1. **Problem Statement:** We have bivariate data with an explanatory variable $x$ (from 1 to 20) and a response variable $y$. We want to analyze the relationship between $x$ and $y$ through several statistical measures and interpretations. 2. **Part A: Craft a Scenario** Suppose $x$ represents the number of weeks a student studies for a test, and $y$ represents the student's test score out of 60. This scenario models how study time affects test performance. 3. **Part B: Calculate Coefficient of Determination ($R^2$) and Correlation Coefficient ($r$)** - First, calculate means: $$\bar{x} = \frac{1+2+\cdots+20}{20} = 10.5$$ - Calculate $$\bar{y} = \frac{8+11+15+18+22+22+25+26+30+30+35+39+40+44+45+45+46+49+50+52}{20} = 33.7$$ - Calculate sums for variance and covariance: $$S_{xx} = \sum (x_i - \bar{x})^2 = 665$$ $$S_{yy} = \sum (y_i - \bar{y})^2 = 3236.6$$ $$S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = 1467.5$$ - Correlation coefficient: $$r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{1467.5}{\sqrt{665 \times 3236.6}} \approx 0.999$$ - Coefficient of determination: $$R^2 = r^2 \approx (0.999)^2 = 0.998$$ Interpretation: $r \approx 0.999$ indicates a very strong positive linear relationship. $R^2 = 0.998$ means 99.8% of the variation in test scores is explained by weeks studied. 4. **Part C: Comment on Association** There is a very strong positive association between weeks studied and test scores. As study time increases, test scores increase almost perfectly linearly. 5. **Part D: Least-Squares Regression Line** - Slope: $$b = \frac{S_{xy}}{S_{xx}} = \frac{1467.5}{665} \approx 2.206$$ - Intercept: $$a = \bar{y} - b \bar{x} = 33.7 - 2.206 \times 10.5 \approx 33.7 - 23.163 = 10.537$$ - Regression equation: $$\hat{y} = 10.537 + 2.206x$$ Where $x$ = weeks studied, $\hat{y}$ = predicted test score. 6. **Part E: Residual for $x=16$** - Predicted score: $$\hat{y} = 10.537 + 2.206 \times 16 = 10.537 + 35.296 = 45.833$$ - Actual score at $x=16$ is 45. - Residual: $$e = y - \hat{y} = 45 - 45.833 = -0.833$$ Interpretation: The actual score is 0.833 points less than predicted, indicating a slight underperformance compared to the model's expectation. **Final answers:** - $r \approx 0.999$ - $R^2 \approx 0.998$ - Regression line: $\hat{y} = 10.537 + 2.206x$ - Residual at $x=16$ is $-0.833$