Question

In a​ poll, adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that ​% of respondents never​ clothes-shop online. Find and interpret a ​% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Answer 1

The question is incomplete. The complete question is :

In a​ poll, 1100 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 43​% of respondents never​ clothes-shop online. Find and interpret a ​95% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Solution :

95% confidence interval for p is :

$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$

0.401 < p < 0.459

Therefore, 95% confidence interval is from 0.401 to 0.459