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Linear Functions: Y = a + bX
X
Y
0 1 2 3 4 5 6 7 8 9 10
2
4
6
8
10
Y = 10 - XY = 1 + 0.8X
Y = 6 + 0.4X
Example 1: Linear functional form
Bt : The per capita consumption of beef in year t (in pounds per person)
Pt : The price of beef in year t (in cents per pound)
Ydt : The per capita disposable income in year t (in thousand of dollars)
Bt = 37.54*** - 0.88***Pt + 11.89***Ydt
se (10.0402) (0.1647) (1.7622)
R2 = 0.6580, N = 28, SER = 6.0806
^
Example 6.7: Double-log functional form
lnBt = 3.5944*** - 0.3444***lnPt + 1.0715***lnYdt
se (0.1413) (0.0622) (0.1485)
R2 = 0.7099, N = 28, SER = 0.0536
Bt : The per capita consumption of beef in year t (in pounds per person)
Pt : The price of beef in year t (in cents per pound)
Ydt : The per capita disposable income in year t (in thousand of dollars)
^
Example 4: Left-side semi-log functional form
lnBt = 3.9970*** - 0.0083***Pt + 0.1139***Ydt
se (0.0945) (0.0015) (0.0166)
R2 = 0.6699, N = 28, SER = 0.0057
Bt : The per capita consumption of beef in year t (in pounds per person)
Pt : The price of beef in year t (in cents per pound)
Ydt : The per capita disposable income in year t (in thousand of dollars)
^
Eg 2 and Eg 5
lnB exp(lnB) R2 = 0.6707
Bt = 227.888*** - 0.804***Pt – 758.093***(1/Ydt)
se (11.7778) (0.0990) (69.9654)
R2 = 0.8306, N = 28, SER = 4.2795
^
lnBt = 3.5944*** - 0.3444***lnPt + 1.0715***lnYdt
se (0.1413) (0.0622) (0.1485)
R2 = 0.7099, N = 28, SER = 0.0536
^
(1) Bt = 37.54*** - 0.88***Pt + 11.89***Ydt R2 = 0.66^
(2) lnBt = 3.59*** - 0.34***lnPt + 1.07***lnYdt R2 = 0.71^
(3) Bt = -71.75*** - 0.87***Pt + 98.87***lnYdt R2 = 0.77^
(4) lnBt = 4.00*** - 0.01***Pt + 0.11***Ydt R2 = 0.67^
(5) Bt = 227.89*** - 0.80***Pt – 758.09***(1/Ydt) R2 = 0.83^
Data: BE4_Tab0604.xlsChild Mortality
CM: Child mortality
FLR: Female literacy rate
PGNP:Per capita GNP in 1980
TFR: Total fertility rate
WAGE EDUC EXPER FEMALE MARRIED
3.10 11 2 1 0
3.24 12 22 1 1
3.00 11 2 0 0
6.00 8 44 0 1
5.40 12 7 0 1
Partial Data for the relation
wage = f(educ, exper, gender, status)
7. The Dummy Variable Approach to the Chow Test
You believe that the data can be classified into two groups, A and B.
The Chow test
can test the hypothesis
cannot tell us the source of the difference.
Yi = 0 + 1X1i + 2X2i + i, i = 1,…,N
Define Di = 1 for group A
Di = 0 otherwise.
Consider the model
Yi = 0 + 0Di + 1X1i + 1(DiX1i)
+ 2X2i + 2(DiX2i) + i
For Di = 0,
Yi = 0 + 1X1i + 2X2i + t
For Di = 1,
Yi = (0 + 0) + (1 + 1)X1i + (2 + 2)X2i + i
Example 17: (HtWt_2008s) The dependent variable is “weight” in Kg. hh = height – 160cm.
Model 1 Model 2 Model 3
Intercept
50.93*** 49.00*** 48.98***
male 10.10*** 10.43***
hh 0.92*** 0.43*** 0.46***
male*hh 0.05
R2 0.63 0.72 0.72
N 61 61 61
RSS 2464.88 1848.68 1849.78
Example 7.9:
** Investment (INV) depends on value of the firm (V) and stock of capital (K).
** INV = 0 + 1 V + 2 K + .
** 2 firms: GE and Westinghouse
** Test whether they have the same investment function using the dummy variable approach?
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Aggregate Disposable Income
Agg
rega
te C
onsu
mtio
n
Scatter Diagram: US Aggregate Consumption, 1940 - 1950
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INC
CO
NS
WAR=0WAR=1
Scatter Diagram: US Aggregate Consumption, 1940 - 1950
