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 0.04 with

95% confidence if

(a) she uses a previous estimate of 0.28?

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

Answer 1

Solution :

Given that,

margin of error = E = 0.04

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

(a)

= 0.28

1 - = 1 - 0.28 = 0.72

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

= (1.96 / 0.04)² * 0.28 * 0.72

= 484.04 = 485

sample size = 485

(b)

= 0.5

1 - = 1 - 0.5 = 0.5

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

= (1.96 / 0.04)² * 0.5 * 0.5

= 600.25 = 601

sample size = 601