Question 8 of 14, Step 1 of 1 An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be \( \$ 32.6 \) for a random sample of 1699 people. Assume the population standard deviation is known to be \( \$ 8.8 \). Construct the \( 95 \% \) confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal

To construct a 95% confidence interval for the mean per capita income, we'll use the formula for the confidence interval: \[ CI = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \(\bar{x} = 32.6\) (sample mean) - \(Z = 1.96\) (Z-score for 95% confidence) - \(\sigma = 8.8\) (population standard deviation) - \(n = 1699\) (sample size) First, we calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{8.8}{\sqrt{1699}} \approx \frac{8.8}{41.23} \approx 0.213 \] Next, we find the margin of error (ME): \[ ME = Z \times SE = 1.96 \times 0.213 \approx 0.417 \] Now, we can construct the confidence interval: \[ CI = 32.6 \pm 0.417 \] Calculating the lower and upper bounds: - Lower bound: \(32.6 - 0.417 \approx 32.2\) - Upper bound: \(32.6 + 0.417 \approx 33.0\) Therefore, the 95% confidence interval for the mean per capita income in thousands of dollars is approximately \((32.2, 33.0)\).