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प्रश्न
Solve the following problem :
Fit a trend line to data in Problem 13 by the method of least squares.
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उत्तर
In the given problem, n = 9 (odd), middle t – value is 1979, h – 1
u = `"t - middle value"/"h" = ("t" - 1979)/(1)` = t – 1979
We obtain the following table.
| Year t |
No. of deaths yt |
u = t – 1979 | u2 | uyt | Trend Value |
| 1975 | 0 | –4 | 16 | 0 | 2.5554 |
| 1976 | 6 | –3 | 9 | –18 | 3.2221 |
| 1977 | 3 | –2 | 4 | –6 | 3.8888 |
| 1978 | 8 | –1 | 1 | –8 | 4.5555 |
| 1979 | 2 | 0 | 0 | 0 | 5.2222 |
| 1980 | 9 | 1 | 1 | 9 | 5.8887 |
| 1981 | 4 | 2 | 4 | 8 | 6.5556 |
| 1982 | 5 | 3 | 9 | 15 | 7.2223 |
| 1983 | 10 | 4 | 16 | 40 | 7.8890 |
| Total | 47 | 0 | 60 | 40 |
From the table, n = 9, `sumy_"t" = 47, sumu = 0, sumu^2 = 60,sumuy_"t" = 40`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 47 = 9a' + b'(0) ...(i) and
40 = a'(0) + b'(60) ...(ii)
From (i), a' = `(47)/(9)` = 5.2222
From (ii), b' = `(40)/(60)` = 0.6667
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 5.2222 + 0.6667 u, where u = t – 1979.
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संबंधित प्रश्न
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| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
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| Production | 3 | 6 | 5 | 1 | 4 | 10 |
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| Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 |
| Production | 0 | 4 | 4 | 2 | 6 | 8 |
| Year | 1982 | 1983 | 1984 | 1985 | 1986 | |
| Production | 5 | 9 | 4 | 10 | 10 |
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| Year | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 | 1969 |
| Production (million barrels) |
0 | 0 | 1 | 1 | 2 | 3 | 4 | 5 |
| Year | 1970 | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 | |
| Production (million barrels) |
6 | 8 | 9 | 9 | 8 | 7 | 10 |
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| Year | 1980 | 1985 | 1990 | 1995 |
| IMR | 10 | 7 | 5 | 4 |
| Year | 2000 | 2005 | 2010 | |
| IMR | 3 | 1 | 0 |
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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 |
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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`
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 |
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10 | 15 | 20 | 25 | 30 |
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| Production xi |
35 | 40 | 45 | 50 | 55 |
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| Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
| IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |
