The document discusses dummy variables, which take on values of 1 and 0 to represent dichotomous categories like male/female or urban/rural. Dummy variables allow for estimating different intercepts for subgroups in a regression. When using two dummy variables to represent a single categorical variable, there is perfect multicollinearity, so only m-1 dummies should be used for a variable with m categories. The document provides examples of using dummy variables to estimate separate regression lines for male and female subgroups.

1. 1
DUMMY
VARIABLES
7
LECTURE

3. Nature of “dummy” variable:
(1)Variables that assume such “1” and “0” values
(2) Variables usually indicates the dichotomized
“presence” or “absence”, “yes” or “no”, etc.
(3) Variables indicates a “quality” or an attribute,
such as “male” or “female”,
“black” or “white”,
“urban” or non-urban”
“before” or “after”
“North” or “south”, “east” or “west”
………..etc.

4. obs
Male=1
Dummy Female=1
Dummy
Salary(K)
Years of
teaching
1 1 0 23 1
2 0 1 19.5 1
3 1 0 24 2
4 0 1 21 2
5 1 0 25 3
6 0 1 22 3
7 1 0 26.5 4
8 0 1 23.1 4
9 0 1 25 5
10 1 0 28 5
11 1 0 29.5 6
12 0 1 26 6
13 0 1 27.5 7
14 1 0 31.5 7
15 0 1 29 6
16 1 0 22 5
17 0 1 19 2
18 1 0 18 2
19 0 1 21.7 5
20 0 1 18.5 2
21 1 0 21 4
22 1 0 20.5 4
23 0 1 17 1
24 0 1 17.5 1
25 1 0 21.2 5

5. Separate male sample
obs Starting salary, Y Years of teaching, X2
1 23 1
3 24 2
5 25 3
7 26.5 4
10 28 5
11 29.5 6
14 31.5 7
16 22 5
18 21.7 5
21 21 4
22 20.5 4
25 21.2 5

6. obs Staring salary, Y Years of teaching, X2
2 19.5 1
4 21 2
6 22 3
8 23.1 4
9 25 5
12 26 6
13 27.5 7
15 29 6
17 19 2
19 18 2
20 18.5 2
23 17 1
24 17.5 1
Separate female sample

7. 10
15
20
25
30
35
0 1 2 3 4 5 6 7 8
Male
Linear (Male)
Salary Y
X
teaching
years
Y = 1 + 2 X (male)
^ ^ ^
Two separate models: Ym = 1 +  2 Xm + um
Yf = ’1 + ’2 Xf + uf
(male)
(female)
Linear (Female)
Female
Y = ’1+ ’2X (female)
^ ^
^

8. Assuming *2 = 2, same slope but different constant
between Yi and Xi.
1st model: Yi = 1 + *1 Di + 2 Xi + ui
Yi = 1 + *1 Di + 2 Xi + *2 DiXi + ui
Yi = annual salary (each obs.)
Xi = years of teaching experience
Di = 1 if male
= 0 otherwise (female)
control
variable
Assuming *2  2, different slope and different constant
between Yi and Xi.
2nd model:

9. Salary Y
X
teaching
years
Y = ’1+ ’2X (female)
^ ^ ^
Y = 1 + 2 X (male)
^ ^ ^
0 1 2 3 4 5 6 7 8
Male
Female
Linear (Male)
Linear (Female)
15
20
25
30
35
10
Y = ”1 + ”2 X (whole)
^ ^ ^
Two separate models: Ym = 1 + 2 Xm + um
Yf = ’1 + ’2 Xf + uf
(male)
(female)

