Economics 405
Due—November 18 at 6:30 pm.
1.
Problem Set #3
A political scientist specified the following population regression model:
voteA =  0 +  1 expendA +  2 expendB +  3 totexpen + u
where voteA is the percentage of the vote captured by candidate A, expendA
is campaign expenditures by candidate A, expendB is campaign
expenditures by candidate B, and totexpen is the sum of campaign
expenditures by both candidates.
a.
Which of the multiple linear regression (MLR) assumptions does this model violate?
Explain why.
The model violates MLR.3 (No Perfect Collinearity). There is an exact linear
relationship between totexpen and expendA and expendB since totexpen = expendA +
expendB. Intuitively, we’re asking the effect of a $1,000 increase in expenditure by
candidate A on voteA, holding total expenditure and expenditure by candidate B
constant. Increase in expenditure by A, with expenditure by B held constant, of
course implies total expenditure does not stay constant. So there is no way for OLS
to estimate  1 (or  2 and  3 for that matter).
b.
Use the Wooldridge data set Vote1 to attempt this regression (you’ll need to generate
the totexpen variable yourself). Attach the regression output and comment on the
outcome.
STATA:
. gen totexpen=expendA+expendB
. regress
voteA expendA expendB totexpen
Source |
SS
df
MS
-------------+-----------------------------Model | 25679.8879
2
12839.944
Residual | 22777.3606
170 133.984474
-------------+-----------------------------Total | 48457.2486
172 281.728189
Number of obs
F( 2,
170)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
173
95.83
0.0000
0.5299
0.5244
11.575
-----------------------------------------------------------------------------voteA |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------expendA | (dropped)
expendB | -.0744583
.0053848
-13.83
0.000
-.0850879
-.0638287
totexpen |
.0383308
.0033868
11.32
0.000
.0316452
.0450165
_cons |
49.619
1.426147
34.79
0.000
46.80376
52.43423
2
EXCEL:
SUMMARY
OUTPUT
Regression Statistics
Multiple R
0.733143139
R Square
0.537498862
Adjusted R Square
0.529288783
Standard Error
11.51575524
Observations
173
ANOVA
df
3
169
172
SS
MS
F
26045.71597 8681.905 65.46817
22411.53258 132.6126
48457.24855
Coefficients
49.60647306
363.9483405
363.8757627
-363.9109519
Standard Error
t Stat
P-value
1.418846846 34.96253 2.82E-79
219.1117063 1.661017 0.098564
219.1128385 1.660678 0.098632
219.1122736 -1.66084 0.098599
Regression
Residual
Total
Intercept
expendA
expendB
totexpen
Note that STATA won’t run the regression I asked it to run. Instead it arbitrarily
drops expendA from the regression. Excel actually runs the regression, but note that
the coefficients on expendA and expendB are virtually identical and that the
coefficient on totexpen is also identical but of the opposite sign. Furthermore,
interpretation of any of the Excel coefficients results in nonsense. For example, the
estimated  1 implies that a $1,000 increase in expenditure by candidate A results in
363.9 percentage point increase in the share of votes for candidate A!
c.
Generate the variables expAsqr (equal to expendA squared) and expBsqr (equal to
expendB squared). Regress voteA on expendA, expAsqr, expendB, and expBsqr.
Write out the fitted model. (Use the fitted model format displayed on p. 127 of the
text, i.e., include coefficient standard errors below estimated coefficients.) Attach
your regression results. Explain why this model is estimable (i.e., explain why the
MLR assumption violated in part b is not violated here).
