1. **State the problem:** We want to analyze the relationship between Daily High Temperature ($X$) and Attendance ($Y$) at town meetings using simple linear regression. 2. **Null hypothesis:** - Formal: $H_0: \beta = 0$ (no linear relationship between temperature and attendance). - Lay terms: Temperature does not predict attendance. 3. **Calculate regression parameters and correlation $r$:** Given data: $$X = [15.3, 24.9, 9.0, 35.0, 36.1, 22.1, 22.7, 24.5]$$ $$Y = [35, 80, 10, 75, 85, 75, 70, 80]$$ Calculate means: $$\bar{X} = \frac{15.3 + 24.9 + 9.0 + 35.0 + 36.1 + 22.1 + 22.7 + 24.5}{8} = 23.44$$ $$\bar{Y} = \frac{35 + 80 + 10 + 75 + 85 + 75 + 70 + 80}{8} = 60$$ Calculate sums for slope $b$ and correlation $r$: $$S_{XY} = \sum (X_i - \bar{X})(Y_i - \bar{Y}) = 1021.94$$ $$S_{XX} = \sum (X_i - \bar{X})^2 = 399.43$$ $$S_{YY} = \sum (Y_i - \bar{Y})^2 = 4625$$ Slope: $$b = \frac{S_{XY}}{S_{XX}} = \frac{1021.94}{399.43} = 2.56$$ Intercept: $$a = \bar{Y} - b\bar{X} = 60 - 2.56 \times 23.44 = 60 - 60.0 = 0.0$$ Correlation: $$r = \frac{S_{XY}}{\sqrt{S_{XX} S_{YY}}} = \frac{1021.94}{\sqrt{399.43 \times 4625}} = 0.75$$ Regression line: $$\hat{Y} = 0.0 + 2.56X$$ 4. **Test significance at $\alpha=0.05$:** Degrees of freedom: $n-2=6$ Calculate $t$-statistic: $$t = r \sqrt{\frac{n-2}{1-r^2}} = 0.75 \sqrt{\frac{6}{1-0.75^2}} = 0.75 \sqrt{\frac{6}{1-0.5625}} = 0.75 \sqrt{13.71} = 2.78$$ Critical $t$ for 6 df at 0.05 two-tailed is about 2.447. Since $2.78 > 2.447$, reject $H_0$; slope is significant. 5. **Explain findings:** - $r=0.75$ indicates a strong positive linear relationship. - $R^2 = r^2 = 0.56$ means 56% of attendance variability is explained by temperature. - Slope $b=2.56$ means for each 1°F increase, attendance increases by about 2.56 people. 6. **95% confidence interval for slope $b$:** Standard error of slope: $$SE_b = \frac{s}{\sqrt{S_{XX}}}$$ Where $s$ is standard error of estimate: $$s = \sqrt{\frac{\sum (Y_i - \hat{Y_i})^2}{n-2}} = 18.3$$ (calculated from residuals) Calculate: $$SE_b = \frac{18.3}{\sqrt{399.43}} = 0.916$$ $t_{0.025,6} = 2.447$ CI: $$b \pm t SE_b = 2.56 \pm 2.447 \times 0.916 = (0.31, 4.81)$$ Lay terms: We are 95% confident the true slope is between 0.31 and 4.81, confirming a positive relationship. 7. **95% confidence interval for $Y$ when $X=25$:** Predicted $Y$: $$\hat{Y} = 0 + 2.56 \times 25 = 64.0$$ Standard error of prediction: $$SE_{pred} = s \sqrt{1 + \frac{1}{n} + \frac{(X_0 - \bar{X})^2}{S_{XX}}} = 18.3 \sqrt{1 + \frac{1}{8} + \frac{(25 - 23.44)^2}{399.43}} = 19.5$$ CI for $Y$: $$64.0 \pm 2.447 \times 19.5 = (15.3, 112.7)$$ Lay terms: We are 95% confident the attendance at 25°F will be between about 15 and 113 people. --- Final regression line: $$\hat{Y} = 2.56X$$