10. D1 + D2 = 1
D1 = 1 - D2
male female
annual
Salary
years of
teaching
obs D2 D1 Y X
1 0 1 23 1
2 1 0 19.5 1
3 0 1 24 2
4 1 0 21 2
5 0 1 25 3
6 1 0 22 3
7 0 1 26.5 4
8 1 0 23.1 4
9 1 0 25 5
10 0 1 28 5
11 0 1 29.5 6
12 1 0 26 6
13 1 0 27.5 7
14 0 1 31.5 7
15 1 0 29 6
16 0 1 22 5
17 1 0 19 2
18 0 1 18 2
19 1 0 21.7 5
20 1 0 18.5 2
21 0 1 21 4
22 0 1 20.5 4
23 1 0 17 1
24 1 0 17.5 1
25 0 1 21.2 5
Each dummy
identify two
different
categories,
but when
sum up two
dummies
it cannot
identify
which is
male or female

11. Caution in the use of Dummy variables
(Dummy variable trap)
If we introduce two dummy variables in one model to
identify two categories of one qualitative variable such as
Yi = 1+ *1 D1i + **1 D2i + 2 Xi + ui
where D1i = 1 if female
= 0 otherwise
where D2i = 1 if male
= 0 otherwise
This model cannot be estimated because of
perfect collinearity between D1 and D2
 D1 = 1 - D2
or D2 = 1 - D1
or D1 + D2 = 1 ( Perfect collinearity )

12. Use two dummy variables to identify two different qualitative
categories in one model will be fall into the
“Trap of perfect multi-collinearity”
General rule : To avoid the perfect multicollinearity
If a qualitative variable has “m” categories,
introduce only “m-1” dummy variables.
1
D1 D2 D3 D4 D5 … Dm-1
age
1 10 20 30 40 m
Categories
dummy =>
Qualitative variable

13. Measure the estimated result for two groups:
Male: ==> Yi = (1 + *1 D2i)+ 2Xi D2i = 1
^ ^ ^ ^
Female: ==> Yi = 1 + 2Xi D2i = 0
^ ^ ^
Now consider different intercepts of two groups:
Model: Yi = 1 + *1 D2i + 2Xi + ui
D2i = 1 if male
= 0 otherwise, (i.e. female)
When a category is assigned the value of zero, this
category is called a control category (or omitted group).
2

14. In order to test whether there is any difference in
the relationships between two categories
Compare: Yi = 1 + 2Xi
^ ^ ^
Yi = (1 + *1 D)+ 2 Xi
^ ^ ^ ^
If t-statistics is significant in *1, there is
different in constant term.
=>same 2 means two categories of X have the
same relationship with Y
^
^
Check the t-statistics

15. H0 : *1 = 0
H1 : *1 > 0 or H1 : *1  0
Appropriate test is the t-test on *0
^
Compare the critical tc
(α/2, n-k) and the estimated t*
If t* > tc ==> reject H0 : *1 = 0
Y = 1 + *1Di+ 2 Xi + *2DiXi
^ ^ ^ ^
^
Check
t-statistics
=
This part is
testing the
difference of
intercept
This part is testing
whether any
difference in slope
of two categories
Check
t-statistics
=

16. Separate Examples for female and male:
Female Male
The two regression results performed differently in slope
and intercept. But are they really statistically different?
We cannot answer from these two separate regression results
unless you test the F*.

17. Set two dummies for the Example: Table 15.1 +15.5
Yi = ( ’1 +”1D1) +2 Xi
^ ^ ^ ^
= (17.937-1.2810) + 1.561X
D1:Female =1
others = 0
D2:Male =1
others = 0
= (16.656+1.2810) + 1.561X
Yi = (1 + *1 D2)+ 2 Xi
^ ^ ^ ^

18. Yi = 1 + 2Xi
^ ^ ^
= 17.095+1.608Xi
Whole Sample

19. D1: Female =1
Male: Y = 1 + 2Xi
^ ^ = 18.689 + 1.373 Xm
Female: Y = (1 + ’1D1)+(2 +’2D1)Xi
^ ^ ^ ^
= 16.255 +1.677 Xf

20. If D2: Male =1
Female: Y = 1 + 2 Xi
^
^ =16.255 + 1.677 X
Male: Y = (1+ ’1 D2)+(2+ ’2D2)X
^ ^ ^ ^ =18.689 + 1.373 X

21. One qualitative variable with more than two categories
(Health care) = 1 + ’1 D2 + ’’1 D3 + 2Income + u
(Y) (X)
D2 = 1 if high school education
= 0 otherwise
D3 = 1 if college education
= 0 otherwise
2

