1. **State the problem:** We want to find the least squares regression line fitting the data with $x$ as the independent variable and $y$ as the dependent variable. 2. **Formula for least squares line:** The regression line is given by $$y = b_0 + b_1 x$$ where $$b_1 = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ and $$b_0 = \frac{\sum y - b_1 \sum x}{n}$$ 3. **Given sums from the table:** $$\sum x = 798$$ $$\sum y = 819$$ $$\sum xy = 66045$$ $$\sum x^2 = 64722$$ $$n = 10$$ 4. **Calculate slope $b_1$:** $$b_1 = \frac{10 \times 66045 - 798 \times 819}{10 \times 64722 - 798^2}$$ Calculate numerator: $$10 \times 66045 = 660450$$ $$798 \times 819 = 653862$$ $$\text{Numerator} = 660450 - 653862 = 6588$$ Calculate denominator: $$10 \times 64722 = 647220$$ $$798^2 = 636804$$ $$\text{Denominator} = 647220 - 636804 = 10416$$ So, $$b_1 = \frac{6588}{10416}$$ Show cancellation: $$b_1 = \frac{\cancel{6588}}{\cancel{10416}} = \frac{549}{868} \approx 0.632$$ 5. **Calculate intercept $b_0$:** $$b_0 = \frac{819 - 0.632 \times 798}{10}$$ Calculate product: $$0.632 \times 798 \approx 504.336$$ Subtract: $$819 - 504.336 = 314.664$$ Divide: $$b_0 = \frac{314.664}{10} = 31.4664$$ 6. **Final regression line:** $$y = 31.4664 + 0.632 x$$ This line minimizes the sum of squared vertical distances (least squares) between the observed $y$ values and the predicted values from the line.