1. **State the problem:** We have data for gas prices (y) over years since 2000 (x): | x | y | |---|-------| | 1 | 1.345 | | 2 | 1.408 | | 3 | 1.537 | | 4 | 1.58 | We want to find the linear regression equation $y = mx + b$ that best fits this data. Then, using this equation, predict the gas price for the year 2005 (which corresponds to $x=6$ since 2005 is 6 years after 2000), and find how much the actual price $2.956$ differs from the predicted price. 2. **Formula for linear regression:** The slope $m$ is given by: $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ The intercept $b$ is: $$b = \frac{\sum y - m \sum x}{n}$$ where $n$ is the number of data points. 3. **Calculate sums:** $n=4$ $\sum x = 1+2+3+4 = 10$ $\sum y = 1.345 + 1.408 + 1.537 + 1.58 = 5.87$ $\sum x^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$ $\sum xy = (1)(1.345) + (2)(1.408) + (3)(1.537) + (4)(1.58) = 1.345 + 2.816 + 4.611 + 6.32 = 15.092$ 4. **Calculate slope $m$:** $$m = \frac{4(15.092) - (10)(5.87)}{4(30) - 10^2} = \frac{60.368 - 58.7}{120 - 100} = \frac{1.668}{20} = 0.0834$$ 5. **Calculate intercept $b$:** $$b = \frac{5.87 - 0.0834 \times 10}{4} = \frac{5.87 - 0.834}{4} = \frac{5.036}{4} = 1.259$$ 6. **Linear regression equation:** $$y = 0.0834x + 1.259$$ 7. **Predict price for 2005 ($x=6$):** $$y = 0.0834 \times 6 + 1.259 = 0.5004 + 1.259 = 1.7594$$ 8. **Calculate difference between actual and predicted price:** Actual price = 2.956 Difference = Actual - Predicted = $2.956 - 1.7594 = 1.1966$ **Answer:** The actual 2005 gas price was approximately 1.197 dollars higher than the predicted price.