Question

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`

Answer 1

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, Σy_t = 30, Σx = 0, Σx² = 28, Σxy_t = – 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