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1. Statistics Homework Help
Statistics Assignment Help
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Statistics Assignment Sample Illustrations and Solutions:
Illustration 1.
Shift the trend origin from 2001 to 2004 in the straight line trend equation, Yc = 25 + 2x,
given that the time unit = 1 year.
Solution
Here, the trend origin is to be shifted forward by 3 years i.e. from 2001 to 2004. Thus k =3.
By the formula of trend shifting se l have,
Yc = a + b (X + K)
Substituting the respective values in the above we get, Yc = 25 + 2 (X + 3)
= 25 + 2X + 6
= 31 + 2X
Thus, the shifted equation is given by
Yc = 31 + 2X
Where, origin of X = 2004, and
X unit = 1 year.
Note. From the above shifted equation, it must be note4d that the value of a only6 has been
changed from 25 to 31, whereas the value of b remains the same 2.

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Illustration 2
The trend equation for a time series is fitted as Yc = 60 + 1.5X shift the trend origin from
2004 ti 1998, given that the time unit is 1 year.
Solution
Here, the trend origin is to be shifte4d back by 5 years i.e. from 2004 to 1999.
Thus k = -5
By the formula of trend shifting we have, Yc = a + b (X –K)
= 60 + 1.5 (X – 5)
= 60 + 1.5X – 7.5
= Yc = 52.5 + 1.5X, where the origin of X = 1999.

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Illustration 3.
Shift the trend origin from 2002 to 2004 in the following trend equation:
Yc = 20.3 + 0.4X + 0.5X2
Given that the time out unit = 1 Year.
Solution
Here, the trend origin is to be shifted forward by 2 years i.e. from 2002 to 2004. Hence k =
2. By the formula of trend shifting we have,
Yc = a + b (X + K) + c (X + K)2
(it is a parabolic equation of second degree)
Substituting the respective values in the above, we get Yc = 20.3 + 0.4 (X + 2) + 0.5 (X +
2)2
= 20.3 + 0.4 X + 0.8 + 0.5 (X2
+ 4X + 4)
= 20.3 + 0.4X + 0.8 + 0.5X2
+ 2X + 2
= Yc = 23.1 + 2.4 X + 0.5X2

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In the above equation, it must be seen that the values of both the a and b have been
changed, whereas the value of c has remained unchanged. This is because; it was the case
of shifting a parabolic equation of second degree.