=(0.25349808,\ 0.32650192)$

$\approx(0.25,\ 0.33)$

Thus, a 99% confidence interval estimate of the proportion of adults who use the Internet for shopping = (0.25,0.33)