1. **Problem 2a: Find the linear regression equation for casino size and revenue.** Given paired data for Size (X) and Revenue (Y): Size: 160, 227, 140, 144, 161, 147, 141 Revenue: 189, 157, 140, 127, 123, 106, 101 2. **Formula for linear regression line:** $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. 3. **Calculate slope $m$:** $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ 4. **Calculate intercept $b$:** $$b = \frac{\sum y - m \sum x}{n}$$ 5. **Calculate sums:** $n=7$ $\sum x = 160 + 227 + 140 + 144 + 161 + 147 + 141 = 1120$ $\sum y = 189 + 157 + 140 + 127 + 123 + 106 + 101 = 943$ $\sum xy = 160\times189 + 227\times157 + 140\times140 + 144\times127 + 161\times123 + 147\times106 + 141\times101 = 30240 + 35639 + 19600 + 18288 + 19803 + 15582 + 14241 = 153393$ $\sum x^2 = 160^2 + 227^2 + 140^2 + 144^2 + 161^2 + 147^2 + 141^2 = 25600 + 51529 + 19600 + 20736 + 25921 + 21609 + 19881 = 184876$ 6. **Calculate slope $m$:** $$m = \frac{7 \times 153393 - 1120 \times 943}{7 \times 184876 - 1120^2} = \frac{1073751 - 1056160}{1294132 - 1254400} = \frac{17591}{39732} \approx 0.4427$$ 7. **Calculate intercept $b$:** $$b = \frac{943 - 0.4427 \times 1120}{7} = \frac{943 - 495.82}{7} = \frac{447.18}{7} \approx 63.88$$ 8. **Linear regression equation:** $$y = 0.4427x + 63.88$$ --- 9. **Problem 2b: Predict revenue for size 200.** Use the regression equation: $$y = 0.4427 \times 200 + 63.88 = 88.54 + 63.88 = 152.42$$ 10. **Interpretation:** The predicted revenue for a casino of size 200 thousand square feet is approximately 152.42. However, since 200 is within the range of observed sizes (140 to 227), the prediction is reasonable but still an estimate. --- 11. **Problem 3a: Find the linear correlation coefficient $r$ for time and height.** Given data: Time (X): 0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8 Height (Y): 0.0, 1.7, 3.1, 3.9, 4.5, 4.7, 4.6, 4.1, 3.3, 2.1 12. **Formula for correlation coefficient:** $$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2 - (\sum y)^2)}}$$ 13. **Calculate sums:** $n=10$ $\sum x = 0 + 0.2 + 0.4 + 0.6 + 0.8 + 1.0 + 1.2 + 1.4 + 1.6 + 1.8 = 7.0$ $\sum y = 0 + 1.7 + 3.1 + 3.9 + 4.5 + 4.7 + 4.6 + 4.1 + 3.3 + 2.1 = 32.0$ $\sum xy = 0\times0 + 0.2\times1.7 + 0.4\times3.1 + 0.6\times3.9 + 0.8\times4.5 + 1.0\times4.7 + 1.2\times4.6 + 1.4\times4.1 + 1.6\times3.3 + 1.8\times2.1 = 0 + 0.34 + 1.24 + 2.34 + 3.6 + 4.7 + 5.52 + 5.74 + 5.28 + 3.78 = 32.34$ $\sum x^2 = 0^2 + 0.2^2 + 0.4^2 + 0.6^2 + 0.8^2 + 1.0^2 + 1.2^2 + 1.4^2 + 1.6^2 + 1.8^2 = 0 + 0.04 + 0.16 + 0.36 + 0.64 + 1.0 + 1.44 + 1.96 + 2.56 + 3.24 = 11.4$ $\sum y^2 = 0^2 + 1.7^2 + 3.1^2 + 3.9^2 + 4.5^2 + 4.7^2 + 4.6^2 + 4.1^2 + 3.3^2 + 2.1^2 = 0 + 2.89 + 9.61 + 15.21 + 20.25 + 22.09 + 21.16 + 16.81 + 10.89 + 4.41 = 123.32$ 14. **Calculate numerator:** $$n\sum xy - \sum x \sum y = 10 \times 32.34 - 7.0 \times 32.0 = 323.4 - 224 = 99.4$$ 15. **Calculate denominator:** $$\sqrt{(10 \times 11.4 - 7.0^2)(10 \times 123.32 - 32.0^2)} = \sqrt{(114 - 49)(1233.2 - 1024)} = \sqrt{65 \times 209.2} = \sqrt{13598} \approx 116.6$$ 16. **Calculate $r$:** $$r = \frac{99.4}{116.6} \approx 0.8529$$ --- 17. **Problem 3b: Conclusion about linear correlation.** Since $r \approx 0.85$, there is a strong positive linear correlation between time and height. --- 18. **Problem 3c: Possible mistake without scatterplot.** Without a scatterplot, one might assume the relationship is perfectly linear and ignore the curvature or peak in the data, leading to incorrect conclusions about the motion (e.g., ignoring that the ball reaches a maximum height and then falls).