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The following are the results needed to test all combinations
of pairs of the coefficients
b1,
b2,
and b3
equal to 0. We shall suppose that
b1,
corresponds to NICO,
b2
corresponds to TAR,
and b3,
corresponds to WEIGHT.
============================================================
LS // Dependent Variable is CARMON
Date: 03/16/98 Time: 23:07
Sample: 1 25
Included observations: 25
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 3.202188 3.461755 0.925019 0.3655
NICO -2.631660 3.900556 -0.674688 0.5072
TAR 0.962574 0.242244 3.973566 0.0007
WEIGHT -0.130480 3.885342 -0.033583 0.9735
============================================================
R-squared 0.918589 Mean dependent var 12.52800
Adjusted R-squared 0.906959 S.D. dependent var 4.739683
S.E. of regression 1.445726 Akaike info criter 0.882869
Sum squared resid 43.89258 Schwarz criterion 1.077890
Log likelihood -42.50933 F-statistic 78.98385
Durbin-Watson stat 2.860469 Prob(F-statistic) 0.000000
============================================================
============================================================
LS // Dependent Variable is CARMON
Date: 03/16/98 Time: 23:15
Sample: 1 25
Included observations: 25
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 1.664666 0.993602 1.675386 0.1074
NICO 12.39541 1.054152 11.75865 0.0000
============================================================
R-squared 0.857378 Mean dependent var 12.52800
Adjusted R-squared 0.851177 S.D. dependent var 4.739683
S.E. of regression 1.828452 Akaike info criter 1.283558
Sum squared resid 76.89449 Schwarz criterion 1.381068
Log likelihood -49.51794 F-statistic 138.2659
Durbin-Watson stat 2.673760 Prob(F-statistic) 0.000000
============================================================
============================================================
LS // Dependent Variable is CARMON
Date: 03/16/98 Time: 23:16
Sample: 1 25
Included observations: 25
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 2.743277 0.675206 4.062875 0.0005
TAR 0.800976 0.050320 15.91759 0.0000
============================================================
R-squared 0.916778 Mean dependent var 12.52800
Adjusted R-squared 0.913160 S.D. dependent var 4.739683
S.E. of regression 1.396721 Akaike info criter 0.744873
Sum squared resid 44.86908 Schwarz criterion 0.842383
Log likelihood -42.78438 F-statistic 253.3698
Durbin-Watson stat 2.892673 Prob(F-statistic) 0.000000
============================================================
============================================================
LS // Dependent Variable is CARMON
Date: 03/16/98 Time: 23:16
Sample: 1 25
Included observations: 25
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C -11.79527 9.721627 -1.213302 0.2373
WEIGHT 25.06820 9.980283 2.511773 0.0195
============================================================
R-squared 0.215258 Mean dependent var 12.52800
Adjusted R-squared 0.181139 S.D. dependent var 4.739683
S.E. of regression 4.288984 Akaike info criter 2.988718
Sum squared resid 423.0939 Schwarz criterion 3.086228
Log likelihood -70.83244 F-statistic 6.309001
Durbin-Watson stat 2.615425 Prob(F-statistic) 0.019481
============================================================
To test
H0:
b2 =
b3 = 0
H1:
b2
¹ 0 and/or
b3
¹
0
We use the first and
second sets of results above to find:
Fq,n-k-1 = {[SSER - SSEUR]/q}/
{SSEUR/n-k-1} =
[(76.89449 - 43.89258)/2]/[43.89258/(25-3-1)] = 7.8947
The table value for F2,21,.05 = 3.47.
Since 7.8947 > 3.47 we reject the null hypothesis at the
a = .05 significance
level. Thus there is sufficient evidence to say that at least one of the two
coefficients is not equal to 0.
This is equivalent to performing the Wald Test:
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(3)=0
C(4)=0
====================================================
F-statistic 7.894729 Probability 0.002775
Chi-square 15.78946 Probability 0.000373
====================================================
To test
H0:
b1 =
b3 = 0
H1:
b1
¹ 0 and/or
b3
¹
0
We use the first and
third sets of results above to find:
Fq,n-k-1 = {[SSER - SSEUR]/q}/
{SSEUR/n-k-1} =
[(44.86908 - 43.89258)/2]/[43.89258/(25-3-1)] = .2336
The table value for F2,21,.05 = 3.47.
Since .2336 < 3.47 we do not reject the null hypothesis at the
a = .05 significance
level. There is not enough evidence to say that at least one of the two
coefficients is not equal to 0.
This is equivalent to the Wald Test:
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(2)=0
C(4)=0
====================================================
F-statistic 0.233600 Probability 0.793708
Chi-square 0.467199 Probability 0.791679
====================================================
To test
H0:
b1 =
b2 = 0
H1:
b1
¹ 0 and/or
b2
¹
0
We use the first and
fourth sets of results above to find:
Fq,n-k-1 = {[SSER - SSEUR]/q}/
{SSEUR/n-k-1} =
[(423.0939 - 43.89258)/2]/[43.89258/(25-3-1)] = 90.7127
The table value for F2,21,.05 = 3.47.
Since 90.7127 > 3.47 we reject the null hypothesis at the
a = .05 significance
level. There is enough evidence to say that at least one of the two
coefficients is not equal to 0.