. gen expAsqr=expendA^2
. gen expBsqr=expendB^2
. regress
voteA expendA expAsqr expendB expBsqr
3
Source |
SS
df
MS
-------------+-----------------------------Model | 36053.1139
4 9013.27847
Residual | 12404.1347
168 73.8341349
-------------+-----------------------------Total | 48457.2486
172 281.728189
Number of obs
F( 4,
168)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
173
122.07
0.0000
0.7440
0.7379
8.5927
-----------------------------------------------------------------------------voteA |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------expendA |
.081002
.005663
14.30
0.000
.0698222
.0921818
expAsqr | -.0000521
5.50e-06
-9.47
0.000
-.0000629
-.0000412
expendB | -.0787343
.0056362
-13.97
0.000
-.0898613
-.0676074
expBsqr |
.0000494
5.34e-06
9.24
0.000
.0000388
.0000599
_cons |
49.26944
1.478249
33.33
0.000
46.3511
52.18778
------------------------------------------------------------------------------
Writing the estimated model out in fitted model format:
^
voteA = 49.27 + .081expendA - .0000521expAsqr - .079expendB + .0000494expBsqr
(1.478) (.0057)
(.0000055)
(.0056)
(.0000053)
2
n = 173 R  .744
The model is estimable since the quadratic terms, though exact functions of the
expenditure variables, are not exact linear functions of the expenditure variables.
d.
What do the quadratic terms added to the model in part c allow for? Explain. Using
the fitted model from part c, what is the impact on predicted voteA of a $1,000
increase in expenditure by candidate A? (In
^
voteA
calculus terms, I’m asking for
.) What is the impact on predicted voteA of
expendA
a $1,000 increase in expenditure by candidate B
^
voteA
(i.e.,
)?
expendB
The quadratic terms allow for possible non-linear relationships between voteA and the
expenditure variables. First note that expAsqr  (expendA)2 and
expBsqr  (expendB)2 . Then
^
voteA
 ˆ1  2ˆ 2 expendA  .081  (2  (.000052))expendA  .081  .000104expendA
expendA
4
The above statement implies that the partial effect of expenditure by candidate A is
positive but diminishing as expenditure by A increases. Taking summary statistics
for expendA:
. summarize expendA
Variable |
Obs
Mean
Std. Dev.
Min
Max
-------------+-------------------------------------------------------expendA |
173
310.611
280.9854
.302
1470.674
Plugging mean expenditure of 310.611 into the partial effect equation above, we get:
^
voteA
 .081  ((.000104)  (310.611))  .049 . Hence, a candidate whose total
expendA
expenditure is at the mean level is predicted to gain almost .05 of a percentage point
of votes cast by raising expenditure by $1,000. On the other hand, a candidate whose
spending level is one standard deviation above the mean level (310.611 + 280.99 =
591.60) is predicted to gain only
^
voteA
 .081  ((.000104)  (591.60))  .019 of a percentage point of votes
expendA
cast by spending another thousand dollars on the campaign. Thus the model predicts
diminishing returns to campaign expenditures by a candidate.
The partial effect of expendB is:
^
voteA
 ˆ3  2ˆ 4 expendB  .079  (2  (.0000494))expendB  .079  .0000998expendB
expendB
Interpreting, expenditure by candidate B has a negative impact on the percentage of
votes cast for candidate A, but this impact becomes less negative the larger is
expendB (by virtue of the second term).
2.
Read the discussion titled “A Partialling Out Interpretation of Multiple Regression” on
pp.78-79 of the text.
a.
Use the Wooldridge dataset Ceosal2 to estimate the following model:
lsalary   0  1comten   2 ceoten   3lsales  u
where lsalary is the log of the CEO’s salary, comten is the total number of
years the CEO has worked for the company in one capacity or
5
another, ceoten is the number of years the CEO has served as CEO of
the company, and lsales is the log of company sales.
Write out the fitted model (again, use the format from p. 127). Interpret each of the
estimated slope coefficients. Attach the regression output as an appendix.