22. Health
care
income
Less than high school education
Y = 1 + 2 X
^ ^ ^
1
^
High school education
Y = (1 + ’1 D2)+ 2X
^ ^ ^ ^
D2 = 1
’1
^
D3 = 1
College education
Y = (1 + 1
” D3)+2 X
^ ^ ^ ^
’’1
^

23. D2 = 1 High school
= 0 otherwise
D3 = 1 College
= 0 otherwise
=========================================
obs Y X D2 D3
=========================================
1 6.000000 40.00000 0.000000 1.000000
2 3.900000 31.00000 1.000000 0.000000
3 1.800000 18.00000 0.000000 0.000000
4 1.900000 19.00000 0.000000 0.000000
5 7.200000 47.00000 0.000000 1.000000
6 3.300000 27.00000 1.000000 0.000000
7 3.100000 26.00000 1.000000 0.000000
8 1.700000 17.00000 0.000000 0.000000
9 6.400000 43.00000 0.000000 1.000000
10 7.900000 49.00000 0.000000 1.000000
11 1.500000 15.00000 0.000000 0.000000
12 3.100000 25.00000 1.000000 0.000000
13 3.600000 29.00000 1.000000 0.000000
14 2.000000 20.00000 0.000000 0.000000
15 6.200000 41.00000 0.000000 1.000000
=========================================

24. Less than high school: Yi = -1.2859 + 0.1722 Xi
^
Yi = (-1.2859 - 0.068 ) + 0.1722 Xi
^
= -1.3539 + 0.1722 X
High school:
If t value of D2 is
statistically significant
Yi = (-1.2859 + 0.447 ) + 0.1722 Xi
^
= -0.8389 + 0.1722 Xi
College:
If t value of D3 is
statistically significant
= -1.2859 + 0.1722 X
= -1.2859 + 0.1722 X
If t-test is not
statistically significant
Measuring the estimated results of different groups:

25. One Qualitative variable with many categories :
Example : An estimate model on three different
age’s medical care expenditure
Yi = 1 + ’1 D1 + ’’1 D2 + 2 Xi + ui
(t-value) (t-value)
where D1 = 1 if 55 > age > 25
= 0 otherwise
D2 = 1 if age > 55
= 0 otherwise
A1 + A2  1
A2 =1
A1 =1
0
25 55

26. measure the estimated models are :
age below 25 Y = 1 + 2 X
^ ^ ^
Y = (1 + ’1D1)+ 2 X
^ ^ ^ ^
25 < age < 55
age > 55 Y = (1 + ’’1D2)+2 X
^ ^ ^ ^
H0 : ’1 = 0, ’’1 = 0 t1
*
H1 : ’1  0, ’’1  0 t2
*
Compare to tc
(α/2, n-k)

27. ’0
^
Y = ( 1 + ’1)+ 2X
^ ^ ^ ^
^
Y = ( 1+ ”1)+2X
^ ^ ^ ^
’’0
In scatter diagram :
0
^
Y
X
Y = (1 ) + 2 X
^ ^ ^

28. One Qualitative variable with many categories :
Example : An estimate model on four different age’s
medical care expenditure
Y = 1 + ’1 D1 + ”1 D2 + ”’1 D3 + 2 X + u
where D1 = 1 if age > 55
= 0 otherwise
D2 = 1 if 35 < age  55
= 0 otherwise
D3 = 1 if 15 < age  35
= 0 otherwise

29. Measure the estimated models are :
age  15 Y = 1 + 2 X
^ ^ ^
15 < age  35 Y = (1 + ’1D3) + 2 X
^ ^ ^ ^
35 < age  55 Y = (1 + ”2D2)+ 2 X
^ ^ ^ ^
age > 55 Y = (1 + ’’’1D1)+ 2 X
^ ^ ^ ^

