1. **Problem Statement:** We have data of temperature (°F) and cups of hot chocolate sold for 11 days. We want to find the linear trend line (line of best fit) and use it to estimate sales at 19°F and 45°F, and describe the association. 2. **Data:** Temperature $T$: 43, 34, 50, 28, 16, 20, 39, 42, 33, 24, 13 Cups sold $C$: 90, 100, 55, 120, 142, 135, 95, 80, 96, 122, 143 3. **Formula for linear regression line:** $$C = mT + b$$ where $m$ is slope and $b$ is y-intercept. 4. **Calculate means:** $$\bar{T} = \frac{43+34+50+28+16+20+39+42+33+24+13}{11} = \frac{342}{11} = 31.09$$ $$\bar{C} = \frac{90+100+55+120+142+135+95+80+96+122+143}{11} = \frac{1178}{11} = 107.09$$ 5. **Calculate slope $m$:** $$m = \frac{\sum (T_i - \bar{T})(C_i - \bar{C})}{\sum (T_i - \bar{T})^2}$$ Calculate numerator and denominator: Numerator: $$(43-31.09)(90-107.09) + (34-31.09)(100-107.09) + \cdots + (13-31.09)(143-107.09) = -2047.27$$ Denominator: $$(43-31.09)^2 + (34-31.09)^2 + \cdots + (13-31.09)^2 = 370.91$$ 6. **Calculate slope:** $$m = \frac{-2047.27}{370.91} = -5.52$$ 7. **Calculate intercept $b$:** $$b = \bar{C} - m \bar{T} = 107.09 - (-5.52)(31.09) = 107.09 + 171.62 = 278.71$$ 8. **Linear trend line equation:** $$C = -5.52T + 278.71$$ 9. **Estimate cups sold at 19°F:** $$C = -5.52(19) + 278.71 = -104.88 + 278.71 = 173.83$$ 10. **Estimate cups sold at 45°F:** $$C = -5.52(45) + 278.71 = -248.4 + 278.71 = 30.31$$ 11. **Association:** Since slope $m$ is negative, the association is **negative**. 12. **Strength:** The data points show a clear downward trend, so the association is **strong**. **Final answers:** - (a) Approximately 174 cups at 19°F. - (b) Approximately 30 cups at 45°F. - (c) Negative association. - (d) Strong association.