. regress
lsalary comten ceoten lsales
Source |
SS
df
MS
-------------+-----------------------------Model |
21.614243
3 7.20474766
Residual | 43.0319701
173 .248739711
-------------+-----------------------------Total | 64.6462131
176 .367308029
Number of obs
F( 3,
173)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
177
28.97
0.0000
0.3343
0.3228
.49874
-----------------------------------------------------------------------------lsalary |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------comten | -.0101184
.0033397
-3.03
0.003
-.0167102
-.0035266
ceoten |
.0170786
.0055813
3.06
0.003
.0060624
.0280949
lsales |
.2481522
.0272305
9.11
0.000
.1944054
.3018989
_cons |
4.880288
.1997791
24.43
0.000
4.48597
5.274606
------------------------------------------------------------------------------
The fitted model is:
^
lsalary  4.880  .010comten  .017ceoten  .248lsales
(.200) (.0033)
(.0056)
(.0272)
2
R  .334
n = 177
The slope coefficient on comten implies that an additional year of tenure with a
company reduces predicted CEO earnings by 1%, other things equal. In other words,
given two CEOs with the same number of years on the job as CEO and whose firm
sales are identical, the CEO with one year more in total tenure with her/his firm is
predicted to earn 1% less than the CEO with one less year of total company tenure.
(Hmmm, interesting result!) The slope coefficient on ceoten implies that an
additional year of tenure as CEO, with total years of tenure at the company and log
sales held constant, increases predicted CEO earnings by 1.7%. Finally, the
coefficient on lsales is an elasticity and implies that a 1% increase in firm sales
increases predicted CEO earnings by .248%, holding total tenure and tenure as CEO
constant.
b.
Confirm the partialling out interpretation for ˆ1 . Do this by first regressing comten
on ceoten and lsales. Attach these regression results. Save the residuals from this
regression (call them r1hat). [Stata users, first run your regression. Your next
command should then be “predict r1hat, residuals.” Excel users, check the residuals
box in the regression dialog.] Now run the simple regression of lsalary on r1hat.
Write out the fitted model. Interpret the estimated coefficient on r1hat. How does
6
the estimated coefficient of r1hat compare with the estimated coefficient on comten
in part a?
. regress
comten ceoten lsales
Source |
SS
df
MS
-------------+-----------------------------Model | 4302.91318
2 2151.45659
Residual | 22301.3354
174 128.168594
-------------+-----------------------------Total | 26604.2486
176 151.160503
Number of obs
F( 2,
174)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
177
16.79
0.0000
0.1617
0.1521
11.321
-----------------------------------------------------------------------------comten |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------ceoten |
.5580039
.1194227
4.67
0.000
.3223003
.7937074
lsales |
2.146732
.5963127
3.60
0.000
.9697944
3.323669
_cons |
2.540943
4.530812
0.56
0.576
-6.401482
11.48337
------------------------------------------------------------------------------
Next:
. predict r1hat, residuals
. regress lsalary r1hat
Source |
SS
df
MS
-------------+-----------------------------Model | 2.28323989
1 2.28323989
Residual | 62.3629732
175 .356359847
-------------+-----------------------------Total | 64.6462131
176 .367308029
Number of obs
F( 1,
175)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
177
6.41
0.0122
0.0353
0.0298
.59696
-----------------------------------------------------------------------------lsalary |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------r1hat | -.0101184
.0039974
-2.53
0.012
-.0180077
-.002229
_cons |
6.582848
.0448702
146.71
0.000
6.494291
6.671404
------------------------------------------------------------------------------
The fitted model is:
^
lsalary = 6.582 - .010r1hat
(.0449) (.0040)
R 2  .035
n = 177
The coefficients on r1hat and on comten are identical as the discussion in the Wooldridge
text indicates they should be. The coefficient on comten in the original regression tells us the
effect on predicted CEO log salary of a one year increase in total tenure with the company,
holding tenure as CEO and firm log sales constant. r1hat is the component of comten that
cannot be explained by variation in ceoten or lsales. In other words, r1hat is comten after
7
netting out the effects of ceoten and lsales. In essence, r1hat also holds comten and lsales
constant and this explains why the coefficient is the same in the two regressions.
3.
Use the data set Hprice2 from the textbook data files to do this problem. The data set
contains information on median housing price in n = 506 communities. Suppose that the
following equation is the true model for log median housing price in a community:
log( price )   0  1 log( nox)   2 rooms  u
where log(price) is the log of median housing price in the community;
log(nox) measures the level of air pollution in the community and is the log of
nitrous oxide concentration measured in parts per million; and
rooms is the average number of rooms per house in the community.
a.
