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 $\beta_1 = 0$ (predictor is not useful). - Alternative hypothesis $H_a$: The slope $\beta_1 \neq 0$ (predictor 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: $$\sum (x_i - \bar{x})^2 = (-4.625)^2 + (0.375)^2 + (-2.625)^2 + (5.375)^2 + (-6.625)^2 + (1.375)^2 + (-0.625)^2 + (7.375)^2$$ $$= 21.3906 + 0.1406 + 6.8906 + 28.8906 + 43.8906 + 1.8906 + 0.3906 + 54.3906 = 157.875$$ Slope: $$b_1 = \frac{-133.5}{157.875} \approx -0.846$$ 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. 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 each $\hat{y}_i$ and residual: - For $x=10$: $\hat{y} = 94.875 - 0.846(10) = 86.415$, residual $= 92 - 86.415 = 5.585$ - For $x=15$: $\hat{y} = 94.875 - 0.846(15) = 82.185$, residual $= 81 - 82.185 = -1.185$ - For $x=12$: $\hat{y} = 94.875 - 0.846(12) = 84.693$, residual $= 84 - 84.693 = -0.693$ - For $x=20$: $\hat{y} = 94.875 - 0.846(20) = 77.955$, residual $= 74 - 77.955 = -3.955$ - For $x=8$: $\hat{y} = 94.875 - 0.846(8) = 88.787$, residual $= 85 - 88.787 = -3.787$ - For $x=16$: $\hat{y} = 94.875 - 0.846(16) = 81.339$, residual $= 80 - 81.339 = -1.339$ - For $x=14$: $\hat{y} = 94.875 - 0.846(14) = 83.031$, residual $= 84 - 83.031 = 0.969$ - For $x=22$: $\hat{y} = 94.875 - 0.846(22) = 75.263$, residual $= 80 - 75.263 = 4.737$ Calculate RSS: $$RSS = 5.585^2 + (-1.185)^2 + (-0.693)^2 + (-3.955)^2 + (-3.787)^2 + (-1.339)^2 + 0.969^2 + 4.737^2$$ $$= 31.204 + 1.404 + 0.480 + 15.642 + 14.337 + 1.793 + 0.939 + 22.435 = 88.234$$ 5. **Calculate standard error of slope:** $$s = \sqrt{\frac{RSS}{n-2}} = \sqrt{\frac{88.234}{6}} = \sqrt{14.7057} = 3.834$$ Standard error of slope: $$SE_{b_1} = \frac{s}{\sqrt{\sum (x_i - \bar{x})^2}} = \frac{3.834}{\sqrt{157.875}} = \frac{3.834}{12.57} = 0.305$$ 6. **Calculate t-statistic:** $$t = \frac{b_1 - 0}{SE_{b_1}} = \frac{-0.846}{0.305} = -2.774$$ 7. **Determine critical t-value:** Degrees of freedom $df = n-2 = 6$. At $\alpha=0.01$ two-tailed test, critical $t_{0.005,6} \approx 3.707$. 8. **Make decision:** Since $|t| = 2.774 < 3.707$, we fail to reject $H_0$. 9. **Conclusion:** At the 0.01 significance level, there is insufficient evidence to conclude that study time is a useful predictor of test score. **Final answer:** The data do not provide sufficient evidence to conclude the predictor variable is useful at $\alpha=0.01$.