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The complicated but efficient method of measuring trend of time series is ______. - Mathematics and Statistics

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Question

The complicated but efficient method of measuring trend of time series is ______.

Options

  • graphical method

  • method of moving average

  • method of least squares

  • method of addition

MCQ
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Solution

The complicated but efficient method of measuring trend of time series is method of least squares.

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Measurement of Secular Trend
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Year 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976
Production
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0 0 1 1 2 3 4 5 6 7 8 9 8 9 10

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Year 1976 1977 1978 1979 1980 1981
Production 0 4 4 2 6 8
Year 1982 1983 1984 1985 1986  
Production 5 9 4 10 10  

Use the method of least squares to fit a trend line to the data given below. Also, obtain the trend value for the year 1975.

Year 1962 1963 1964 1965 1966 1967 1968 1969
Production
(million barrels)
0 0 1 1 2 3 4 5
Year 1970 1971 1972 1973 1974 1975 1976  
Production
(million barrels)
6 8 9 9 8 7 10  

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

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Year Production (y) x x2 xy
2006 19 – 9 81 – 171
2007 20 – 7 49 – 140
2008 14 – 5 25 – 70
2009 16 – 3 9 – 48
2010 17 – 1 1 – 17
2011 16 1 1 16
2012 18 3 9 54
2013 17 5 25 85
2014 21 7 49 147
2015 19 9 81 171
Total 177 0 330 27

Let the equation of trend line be y = a + bx   .....(i)

Here n = `square` (even), two middle years are `square` and 2011, and h = `square`

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As Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx2

As Σx = 0, b = `square`

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y = `square`


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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
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1981 4 2 4 8
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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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