What are the probable signs of  1 and  2 ? Explain.
1  0 since a greater level of pollution should reduce home values.
 2  0 since larger homes should be worth more.
b.
What is the interpretation of  1 ? What is the interpretation of  2 ?
 1 is the percent change in expected home value due to a one percent increase in
nitrous oxide concentration, holding the average number of rooms per house constant.
 2 is the percent change in expected home value due to a one room increase in
average number of rooms, holding log(nox) constant.
c.
Run a simple regression of rooms on log(nox). Write out the fitted model. Attach the
regression output as an appendix to the problem set. Interpret the slope coefficient on
log(nox).
. regress rooms
lnox
Source |
SS
df
MS
-------------+-----------------------------Model | 23.1881055
1 23.1881055
Residual | 226.099087
504
.4486093
-------------+-----------------------------Total | 249.287193
505 .493638005
Number of obs
F( 1,
504)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
506
51.69
0.0000
0.0930
0.0912
.66978
-----------------------------------------------------------------------------rooms |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------lnox | -1.063912
.1479814
-7.19
0.000
-1.354648
-.7731754
_cons |
8.085351
.2523091
32.05
0.000
7.589644
8.581058
8
The fitted model is:
~
rooms = 8.085 – 1.064lnox
(.252) (.148)
R 2  .093
n = 506
The estimated model here is a level-log model. As always the slope coefficient tells
us the change in predicted dependent variable due to a 1-unit change in the regressor.
Since the regressor is in logs, a 1-unit change is a “big” change (i.e., a 100 percent
change). Usually, we’re interested in what happens given more marginal change in
the regressor, for example, a 1 percent change (or a .01 change in proportional terms).
To find this effect, take ˆ1 / 100  1.064 / 100  .01064 . Hence the interpretation
of the slope coefficient on lnox is that a 1% increase in nitrous oxide concentration is
associated with a decrease in predicted rooms of .01, a negative but relatively small
effect. (See Wooldridge, p. 46, for interpretation of the regression coefficient in the
level-log model.)
d.
Run the simple regression of log(price) on log(nox). Write out the fitted model.
Interpret the model’s slope coefficient. Attach the regression output as an appendix.
. regress lprice lnox
Source |
SS
df
MS
-------------+-----------------------------Model | 22.2916542
1 22.2916542
Residual | 62.2905708
504 .123592402
-------------+-----------------------------Total |
84.582225
505 .167489554
Number of obs
F( 1,
504)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
506
180.36
0.0000
0.2636
0.2621
.35156
-----------------------------------------------------------------------------lprice |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------lnox | -1.043144
.0776728
-13.43
0.000
-1.195746
-.8905413
_cons |
11.70719
.1324325
88.40
0.000
11.44701
11.96738
------------------------------------------------------------------------------
~
log(price) = 11.707 - 1.043lnox
(.132) (.0777)
R 2  .264
n = 506
Since both variables are in logarithms, the slope coefficient is an elasticity.
Interpreting, the slope coefficient implies that a 1% increase in nitrous oxide
concentration in a community is associated with a 1.043% reduction in predicted
home value. Note, however, that the effect of home size, as measured by average
number of rooms, is not controlled for in this model.
9
e.
. regress
Run the multiple regression of log(price) on log(nox) and rooms. Write out the fitted
model. Interpret the slope coefficient of each regressor. Attach the regression output
as an appendix.
lprice
lnox rooms
Source |
SS
df
MS
-------------+-----------------------------Model | 43.4513652
2 21.7256826
Residual | 41.1308598
503 .081771093
-------------+-----------------------------Total |
84.582225
505 .167489554
Number of obs
F( 2,
503)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
506
265.69
0.0000
0.5137
0.5118
.28596
-----------------------------------------------------------------------------lprice |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------lnox | -.7176736
.0663397
-10.82
0.000
-.8480106
-.5873366
rooms |
.3059183
.0190174
16.09
0.000
.268555
.3432816
_cons |
9.233738
.1877406
49.18
0.000
8.864885
9.60259
The fitted model is:
^
log(price) = 9.234 - .718lnox + .306rooms
(.188) (.0663)
(.019)
2
R  .514
n = 506
Since both log(price) and lnox are logarithmic, the coefficient of lnox is an elasticity.
