Question

Solve the following problem :

Following tables shows the wheat yield (‘000 tonnes) in India for years 1959 to 1968.

Year	1959	1960	1961	1962	1963	1964	1965	1966	1967	1968
Yield	0	1	2	3	1	0	4	1	2	10

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

Answer 1

In the given problem, n = 10 (even), two middle t – value are 1963 and 1964, h = 1

u = `"t - mean of two middle values"/("h"/2) = ("t" - 1963.5)/(1/2)` = 2(t – 1963.5)

We obtain the following table.

Year t	Yield (in '000 tonnes)  yt	u = 2(t – 1963.5)	u2	uyt	Trend Value
1959	0	–9	81	0	–0.1632
1960	1	–7	49	–7	0.4064
1961	2	–5	25	–10	0.9760
196	3	–3	9	–9	1.5456
1963	1	–1	1	–1	2.1152
1964	0	1	1	0	2.6848
1965	4	3	9	12	3.2544
1966	1	5	25	5	3.8240
1967	2	7	49	14	4.3936
1968	10	9	81	90	4.9632
Total	24	0	330	94

From the table, n = 10, `sumy_"t" = 24, sumu = 0, sumu^2 = 330,sumuy_"t" = 94`

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

∴ 24 = 10a' + b'(0)               ...(i)   and
94 = a'(0) + b'(330)              ...(ii)

From (i), a' = `(24)/(10)` = 2.4

From (ii), b' = `(94)/(330)` = 0.2848
∴  The equation of the trend line is y_t = a' + b'u
i.e., y_t = 2.4 + 0.2848 u, where u = 2(t – 1963.5).