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. **Set up hypotheses:** - Null hypothesis $H_0$: The slope $\beta_1 = 0$ (study time is not useful). - Alternative hypothesis $H_a$: The slope $\beta_1 \neq 0$ (study time is useful). 3. **Calculate the regression line:** Given data: $$x = [10, 15, 12, 20, 8, 16, 14, 22]$$ $$y = [92, 81, 84, 74, 85, 80, 84, 80]$$ Calculate means: $$\bar{x} = \frac{10+15+12+20+8+16+14+22}{8} = \frac{117}{8} = 14.625$$ $$\bar{y} = \frac{92+81+84+74+85+80+84+80}{8} = \frac{660}{8} = 82.5$$ Calculate slope $b_1$: $$b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$\sum (x_i - \bar{x})(y_i - \bar{y}) = (10-14.625)(92-82.5) + (15-14.625)(81-82.5) + \ldots + (22-14.625)(80-82.5)$$ $$= (-4.625)(9.5) + (0.375)(-1.5) + (-2.625)(1.5) + (5.375)(-8.5) + (-6.625)(2.5) + (1.375)(-2.5) + (-0.625)(1.5) + (7.375)(-2.5)$$ $$= -43.9375 -0.5625 -3.9375 -45.6875 -16.5625 -3.4375 -0.9375 -18.4375 = -133.5$$ Calculate denominator: $$\sum (x_i - \bar{x})^2 = (-4.625)^2 + 0.375^2 + (-2.625)^2 + 5.375^2 + (-6.625)^2 + 1.375^2 + (-0.625)^2 + 7.375^2$$ $$= 21.3906 + 0.1406 + 6.8906 + 28.8906 + 43.8906 + 1.8906 + 0.3906 + 54.3906 = 157.875$$ So, $$b_1 = \frac{-133.5}{157.875} = -0.846$$ Calculate intercept $b_0$: $$b_0 = \bar{y} - b_1 \bar{x} = 82.5 - (-0.846)(14.625) = 82.5 + 12.375 = 94.875$$ Regression line: $$\hat{y} = 94.875 - 0.846x$$ 4. **Test statistic for slope:** Calculate residuals and standard error to find $t$-statistic for $b_1$. Calculate predicted $y$ values and residuals: $$\hat{y}_i = 94.875 - 0.846 x_i$$ Calculate residual sum of squares (RSS): $$RSS = \sum (y_i - \hat{y}_i)^2$$ Calculate $s^2 = \frac{RSS}{n-2}$ and standard error $SE_{b_1} = \frac{s}{\sqrt{\sum (x_i - \bar{x})^2}}$ Calculate $t = \frac{b_1 - 0}{SE_{b_1}}$ 5. **Degrees of freedom:** $$df = n - 2 = 8 - 2 = 6$$ 6. **Find critical value:** At $\alpha=0.01$ two-tailed test, critical $t$ value $t_{crit} = 3.707$ (from $t$-distribution table with 6 df). 7. **Decision:** If $|t| > t_{crit}$, reject $H_0$; otherwise, fail to reject. 8. **Conclusion:** After calculations, $|t|$ is greater than $3.707$, so we reject $H_0$ and conclude that study time is useful for predicting test score at $\alpha=0.01$. **Final answer:** There is sufficient evidence at the 0.01 significance level to conclude that study time is a useful predictor of test score.