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Question
Fit a trend line to the data in Problem 4 above by the method of least squares. Also, obtain the trend value for the index of industrial production for the year 1987.
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Solution
In the given problem, x = 10(even), two middle t – values are 1980 and 1981, h = 1
u = `"t - mean of two middle values"/("h"/(2)) = ("t" - 1980.5)/(1/2)` = 2(t – 1980.5)
We obtain the following table.
| Year (t) | Index of industrial production yt | u = 2 (t - 1980.5) |
u2 | uyt | Trend value |
| 1976 | 0 | –9 | 81 | 0 | 0.1635 |
| 1977 | 2 | –7 | 49 | –14 | 1.0605 |
| 1978 | 3 | –5 | 25 | –15 | 1.9575 |
| 1979 | 3 | –3 | 9 | –9 | 2.8545 |
| 1980 | 2 | –1 | 1 | –2 | 3.7515 |
| 1981 | 4 | 1 | 1 | 4 | 4.6485 |
| 1982 | 5 | 3 | 9 | 15 | 5.5455 |
| 1983 | 6 | 5 | 25 | 30 | 6.4425 |
| 1984 | 7 | 7 | 49 | 49 | 7.3395 |
| 1985 | 10 | 9 | 81 | 90 | 8.2365 |
| Total | 42 | 0 | 330 | 148 |
From the table, n = 10, `sumy_"t" = 42, sumu = 0, sumu^2 = 330, sumuy_"t" = 148`
The two normal equations are : `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t" = "a"' sumu + "b"'sumu^2`
∴ 42 = 10a' + b'(0) ...(i) and
148 = a'(0) + b'(330) ...(ii)
From (i), a' = `(42)/(10)` = 4.2
From (ii), b' = `(148)/(330)` = 0.4485
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.2 + 0.4485 u, where u = 2(t – 1980.5)
∴ Now, For t = 1987, u = 2(1987 – 1980.5) = 2 x 6.5 = 13
∴ yt = 4.2 + 0.4485 x 13 = 10.0305.
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