1. **State the problem:** We want to determine if the predictor variable (study time $x$) is useful for predicting the response variable (test score $y$) at significance level $\alpha=0.01$. 2. **Formula and assumptions:** We use linear regression inference. The null hypothesis $H_0$: slope $\beta_1=0$ (no relationship). The alternative hypothesis $H_a$: slope $\beta_1\neq 0$ (predictor is useful). 3. **Calculate regression parameters:** - Data points: $x = [10,15,12,20,8,16,14,22]$, $y = [92,81,84,74,85,80,84,80]$ - Calculate means: $$\bar{x} = \frac{10+15+12+20+8+16+14+22}{8} = 14.625$$ $$\bar{y} = \frac{92+81+84+74+85+80+84+80}{8} = 82.5$$ 4. **Calculate slope $b_1$ and intercept $b_0$:** $$b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$(10-14.625)(92-82.5) + (15-14.625)(81-82.5) + \cdots + (22-14.625)(80-82.5) = -111.75$$ Calculate denominator: $$(10-14.625)^2 + (15-14.625)^2 + \cdots + (22-14.625)^2 = 110.875$$ So, $$b_1 = \frac{-111.75}{110.875} \approx -1.007$$ Calculate intercept: $$b_0 = \bar{y} - b_1 \bar{x} = 82.5 - (-1.007)(14.625) = 82.5 + 14.73 = 97.23$$ 5. **Form regression equation:** $$\hat{y} = 97.23 - 1.007x$$ 6. **Test significance of slope:** Calculate standard error of slope $SE_{b_1}$ and t-statistic: - Calculate residuals and sum of squared errors (SSE): $$SSE = \sum (y_i - \hat{y_i})^2 = 178.5$$ - Calculate standard error: $$SE_{b_1} = \sqrt{\frac{SSE}{(n-2)\sum (x_i - \bar{x})^2}} = \sqrt{\frac{178.5}{6 \times 110.875}} \approx 0.164$$ - Calculate t-statistic: $$t = \frac{b_1 - 0}{SE_{b_1}} = \frac{-1.007}{0.164} \approx -6.14$$ 7. **Decision:** Degrees of freedom $df = n-2 = 6$. Critical t-value for two-tailed test at $\alpha=0.01$ is approximately $\pm 3.707$. Since $|t|=6.14 > 3.707$, we reject $H_0$. 8. **Conclusion:** There is sufficient evidence at the 0.01 significance level to conclude that study time is useful for predicting test score. **Final answer:** The predictor variable (study time) is useful for predicting the response variable (test score) at $\alpha=0.01$.