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Question
The annual production of a commodity is given as follows:
| Year | production (in tones) |
| 1995 | 155 |
| 1996 | 162 |
| 1997 | 171 |
| 19988 | 182 |
| 1999 | 158 |
| 2000 | 880 |
| 2001 | 178 |
Fit a straight line trend by the method of least squares
Solution
Computation of trend values by the method of least squares. (ODD years)
| Year (x) | Production (in tones) (Y) |
X = x – 1998 | X2 | XY |
Trend values |
| 1995 | 155 | – 3 | 9 | – 465 | 159.57 |
| 1996 | 162 | – 2 | 4 | – 324 | 162.86 |
| 1997 | 171 | – 1 | 1 | – 171 | 166.14 |
| 1998 | 182 | 0 | 0 | 0 | 169.43 |
| 1999 | 158 | 1 | 1 | 158 | 172.72 |
| 2000 | 180 | 2 | 4 | 360 | 176.00 |
| 2001 | 178 | 3 | 9 | 534 | 179.29 |
| N = 7 | `sum"Y"` = 1186 | `sum"X"` = 0 | `sum"X"^2` = 28 | `sum"XY"` = 92 | `sum"Y"_"t"` = 1186.01 |
a =`(sum"Y")/"N" = 1186/7` = 169.429
b = `(sum"XY")/(sum"X"^2) = 92/28` = 3.286
Therefore, the required equation of the straight-line trend is given by Y = a + bX
i. Y = 169.429 + 3.286 X or Y = 169.429 + 3.286 (x – 1998)
The trends values are obtained by
When x = 1995, Yt = 169.429 + 3.286(1995 – 1998) = 159.57
When x = 1996, Yt = 169.429 + 3.286(1996 – 1998) = 162.86
When x = 1997, Yt = 169.429 + 3.286(1997 – 1998) = 166.14
When x = 1998, Yt = -169.429 + 3.286(1998 – 1998) = 169.43
When x = 1999, Yt = 169.429 + 3.286(1999 – 1998) = 172.72
When x = 2000, Yt = 169.429 + 3.286(2000 – 1998) = 176.00
When x = 2001, Yt = 169.429 + 3.286(2001 – 1998) = 179.29
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