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 90​% confidence if ​

(a) she uses a previous estimate of 0.38​? ​n=

(b) she does not use any prior​ estimate? n=

Answer 1

Solution:

Given that,

a ) A previous estimate of 0.38​

= 0.38

1 - = 1 - 0.38 = 0.62

margin of error = E = 0.04

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= (1.645 / 0.04)² * 0.38 * 0.62

                         = 398.46

                        = 398

n = sample size = 398

b ) A prior estimate of 0.38​

= 0.50

1 - = 1 - 0.50 = 0.50

margin of error = E = 0.04

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= (1.645 / 0.04)² * 0.50 * 0.50

                         = 422.81

                        = 423

n = sample size = 423