Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $31.3$⁢31.3, and the standard deviation is known to be $9.4$⁢9.4. How large of a sample would be required in order to estimate the mean per capita income at the 85%85% level of confidence with an error of at most $0.46$⁢0.46? Round your answer up to the next integer.

Answer 1

Solution

standard deviation = =$9.4

Margin of error = E = $0.46

At 5% confidence level the z is ,

= 1 - 85% = 1 - 0.5 = 0.15

/ 2 = 0.075

Z/2 = Z0.075 = 1.44( Using z table  

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

n = ( 1.44* 9.4/ 0.46 )2

n =865.89

Sample size = n =866