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प्रश्न
Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010
Year | 1980 | 1985 | 1990 | 1995 |
IMR | 10 | 7 | 5 | 4 |
Year | 2000 | 2005 | 2010 | |
IMR | 3 | 1 | 0 |
Fit a trend line by the method of least squares
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 |
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`
उत्तर
Let the equation of trend line be y = a + bx .....(i)
Here n = 7(odd), middle year is 1995 and h = 5
x = `("t" - "middle year")/"h"`
= `("t" - 1995)/5`
We obtain the following table:
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 |
2010 | 0 | 3 | 9 | 0 |
Total | 30 | 0 | 28 | – 44 |
From the table, n = 7, Σyt = 30, Σx = 0, Σx2 = 28, Σxyt = – 44
The normal equations are
Σy = na + bΣx
∴ 30 = 7a + bΣx
As, Σx = 0, a = `30/7` = 4.2857
Also, Σxy = aΣx + bΣx2
∴ – 44 = aΣx + b′(28)
As, Σx = 0, b =`(-44)/28` = – 1.5714
∴ The equation of trend line is y = a + bx
∴ The equation of trend line is y = 4.2857 – 1.5714x
APPEARS IN
संबंधित प्रश्न
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Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
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Year | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 |
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The following table shows gross capital information (in Crore ₹) for years 1966 to 1975:
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 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |
Number of deaths | 0 | 6 | 3 | 8 | 2 | 9 | 4 | 5 | 10 |
Solution:
Here n = 9. We transform year t to u by taking u = t - 1979. We construct the following table for calculation :
Year t | Number of deaths xt | u = t - 1979 | u2 | uxt |
1975 | 0 | - 4 | 16 | 0 |
1976 | 6 | - 3 | 9 | - 18 |
1977 | 3 | - 2 | 4 | - 6 |
1978 | 8 | - 1 | 1 | - 8 |
1979 | 2 | 0 | 0 | 0 |
1980 | 9 | 1 | 1 | 9 |
1981 | 4 | 2 | 4 | 8 |
1982 | 5 | 3 | 9 | 15 |
1983 | 10 | 4 | 16 | 40 |
`sumx_t` =47 | `sumu`=0 | `sumu^2=60` | `square` |
The equation of trend line 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 = 47, sumu= 0, sumu^2 = 60`
By putting these values in normal equations, we get
47 = 9a' + b' (0) ...(3)
40 = a'(0) + b'(60) ...(4)
From equation (3), we get a' = `square`
From equation (4), we get b' = `square`
∴ the equation of trend line is xt = `square`