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Question
Determine the equation of a straight line which best fits the following data
| Year | 2000 | 2001 | 2002 | 2003 | 2004 |
| Sales (₹ '000) | 35 | 36 | 79 | 80 | 40 |
Compute the trend values for all years from 2000 to 2004
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Solution
Computation of trend values by the method of least squares. (ODD years)
| Year (x) |
Sales (y) (₹ 000) |
X = (x - 2002) | X2 | XY | Tred Value (Yt) |
| 2000 | 335 | – 2 | 4 | – 70 | 43.2 |
| 2001 | 36 | – 1 | 1 | – 36 | 48.6 |
| 2002 | 79 | 0 | 0 | 0 | 54 |
| 2003 | 80 | 1 | 1 | 80 | 59.4 |
| 2004 | 40 | 2 | 4 | 80 | 64.8 |
| N = 5 | `sum"Y"` = 270 | `sum"X"` = 0 | `sum"X"^2` = 10 | `sum"XY"` = 54 | `sum"Y""t"` 270 |
a = `(sum"Y")/"n" = 270/5` = 54
b = `(sum"XY")/(sum"X"^2) = 54/10` = 5.4
Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 54 + 5.4X
Y = 54 + 5.4(x – 2002)
The trend value can be obtained by when x = 2000
Yt = 54 + 5.4(2000 – 2002)
Y = 54+ 5.4(– 2)
= 54 – 10.8
= 43.2
When x = 2001
Yt = 54 + 5.4(2001 – 2002)
Y = 54 + 5.4(– 1)
= 54 – 5.4
= 48.6
When x = 2002
Yt = 54 + 5.4(2002 – 2002)
V = 54 + 5.4(0)
= 54
When x = 2003
Yt = 54 + 5.4(2003 – 2002)
Y = 54 + 5.4(1)
= 54 + 5.4
= 59.4
When x = 2004
Yt = 54 + 5.4(2004 – 2002)
Y = 54 + 5.4(2)
= 54 + 10.8
= 64.8
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