1. **Problem Statement:** We are given sums and means of two variables $x$ (catalyst amount) and $y$ (chemical process yield) and need to analyze their relationship, likely by finding the linear regression line. 2. **Given Data:** - $\sum x_i = 13.8$ - $\sum y_i = 523.3$ - $\sum x_i y_i = 911.58$ - $\sum x_i^2 = 25.16$ - $\sum y_i^2 = 34356.85$ - $\bar{x} = 1.725$ - $\bar{y} = 65.412$ 3. **Formula for Linear Regression Line:** The regression line of $y$ on $x$ is given by: $$y = a + bx$$ where $$b = \frac{\sum (x_i y_i) - n \bar{x} \bar{y}}{\sum (x_i^2) - n \bar{x}^2}$$ $$a = \bar{y} - b \bar{x}$$ 4. **Calculate $n$ (number of data points):** Since $\bar{x} = \frac{\sum x_i}{n}$, then $$n = \frac{\sum x_i}{\bar{x}} = \frac{13.8}{1.725} = 8$$ 5. **Calculate slope $b$:** $$b = \frac{911.58 - 8 \times 1.725 \times 65.412}{25.16 - 8 \times (1.725)^2}$$ Calculate numerator: $$911.58 - 8 \times 1.725 \times 65.412 = 911.58 - 902.68 = 8.9$$ Calculate denominator: $$25.16 - 8 \times 2.976 = 25.16 - 23.81 = 1.35$$ So, $$b = \frac{8.9}{1.35} \approx 6.59$$ 6. **Calculate intercept $a$:** $$a = 65.412 - 6.59 \times 1.725 = 65.412 - 11.36 = 54.05$$ 7. **Final regression equation:** $$y = 54.05 + 6.59x$$ This means for each additional liter of catalyst, the yield increases by approximately 6.59%.