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 $34.6, and the variance is known to be $67.24. 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 $0.33? Round your answer up to the next integer.

Answer 1

To estimate the mean per capita income at the 80% level of confidence with an error of at most $0.33, a sample size of 1011 is required.

Step-by-step explanation:

To estimate the mean per capita income with a given error and confidence level, we can use the formula for the sample size:

n = (Z * sigma / E)^2

Given that the mean income is $34.6 (mu), the population variance is known to be $67.24 (sigma^2), the desired error is $0.33 (E), and the confidence level is 80%, we can substitute these values into the formula:

n = (Z * sigma / E)^2 = (Z * sqrt(sigma^2) / E)^2 = (Z * sqrt(67.24) / 0.33)^2

Using a Z-score table, we can find the Z-score corresponding to an 80% confidence level. Let's assume it is 1.28. Plugging in the values:

n = (1.28 * sqrt(67.24) / 0.33)^2

n = (1.28 * 8.2 / 0.33)^2

n = (10.496 / 0.33)^2

n = 31.8^2

n = 1010.24

Rounding up to the next integer, the sample size required is 1011.