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 .03 with 99% confidence if a) she uses a previous estimate of .58? b) she does not use any prior estimate?

Answer 1

Solution :

Given that,

a)

= 0.58

1 - = 1 - 0.58 = 0.42

margin of error = E = 0.03

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z _0.005 = 2.576

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

= (2.576 / 0.03)² * 0.58 * 0.42

= 1796

sample size = 1796

b)

= 1 - = 0.5

margin of error = E = 0.03

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z _0.005 = 2.576

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

= (2.576 / 0.03)² * 0.5 * 0.5

= 1843

sample size = 1843