1. **Problem Statement:** We have petrochemical export data for years 2010 to 2014 with values 17, 18, 20, 22, and 23.5 billion respectively. We use two models: $$y = 1.4x + 2.35$$ and $$y = 2x - 3$$ where $x$ corresponds to the year minus 2000 (so $x=10$ for 2010). We need to find residuals and their sums for each model, sum of squares of residuals, and decide which model fits better. 2. **Calculate residuals for each year and model:** For each year, calculate predicted $y$ using the model, then residual = actual $y$ - predicted $y$. Years and $x$ values: 2010: $x=10$ 2011: $x=11$ 2012: $x=12$ 2013: $x=13$ 2014: $x=14$ **Model 1: $y=1.4x + 2.35$** - 2010: $1.4(10)+2.35=14+2.35=16.35$, residual $=17-16.35=0.65$ - 2011: $1.4(11)+2.35=15.4+2.35=17.75$, residual $=18-17.75=0.25$ - 2012: $1.4(12)+2.35=16.8+2.35=19.15$, residual $=20-19.15=0.85$ - 2013: $1.4(13)+2.35=18.2+2.35=20.55$, residual $=22-20.55=1.45$ - 2014: $1.4(14)+2.35=19.6+2.35=21.95$, residual $=23.5-21.95=1.55$ Sum of residuals Model 1: $$0.65 + 0.25 + 0.85 + 1.45 + 1.55 = 4.75$$ **Model 2: $y=2x - 3$** - 2010: $2(10)-3=20-3=17$, residual $=17-17=0$ - 2011: $2(11)-3=22-3=19$, residual $=18-19=-1$ - 2012: $2(12)-3=24-3=21$, residual $=20-21=-1$ - 2013: $2(13)-3=26-3=23$, residual $=22-23=-1$ - 2014: $2(14)-3=28-3=25$, residual $=23.5-25=-1.5$ Sum of residuals Model 2: $$0 + (-1) + (-1) + (-1) + (-1.5) = -4.5$$ 3. **Sum of squares of residuals:** Model 1: $$0.65^2 + 0.25^2 + 0.85^2 + 1.45^2 + 1.55^2 = 0.4225 + 0.0625 + 0.7225 + 2.1025 + 2.4025 = 5.7125$$ Rounded to two decimals: 5.71 Model 2: $$0^2 + (-1)^2 + (-1)^2 + (-1)^2 + (-1.5)^2 = 0 + 1 + 1 + 1 + 2.25 = 5.25$$ 4. **Conclusion:** The model with the smaller sum of squares of residuals fits better. Since $5.25 < 5.71$, the model $y=2x - 3$ is the better fit. **Final answers:** (a) Residuals Model 1: [0.65, 0.25, 0.85, 1.45, 1.55], sum = 4.75 Residuals Model 2: [0, -1, -1, -1, -1.5], sum = -4.5 (b) Sum of squares Model 1: 5.71 Sum of squares Model 2: 5.25 (c) The model $y=2x - 3$ is the better fit because it has the smaller sum of squares of residuals.