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प्रश्न
Fit a trend line to the data in Problem 7 by the method of least squares. Also, obtain the trend value for the year 1990.
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उत्तर
In the given problem, n = 11 (odd), middle t- values is 1981, h = 1
u = `"t - middle value"/"h" = ("t" - 1981)/(1)` = t – 1981
We obtain the following table.
| Year t |
Production yt |
u = t–1981 | u2 | uyt | Trend Value |
| 1976 | 0 | –5 | 25 | 0 | 1.6819 |
| 1977 | 4 | –4 | 16 | –16 | 2.4728 |
| 1978 | 4 | –3 | 9 | –12 | 3.2637 |
| 1979 | 2 | –2 | 4 | –4 | 4.0546 |
| 1980 | 6 | –1 | 1 | –6 | 4.8455 |
| 1981 | 8 | 0 | 0 | 0 | 5.6364 |
| 1982 | 5 | 1 | 1 | 5 | 6.4273 |
| 1983 | 9 | 2 | 4 | 18 | 7.2182 |
| 1984 | 4 | 3 | 9 | 12 | 8.0091 |
| 1985 | 10 | 4 | 16 | 40 | 8.8 |
| 1986 | 10 | 5 | 25 | 50 | 9.5909 |
| Total | 62 | 0 | 110 | 87 |
From the table, n = 11, `sumy_"t" = 62, sumu = 0, sumu^2 = 110, sumuy_"t" = 87`
The two normal equations are : `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t" = "a"' sumu + "b"'sumu^2`
∴ 62 = 11a' + b'(0) ...(i) and
87 = a'(0) + b'(110) ...(ii)
From (i), a' = `(62)/(11)` = 5.6364
From (ii), b' = `(87)/(110)` = 0.7909
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 5.6364 + 0.7909 u, where u = t – 1981
∴ Now, For t = 1990, u = 1990 – 1981= 9
∴ yt = 5.6364 + 0.7909 x 9 = 12.7545.
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| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 |
| Production | 0 | 4 | 4 | 2 | 6 | 8 |
| Year | 1982 | 1983 | 1984 | 1985 | 1986 | |
| Production | 5 | 9 | 4 | 10 | 10 |
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Solution:
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| 1980 | 10 | – | – |
| 1985 | 7 | `square` | 7.33 |
| 1990 | 5 | 16 | `square` |
| 1995 | 4 | 12 | 4 |
| 2000 | 3 | 8 | `square` |
| 2005 | 1 | `square` | 1.33 |
| 2010 | 0 | – | – |
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| Production xi |
10 | 15 | 20 | 25 | 30 |
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35 | 40 | 45 | 50 | 55 |
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