30. Two qualitative variables
(Y) Salary = 1 + ’1D1 + ”1 D2 + 2X + u
or Y = 1+ ’1D1+ ”1D2 + 2X + ’2D1*X + ”2D2*X + u’
D1 = 1 if male
= 0 otherwise
sex
D2 = 1 if white
= 0 otherwise race
(1) Mean salary for “non-white” female teacher:
Y = 1 + 2X that is D1 = 0, D2 = 0
^ ^ ^
(2) Mean salary for “non-white” male teacher:
Y = (1 + ’1 D1) + (2+ ’2D1)X that is D1 = 1, D2 = 0
^ ^ ^ ^ ^

31. (3) Mean salary for “white” female teacher:
Y = (1 + ’’1 D2) + 2 X + ”2D2X that is D1 = 0, D2 = 1
^ ^ ^ ^
(4) Mean salary for “white” male teacher:
Y = (1 + ’1 D1 +”1D2)+ (2+ ’2D1+ ”2D2)X that is D1 = 1,
D2 = 1
^ ^ ^ ^ ^ ^ ^

32. D = 1 if 1970-1981
= 0 otherwise
(1982-1995)
Different types of dummy regression:
Y = 1 + 2 X + ’1D + ’2D*X
H0 : ’1 = 0 and ’2 = 0
Y = 1 + 2 X + ’1 D + ’2D*X
H0 : ’1 = 0
Y = 1 + 2 X + ’1D + ’2D*X
H0 : ’1  0 and ’2  0
Y = 1 + 2 X + ’1 D + ’2D*X
H0 : ’2 = 0
1.
2
3
4

33. (1970-1981): Yt = A1 + A2 Xt + u1t
(1982-1995: Yt = B1 + B2 Xt + u2t
Y
X
A1 = B1
1
A2 = B2
Identical regressions
Y
X
A1
1
A2
Parallel regressions
A1  B1, A2 = B2
B2
1
B1

34. Y
X
A0 = B0
1
B1
Concurrent regressions
A1
1
A0 = B0, A1  B1
Y
X
A0
1
A1
dissimilar regressions
A0  B0, A1  B1
B0
1
B1

35. Interactive effects between the two qualitative variables
Spending(Y) = 1 + ’1 D1 + ”1 D2 + 2 income(X) + u
D1 = 1 if female
= 0 otherwise
sex
D2 = 1 if college graduate
= 0 otherwise
education
Spending(Y) = 1 + ’1D1 + ”1D2 + ’”1D1*D2 + 2income(X) + u
Interaction effect:
’1 = different effect of being a female
”1= different effect of being a college graduate
”’1 = different effect of being a female with college graduate

36. Example : how can we test the hypothesis that the gasoline
spending is different between a new car and a used car ?
Let us assume that at the begin mile, there is no
different between used car and new car.
gas spending
miles running
Y
X
0
^
New car Y = 1 + 2 X
^ ^ ^
used car Y = 1+ 2X
^
Y = 1+ (2 +  ’2)X
^ ^ ^ ^
^ ^

37. The estimated relations are :
used car : Yi = 1 + (2 +  ’2D) Xi where D = 1
^ ^ ^ ^
new car : Yi = 1 + 2 Xi
^ ^ ^
==
Yi = 1 + 2 Xi
==
^ ^ ^
or
If ’2 0, means the estimated slopes for cars is different.
^
Let 2= 2 + ’2 D where D = 1 if used car
= 0 otherwise
Now in one model :
multiplicative
dummy variable
Yi = 1 + (2 + ’2 D) Xi + ui
= 1 + 2 Xi + ’2 D*Xi +ui
= 1 + 2 Xi + ’2 Zi + ui

