1. **Problem Statement:** We have data on malpractice insurance rates (as % of income) and doctors per 10,000 population for 10 U.S. states. We want to analyze the relationship using simple linear regression. 2. **(a) Null Hypothesis:** - Formal: $H_0: \beta = 0$ (no linear relationship between doctors per 10,000 population and malpractice rate). - Lay terms: There is no relationship between the number of doctors and malpractice insurance rates. 3. **(b) Calculate correlation $r$ and regression line $y = a + bx$:** - Data points: $(x_i, y_i)$ where $x_i$ = doctors per 10,000, $y_i$ = malpractice rate. - Calculate means: $\bar{x} = \frac{30+20+50+20+5+8+23+25+5+7}{10} = 19.3$ $\bar{y} = \frac{2+3+1+5+6+6+4+2+9+8}{10} = 4.6$ - Calculate sums: $S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = -178.1$ $S_{xx} = \sum (x_i - \bar{x})^2 = 1322.1$ $S_{yy} = \sum (y_i - \bar{y})^2 = 52.4$ - Correlation coefficient: $$r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{-178.1}{\sqrt{1322.1 \times 52.4}} = -0.676$$ - Regression slope: $$b = \frac{S_{xy}}{S_{xx}} = \frac{-178.1}{1322.1} = -0.135$$ - Regression intercept: $$a = \bar{y} - b \bar{x} = 4.6 - (-0.135)(19.3) = 7.2$$ - Regression line: $$y = 7.2 - 0.135x$$ - Test significance of slope at $\alpha=0.05$: - Standard error of slope: $$SE_b = \sqrt{\frac{SSE}{(n-2)S_{xx}}}$$ where $SSE = S_{yy} - b S_{xy} = 52.4 - (-0.135)(-178.1) = 28.5$ $$SE_b = \sqrt{\frac{28.5}{8 \times 1322.1}} = 0.052$$ - Test statistic: $$t = \frac{b}{SE_b} = \frac{-0.135}{0.052} = -2.60$$ - Degrees of freedom: $n-2=8$, critical $t_{0.025,8} \approx 2.306$ Since $|t|=2.60 > 2.306$, reject $H_0$. The slope is significantly different from zero. 4. **(c) Interpretation:** - $r = -0.676$ indicates a moderate negative correlation: as doctors per 10,000 increase, malpractice rate tends to decrease. - $R^2 = r^2 = 0.457$ means about 45.7% of the variation in malpractice rate is explained by the number of doctors. - The slope $b = -0.135$ means for each additional doctor per 10,000 population, the malpractice rate decreases by 0.135% of income on average. 5. **(d) 95% confidence interval for slope $b$:** - Using $t_{0.025,8} = 2.306$ and $SE_b=0.052$: $$b \pm t SE_b = -0.135 \pm 2.306 \times 0.052 = (-0.255, -0.015)$$ - Lay terms: We are 95% confident the true slope is between -0.255 and -0.015, confirming a negative relationship. 6. **(e) 95% confidence interval for predicted $Y$ when $X=5$:** - Predicted $\hat{y} = 7.2 - 0.135 \times 5 = 6.525$ - Standard error of prediction: $$SE_{pred} = \sqrt{MSE \left(1 + \frac{1}{n} + \frac{(x_0 - \bar{x})^2}{S_{xx}}\right)}$$ where $MSE = \frac{SSE}{n-2} = \frac{28.5}{8} = 3.56$ $$SE_{pred} = \sqrt{3.56 \left(1 + \frac{1}{10} + \frac{(5 - 19.3)^2}{1322.1}\right)} = \sqrt{3.56 \times 1.17} = 2.04$$ - Confidence interval: $$6.525 \pm 2.306 \times 2.04 = (1.8, 11.25)$$ - Lay terms: When there are 5 doctors per 10,000 population, we predict the malpractice rate to be about 6.5%, but it could reasonably be as low as 1.8% or as high as 11.25%. Final answers: - Regression line: $y = 7.2 - 0.135x$ - Correlation: $r = -0.676$ - Slope 95% CI: $(-0.255, -0.015)$ - Prediction 95% CI at $x=5$: $(1.8, 11.25)$