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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.
APPEARS IN
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| 1934 | 2 | 1940 | 1 |
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Complete the following activity to fit a trend line to the following data by the method of least squares.
| 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`
