Question

Solve the following problem :

Fit a trend line to data in Problem 4 by the method of least squares.

Answer 1

In the given problem, n = 12 (odd), middle t – value is 1976, h = 1

u = `"t - middle value"/("h"/2) = ("t" - 1976.5)/(1/2)` = 2(t – 1976.5)

We obtain the following table.

Year t	Production yt	u = 2(t – 1976.5)	u2	uyt	Trend Value
1971	1	–11	121	–11	0.0900
1972	0	–9	81	0	0.6494
1973	1	–7	49	–7	1.2088
1974	2	–5	25	–10	1.7682
1975	3	–	9	–9	2.3276
1976	2	–1	1	–2	2.8870
1977	3	1	1	3	3.4464
1978	6	3	9	18	4.0058
1979	5	5	25	25	4.5652
1980	1	7	49	7	5.1246
1981	4	9	81	36	5.6840
1982	10	11	121	110	6.243
Total	38	0	572	160

From the table, n = 12, `sumy_"t" = 38, sumu = 0, sumu^2 = 572,sumuy_"t" = 160`

The two normal equations are: `sumy_"t" = "na"' + "b"' sumu  "and" sumuy_"t", = a'sumu + b'sumu^2`

∴ 38 = 12a' + b'(0)            ...(i)   and
160 = a'(0) + b'(572)         ...(ii)

From (i), a' = `(38)/(12)` = 3.1667

From (ii), b' = `(160)/(572)` = 0.2797
∴  The equation of the trend line is y_t = a' + b'u
i.e., y_t = 3.1667 + 0.2797 u, where u = 2(t – 1976.5).