Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $21.3, and the variance is known to be 33.91. How large of a sample would be required in order to estimate the mean per capita income at the 80% level of confidence with an error of at most $0.5? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

Variance = 33.91

Population standard deviation = = 33.91 = 5.82

Margin of error = E = 0.5

At 80% confidence level the z is,

= 1 - 80%

= 1 - 0.80 = 0.20

/2 = 0.10

Z/2 = Z0.10 = 1.282  

sample size = n = [Z/2* / E] 2

n = [1.282 * 5.82 / 0.5 ]2

n = 222.68

Sample size = n = 223