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प्रश्न
Solve the following problem :
Fit a trend line to data by the method of least squares.
| Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
| Number of boxes (in ten thousands) | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
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उत्तर
In the given problem, n = 8 (even), two middle t – values are 1980 and 1981, h – 1
u = `"t - mean of two middle values"/("h"/2) = ("t" - 1980.5)/(1/2)` = 2(t – 1980.5)
We obtain the following table.
| Year t |
No. of boxes (in ten thousands) yt |
u = 2(t – 1980.5) | u2 | uyt | Trend Value |
| 1977 | 1 | –7 | 49 | –7 | 1.5836 |
| 1978 | 0 | –5 | 25 | 0 | 2.5240 |
| 1979 | 3 | –3 | 9 | –9 | 3.4644 |
| 1980 | 8 | –1 | 1 | –8 | 4.4048 |
| 1981 | 10 | 1 | 1 | 10 | 5.3452 |
| 1982 | 4 | 3 | 9 | 12 | 6.2856 |
| 1983 | 5 | 5 | 25 | 25 | 7.2260 |
| 1984 | 8 | 7 | 49 | 56 | 8.1664 |
| Total | 39 | 0 | 168 | 79 |
From the table, n = 8, `sumy_"t" = 39, sumu = 0, sumu^2 = 168,sumuy_"t" = 79`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 39 = 8a' + b'(0) ...(i) and
79 = a'(0) + b'(168) ...(ii)
From (i), a' = `(39)/(8)` = 4.875
From (ii), b' = `(79)/(168)` = 0.4702
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.875 + 0.4702 u, where u = 2(t – 1980.5).
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संबंधित प्रश्न
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| Year | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 | 1969 | 1970 | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production (Million Barrels) |
0 | 0 | 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 8 | 9 | 10 |
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| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
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| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
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| Year | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |
| No. of deaths | 0 | 6 | 3 | 8 | 2 | 9 | 4 | 5 | 10 |
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| Year | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 |
| Production | 1 | 0 | 1 | 2 | 3 | 2 |
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| Production | 4 | 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 | 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
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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`
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As, Σx = 0, b =`square`
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| 1980 | 10 | – | – |
| 1985 | 7 | `square` | 7.33 |
| 1990 | 5 | 16 | `square` |
| 1995 | 4 | 12 | 4 |
| 2000 | 3 | 8 | `square` |
| 2005 | 1 | `square` | 1.33 |
| 2010 | 0 | – | – |
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| 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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| No. of subscribers (in millions) |
12 | 11 | 19 | 17 |
| Years | 1980 | 1981 | 1982 | 1983 |
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