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

0.040.04

with

9999​%

confidence if

​(a) she uses a previous estimate of

0.560.56​?

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

Answer 1

(A)Solution :

Given that,

= 0.56

1 - = 1 - 0.56= 0.44

margin of error = E =0.04

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576 ( Using z table )

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

= (2.576 / 0.04)2 * 0.56 * 0.44

=1021.9

Sample size = 1022

(B)Solution :

Given that,

= 0.5

1 - = 1 - 0.5= 0.5

margin of error = E =0.04

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576 ( Using z table )

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

= (2.576 / 0.04)2 * 0.5* 0.5

=1036.84

Sample size = 1037