The coefficient on lnox implies that a 1% increase in nitrous oxide concentration
reduces home value by .718%, with average number of rooms held constant. Note
that once we control for number of rooms, the estimated impact of pollution becomes
considerably smaller. The omission of number of rooms from the regression of part d
resulted in downward bias away from zero.
The coefficient on rooms implies that a one room increase in the average number of
rooms raises predicted home value by 30.6%, with pollution level held constant.
f.
~
~
Verify mathematically that 1  ˆ1  ˆ 2 1 .
~
From part d, 1  1.043 . From part e, ˆ1  .718 and ˆ 2  .306 . From part c,
~
 1  1.064 . Plugging these values into the statement,
-1.043 = -.718 + [(.306)(-1.064)] = -1.043. The statement is verified.
Intuitively, the negative impact of pollution is over-estimated in part d because more
polluted communities also tend to have smaller homes. Smaller homes are worth less
in the real estate market. Ignoring house size in the log(price) regression means we
are not holding home size constant in evaluating the impact of pollution on home
value. Since more polluted communities also tend to have smaller homes, some of
the adverse impact of pollution on home price found in the simple regression will in
10
actuality be the result of smaller homes being found in those communities. To see the
true impact of pollution on home price, we need to hold home size constant.
4.
Suppose we want to estimate the ceteris paribus effect of x1 on y. Data for two additional
~
control variables, x2 and x3, are collected. Let  1 be the simple regression estimate from y on
x1 and let ˆ be the multiple regression estimate of y on x1, x2, and x3.
1
a.
If x1 is highly correlated with x2 and x3 in the sample, and x2 and x3 have large partial
~
effects on y, would you expect  1 and ˆ1 to be similar or very different? Explain.
Because x1 is highly correlated with x2 and x3 , and these latter variables have large
partial effects on y, the simple and multiple regression coefficients on x1 can differ
by large amounts. Intuitively, think in terms of equation (3.45) from the book, i.e.,
~
~
E ( 1 )  1   2 1 , where the betas are the true partial effects of the relevant
variables and the delta term is the regression coefficient on x1 in a regression of the
omitted variable on the other explanatory variables in the model. In the case at hand,
 2 and  3 will both be large (“x2 and x3 have large partial effects on y”) and
~
 1 terms will also be large (“x1 is highly correlated with x2 and x3”).
b.
If x1 is almost uncorrelated with x2 and x3, but x2 and x3 are highly correlated with
~
each other, would you expect  1 and ˆ1 to be similar or very different? Explain.
Here we would expect 1 and ̂1 to be similar. The amount of correlation
between x2 and x3 does not directly affect the multiple regression
~
estimate on x1 if x1 is essentially uncorrelated with x2 and x3 (i.e., the  1
terms are approximately equal to 0).
c.
If x1 is highly correlated with x2 and x3, but x2 and x3 have small partial effects on y,
~
would you expect se( 1 ) or se( ˆ1 ) to be smaller? Explain.
In this case we are (unnecessarily) introducing multicollinearity into the
regression: x2 and x3 have small partial effects on y and yet x2 and x3 are
highly correlated with x1 . Adding x2 and x3 likely increases the standard
error of the coefficient on x1 substantially, so se( ̂1 ) is likely to be much
larger than se( 1 ).
d.
If x1 is almost uncorrelated with x2 and x3, x2 and x3 have large partial effects on y,
~
and x2 and x3 are highly correlated, would you expect se( 1 ) or se( ˆ1 ) to be
smaller? Explain.
11
In this case, adding x2 and x3 will decrease the residual variance without
causing much collinearity (because x1 is almost uncorrelated with x2
and x3 ), so we should see se( ̂1 ) smaller than se( 1 ). The amount of
correlation between x2 and x3 does not directly affect se( ̂1 ).
5.
a.
b.
What is the Gauss-Markov Theorem?
