1. **State the problem:** We are given two variables, $Y$ (Sales) and $x$ (Adv), and we want to find the estimated linear regression equation of the form $$Y = b^0 + b^1 x$$ where $b^0$ is the intercept and $b^1$ is the slope. 2. **Formulas used:** - Slope: $$b^1 = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ - Intercept: $$b^0 = \bar{y} - b^1 \bar{x}$$ where $n$ is the number of data points, $\sum xy$ is the sum of products of $x$ and $y$, $\sum x$ and $\sum y$ are sums of $x$ and $y$ values respectively, and $\bar{x}$ and $\bar{y}$ are the means of $x$ and $y$. 3. **Calculate sums and means:** - $n = 10$ - $\sum x = 7.4 + 8.4 + 7.8 + 10.7 + 8.8 + 6.4 + 6.3 + 8.5 + 8.3 + 8.8 = 81.4$ - $\sum y = 88.9 + 67.4 + 74.8 + 83.9 + 57.1 + 37.4 + 80.6 + 66.3 + 70.1 + 60.8 = 686.3$ - $\sum x^2 = 7.4^2 + 8.4^2 + 7.8^2 + 10.7^2 + 8.8^2 + 6.4^2 + 6.3^2 + 8.5^2 + 8.3^2 + 8.8^2 = 676.14$ - $\sum xy = (7.4)(88.9) + (8.4)(67.4) + (7.8)(74.8) + (10.7)(83.9) + (8.8)(57.1) + (6.4)(37.4) + (6.3)(80.6) + (8.5)(66.3) + (8.3)(70.1) + (8.8)(60.8) = 5657.68$ - $\bar{x} = \frac{81.4}{10} = 8.14$ - $\bar{y} = \frac{686.3}{10} = 68.63$ 4. **Calculate slope $b^1$:** $$b^1 = \frac{10 \times 5657.68 - 81.4 \times 686.3}{10 \times 676.14 - (81.4)^2} = \frac{56576.8 - 55856.82}{6761.4 - 6626.0} = \frac{719.98}{135.4} \approx 5.32$$ 5. **Calculate intercept $b^0$:** $$b^0 = 68.63 - 5.32 \times 8.14 = 68.63 - 43.31 = 25.32$$ 6. **Final regression equation:** $$Y = 25.32 + 5.32 x$$ This means for each unit increase in advertising, sales increase by approximately 5.32 units, starting from a base sales of 25.32 when advertising is zero.