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Question
Solve the following problem :
Fit a trend line to data in Problem 4 by the method of least squares.
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Solution
In the given problem, n = 12 (odd), middle t – value is 1976, h = 1
u = `"t - middle value"/("h"/2) = ("t" - 1976.5)/(1/2)` = 2(t – 1976.5)
We obtain the following table.
| Year t |
Production yt |
u = 2(t – 1976.5) | u2 | uyt | Trend Value |
| 1971 | 1 | –11 | 121 | –11 | 0.0900 |
| 1972 | 0 | –9 | 81 | 0 | 0.6494 |
| 1973 | 1 | –7 | 49 | –7 | 1.2088 |
| 1974 | 2 | –5 | 25 | –10 | 1.7682 |
| 1975 | 3 | – | 9 | –9 | 2.3276 |
| 1976 | 2 | –1 | 1 | –2 | 2.8870 |
| 1977 | 3 | 1 | 1 | 3 | 3.4464 |
| 1978 | 6 | 3 | 9 | 18 | 4.0058 |
| 1979 | 5 | 5 | 25 | 25 | 4.5652 |
| 1980 | 1 | 7 | 49 | 7 | 5.1246 |
| 1981 | 4 | 9 | 81 | 36 | 5.6840 |
| 1982 | 10 | 11 | 121 | 110 | 6.243 |
| Total | 38 | 0 | 572 | 160 |
From the table, n = 12, `sumy_"t" = 38, sumu = 0, sumu^2 = 572,sumuy_"t" = 160`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 38 = 12a' + b'(0) ...(i) and
160 = a'(0) + b'(572) ...(ii)
From (i), a' = `(38)/(12)` = 3.1667
From (ii), b' = `(160)/(572)` = 0.2797
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 3.1667 + 0.2797 u, where u = 2(t – 1976.5).
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Let the equation of trend line be y = a + bx .....(i)
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| 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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| 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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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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| Years | 1966 | 1967 | 1968 | 1969 | 1970 |
| Gross Capital information | 20 | 25 | 25 | 30 | 35 |
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| Gross Capital information | 30 | 45 | 40 | 55 | 65 |
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| 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`
