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 $20, and the variance is known to be $123.21. How large of a sample would be required in order to estimate the mean per capita income at the 99% level of confidence with an error of at most $0.56? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

Population standard deviation = = 11.1

Margin of error = E = 0.56

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z_/2* / E) ²

n = (2.576 *11.1 / 0.56)²

n = 2607.12

n = 2608

Sample size = 2608