38. Test whether  ’2 = 0 or not ?
^
(ii) use t-test on  ’2:Y = 1 +2 Xi +  ’2 Z
^
^ ^ ^ ^
H0 : ’2 = 0
^
H1 : ’2 > 0 Used car is spending more gasoline per mile
^
compare tc
(α, N-3) and t*
If t* > tc
（α, N-3） reject H0
or H1: ’2  0 If t* > tc
(α/2, n-3)  reject Ho
(i) Compare two separate models: (a) Y = 1 + 2 X
(b) Y = ’1 +  2X
^ ^ ^
^ ^ ^

39. Check the t-value
…...
…...
…...
…...
…...
Y = 1 + 2 Xi + ’2 Zi
obs Yi Xi Di (Di Xi) = Zi
1 210 100 0 0
2 250 110 1 110
3 340 150 1 150
4 305 120 1 120
^ ^ ^ ^

40. Shifts in both intercept and slope
E = 1 + 2 T + u
E : electricity consumption
T : temperature
To capture effect of seasonal factors
E = 1 + ’1D1 +”1D2 + ’’’1 D3 + 2T + u
where D1 = 1 if winter
0 otherwise
D2 = 1 if spring
0 otherwise
D3 = 1 if summer
0 otherwise
spring summer fall winter
Q1 Q2 Q3 Q4

41. ’’’1
E=(1+ ”’1)+ 2T (Summer)
^
^
^
^
^
1
^
T
E
E = 1 + 2T (Fall)
^ ^
^
E = (1 + ”1) + 2 T (Spring)
^ ^
^
^
’’1
^
E = (1 + ’1) + 2 T (winter)
^ ^ ^
’1
^
^
Measure the basic difference of four seasonal results :
Fall E = 1 + 2 T
^ ^ ^
Spring E = (1 + ”1)+ 2 T
^ ^ ^ ^
Winter E = (1 + ’1)+ 2 T
^ ^ ^ ^
Summer E = ( 1+”’1)+ 2 T
^ ^ ^ ^

42. Also consider the slope in different seasons
Let *2 = 2 + ’2D1 + ’’2 D2 + ’’’2 D3
Thus, the full general specification is
E = [1+ ’1D1 + ”1D2+ ”’1D3] + 2T + ’2 D1 T + ”2D2 T
+ ”’2D3 T + u Z1 Z2
Z3

43. Measure the four seasonal results :
Fall E = 1 + 2 T
^ ^ ^
Spring E = (1 + ”1)+(2 + ”1) T
^ ^ ^ ^
^
Winter E = (1 + ’1)+ (2 + ’2) T
^ ^ ^ ^
^
Summer E = ( 1+”’1)+ (2 + ”’2) T
^ ^ ^ ^
^
1
^
T
E
E = 1 + 2T (Fall)
^ ^
^
E = (1 + ’1)+(2 +’2)T(winter)
^ ^ ^ ^ ^
’1
^
E = (1 + ”1)+(2 + ”2)T (Spring)
^
^
^
^
’’1
^
’’’1
E=(1+ ”’1)+(2+ ”’2)T(Summer)
^
^
^
^
^
^

44. Quarterly effect is same as seasonal effect
D1 = 1 1st Quarter
= 0 otherwise
D2 = 1 2nd Quarter
= 0 otherwise
D3 = 1 3rd Quarter
= 0 otherwise
Control quarter is the 4th quarter

45. 1. Set the seasonal dummy01= 1 if there is the 1st quarter
= 0 otherwise

46. How
does the
quarterly
dummy
variable
look
like?

47. Basic model
Yt = 1 + 2 Xt + ut
1974
1960
1989
Define a dummy variable : D = 1 for the period
1974 onward
= 0 otherwise
To test whether the structures of two periods are
different, the specification must assume that
*1 = 1 + ’1 D
*2 = 2 + ’2 D
Dummy regression:
Yt = 1 + ’1 D + 2 Xt + ’2D Xt + ut
(2)

