1. **Problem Statement:** Given advertising expenditure $X$ (in $100$ units) and sales revenue $Y$ (in $1000$ units) as follows: $X: 1, 2, 3, 4, 5$ $Y: 3, 3, 5, 5, 6$ Calculate: a) The coefficient of correlation between $X$ and $Y$. b) The linear regression equation $y = a + bx$. c) The reliability of the model. 2. **Formulas and Important Rules:** - Mean of $X$: $\bar{X} = \frac{\sum X_i}{n}$ - Mean of $Y$: $\bar{Y} = \frac{\sum Y_i}{n}$ - Covariance: $S_{XY} = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{n}$ - Variance of $X$: $S_X^2 = \frac{\sum (X_i - \bar{X})^2}{n}$ - Variance of $Y$: $S_Y^2 = \frac{\sum (Y_i - \bar{Y})^2}{n}$ - Coefficient of correlation: $r = \frac{S_{XY}}{S_X S_Y}$ - Regression coefficients: $$b = \frac{S_{XY}}{S_X^2}, \quad a = \bar{Y} - b \bar{X}$$ - Reliability of the model (coefficient of determination): $r^2$ 3. **Calculations:** - $n = 5$ - Calculate sums: $$\sum X = 1+2+3+4+5 = 15$$ $$\sum Y = 3+3+5+5+6 = 22$$ $$\sum X^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$$ $$\sum Y^2 = 3^2 + 3^2 + 5^2 + 5^2 + 6^2 = 9 + 9 + 25 + 25 + 36 = 104$$ $$\sum XY = (1)(3) + (2)(3) + (3)(5) + (4)(5) + (5)(6) = 3 + 6 + 15 + 20 + 30 = 74$$ - Means: $$\bar{X} = \frac{15}{5} = 3$$ $$\bar{Y} = \frac{22}{5} = 4.4$$ - Calculate covariance: $$S_{XY} = \frac{\sum XY}{n} - \bar{X} \bar{Y} = \frac{74}{5} - (3)(4.4) = 14.8 - 13.2 = 1.6$$ - Calculate variance of $X$: $$S_X^2 = \frac{\sum X^2}{n} - \bar{X}^2 = \frac{55}{5} - 3^2 = 11 - 9 = 2$$ - Calculate variance of $Y$: $$S_Y^2 = \frac{104}{5} - 4.4^2 = 20.8 - 19.36 = 1.44$$ - Calculate standard deviations: $$S_X = \sqrt{2} \approx 1.414$$ $$S_Y = \sqrt{1.44} = 1.2$$ - Calculate coefficient of correlation: $$r = \frac{1.6}{1.414 \times 1.2} = \frac{1.6}{1.6968} \approx 0.9428$$ - Calculate regression coefficients: $$b = \frac{S_{XY}}{S_X^2} = \frac{1.6}{2} = 0.8$$ $$a = \bar{Y} - b \bar{X} = 4.4 - 0.8 \times 3 = 4.4 - 2.4 = 2$$ - Regression equation: $$y = 2 + 0.8x$$ - Reliability of the model: $$r^2 = (0.9428)^2 \approx 0.889$$ 4. **Interpretation:** - The coefficient of correlation $r \approx 0.943$ indicates a strong positive linear relationship between advertising expenditure and sales revenue. - The regression line $y = 2 + 0.8x$ can be used to predict sales revenue based on advertising expenditure. - The reliability $r^2 \approx 0.889$ means about 88.9% of the variation in sales revenue is explained by the advertising expenditure. **Final answers:** a) $r \approx 0.943$ b) $y = 2 + 0.8x$ c) Reliability $r^2 \approx 0.889$