1. **Problem Statement**: We have Physics and Chemistry marks of 8 students and want to analyze the relation with regression and correlation. 2. **Data**: Physics ($x$): 70, 36, 56, 56, 58, 45, 67, 72 Chemistry ($y$): 78, 49, 55, 50, 75, 50, 60, 57 3. **(a) Scatter Diagram**: Plot points $(x,y)$: (70,78), (36,49), (56,55), (56,50), (58,75), (45,50), (67,60), (72,57). 4. **(b) Regression Line Calculation**: Calculate means: $$\bar{x} = \frac{70+36+56+56+58+45+67+72}{8} = \frac{460}{8} = 57.5$$ $$\bar{y} = \frac{78+49+55+50+75+50+60+57}{8} = \frac{474}{8} = 59.25$$ Calculate sums: $$\sum{(x_i - \bar{x})(y_i - \bar{y})} = \sum{x_i y_i} - n \bar{x} \bar{y}$$ First compute $\sum{x_i y_i}$: $$70\times78 + 36\times49 + 56\times55 + 56\times50 + 58\times75 + 45\times50 + 67\times60 + 72\times57 =$$ $$5460 + 1764 + 3080 + 2800 + 4350 + 2250 + 4020 + 4104 = 27828$$ Also compute $\sum{x_i^2}$: $$70^2 + 36^2 + 56^2 + 56^2 + 58^2 + 45^2 + 67^2 + 72^2 =$$ $$4900 + 1296 + 3136 + 3136 + 3364 + 2025 + 4489 + 5184 = 27530$$ Compute covariance numerator: $$S_{xy} = 27828 - 8 \times 57.5 \times 59.25 = 27828 - 27240 = 588$$ Compute variance of $x$: $$S_{xx} = 27530 - 8 \times 57.5^2 = 27530 - 8 \times 3306.25 = 27530 - 26450 = 1080$$ Slope $b$: $$b = \frac{S_{xy}}{S_{xx}} = \frac{588}{1080} = 0.5444$$ Intercept $a$: $$a = \bar{y} - b \bar{x} = 59.25 - 0.5444 \times 57.5 = 59.25 - 31.3 = 27.95$$ Regression line equation: $$y = 27.95 + 0.5444x$$ 5. **(c) Hypothesis test for positive correlation at 5% level**: Null hypothesis $H_0$: $\rho = 0$ (no correlation) Alternative hypothesis $H_a$: $\rho > 0$ (positive correlation) Calculate Pearson correlation coefficient $r$: $$r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}$$ Calculate $S_{yy}$: $$\sum y_i^2 = 78^2 + 49^2 + 55^2 + 50^2 + 75^2 + 50^2 + 60^2 + 57^2 =$$ $$6084 + 2401 + 3025 + 2500 + 5625 + 2500 + 3600 + 3249 = 28984$$ $$S_{yy} = 28984 - 8 \times 59.25^2 = 28984 - 8 \times 3509.06 = 28984 - 28072 = 912$$ Compute $r$: $$r = \frac{588}{\sqrt{1080 \times 912}} = \frac{588}{\sqrt{985,000}} = \frac{588}{992.46} = 0.5925$$ Test statistic: $$t = r \sqrt{\frac{n-2}{1-r^2}} = 0.5925 \sqrt{\frac{6}{1 - 0.5925^2}} = 0.5925 \sqrt{\frac{6}{1 - 0.351}} = 0.5925 \sqrt{\frac{6}{0.649}} = 0.5925 \times 3.04 = 1.8$$ Degrees of freedom = 6. Critical value (one-tailed, 5%) $t_{0.05,6} = 1.943$. Decision: $t=1.8 < 1.943$ not significant. Conclusion: insufficient evidence at 5% to conclude a positive correlation. Final answers: - Regression line: $$y = 27.95 + 0.5444x$$ - Correlation coefficient: $$r = 0.5925$$ - Test statistic: $$t = 1.8$$, fail to reject null at 5%.