1. **State the problem:** We are given data on advertising costs (X) and sales (Y) and asked to find the regression equation, correlation coefficient, estimate sales for X=15, calculate the standard error of estimate, and find the 99% confidence range for the estimate. 2. **Regression equation:** The linear regression equation is $$\hat{Y} = a + bX$$ where $$b = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2}$$ $$a = \frac{\sum Y - b\sum X}{n}$$ 3. **Calculate slope $b$:** $$b = \frac{7(113,100) - 245 \times 3,125}{7(9,275) - 245^2} = \frac{791,700 - 765,625}{64,925 - 60,025} = \frac{26,075}{4,900} \approx 5.3214$$ 4. **Calculate intercept $a$:** $$a = \frac{3,125 - 5.3214 \times 245}{7} = \frac{3,125 - 1,303.743}{7} = \frac{1,821.257}{7} \approx 260.1796$$ 5. **Regression equation:** $$\hat{Y} = 260.18 + 5.32X$$ 6. **Calculate Pearson correlation coefficient $r$:** $$r = \frac{n\sum XY - (\sum X)(\sum Y)}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$$ Calculate denominator parts: $$n\sum Y^2 - (\sum Y)^2 = 7 \times 1,417,275 - 3,125^2 = 9,920,925 - 9,765,625 = 155,300$$ $$r = \frac{26,075}{\sqrt{4,900 \times 155,300}} = \frac{26,075}{\sqrt{760,970,000}} = \frac{26,075}{27,585.68} \approx 0.9452$$ 7. **Interpretation:** $r \approx 0.95$ indicates a very strong positive linear relationship. 8. **Estimate sales for $X=15$:** $$\hat{Y} = 260.18 + 5.32 \times 15 = 260.18 + 79.8 = 339.98$$ Sales estimate is 339.98 (thousands). 9. **Calculate standard error of estimate $s_e$:** $$s_e = \sqrt{\frac{\sum Y^2 - a\sum Y - b\sum XY}{n-2}}$$ Plug in values: $$s_e = \sqrt{\frac{1,417,275 - 260.1796 \times 3,125 - 5.3214 \times 113,100}{5}} = \sqrt{\frac{1,417,275 - 813,061.25 - 601,850.34}{5}} = \sqrt{\frac{2,363.41}{5}} = \sqrt{472.682} \approx 21.741$$ 10. **Calculate 99% confidence range for estimate at $X=15$:** Degrees of freedom $df = 7 - 2 = 5$, critical t-value $t_{0.005,5} = 4.032$ Mean of X: $$\bar{X} = \frac{245}{7} = 35$$ Margin of error $E$: $$E = t \times s_e \times \sqrt{1 + \frac{1}{n} + \frac{(X_0 - \bar{X})^2}{\sum X^2 - \frac{(\sum X)^2}{n}}}$$ Calculate denominator inside root: $$\sum X^2 - \frac{(\sum X)^2}{n} = 9,275 - \frac{245^2}{7} = 9,275 - 8,575 = 700$$ Calculate root term: $$\sqrt{1 + \frac{1}{7} + \frac{(15 - 35)^2}{700}} = \sqrt{1 + 0.1429 + \frac{400}{700}} = \sqrt{1 + 0.1429 + 0.5714} = \sqrt{1.7143} \approx 1.3093$$ Calculate $E$: $$E = 4.032 \times 21.741 \times 1.3093 = 87.660 \times 1.3093 \approx 114.77$$ 11. **Confidence interval:** Lower limit: $$339.98 - 114.77 = 225.21$$ Upper limit: $$339.98 + 114.77 = 454.75$$ **Conclusion:** At 99% confidence, sales for advertising cost 15 will be between 225.21 and 454.75 (thousands), which matches the user's calculation and is correct.