# 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. - Mathematics and Statistics

Sum

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.

#### Solution

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.

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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