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प्रश्न
Fit a straight line trend by the method of least squares to the following data
| Year | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 | 1987 |
| Sales | 50.3 | 52.7 | 49.3 | 57.3 | 56.8 | 60.7 | 62.1 | 58.7 |
उत्तर
| Year (x) |
Sales (y) |
x = `((x - 1983.5))/0.5` | xy | x2 | Trend values (yt) |
| 1980 | 50.3 | – 7 | – 352.1 | 49 | 50.1775 |
| 1981 | 52.7 | – 5 | – 263.5 | 25 | 51.8375 |
| 1982 | 49.3 | – 3 | – 147.9 | 9 | 53.4975 |
| 1983 | 57.3 | – 1 | – 57.3 | 1 | 55.1575 |
| 1984 | 56.8 | 1 | 56.8 | 1 | 56.8175 |
| 1985 | 60.7 | 3 | 182.1 | 9 | 58.4775 |
| 1986 | 62.1 | 5 | 310.5 | 25 | 60.1375 |
| 1987 | 58.7 | 7 | 410.9 | 49 | 61.7975 |
| N = 8 | `sum` = 447.9 | `sumx` = 0 | `sumxy` = 139.5 | `sumx^2` = 168 | `sumy"t"` = 447.9 |
a = `(sumy)/"n" = 447.9/8` = 55.9875
b = `(sumxy)/(sumx^2) = 139.5/168` = 0.830
Therefore, the required equation of the straight line trend is given by
y = a + bx
y = 55.9875 + 0.830x
⇒ y = 55.9875 + 0.83`((x - 1983.5)/0.5)`
The trend values can be obtained by
When x = 1980
y = 55.9875 + 0.83 `((1980 - 1983.5)/0.5)`
= 55.9875 + 0.83(– 7)
= 55.9875 – 5.81
= 50.1775
When x = 1981
y = 55.9875 + 0.83 `((1981 - 1983.5)/0.5)`
= 55.9875 + 0.83(– 5)
= 55.9875 – 4.15
= 51.8375
When x = 1982
y = 55.9875 + 0.83 `((1981 - 1983.5)/0.5)`
= 55.9875 + 0.83(– 3)
= 55.9875 – 2.49
= 53.4975
When x = 1983
y = 55.9875 + 0.83 `((1983 - 1983.5)/0.5)`
= 55.9875 + 0.83(– 1)
= 55.9875 – 0.83
= 55.1575
When x = 1984
y = 55.9875 + 0.83 `((1984 - 1983.5)/0.5)`
= 55.9875 + 0.83(1)
= 56.8175
When x = 1985
y = 55.9875 + 0.83 `((1985 - 1983.5)/0.5)`
= 55.9875 + 0.83(3)
= 55.9875 + 2.49
= 58.4775
When x = 1986
y = 55.9875 + 0.83 `((1986 - 1983.5)/0.5)`
= 55.9875 + 0.83(5)
= 55.9875 + 4.15
= 60.1375
When x = 1987
y = 55.9875 + 0.83 `((1987 - 1983.5)/0.5)`
= 55.9875 + 0.83(7)
= 55.9875 + 5.81
= 61.7975
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