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The following table gives the production of steel (in millions of tons) for years 1976 to 1986.
Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 |
Production | 0 | 4 | 4 | 2 | 6 | 8 | 5 | 9 | 4 | 10 | 10 |
Obtain the trend value for the year 1990
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Solution
In the given problem, n = 11 (odd), middle t- values is 1981, h = 1
u = `("t" - "middle t value")/"h"`
= `("t" - 1981)/1`
= t – 1981
We obtain the following table:
Year t |
Production y_{t} |
u = t − 1981 | u^{2} | uy_{t} | 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 | 87 |
From the table, n = 11, ∑y_{t} = 62, ∑u = 0, ∑u^{2} = 110, ∑uy_{t} = 87
The two normal equations are:
∑y_{t} = na' + b'∑u and ∑uy_{t} = a'∑u + b'∑u^{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 y_{t} = a′ + b′u
i.e., y_{t} = 5.6364+ 0.7909 u,
where u = t – 1981
Now, for t = 1990,
u = 1990 – 1981
= 9
∴ y_{t} = 5.6364 + 0.7909 × 9
= 12.7545
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