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 $20.4, and the standard deviation is known to be $8.8. How large of a sample would be required in order to estimate the mean per capita income at the 98% level of confidence with an error of at most $0.54$? Round your answer up to the next integer.

Accepted Solution

Answer:1442Step-by-step explanation:Given,Mean is,[tex]\mu = 20.4[/tex]Standard deviation,[tex]\sigma = 8.8[/tex],Margin of error(E) = 0.54,For 98% level of confidence, [tex]z_{\frac{\alpha}{2}} = 2.33[/tex]Thus, the sample size would be,[tex]n=[\frac{z_{\frac{\alpha}{2}}\times \sigma}{E}]^2[/tex][tex]=[\frac{2.33\times 8.8}{0.54}]^2[/tex][tex]=1441.74902606[/tex][tex]\approx 1442[/tex]