Question

In a recent Dew Research Poll, 1000 U.S. adults were asked about their online versus in-store clothes shopping. One finding was that 32% of respondents never clothes-shop online. Construct and interpret a 95% confidence interval for the proportion of all U.S. adults who never clothes-shop online.

Answer 1

Solution :

Given that,

n = 1000

Point estimate = sample proportion = = 32%=0.32

1 -   = 1- 0.32 =0.68

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.32*0.68) /1000 )

E = 0.029

A 95% confidence interval is ,

- E < p < + E

0.32 - 0.029 < p < 0.32 +0.029

0.291< p < 0.349