Question

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%

Answer 1

Solution :

Given that,

= 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

=317.5

Sample size = 318 rounded