Question

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

Answer 1

Solution :

Given that,

= 0.32

1 - = 1 - 0.32 = 0.68

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 = Z0.025 = 1.96 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.03)2 * 0.32* 0.68

= 929

Sample size =929

(B)

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

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 = Z0.025 = 1.96 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.03)2 * 0.5* 0.5

= 1067.111

Sample size =1068