48. The Chow test on the Unemployment rate-capacity utilization rate
Dependent Var. Constant CAPt R2 F RSS n
_
unemplt 30.0 -0.293 0.761 93.6 17.15 30
(12.1) (9.7) RSSR
^
unemplt 19.64 -0.175 0.59 19.7 4.69 14
(5.9) (4.4) RSS1
^
unemplt 30.63 -0.296 0.871 102.1 3.29 16
(13.1) (10.1) RSS2
^
Note : t-values are in parentheses

49. H0 : No structural change
H1 : yes
For the unrestricted model :
RSSu = RSS1 + RSS2 = 4.69 + 3.29
= 7.98
F* =
(RSSR - RSSu) / k
RSSu / (T - 2k)
=
(17.15 - 7.98) / 2
7.98 / (30 - 4)
= 14.9
F* > Fc ==> reject H0
Fc
0.01, k, T -2k = Fc
0.01 = 5.53
0.05 0.05, 2, 26 = 3.37
Restriction F-test procedures:

50. Sample : 1960 - 1989
Dt = 1 1974 to 1980
= 0 prior to 1974
unempl = 19.6 + 11.0 Dt - 0.175 CAPt - 0.121 (Dt*CAPt)
^
(6.7) (2.7) (5.0) (2.5)
R2 = 0.88 SEE = 0.554 F = 72.2 n = 30
_
The estimated of 1974-1980:
unempl = (19.6+11.0) - (0.175+0.121)CAP
= 30.6 - 0.296 CAP
^
^
The estimated of 1960-1973:
unempl = 19.6 - 0.175 CAP
Using the dummy variable to identify the structural change

51. D = 1 if t  74
= 0 otherwise
Observed data
Year Ut CAPt Dt Dt*CAPt
60 4.20 5.70 0 0
61 0 0
62 0 0
63 0 0
… … ...
68 0 0
69 0 0
70 0 0
71 0 0
72 0 0
73 0 0
74 1
75 1
76 1
77 1
... 1
... 1
... 1
89 1
…
………….……
…
…...….……....
10.5
11.2
10.5
11.2
Ut = 1 + 2 CAPt + ’1Dt + ’2 Dt*CAPt

52. The estimated models are :
1974 : 1 and onward Y = 1 + 2 X
^ ^ ^
Now the basic model becomes
Yt = 1 + ’1 D + 2 Xt + ’2 D Xt + ut
Yt = 1 + ’1 D + 2 Xt + ’2 X*
t + ut
==>
=== ==
Check t* > 2.0
1974 : 1
1950
1995
Prior to 1974 : 1 Y = ( 1 + ’1D)+(2 + ’2D) X
^ ^ ^ ^ ^
* *
(4)

53. GENR DUMMY = 1 (sample 1970 - 1980)
GENR DUMMY = 0 (sample 1981 - 1991)
=================================================
obs SAVINGS INCOME DUMMY D*INCOME
=================================================
1970 57.50000 831.0000 1.000000 831.0000
1971 65.40000 893.5000 1.000000 893.5000
1972 59.70000 980.5000 1.000000 980.5000
1973 86.10000 1098.700 1.000000 1098.700
1974 93.40000 1205.700 1.000000 1205.700
1975 100.3000 1307.300 1.000000 1307.300
1976 93.00000 1446.300 1.000000 1446.300
1977 87.90000 1601.300 1.000000 1601.300
1978 107.8000 1807.900 1.000000 1807.900
1979 123.3000 2033.100 1.000000 2033.100
1980 153.8000 2265.400 1.000000 2265.400
1981 191.8000 2534.700 0.000000 0.000000
1982 199.5000 2690.900 0.000000 0.000000
1983 168.7000 2862.500 0.000000 0.000000
1984 222.0000 3154.600 0.000000 0.000000
1985 189.3000 3379.800 0.000000 0.000000
1986 187.5000 3590.400 0.000000 0.000000
1987 142.0000 3802.000 0.000000 0.000000
1988 155.7000 4075.900 0.000000 0.000000
1989 175.6000 4664.200 0.000000 0.000000
1990 175.6000 4664.200 0.000000 0.000000
1991 199.6000 4828.300 0.000000 0.000000
=================================================

