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 $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.

Answer 1

Answer:

1442

Step-by-step explanation:

Given,

Mean is,

$\mu = 20.4$

Standard deviation,

$\sigma = 8.8$,

Margin of error(E) = 0.54,

For 98% level of confidence, $z_{\frac{\alpha}{2}} = 2.33$

Thus, the sample size would be,

$n=[\frac{z_{\frac{\alpha}{2}}\times \sigma}{E}]^2$

$=[\frac{2.33\times 8.8}{0.54}]^2$

$=1441.74902606$

$\approx 1442$