Answer: (0.25,0.33)
Step-by-step explanation:
A 99% confidence interval for population proportion is given by:-
[tex]\hat{p}\pm 2.576\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex], where [tex]\hat{p}[/tex] = sample proportion, [tex]n[/tex] = sample size.
Given: [tex]n=1025, \hat{p}=0.29[/tex]
A 99% confidence interval estimate of the proportion of adults who use the Internet for shopping:
[tex]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)[/tex]
[tex]\approx(0.25,\ 0.33)[/tex]
Thus, a 99% confidence interval estimate of the proportion of adults who use the Internet for shopping = (0.25,0.33)
