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

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Sum

Solve the following problem :

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

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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
  Is there an error in this question or solution?
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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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