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Obtain the trend line for the above data using 5 yearly moving averages. - Mathematics and Statistics

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Question

Obtain the trend line for the above data using 5 yearly moving averages.

Sum
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Solution

Construct the following table to obtain 5 yearly moving averages in data of problem 1.

Year T Production yt
(in' 000 tonnes)
5 – yearly moving total 5 – yearly moving averages trend value
1962 0
1963 0
1964 1 6 1.2
1965 1 8 1.6
1966 4 12 2.4
1967 2 20 4
1968 4 26 5.2
1969 9 32 6.4
1970 7 38 7.6
1971 10
1972 8
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Measurement of Secular Trend
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Chapter 4: Time Series - Exercise 4.1 [Page 66]

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Solution: Let us fit equation of trend line for above data.

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

Here n = 7(odd), middle year is `square` and h = 5

Year IMR (y) x x2 x.y
1980 10 – 3 9 – 30
1985 7 – 2 4 – 14
1990 5 – 1 1 – 5
1995 4 0 0 0
2000 3 1 1 3
2005 1 2 4 2
2010 0 3 9 0
Total 30 0 28 – 44

The normal equations are

Σy = na + bΣx

As, Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx2

As, Σx = 0, b =`square`

∴ The equation of trend line is y = `square`


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Solution:

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

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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
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  `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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