1. **State the problem:** We want to find a 90% confidence interval for the mean weekend customer spend amount at Forest City Casino based on the given sample data. 2. **Identify the type of problem:** Since we are estimating the average amount spent (a numerical value), this is a problem involving the **mean**. 3. **Formula for confidence interval for the mean:** $$\text{CI} = \bar{x} \pm t^* \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $s$ is the sample standard deviation - $n$ is the sample size - $t^*$ is the critical value from the t-distribution for 90% confidence and $n-1$ degrees of freedom 4. **Calculate sample size $n$:** Count the number of data points. There are 150 data points. 5. **Calculate sample mean $\bar{x}$:** Sum all values and divide by 150. 6. **Calculate sample standard deviation $s$:** Use the formula $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ 7. **Find $t^*$ value:** For 90% confidence and $df = 149$, $t^* \approx 1.655$ (from t-tables or software). 8. **Compute margin of error:** $$\text{ME} = t^* \times \frac{s}{\sqrt{n}}$$ 9. **Calculate confidence interval:** $$\text{Lower bound} = \bar{x} - \text{ME}$$ $$\text{Upper bound} = \bar{x} + \text{ME}$$ 10. **Perform calculations with the data:** - Sum of all spends $= 83075$ - Sample mean $\bar{x} = \frac{83075}{150} = 553.83$ - Sample standard deviation $s \approx 1703.44$ - Margin of error $= 1.655 \times \frac{1703.44}{\sqrt{150}} \approx 229.88$ 11. **Final 90% confidence interval:** $$553.83 - 229.88 = 323.95$$ $$553.83 + 229.88 = 783.71$$ **Interpretation:** We are 90% confident that the true mean weekend customer spend amount is between $323.95 and $783.71. **Answers:** - Blank #1: MEAN - Blank #2: 323.95 - Blank #3: 783.71