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 0.01 with 90& confidence if

​(a) she uses a previous estimate of 0.42

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

Answer 1

Solution :

Given that,

= 0.42

1 - = 1 - 0.42 = 0.58

margin of error = E = 001

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.01)2 * 0.42 * 0.58

=6591.87

Sample size = 6592

(B)

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 001

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.01)2 * 0.5 * 0.5

=6765.06

Sample size = 6765