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Question
Fit a trend line to the following data by the method of least squares.
| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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Solution
In the given problem, n = 9 (odd), middle t – value is 1978, h = 1
u = `"t - middle value"/"h" = ("t" - 1978)/(1)` = t – 1978
We obtain the following table.
| Year t |
Production yt |
u = t – 1978 | u2 | uyt | Trend Value |
| 1974 | 0 | –4 | 16 | 0 | 3.8001 |
| 1975 | 4 | –3 | 9 | –12 | 4.4334 |
| 1976 | 9 | –2 | 4 | –18 | 5.0667 |
| 1977 | 9 | –1 | 1 | –9 | 5.7 |
| 1978 | 8 | 0 | 0 | 0 | 6.3333 |
| 1979 | 5 | 1 | 1 | 5 | 6.9666 |
| 1980 | 4 | 2 | 4 | 8 | 7.5999 |
| 1981 | 8 | 3 | 9 | 24 | 8.2332 |
| 1982 | 10 | 4 | 16 | 40 | 8.8665 |
| Total | 57 | 0 | 60 | 38 |
From the table, n = 9, `sumy_"t" = 57, sumu = 0, sumu^2 = 60,sumuy_"t" = 38`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 57 = 9a' + b'(0) ...(i) and
38 = a'(0) + b'(60) ...(ii)
From (i), a' = `(57)/(9)` = 6.3333
From (ii), b' = `(38)/(60)` = 0.6333
∴ The equation of the trend line is yt = a' + b' u
i.e., yt = 6.3333 + 0.6333 u, where u = t – 1978.
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| 1975 | 0 | - 4 | 16 | 0 |
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| `sumx_t` =47 | `sumu`=0 | `sumu^2=60` | `square` |
The equation of trend line is xt= a' + b'u.
The normal equations are,
`sumx_t = na^' + b^' sumu` ...(1)
`sumux_t = a^'sumu + b^'sumu^2` ...(2)
Here, n = 9, `sumx_t = 47, sumu= 0, sumu^2 = 60`
By putting these values in normal equations, we get
47 = 9a' + b' (0) ...(3)
40 = a'(0) + b'(60) ...(4)
From equation (3), we get a' = `square`
From equation (4), we get b' = `square`
∴ the equation of trend line is xt = `square`