Discuss the importance/relevance of the Gauss-Markov Theorem.
Read pp. 102-104 of Wooldridge for answers to both a and b.
6.
Assume that MLR.1-MLR.6 are valid for a k regressor model:
a.
Draw the sampling distribution for ˆ j .
f ( ˆ j )
j
b.
ˆ j
Explain why the graph looks as you have drawn it.
By Theorem 4.1, when MLR.1 – MLR.6 are valid for the k regressor model, the
sampling distribution of ˆ j is ˆ j ~ N (  j ,Var ( ˆ j )) . The distribution
12
characterized in the graph of part a is normal and centered at  j per the distribution
statement.
c.
If the sample size were to double, how (if at all) would that affect your drawing?
Illustrate and explain.
Var ( ˆ j ) 
2
. As n increases, SSTj increases, hence Var ( ˆ j ) decreases.
SST j  [1  R ]
The implication for the sampling distribution is that it becomes less dispersed about
 j as in the graph below.
f ( ˆ )
2
j
j
Distribution becomes
less dispersed as n
increases
j
ˆ j
13
7.
Use the textbook data file Lawsch85 to estimate:
log( salary )   0  1 LSAT   2 GPA   3 log( libvol )   4 log( cost )   5 rank  u
where log(salary) is the log of median starting salary for a law school’s graduates; LSAT is
the median LSAT score for the law school’s admits; GPA is the median
undergraduate GPA of the law school’s admits; libvol is the number of volumes in the
school’s library (in thousands); cost is the school’s tuition cost; and rank is the law
school’s ranking (where 1 is the best and 175 is the worst).
a.
. regress
Attach your regression output as an appendix.
lsalary LSAT GPA llibvol lcost rank
Source |
SS
df
MS
-------------+-----------------------------Model | 8.73362207
5 1.74672441
Residual | 1.64272974
130 .012636383
-------------+-----------------------------Total | 10.3763518
135 .076861865
Number of obs
F( 5,
130)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
136
138.23
0.0000
0.8417
0.8356
.11241
-----------------------------------------------------------------------------lsalary |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------LSAT |
.0046965
.0040105
1.17
0.244
-.0032378
.0126308
GPA |
.2475239
.090037
2.75
0.007
.0693964
.4256514
llibvol |
.0949932
.0332543
2.86
0.005
.0292035
.160783
lcost |
.0375538
.0321061
1.17
0.244
-.0259642
.1010718
rank | -.0033246
.0003485
-9.54
0.000
-.004014
-.0026352
_cons |
8.343226
.5325192
15.67
0.000
7.2897
9.396752
------------------------------------------------------------------------------
b.
What is the estimated ceteris paribus effect of a 1-point increase in GPA on salary?
What has truly been “held constant” when you make this statement?
The model predicts that a 1-point increase in GPA increases predicted earnings by
24.8%. LSAT scores, library size (measured by books), law school cost, and law
school rank have been controlled for in the model. That is, given two law schools
with the same median LSAT score for admits, the same library size, the same tuition
cost, and the same rank, predicted earnings are 24.8% higher for the school for
which undergraduate GPA is 1 point higher.
c.
Use Stata or Excel to calculate the standard deviation of GPA (attach the results). Is a
1 point change in GPA from part b a relatively “big” change in GPA or a relatively
“small” change in GPA? Explain.
The Lawsch85 data set has 156 observations in total, but there are missing values in
both the dependent variable and in some of the regressors. As a consequence, there
are only 136 observations that can be used in the log(salary) regression (see part a).
14
In doing the summary statistics, I first restricted the calculations to observations in the
regression sample. The command to do this in Stata is:
. keep if e(sample)
This command retains only observations that were in the regression sample.
. summarize GPA
Variable |
Obs
Mean
Std. Dev.
Min
Max
-------------+-------------------------------------------------------GPA |
136
3.309632
.1972305
2.8
3.82
Given that the standard deviation of GPA is only .197, we can conclude that a 1point change in GPA across these law schools represents a huge (unrealistically
large) change in GPA. So a more interesting question to ask might be what is the
ceteris paribus effect of a one standard deviation change in GPA on salary?
d.
