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 $\$ 30.8$, and the standard deviation is known to be $\$ 8.2$. How large of a sample would be required in order to estimate the mean per capita income at the $95 \%$ level of confidence with an error of at most $\$ 0.39$ ? Round your answer up to the next integer. Answer How to enter your answer (opens in new window) Tables Keypad Keyboard Shortcuts Submit Answer O 2023 Hankes Leaming

Step 1 ：The problem is asking for the sample size needed to estimate the mean per capita income with a certain level of confidence and margin of error. This is a problem of sample size determination for estimating a population mean. The formula for this is: \(n = (Z*σ/E)^2\), where: n is the sample size, Z is the Z-score corresponding to the desired level of confidence, σ is the standard deviation of the population, E is the desired margin of error.

Step 2 ：In this case, we know that: The level of confidence is 95%, so the Z-score is approximately 1.96 (this value can be found in a standard Z-table), The standard deviation σ is $8.2, The desired margin of error E is $0.39.

Step 3 ：We can plug these values into the formula to find the required sample size. Since we can't have a fraction of a sample, we'll need to round up to the next whole number.

Step 4 ：Z = 1.96, σ = 8.2, E = 0.39, n = 1699

Step 5 ：Final Answer: The required sample size to estimate the mean per capita income at the 95% level of confidence with an error of at most $0.39 is \(\boxed{1699}\).