Question

In a​ poll,1000 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 28​% 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.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = 0.28

1 - = 1 - 0.28 = 0.72

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.28 * 0.72) / 1000)

= 0.028

A 95% confidence interval for population proportion p is ,

± E

= 0.28  ± 0.028

= ( 0.252, 0.308 )

We are 95% confident that the true proportion of all adults in the region who never​ clothes-shop online between 0.252 and 0.308.