Question

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within

0.01 with 95% confidence if:

​(a) she uses a previous estimate of .42?

​(b) she does not use any prior​ estimates?

Answer 1

Solution :

Given that,

a)

= 0.42

1 - = 1 - 0.42 = 0.58

margin of error = E = 0.01

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.01)² * 0.42 * 0.58

= 9358.14

sample size = 9359

b)

= 0.50

1 - = 0.50

margin of error = E = 0.01

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.01)² * 0.50 * 0.50

= 9604

sample size = 9604