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Solve the following problem : Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010.Fit a trend line to the above data by graphical method. - Mathematics and Statistics

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Question

Solve the following problem :

Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010.

Year 1980 1985 1990 1995 2000 2005 2010
IMR 10 7 5 4 3 1 0

Fit a trend line to the above data by graphical method.

Graph

Solution

Taking year on X-axis and mortality rate on Y-axis, we plot the points for mortality rate corresponding to years. Joining these points we get the graph of time series. We fit the trend line as shown in the following graph.

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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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Obtain trend values for data, using 3-yearly moving averages
Solution:

Year IMR 3 yearly
moving total
3-yearly moving
average

(trend value)
1980 10
1985 7 `square` 7.33
1990 5 16 `square`
1995 4 12 4
2000 3 8 `square`
2005 1 `square` 1.33
2010 0

Complete the following activity to fit a trend line to the following data by the method of least squares.

Year 1975 1976 1977 1978 1979 1980 1981 1982 1983
Number of deaths 0 6 3 8 2 9 4 5 10

Solution:

Here n = 9. We transform year t to u by taking u = t - 1979. We construct the following table for calculation :

Year t Number of deaths xt u = t - 1979 u2 uxt
1975 0 - 4 16 0
1976 6 - 3 9 - 18
1977 3 - 2 4 - 6
1978 8 - 1 1 - 8
1979 2 0 0 0
1980 9 1 1 9
1981 4 2 4 8
1982 5 3 9 15
1983 10 4 16 40
  `sumx_t` =47 `sumu`=0 `sumu^2=60` `square`

The equation of trend line is xt= a' + b'u.

The normal equations are,

`sumx_t = na^' + b^' sumu`              ...(1)

`sumux_t = a^'sumu + b^'sumu^2`      ...(2)

Here, n = 9, `sumx_t = 47, sumu= 0, sumu^2 = 60`

By putting these values in normal equations, we get

47 = 9a' + b' (0)       ...(3)

40 = a'(0) + b'(60)      ...(4)

From equation (3), we get a' = `square`

From equation (4), we get b' = `square`

∴ the equation of trend line is xt = `square`


Following table gives the number of road accidents (in thousands) due to overspeeding in Maharashtra for 9 years. Complete the following activity to find the trend by the method of least squares.

Year 2008 2009 2010 2011 2012 2013 2014 2015 2016
Number of accidents 39 18 21 28 27 27 23 25 22

Solution:

We take origin to 18, we get, the number of accidents as follows:

Year Number of accidents xt t u = t - 5 u2 u.xt
2008 21 1 -4 16 -84
2009 0 2 -3 9 0
2010 3 3 -2 4 -6
2011 10 4 -1 1 -10
2012 9 5 0 0 0
2013 9 6 1 1 9
2014 5 7 2 4 10
2015 7 8 3 9 21
2016 4 9 4 16 16
  `sumx_t=68` - `sumu=0` `sumu^2=60` `square`

The equation of trend is xt =a'+ b'u.

The normal equations are,

`sumx_t=na^'+b^'sumu             ...(1)`

`sumux_t=a^'sumu+b^'sumu^2      ...(2)`

Here, n = 9, `sumx_t=68,sumu=0,sumu^2=60,sumux_t=-44`

Putting these values in normal equations, we get

68 = 9a' + b'(0)     ...(3)

∴ a' = `square`

-44 = a'(0) + b'(60)          ...(4)

∴ b' = `square`

The equation of trend line is given by

xt = `square`


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