1. **Problem Statement:** We are given a regression equation $\hat{y} = 1 - 2x$ and data points for $x$ and $y$. We need to: a. Test at the 10% significance level if $x$ is useful for predicting $y$. b. Find a 90% confidence interval for the slope of the population regression line. 2. **Hypothesis Test for Slope:** - Null hypothesis $H_0$: slope $\beta = 0$ (x is not useful) - Alternative hypothesis $H_a$: slope $\beta \neq 0$ (x is useful) - Significance level $\alpha = 0.10$ 3. **Given Regression Equation:** $$\hat{y} = 1 - 2x$$ Here, slope $b = -2$. 4. **Test Statistic:** The test statistic for slope is $$t = \frac{b - 0}{SE_b}$$ where $SE_b$ is the standard error of the slope. 5. **Data Provided:** From the problem, $x = [3,1,2]$ and $y = [-4,0,-5]$. 6. **Calculate necessary statistics:** - Mean of $x$: $\bar{x} = \frac{3+1+2}{3} = 2$ - Mean of $y$: $\bar{y} = \frac{-4+0-5}{3} = -3$ 7. **Calculate residuals and standard error:** Predicted $\hat{y}$ values: - For $x=3$: $1 - 2(3) = 1 - 6 = -5$ - For $x=1$: $1 - 2(1) = 1 - 2 = -1$ - For $x=2$: $1 - 2(2) = 1 - 4 = -3$ Residuals $e = y - \hat{y}$: - $-4 - (-5) = 1$ - $0 - (-1) = 1$ - $-5 - (-3) = -2$ 8. **Calculate residual sum of squares (RSS):** $$RSS = 1^2 + 1^2 + (-2)^2 = 1 + 1 + 4 = 6$$ 9. **Calculate standard error of estimate:** Degrees of freedom $df = n - 2 = 3 - 2 = 1$ $$s = \sqrt{\frac{RSS}{df}} = \sqrt{6} \approx 2.449$$ 10. **Calculate $S_{xx}$:** $$S_{xx} = \sum (x_i - \bar{x})^2 = (3-2)^2 + (1-2)^2 + (2-2)^2 = 1 + 1 + 0 = 2$$ 11. **Calculate standard error of slope:** $$SE_b = \frac{s}{\sqrt{S_{xx}}} = \frac{2.449}{\sqrt{2}} = \frac{2.449}{1.414} \approx 1.732$$ 12. **Calculate t-statistic:** $$t = \frac{-2 - 0}{1.732} = \frac{-2}{1.732} \approx -1.1547$$ 13. **Critical t-value:** For $df=1$ and two-tailed $\alpha=0.10$, critical $t_{0.05,1} \approx 6.314$ (from t-distribution table). 14. **Decision:** Since $|t| = 1.1547 < 6.314$, we fail to reject $H_0$. There is insufficient evidence at 10% level to conclude $x$ is useful for predicting $y$. 15. **Confidence Interval for Slope:** $$b \pm t_{\alpha/2, df} \times SE_b = -2 \pm 6.314 \times 1.732$$ $$= -2 \pm 10.93$$ $$= (-12.93, 8.93)$$ 16. **Interpretation:** The 90% confidence interval for the slope is very wide and includes 0, consistent with the hypothesis test result. **Final answers:** - a. Insufficient evidence to conclude $x$ is useful at 10% significance. - b. 90% confidence interval for slope: $(-12.93, 8.93)$