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 99% confidence if

​(a) she uses a previous estimate of 0.28​?

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

Answer 1

Solution :

Given that,

= 0.28

1 - = 1 - 0.28 = 0.72

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 = 2.58    ( Using z table )

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

= (2.58 / 0.04)2 * 0.28 * 0.72

=839

Sample size = 839

B

Given that,

=0.5

1 - = 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 = 2.58    ( Using z table )

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

= (2.58 / 0.04)2 * 0.5 * 0.5

=1040

Sample size = 1040