The models below have been estimated using monthly data for the period 2000: 1-2020: 12. The natural logarithm of the variables used in this model is taken and the transaction is made. D08 is the dummy variable that takes a value of 1 in 2008 and after and 0 in other periods. s.e. indicates the standard error.Comment on the coefficients of variables D08 and (X2t * D08) in the model (3) above. Y = 0.036 -1.102*X1, + 0.412*X2,(1)R? = 0.853; RSS = 0.228Y; = 0.046 - 1.074*X1; + 0.681*X2, +0.565*X3; + 0.452*X4(2)R? = 0.936; RSS = 0.142= 0.042 + 1.130*X1, +0.415*X2, 0.150*D08 + 0.113*(X2, *D08)s.e. (0.007) (0.115)R? = 0.963; RSS = 0.0136Y,(3)(0.005)(0.019)(0.005)& = -0.041 + 0.082X1, - 0.071X1-1 + 0.046&-1+ 0.069&-2 + 0.265 &-10(0.151)(0.38 0) (0.150)(0.001 )(0.014)(0.008)R? = 0.280(4)E? = 0.648+ 0.0675£² + 0.092ê?, + 0,765 £?, + 0.662 £?,(0.38 0)R? = 0.550(0.014)(0.085)(0.251)(0.214)(5)? = 0.648+ 0.252X,(0.38 0) (0.015)R? = 0.6890.138X2t(0.056)1t(6)

Yt=0.042+1.130*X1t+0.415X2t 0.150*D08+0.113*(X2t*D08) Variable DO8 is a dummy variable that…

Answered: Y, = 0.042 + 1.130*X1, +0.415*X2,…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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The models below have been estimated using monthly data for the period 2000: 1-2020: 12. The natural logarithm of the variables used in this model is taken and the transaction is made. D08 is the dummy variable that takes a value of 1 in 2008 and after and 0 in other periods. s.e. indicates the standard error.

Comment on the coefficients of variables D08 and (X2t * D08) in the model (3) above.

Y = 0.036 -1.102*X1, + 0.412*X2,
(1)
R? = 0.853; RSS = 0.228
Y; = 0.046 - 1.074*X1; + 0.681*X2, +0.565*X3; + 0.452*X4
(2)
R? = 0.936; RSS = 0.142
= 0.042 + 1.130*X1, +0.415*X2, 0.150*D08 + 0.113*(X2, *D08)
s.e. (0.007) (0.115)
R? = 0.963; RSS = 0.0136
Y,
(3)
(0.005)
(0.019)
(0.005)
& = -0.041 + 0.082X1, - 0.071X1-1 + 0.046&-1+ 0.069&-2 + 0.265 &-10
(0.151)
(0.38 0) (0.150)
(0.001 )
(0.014)
(0.008)
R? = 0.280
(4)
E? = 0.648+ 0.0675£² + 0.092ê?, + 0,765 £?, + 0.662 £?,
(0.38 0)
R? = 0.550
(0.014)
(0.085)
(0.251)
(0.214)
(5)
? = 0.648+ 0.252X,
(0.38 0) (0.015)
R? = 0.689
0.138X2t
(0.056)
1t
(6)

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