Question

Among 500 randomly selected adults were surveyed, there are 25% of them said that they used the Internet for shopping at least a few times a year. Construct a 99% confidence interval estimate of the percentage of adults who use the Internet for shopping.

Answer 1

Solution :

Given that,

n = 500

Point estimate = sample proportion = = 0.25

1 -   = 1- 0.25 =0.75

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.25*0.75) / 500)

E =0.0499

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.25-0.0499  < p <0.25+ 0.0499

0.2001< p < 0.2999

The 99% confidence interval for the population proportion p is : 0.2001, 0.2999