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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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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 |
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| Year | Production | 4 yearly moving total |
4 yearly centered total |
4 yearly centered moving average (trend values) |
| 2006 | 19 | – | – | |
| `square` | ||||
| 2007 | 20 | – | `square` | |
| 72 | ||||
| 2008 | 17 | 142 | 17.75 | |
| 70 | ||||
| 2009 | 16 | `square` | 17 | |
| `square` | ||||
| 2010 | 17 | 133 | `square` | |
| 67 | ||||
| 2011 | 16 | `square` | `square` | |
| `square` | ||||
| 2012 | 18 | 140 | 17.5 | |
| 72 | ||||
| 2013 | 17 | 147 | 18.375 | |
| 75 | ||||
| 2014 | 21 | – | – | |
| – | ||||
| 2015 | 19 | – | – |
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10 | 15 | 20 | 25 | 30 |
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| Production xi |
35 | 40 | 45 | 50 | 55 |
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| 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`
