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. - Mathematics and Statistics

Sum

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.

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.

Concept: Measurement of Secular Trend
Is there an error in this question or solution?
Chapter 4: Time Series - Exercise 4.1 [Page 66]

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