# Solve the following problem : Fit a trend line to data in Problem 4 by the method of least squares. - Mathematics and Statistics

Sum

Solve the following problem :

Fit a trend line to data in Problem 4 by the method of least squares.

#### Solution

In the given problem, n = 12 (odd), middle t – value is 1976, h = 1

u = "t - middle value"/("h"/2) = ("t" - 1976.5)/(1/2) = 2(t – 1976.5)

We obtain the following table.

 Year t Production yt u = 2(t – 1976.5) u2 uyt Trend Value 1971 1 –11 121 –11 0.0900 1972 0 –9 81 0 0.6494 1973 1 –7 49 –7 1.2088 1974 2 –5 25 –10 1.7682 1975 3 – 9 –9 2.3276 1976 2 –1 1 –2 2.8870 1977 3 1 1 3 3.4464 1978 6 3 9 18 4.0058 1979 5 5 25 25 4.5652 1980 1 7 49 7 5.1246 1981 4 9 81 36 5.6840 1982 10 11 121 110 6.243 Total 38 0 572 160

From the table, n = 12, sumy_"t" = 38, sumu = 0, sumu^2 = 572,sumuy_"t" = 160

The two normal equations are: sumy_"t" = "na"' + "b"' sumu  "and" sumuy_"t", = a'sumu + b'sumu^2

∴ 38 = 12a' + b'(0)            ...(i)   and
160 = a'(0) + b'(572)         ...(ii)

From (i), a' = (38)/(12) = 3.1667

From (ii), b' = (160)/(572) = 0.2797
∴  The equation of the trend line is yt = a' + b'u
i.e., yt = 3.1667 + 0.2797 u, where u = 2(t – 1976.5).

Concept: Measurement of Secular Trend
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Chapter 4: Time Series - Miscellaneous Exercise 4 [Page 69]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Time Series
Miscellaneous Exercise 4 | Q 4.05 | Page 69
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