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Fit equation of trend line for the data given below. 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 16 201 - Mathematics and Statistics

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Question

Fit equation of trend line for the data given below.

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`

The normal equations are Σy = na + bΣx

As Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx2

As Σx = 0, b = `square`

Substitute values of a and b in equation (i) the equation of trend line is `square`

To find trend value for the year 2016, put x = `square` in the above equation.

y = `square`

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Solution

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

Here n = 10 (even), two middle years are 2010 and 2011, and h = 2 

The normal equations are Σy = na + bΣx

As Σx = 0,

∴ 177 = 10a + b(0)

∴ a = `177/10`

a = 17.7 

Also, Σxy = aΣx + bΣx2

As Σx = 0, 

27 = a(0) + b(330)

∴ b = `27/330`

= 0.08

b = 0.1

Substitute values of a and b in equation (i) the equation of trend line is y = 17.7 + 0.1x

To find trend value for the year 2016, put x = 11 in the above equation.

y = 17.7 + 1.1

y = 18.8

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Measurement of Secular Trend
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Chapter 2.4: Time Series - Q.5

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Solution:

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

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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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