What is the estimated ceteris paribus effect of a one standard deviation change in
GPA on salary? Show your work.
^
Recall yˆ  ̂ j  x j . Therefore:  log( salary )  (.248)  (.197)  .049 Hence, the
predicted median salary, other factors held constant, is about 5% higher for a school
for which the undergrad GPA of its students is a standard deviation higher.
e.
State and interpret the value of the regression’s ̂ .
The standard error of the regression is .112. Therefore, the typical amount of
unexplained difference between the actual value of log(salary) and its predicted
value is .112 log points. Since the dependent variable here is measured in logs, we
can easily convert to percentages by multiplying by 100. Therefore, the typical
amount of unexplained difference between actual and predicted salary is 11.2%.
f.
Determine SSTGPA (you should be able to figure this out based on your work for part
2
2
c) and RGPA
(where RGPA
is the R2 from a regression of GPA on the other explanatory
2
variables in the model). Attach your results. Using ̂ , SSTGPA , and RGPA
, verify the
computer’s calculation of se(ˆ ) from part a. Show your work. Evaluate and
GPA
discuss the degree of multicollinearity in ̂ GPA .
15
2
 .0388998 .
From part c, sGPA  .1972305  sGPA
2
sGPA

. regress
SSTGPA
SSTGPA
 .0388998 
 SSTGPA  135  .0388998  5.251473 .
n 1
136  1
GPA LSAT llibvol lcost rank
Source |
SS
df
MS
-------------+-----------------------------Model | 3.69271787
4 .923179469
Residual | 1.55876379
131
.01189896
-------------+-----------------------------Total | 5.25148166
135 .038899864
Number of obs
F( 4,
131)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
136
77.58
0.0000
0.7032
0.6941
.10908
-----------------------------------------------------------------------------GPA |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------LSAT |
.0266461
.0031186
8.54
0.000
.0204768
.0328154
llibvol |
.0151729
.0322422
0.47
0.639
-.0486099
.0789556
lcost | -.1460243
.0284231
-5.14
0.000
-.2022521
-.0897966
rank | -.0014423
.0003138
-4.60
0.000
-.002063
-.0008215
_cons |
.4847979
.5150086
0.94
0.348
-.534012
1.503608
-----------------------------------------------------------------------------2
 .7032 , i.e., only
From the regression results table above, we observe that RGPA
about 30% of the variation in GPA is independent of the other explanatory variables.
ˆ
.11241
Estimated se( ˆ j ) 

 .090039 , which
2
(
5
.
251473
)
*
(
1

.
7032
)
SST j  [1  R j ]
is the same as the standard error value reported for GPA’s coefficient in the
regression results table of part a.
2
The high RGPA
suggests a considerable degree of collinearity between GPA and the
other regressors. Perhaps this is not too surprising. For example, one would expect
average LSAT and average undergraduate GPA to be highly correlated. The effect
that collinearity has on the variance of the estimated regression coefficient can be
seen by re-writing the estimated variance as
 1  ˆ 2
ˆ 2
Var ( ˆ j ) 



SST j  [1  R 2j ] 1  R 2j  SST j
The second term to the right of the last equality is like estimated coefficient variance
in simple regression. The first term to the right of the last equality is > 1 and
measures the degree by which the coefficient variance is inflated as a result of
correlation between the regressor in question and the other regressors in the model.
If the regressor in question is uncorrelated with the other variables in the model, then
R 2j  0 and no inflation of the variance occurs. The larger is R 2j , however, the
larger is
1
1
 3.369 ,
. In GPA’s case, its variance inflation factor is
2
1  .7032
1 Rj
16
hence the variance of GPA’s coefficient is much larger than would be the case in the
absence of collinearity. With all of this said, however, the large variance inflation
factor hasn’t rendered GPA statistically insignificant (its t-ratio is 2.75). What has
saved the day? Probably the large sample size as this makes SSTj large,
counteracting the effect of the collinearity.