1. **Problem Statement:** A farmer collected data on fertilizer amount $x$ (hundreds of kgs per acre) and soya bean yield $y$ (bushels per acre). We analyze the relationship between $x$ and $y$. 2. **Scatter Diagram and Correlation:** Plot points $(x,y)$: (1,25), (2.5,32), (3,35), (3,32), (3.4,35), (4,39), (4,41), (4.5,40). The points generally rise as $x$ increases, indicating a **positive correlation**. 3. **Linear Regression Model:** The linear model is $y = a + bx$ where $b$ is slope and $a$ is intercept. Calculate means: $$\bar{x} = \frac{1+2.5+3+3+3.4+4+4+4.5}{8} = 3.425$$ $$\bar{y} = \frac{25+32+35+32+35+39+41+40}{8} = 34.875$$ Calculate slope: $$b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} = \frac{(1-3.425)(25-34.875)+...+(4.5-3.425)(40-34.875)}{(1-3.425)^2+...+(4.5-3.425)^2} = 5.44$$ Calculate intercept: $$a = \bar{y} - b\bar{x} = 34.875 - 5.44 \times 3.425 = 16.33$$ So regression line: $$y = 16.33 + 5.44x$$ Meaning: For each additional hundred kgs of fertilizer, bushels increase by about 5.44. When fertilizer is zero, predicted yield is 16.33 bushels. 4. **Prediction for $x=7.5$:** $$y = 16.33 + 5.44 \times 7.5 = 16.33 + 40.8 = 57.13$$ Predicted bushels per acre is approximately 57.13. 5. **Correlation Coefficient $r$:** Calculate: $$r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} = 0.91$$ This shows a strong positive linear relationship. 6. **Coefficient of Determination $r^2$:** $$r^2 = (0.91)^2 = 0.83$$ Interpretation: About 83% of the variation in bushels is explained by fertilizer amount.