1. **State the problem:** We have data for number of absences $x$ and final grades $y$. We want to find the linear regression equation $\hat{y} = mx + b$ that predicts final grade from absences. 2. **Formula and rules:** The regression line is $\hat{y} = mx + b$ where $m$ is slope and $b$ is y-intercept. 3. **Calculate means:** $$\bar{x} = \frac{0+1+2+3+4+5+6+7+8+9}{10} = 4.5$$ $$\bar{y} = \frac{88.8+85.9+82.9+80.4+77.4+73.0+63.4+67.9+64.9+62.0}{10} = 74.16$$ 4. **Calculate slope $m$:** $$m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$\sum (x_i - 4.5)(y_i - 74.16) = (0-4.5)(88.8-74.16) + (1-4.5)(85.9-74.16) + \cdots + (9-4.5)(62.0-74.16) = -370.65$$ Calculate denominator: $$\sum (x_i - 4.5)^2 = (0-4.5)^2 + (1-4.5)^2 + \cdots + (9-4.5)^2 = 82.5$$ So, $$m = \frac{-370.65}{82.5} = -4.494$$ 5. **Calculate intercept $b$:** $$b = \bar{y} - m\bar{x} = 74.16 - (-4.494)(4.5) = 74.16 + 20.22 = 94.38$$ 6. **Regression equation:** $$\hat{y} = -4.494x + 94.38$$ 7. **Interpretations:** - A. For a final score of zero, predicted absences is not directly from this regression (predicts $y$ from $x$). - B. For every unit change in final grade, absences fall by about $\frac{1}{4.494} \approx 0.222$ days (inverse slope). - C. For zero absences, predicted final score is $94.38$. - D. For every day absent, final grade falls by about $4.494$ points. - E. It is appropriate to interpret the slope here as the data shows a linear trend. 8. **Predicted final grade for given $x$:** Use the formula $\hat{y} = -4.494x + 94.38$. **Final answers:** - Regression line: $\hat{y} = -4.494x + 94.38$ - For zero absences, predicted grade: $94.4$ - For every day absent, grade falls by $4.494$ - For every unit change in grade, absences change by about $0.222$ days