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 \( \$ 21.5 \), and the variance is known to be \( \$ 30.25 \). 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 \( \$ 0.27 \) ? Round your answer up to the next integer.

To determine the sample size required to estimate the mean per capita income at a \( 98\% \) confidence level with a specified error, you can use the formula for sample size in estimating a population mean: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Where: - \( Z \) is the Z-score corresponding to the desired confidence level, - \( \sigma \) is the standard deviation (the square root of variance), - \( E \) is the margin of error. 1. For a \( 98\% \) confidence level, the Z-score is approximately \( 2.33 \). 2. Given variance \( \sigma^2 = 30.25 \), the standard deviation \( \sigma = \sqrt{30.25} \approx 5.5 \). 3. The margin of error \( E = 0.27 \). Now, plug these values into the sample size formula: \[ n = \left( \frac{2.33 \cdot 5.5}{0.27} \right)^2 \] Calculating this gives: \[ n = \left( \frac{12.815}{0.27} \right)^2 \approx (47.52)^2 \approx 2255.23 \] Finally, rounding up to the next integer, the required sample size is \( n = 2256 \). So, you would need a sample of at least 2256 people to estimate the mean per capita income with the desired confidence and error margin.