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Solve the following problem : Fit a trend line to the data in Problem 7 by the method of least squares. - Mathematics and Statistics

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Question

Solve the following problem :

Fit a trend line to the data in Problem 7 by the method of least squares.

Sum

Solution

In the given problem, n = 10 even), middle t – value is 1980 and 1985, h = 5

u = `"t - mean of two middle values"/("h"/2) = ("t" - 1985.5)/(5/2) = (2("t" – 1985.5))/(5)`

We obtain the following table.

Year
t
Percentage of enrolment 
yt
u = `(2("t" – 1982.5))/(5)` u2 uyt Trend Value
1960 0 –9 81 0 0.8187
1965 3 –7 49 –21 1.7701
1970 3 –5 25 –15 2.7215
1975 4 –3 9 –12 3.6729
1980 4 –1 1 –4 4.6243
1985 5 1 1 5 5.5757
1990 6 3 9 18 6.5271
1995 8 5 25 40 7.4785
2000 8 7 49 56 8.4299
2005 10 9 81 90 9.3813
Total 51 0 330 157  

From the table, n = 10, `sumy_"t" = 51, sumu = 0, sumu^2 = 330,sumuy_"t" = 157`

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

∴ 51 = 10a' + b'(0)            ...(i)   and
157 = a'(0) + b'(330)         ...(ii)

From (i), a' = `(51)/(10)` = 5.1

From (ii), b' = `(157)/(330)` = 0.4757
∴  The equation of the trend line is yt = a' + b'u
i.e., yt = 5.1 + 0.4757 u, where u = `(2("t" – 1985.5))/(5)`.

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

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