# Solve the following problem : Fit a trend line to data in Problem 16 by the method of least squares. - Mathematics and Statistics

Sum

Solve the following problem :

Fit a trend line to data in Problem 16 by the method of least squares.

#### Solution

In the given problem, n = 7 (odd), middle t – value is 1995, h = 5

u = "t - middle value"/"h" = ("t" - 1995)/(1)

We obtain the following table.

 Year t Infant mortality rate yt u = ("t" - 1995)/(5) u2 uyt Trend Value 1980 10 –3 9 –30 8.9999 1985 7 –2 4 –14 7.4285 1990 5 –1 1 –5 5.8571 1995 4 0 0 0 4.2857 2000 3 1 1 3 2.7143 2005 1 2 4 2 1.1429 2010 0 3 9 0 –0.4285 Total 30 0 28 –44

From the table, n = 7, sumy_"t" = 10, sumu = 0, sumu^2 = 28,sumuy_"t" = – 44

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

∴ 30 = 7a' + b'(0)               ...(i)   and
– 44 = a'(0) + b'(28)           ...(ii)

From (i), a' = (30)/(7) = 4.2857

From (ii), b' = (-44)/(28) = 1.5714
∴  The equation of the trend line is yt = a' + b'u
i.e., yt = 4.2857 – 1.5714 u, where u = ("t" - 1995)/(5).

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Time Series
Miscellaneous Exercise 4 | Q 4.17 | Page 70