54. Savings = 1 + 2 Income + ’1D + ’2D*Income + u
D = 1 1970--1980
= 0 1981--1991
Estimated for 1970 - 1980 : D = 1
Savings = (1 + ’1) +(2 + ’2) Income
^ ^ ^ ^
1
Estimated for 1981 - 1991 : D = 0
Savings = 1 + 2 Income
^ ^
2
Dummy Regression Results: 1970 - 1991 :
Savings = 217.81 - 203.19 D - 0.010 Income + 0.066 D*Income
(7.96) (-6.19) (-1.39) (4.63)

55. 70 - 80:
Savings = (217.81 - 203.19) + (-0.010 + 0.066) Income
= 14.62 + 0.056 Income
81 - 91:
Savings = 217.81 - 0.010 Income
1970 - 1991 :
Savings = 57.63 + 0.031 Income
(3.86) (5.95)
1970 - 1980 :
Savings = 14.61 + 0.056 Income
(1.40) (7.93)
1981 - 1991 :
Savings = 217.81 + 0.010 Income
(6.16) (-1.08)

56. Savings = -1.250 + 0.091 Dummy + 0.125 Income
(-3.42) (0.506) (7.04)
LS // Dependent Variable is SAVINGS
Date: 03/02/99 Time: 22:23
Sample: 1946 1963
Number of observations: 18
=====================================================
Variable Coefficient Std. Error t-Statistic. Prob.
=====================================================
C -1.250957 0.364879 -3.428419 0.0037
DUMMY 0.091857 0.181244 0.506816 0.6197
INCOME -0.125655 0.017837 -7.044517 0.0000
=====================================================
R-squared 0.919909 Mean dependent var 0.773333
Adjusted R-squared 0.909230 S.D. dependent var 0.642806
S.E. of likelihood 0.193665 Akaike info criterion -3.132238
Sum squared resid 0.562593 Schwarz criterion -2.983843
Log likelihood 5.649250 F-statistic 86.14326
Durbin-Watson stat 0.976197 Prob(F-statistic) 0.000000
=====================================================
Only consider the difference in intercept

57. LS // Dependent Variable is SAVINGS
Date: 03/02/99 Time: 22:23
Sample: 1946 1963
Number of observations: 18
=====================================================
Variable Coefficient Std. Error t-Statistic. Prob.
=====================================================
C -1.750172 0.331888 -5.273377 0.0001
DUMMY 1.483923 0.470362 3.154852 0.0070
INCOME 0.150450 0.016286 9.238172 0.0000
DINCOME -0.103422 0.033260 -3.109471 0.0077
=====================================================
R-squared 0.952626 Mean dependent var 0.773333
Adjusted R-squared 0.942475 S.D. dependent var 0.642806
S.E. of likelihood 0.154173 Akaike info criterion -3.546228
Sum squared resid 0.332771 Schwarz criterion -3.348367
Log likelihood 10.37516 F-statistic 93.84109
Durbin-Watson stat 1.468099 Prob(F-statistic) 0.000000
=====================================================
Whether intercept and slope change?

58. 1946 - 1954 : = -0.2662 + 0.047 Income D1 = 1
1955 - 1963 : = -1.750 + 0.150 Income D1 = 0
Savings = -1.750 + 1.483 D + 0.150 Income - 0.103 (Income*D)
(-5.273) (3.154) (9.238) (-3.109)
^

59. ln Y = 1 + 2 X + ’1 D
(Salary) (years of teaching)
D1 = 1 for male
= 0 otherwise
ln Y = 2.9298 + 0.0546 X2 + 0.1341 D
^
t=(481.5) (48.3) (27.2)
R2 = 0.995 DW = 2.51
exp(0.1341)=1.1435
Exp(0)=1
This means the starting salary of male teacher is higher than the
female teacher by 14.35 percent.
The estimated male teacher salary :
ln Y = (2.9298 + 0.1341) + 0.0546 X
ln Y = 3.0639 + 0.0546 X
^
^