A survey of 66 randomly selected students finds that they spend a mean of \( \$ 395 \) per semester. Assume that the population standard deviation is \( \$ 54 \) per month. Confidence Interval: What is the \( 90 \% \) confidence interval to estimate the population mean? Round your answers to two decimal places. \[ \text {. } \]

**Step 1. Identify the given values** We are given: - Sample size: \( n = 66 \) - Sample mean: \( \bar{x} = 395 \) - Population standard deviation: \( \sigma = 54 \) - Confidence level: \( 90\% \) which implies a critical value \( z_{\alpha/2} \) corresponding to \( \alpha = 0.10 \) (i.e. \( \alpha/2 = 0.05 \)). The critical value is \( z_{0.05} \approx 1.645 \). **Step 2. Compute the standard error of the mean** The standard error (SE) is given by: \[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{54}{\sqrt{66}} \] Calculating: \[ \sqrt{66} \approx 8.124 \] \[ \text{SE} \approx \frac{54}{8.124} \approx 6.646 \] **Step 3. Compute the margin of error** The margin of error (ME) is calculated as: \[ \text{ME} = z_{\alpha/2} \times \text{SE} = 1.645 \times 6.646 \approx 10.935 \] **Step 4. Construct the confidence interval** The \( 90\% \) confidence interval for the population mean is: \[ \bar{x} \pm \text{ME} \] Thus, the lower limit is: \[ 395 - 10.935 \approx 384.07 \] and the upper limit is: \[ 395 + 10.935 \approx 405.94 \] **Step 5. Write the final answer** The \( 90\% \) confidence interval to estimate the population mean is: \[ (384.07, \, 405.94) \]