Question

Use the method of least squares to fit a trend line to the data given below. Also, obtain the trend value for the year 1975.

Year	1962	1963	1964	1965	1966	1967	1968	1969
Production (million barrels)	0	0	1	1	2	3	4	5
Year	1970	1971	1972	1973	1974	1975	1976
Production (million barrels)	6	8	9	9	8	7	10

Answer 1

In the given problem, n = 15 (odd), middle t – values is 1969, h = 1

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

= `("t" - 1969)/1`

= t – 1969

We obtain the following table:

Year  t	Production yt	u = t − 1969	u2	uyt	Trend Value
1962	0	−	49	0	− 0.6
1963	0	− 6	36	0	0.2
1964	1	− 5	25	− 5	1
1965	1	− 4	16	− 4	1.8
1966	2	− 3	9	− 6	2.6
1967	3	− 2	4	− 6	3.4
1968	4	− 1	1	− 4	4.2
1969	5	0	0	0	5
1970	6	1	1	6	5.8
1971	8	2	4	16	6.6
1972	9	3	9	27	7.4
1973	9	4	16	36	8.
1974	8	5	25	40	9
1975	9	6	36	54	9.8
1976	10	7	49	70	10.6
Total	75	0	280	224

From the table, n = 15, ∑y_t = 75, ∑u = 0, ∑u² = 280, ∑uy_t = 224

The two normal equations are:

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

∴ 75 = 15a' + b'(0)   ......(i)

and

224 = a′(0) + b′(280)  .....(ii)

From (i), a′ = `75/15` = 5

From (ii), b′= `224/280` = 0.8

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

i.e., y_t = 5 + 0.8 u, where u = t – 1969

Now, for t = 1975, u = 1975 – 1969 = 6

∴  y_t = 5 + 0.8 × 6 = 9.8