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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 Copyright Š 2012-2015 Economicshelpdesk.com, All rights reserved
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.
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