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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 | 2000 | 2005 | 2010 |
| IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |
Fit the trend line to the above data by the method of least squares.
उत्तर
Let Yt = a + bu be the trend equation where a and b are constants obtained by solving the normal equations.
Here n = 7 and interval h = 5,
middle year = 1995.
u = `("t" - "middle year")/"interval "`
u = `("t" - 1995)/5`
We obtain the following table:
| Year (t) | IMR (y) | u = `bb((("t" - 1995)/5))` | u2 | u.y | yt |
| 1980 | 10 | – 3 | 9 | – 30 | 8.9999 |
| 1985 | 7 | – 2 | 4 | – 14 | 7.4285 |
| 1990 | 5 | – 1 | 1 | – 5 | 5.8571 |
| 1995 | 4 | 0 | 0 | 0 | 4.2857 |
| 2000 | 3 | 1 | 1 | 3 | 2.7143 |
| 2005 | 1 | 2 | 4 | 2 | 1.1429 |
| 2010 | 0 | 3 | 9 | 0 | – 0.4285 |
| Total | 30 | 0 | 28 | – 44 |
From the table, n = 7, Σy = 30, Σu = 0,
Σu2 = 28 and Σuy = – 44
The two normal equations are
Σy = na + bΣu
and Σuy = aΣu + Σu2
Substituting the values, we will get
∴ 30 = 7a and – 44 = 28b
a = `30/7` and b = `(– 44)/28`
a = 4.2857 and b = –1.5714
∴ Equation of the trend line yt = a + bu is
yt = 4.2857 + (–1.5714)u
yt = 4.2857 – 1.5714u ... (i)
Substituting u = –3, –2, –1, 0, 1, 2, 3 in equation (i), we will get the trend values of y 1 and tabulate in the last column.
When u = –3,
yt = 4.2857 – 1.5741 x (–3)
= 4.2857 + 4.7142
= 8.9999
When u = –2,
yt = 4.2857 – 1.5741 x (–2)
= 7.4285
When u = –1,
yt = 4.2857 – 1.5741 x (–1)
= 5.8571
When u = 0,
yt = 4.2857 – 1.5741 x (0)
= 4.2857
When u = 1, yt = 2.7143
When u = 2, yt = 1.1429
When u = 3, yt = – 0.4285
Tabulate the trend values yt in the last column.
