Question

In a Pew Research Center poll, 2559 of 3011 adults surveyed said that they use the Internet. Construct a 95% confidence interval estimate of the proportion of all adults who use the Internet.

Answer 1

Solution :

Given that,

n = 3011

x = 2559

Point estimate = sample proportion = = x / n = 2559/3011=0.850

1 -   = 1-0.850 =0.15

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 (((0.85*0.15) /3011 )

= 0.013

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.85-0.013 < p < 0.85+0.013

0.837 < p <0.863