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 :
[tex]$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$[/tex]
[tex]$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$[/tex]
0.401 < p < 0.459
Therefore, 95% confidence interval is from 0.401 to 0.459
