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.

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