In: Statistics and Probability
Find the sample size needed to estimate the percentage of Virginia residents who are left-handed. Use a Margin of Error of 3% and use a confidence level of 90% for both situations as follows.
a. Assume phat is unknown
b. Based on prior studies, use phat= 12%
Solution :
Given that,
= 0.5 ( assume 0.5)
1 - = 1 - 0.5= 0.5
margin of error = E = 3% = 0.03
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 / 0.03)2 * 0.5 * 0.5
=751.67
Sample size = 752
(B)
Solution :
Given that,
=0.12
1 - = 1 - 0.12= 0.88
margin of error = E = 3% = 0.03
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 / 0.03)2 * 0.12 * 0.88
=317.5069
Sample size = 318