1. **Problem Statement:** Use the least squares method to develop a linear trend equation for the annual aluminum shipments data from 1989 to 1999 and forecast the shipment for the year 2004. 2. **Formula and Explanation:** The linear trend equation is given by: $$y = a + bx$$ where $y$ is the shipment, $x$ is the time variable (year), $a$ is the intercept, and $b$ is the slope. The slope $b$ and intercept $a$ are calculated by: $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$a = \frac{\sum y - b \sum x}{n}$$ where $n$ is the number of data points. 3. **Assigning Values:** Let $x=0$ correspond to 1989, so $x$ runs from 0 to 10 for years 1989 to 1999. 4. **Data Table:** | Year | x | y (Shipments) | xy | x^2 | |------|---|---------------|----|-----| | 1989 | 0 | 2 | 0 | 0 | | 1990 | 1 | 3 | 3 | 1 | | 1991 | 2 | 6 | 12 | 4 | | 1992 | 3 | 10 | 30 | 9 | | 1993 | 4 | 8 | 32 | 16 | | 1994 | 5 | 7 | 35 | 25 | | 1995 | 6 | 12 | 72 | 36 | | 1996 | 7 | 14 | 98 | 49 | | 1997 | 8 | 14 | 112| 64 | | 1998 | 9 | 18 | 162| 81 | | 1999 | 10| 19 | 190| 100 | Calculate sums: $$\sum x = 0+1+2+...+10 = 55$$ $$\sum y = 2+3+6+10+8+7+12+14+14+18+19 = 113$$ $$\sum xy = 0+3+12+30+32+35+72+98+112+162+190 = 646$$ $$\sum x^2 = 0+1+4+9+16+25+36+49+64+81+100 = 385$$ $$n = 11$$ 5. **Calculate slope $b$:** $$b = \frac{11 \times 646 - 55 \times 113}{11 \times 385 - 55^2} = \frac{7106 - 6215}{4235 - 3025} = \frac{891}{1210} \approx 0.7364$$ 6. **Calculate intercept $a$:** $$a = \frac{113 - 0.7364 \times 55}{11} = \frac{113 - 40.502}{11} = \frac{72.498}{11} \approx 6.5907$$ 7. **Linear trend equation:** $$y = 6.5907 + 0.7364x$$ 8. **Forecast for 2004:** Since $x=0$ for 1989, $x$ for 2004 is $2004 - 1989 = 15$. $$y_{2004} = 6.5907 + 0.7364 \times 15 = 6.5907 + 11.046 = 17.6367$$ **Final forecasted shipment for 2004 is approximately 17.64 tons.**