# 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 - Mathematics and Statistics

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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

#### 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 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 87

From the table, n = 11, ∑yt = 62, ∑u = 0, ∑u2 = 110, ∑uyt = 87

The two normal equations are:

∑yt = na' + b'∑u and ∑uyt = a'∑u + b'∑u2

∴ 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 × 9

= 12.7545

Concept: Measurement of Secular Trend
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Chapter 2.4: Time Series - Q.4
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