Question

The following table gives the production of steel (in millions of tons) for years 1976 to 1986:

Year	1976	1977	1978	1979	1980	1981	1982	1983	1984	1985	1986
Production	0	4	4	2	6	8	5	9	4	10	10

Fit a trend line to the above data by the method of least squares.

Answer 1

Here, n = 11. We transform year t to u by taking u = t − 1981.

We construct the following table for calculation:

Year (t)	Production (xt)	u = t − 1981	u2	uxt
1976	0	−5	25	0
1977	4	−4	16	−16
1978	4	−3	9	−12
1979	2	−2	4	−4
1980	6	−1	1	−6
1981	8	0	0	0
1982	5	1	1	5
1983	9	2	4	18
1984	4	3	9	12
1985	10	4	16	40
1986	10	5	25	50
Total	∑xt = 62	∑u = 0	∑u2 = 110	125 − 38 =  ∑uxt = 87

The equation of trend line is x_t = a′ + b′u

The normal equations are

Σx_t = na′ + b′Σu    ...(1)

Σux_t = a′Σu + b′Σu²    ...(2)

Here, n = 11, Σx_t = 62, Σu = 0, Σu² = 110, Σux_t = 87

Putting these values in normal equations, we get

62 = 11a′ + b′(0)    ...(3)

87 = a′(0) + b′(110)    ...(4)

From equation (3), we get

a' = `62/11 = 5.6364`

From equation (4), we get

b' = `87/110 = 0.7909`

Putting a′ = 5.6364 and b′ = 0.7909 in the equation

x_t = a′ + b′u, we get, the equation of trend line as

x_t = 5.6364 + 0.7909 u.