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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.
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Solution
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 yt = a' + b'u
i.e., yt = 2.4 + 0.2848 u, where u = 2(t – 1963.5).
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