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Solve the following problem : Fit a trend line to data by the method of least squares. Year 1977 1978 1979 1980 1981 1982 1983 1984 Number of boxes (in ten thousands) 1 0 3 8 10 4 5 8

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Question

Solve the following problem :

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

Year 1977 1978 1979 1980 1981 1982 1983 1984
Number of boxes (in ten thousands) 1 0 3 8 10 4 5 8
Sum
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Solution

In the given problem, n = 8 (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
No. of boxes (in ten thousands) 
yt
u = 2(t –  1980.5) u2 uyt Trend Value
1977 1 –7 49 –7 1.5836
1978 0 –5 25 0 2.5240
1979 3 –3 9 –9 3.4644
1980 8 –1 1 –8 4.4048
1981 10 1 1 10 5.3452
1982 4 3 9 12 6.2856
1983 5 5 25 25 7.2260
1984 8 7 49 56 8.1664
Total 39 0 168 79  

From the table, n = 8, `sumy_"t" = 39, sumu = 0, sumu^2 = 168,sumuy_"t" = 79`

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

∴ 39 = 8a' + b'(0)            ...(i)   and
79 = a'(0) + b'(168)         ...(ii)

From (i), a' = `(39)/(8)` = 4.875

From (ii), b' = `(79)/(168)` = 0.4702
∴  The equation of the trend line is yt = a' + b'u
i.e., yt = 4.875 + 0.4702 u, where u = 2(t – 1980.5).

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Measurement of Secular Trend
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Chapter 4: Time Series - Miscellaneous Exercise 4 [Page 70]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 4 Time Series
Miscellaneous Exercise 4 | Q 4.11 | Page 70

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