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प्रश्न
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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उत्तर
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: 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 |
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| 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`
