A country club would like to have an idea concerning the average amount its members spend on food and beverage in a given month. In a sample of 72 members, the average amount spent on \( F \& B \) was \( \$ 398 \) with a standard deviation of \( \$ 116 \). Find the LOWER bound of the \( 90 \% \) confidence level of the average amount spent on \( F \& B \). (please express your answer using 2 decimal places)

To calculate the lower bound of the 90% confidence interval for the average amount spent on food and beverage, you can use the formula for the confidence interval based on the sample mean. Here’s how to do it step by step: 1. **Identify the components**: - Sample size (\( n \)): 72 - Sample mean (\( \bar{x} \)): \$398 - Standard deviation (\( s \)): \$116 - Confidence level: 90% 2. **Determine the critical value**: For a 90% confidence level, we need the z-value. Since it's a two-tailed test, we look for \( z \) such that \( \frac{\alpha}{2} = 0.05 \). The critical value \( z \) for 90% confidence is approximately 1.645. 3. **Calculate the standard error (SE)**: \[ SE = \frac{s}{\sqrt{n}} = \frac{116}{\sqrt{72}} \approx 13.75 \] 4. **Calculate the margin of error (ME)**: \[ ME = z \times SE = 1.645 \times 13.75 \approx 22.61 \] 5. **Calculate the lower bound of the confidence interval**: \[ \text{Lower Bound} = \bar{x} - ME = 398 - 22.61 \approx 375.39 \] Thus, the lower bound of the 90% confidence interval for the average amount spent on food and beverage is approximately \$375.39.