1. **State the problem:** We are given data points for % level of engagement ($x$) and % volume of sales ($y$). We want to find the linear regression line $y = bx + a$ that best fits the data. 2. **Formula used:** The formulas for the slope $b$ and intercept $a$ in linear regression are: $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$a = \frac{\sum y - b \sum x}{n}$$ where $n$ is the number of data points. 3. **Given data:** \begin{align*} x &: 69, 63, 34, 41, 81 \\ y &: 73, 67, 38, 43, 84 \end{align*} 4. **Calculate sums:** \begin{align*} \sum x &= 69 + 63 + 34 + 41 + 81 = 288 \\ \sum y &= 73 + 67 + 38 + 43 + 84 = 305 \\ \sum xy &= (69)(73) + (63)(67) + (34)(38) + (41)(43) + (81)(84) \\ &= 5037 + 4221 + 1292 + 1763 + 6804 = 19117 \\ \sum x^2 &= 69^2 + 63^2 + 34^2 + 41^2 + 81^2 \\ &= 4761 + 3969 + 1156 + 1681 + 6561 = 18128 \\ n &= 5 \end{align*} 5. **Calculate slope $b$:** $$b = \frac{5 \times 19117 - 288 \times 305}{5 \times 18128 - 288^2} = \frac{95585 - 87840}{90640 - 82944} = \frac{6745}{7696} \approx 0.8767$$ 6. **Calculate intercept $a$:** $$a = \frac{305 - 0.8767 \times 288}{5} = \frac{305 - 252.5}{5} = \frac{52.5}{5} = 10.5$$ 7. **Final linear regression equation:** $$y = 0.8767x + 10.5$$ This means for each 1% increase in engagement, sales increase by approximately 0.8767%, starting from a base of 10.5% sales when engagement is zero.