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

Obtain the trend value for the year 1990

Answer 1

In the given problem, n = 11 (odd), middle t- values is 1981, h = 1

u = `("t" - "middle t value")/"h"`

= `("t" - 1981)/1`

= t – 1981

We obtain the following table:

Year  t	Production yt	u = t − 1981	u2	uyt	Trend Value
1976	0	− 5	25	0	1.6819
1977	4	− 4	16	− 16	2.4728
1978	4	− 3	9	− 12	3.2637
1979	2	− 2	4	− 4	4.0546
1980	6	− 1	1	− 6	4.8455
1981	8	0	0	0	5.6364
1982	5	1	1	5	6.4273
1983	9	2	4	18	7.2182
1984	4	3	9	12	8.0091
1985	10	4	16	40	8.8
1986	10	5	25	50	9.5909
Total	62	0	110	87	87

From the table, n = 11, ∑y_t = 62, ∑u = 0, ∑u² = 110, ∑uy_t = 87

The two normal equations are:

∑y_t = na' + b'∑u and ∑uy_t = a'∑u + b'∑u²

∴ 62 = 11a' + b'(0)   .....(i)

and

87 = a'(0) + b'(110)  .....(ii)

From (i), a′ = `62/11` = 5.6364

From (ii), b′ = `87/110` = 0.7909

∴ The equation of the trend line is y_t = a′ + b′u

i.e., y_t = 5.6364+ 0.7909 u,

where u = t – 1981

Now, for t = 1990,

u = 1990 – 1981

= 9

∴ y_t = 5.6364 + 0.7909 × 9

= 12.7545