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? N=

b. She does not use any prior estimates? N=

Answer 1

Solution :

Given that,

= 0.42

1 - = 1 - 0.42= 0.58

margin of error = E = 0.01

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )   

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

= (1.645 / 0.01)2 * 0.42 * 0.58

=6592

Sample size = 6592

(b)

Solution :

Given that,

= 0.5 ( assume 0.5)

1 - = 1 - 0.5= 0.5

margin of error = E = 0.01

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )   

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

= (1.645 / 0.01)2 * 0.5 * 0.5

=6765

Sample size = 6765