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)

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To find the lower bound of the 90% confidence interval for the average amount spent on food and beverage ($$ F \& B $$) by the country club members, we'll follow these steps:

### Given Data:
- Sample size ($$ n $$): 72
- Sample mean ($$ \bar{x} $$): \$398
- Sample standard deviation ($$ s $$): \$116
- Confidence level: 90%

### Steps to Calculate the Lower Bound:

1. **Calculate the Standard Error (SE):**
   $$
   SE = \frac{s}{\sqrt{n}} = \frac{116}{\sqrt{72}} \approx \frac{116}{8.485} \approx 13.67
   $$

2. **Determine the Critical Value ($$ t^* $$) for 90% Confidence:**
   - For a 90% confidence interval with $$ n - 1 = 71 $$ degrees of freedom, the critical value $$ t^* $$ is approximately 1.669. (This value can be found using a t-distribution table or calculator.)

3. **Calculate the Margin of Error (ME):**
   $$
   ME = t^* \times SE = 1.669 \times 13.67 \approx 22.82
   $$

4. **Find the Lower Bound of the Confidence Interval:**
   $$
   \text{Lower Bound} = \bar{x} - ME = 398 - 22.82 \approx 375.18
   $$

### Conclusion:
The **lower bound** of the 90% confidence interval for the average amount spent on $$ F \& B $$ is **\$375.18**.

**Answer:**  
The lower bound is $375.18
The lower bound of the 90% confidence interval for the average amount spent on $$ F \& B $$ is \$375.18.

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