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 0.03 with 95 % confidence if

(a) she uses a previous estimate of 0.34 ?

(b) she does not use any prior estimates?

Answer 1

Solution :

Given that,

Margin of error = E = 0.03

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)

= 0.34

1 - = 1 - 0.34 = 0.66

Sample size = (  Z_/2 / E)² * * (1 - )

= (1.96 / 0.03)² * 0.34 * 0.66

= 957.83 = 958

Sample size = n = 958

(b)

= 0.5

1 - = 1 - 0.5 = 0.5

Sample size = (  Z_/2 / E)² * * (1 - )

= (1.96 / 0.03)² * 0.5 * 0.5

= 1067.11 = 1068

Sample size = n = 1068