1. **State the problem:** We want to analyze if the air contamination levels differ with distance from the factory complex and understand the relationship between distance and contamination levels. 2. **Identify variables:** - Independent variable: Distance from the factory (miles). - Dependent variable: Air contamination levels (parts per billion). 3. **Hypotheses for ANOVA test:** - Null hypothesis $H_0$: There is no difference in contamination levels at different distances. - Alternative hypothesis $H_a$: There is a difference in contamination levels at different distances. 4. **ANOVA table given:** \begin{align*} \text{Source} & \quad SS & \quad df & \quad MS & \quad F & \quad p \\ \text{Accountable} & 584 & 9 & 80.89 & 1.24 & \text{(not given)} \\ \text{Unaccountable} & 130 & 2 & 65 & & \\ \text{Total} & 858 & 11 & & & \end{align*} 5. **Interpretation:** The F-value is 1.24, which is relatively low, suggesting weak evidence against $H_0$. Without the p-value, we cannot conclusively reject $H_0$. 6. **Scatterplot and slope:** - Data points are plotted at distances 1, 3, and 8 miles with corresponding contamination levels. - The freehand line drawn through the points slopes downward from left to right. - This indicates a **negative slope** because contamination levels decrease as distance increases. 7. **Correlation coefficient $r$:** - Given $r = -0.895$, which is a strong negative correlation. - This confirms the negative relationship between distance and contamination levels. 8. **Calculations for means and variances:** - Mean distance $\bar{x} = \frac{1 + 3 + 8}{3} = \frac{12}{3} = 4$ - Variance of $x$: $\frac{(1-4)^2 + (3-4)^2 + (8-4)^2}{3-1} = \frac{9 + 1 + 16}{2} = \frac{26}{2} = 13$ - Mean contamination $\bar{y} = \frac{27 + 20 + 10}{3} = \frac{57}{3} = 19$ - Variance of $y$: $\frac{(27-19)^2 + (20-19)^2 + (10-19)^2}{3-1} = \frac{64 + 1 + 81}{2} = \frac{146}{2} = 73$ - Standard deviation of $y = \sqrt{73} \approx 8.54$ 9. **Summary:** The negative correlation and downward slope indicate contamination decreases as distance increases. The ANOVA suggests no strong evidence of difference in means at the given significance level.