This is equivalent to the Wald Test:
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(2)=0
C(3)=0
====================================================
F-statistic 90.71269 Probability 0.000000
Chi-square 181.4254 Probability 0.000000
====================================================
The results from the three F-tests seem to indicate that
b1 and
b3 are
not significantly different from 0 either by themselves or jointly. Both
F-tests that included
b2
were rejected, but that should not be surprising
because b2
itself was significant in the model that included all the
coefficients. The proper choice of the model would be that using only the
variable TAR and an intercept to predict the level. Notice, in fact,
that in
this model the adjusted R2 is higher in the model with only the
variable TAR than it is in the unrestricted model.
The test
H0:
b2 = 1
H1:
b2
¹
1
can be done using either the
unrestricted model or the restricted model with only the variable TAR in
it. However based on what was said above, it should really be done on
the model with only TAR and the intercept.
For the unrestricted case:
Test Statistic = (b2 - 1)/SE{b2} =
(.962574 - 1)/.242244 = -.1545. Now, since
t21,.025 = 2.080, we would fail to reject the null
hypothesis that
b2 = 1
at the .05 significance level.
The equivalent Wald Test is (note that the F = t2 here):
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(3)=1
====================================================
F-statistic 0.023870 Probability 0.878692
Chi-square 0.023870 Probability 0.877217
====================================================
For the restricted case:
Test Statistic = (b2 - 1)/SE{b2} =
(.800976 - 1)/.050320 = -3.9552. Now since
t23,.025 = 2.069,
we reject the null hypothesis
b2 = 1
at the .05 significance level.
The equivalent Wald Test is (note that the F = t2 here):
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(2)=1
====================================================
F-statistic 15.64324 Probability 0.000629
Chi-square 15.64324 Probability 0.000076
====================================================
In summary, the model that uses only the variable TAR and the
intercept to predict the amount of carbon monoxide should be used as
there is no evidence that either of the other variables have any
predictive power over the amount of carbon monoxide given off by a
cigarette.
The model which predicts the number of wildcat wells from the
variables OILCON, PCI, PRICEOK and VEHICLE is shown below:
============================================================
LS // Dependent Variable is WILDCT2
Date: 03/20/98 Time: 22:23
Sample: 1936 1987
Included observations: 52
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 29855.87 7030.455 4.246649 0.0001
OILCON 53073.31 248862.9 0.213263 0.8320
PCI -1875.729 1549.480 -1.210554 0.2321
PRICEOK 2100.154 406.8240 5.162317 0.0000
VEHICLE 9139.570 39587.01 0.230873 0.8184
============================================================
R-squared 0.623885 Mean dependent var 43583.69
Adjusted R-squared 0.591875 S.D. dependent var 16546.58
S.E. of regression 10570.73 Akaike info criter 18.62290
Sum squared resid 5.25E+09 Schwarz criterion 18.81052
Log likelihood -552.9802 F-statistic 19.49042
Durbin-Watson stat 0.403879 Prob(F-statistic) 0.000000
============================================================
The results of this model show that individually the only variable that
is significant in predicting the number wildcat oil wells drilled is the price
of oil, PRICEOK. The coefficient is positive and says that as the price of a
barrel of oil increases by one dollar the expected number of wells drilled
will increase by 2100 or so. The other variables are not individually
significant. The result makes sense if one believes that the major cause for
drilling more wells is financial. It could be that the demand for oil, or the
number of vehicles on the road, or even the per capita income are exogenous
and do not determine whether it is actually cost effective to drill wells. In
fact, this is what the the model shown here seems to imply.
The regression below is used to perform the
joint hypothesis test on the other coefficients.
============================================================
LS // Dependent Variable is WILDCT2
Date: 03/20/98 Time: 22:26
Sample: 1936 1987
Included observations: 52
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 20900.92 2991.614 6.986503 0.0000
PRICEOK 1820.533 209.4608 8.691520 0.0000
============================================================
R-squared 0.601729 Mean dependent var 43583.69
Adjusted R-squared 0.593763 S.D. dependent var 16546.58
S.E. of regression 10546.25 Akaike info criter 18.56475
Sum squared resid 5.56E+09 Schwarz criterion 18.63980
Log likelihood -554.4684 F-statistic 75.54251
Durbin-Watson stat 0.343020 Prob(F-statistic) 0.000000
============================================================
To test
H0:
boilcon =
bpci =
bvehicle =
0
H1:
boilcon
¹ 0 and/or
bpci
¹ 0 and/or
bvehicle
¹
0
Fq,n-k-1 = {[SSER - SSEUR]/q}/
{SSEUR/n-k-1} =
[(5.56E09 - 5.25E09)/3]/[5.25E09/(52-4-1)] = .9229
F3,47,.05 = 2.80.
Since .9251 < 2.80 we do not reject the null hypothesis. We do not
have sufficient evidence to say that at least one of the three coefficients
is not equal to 0. Note again (as in problem one) the adjusted R2
is higher when we restrict the coefficients of OILCON, PCI and
VEHICLE to be 0.
The equivalent Wald Test is:
====================================================
Wald Test:
Equation: Untitled
====================================================
Null HypothesisC(2)=0
C(3)=0
C(5)=0
====================================================
F-statistic 0.922888 Probability 0.437107
Chi-square 2.768665 Probability 0.428684
====================================================
The results of fitting the model described in the homework are shown
below. The variables were all transformed by taking logs.
============================================================
LS // Dependent Variable is LWILDCT2
Date: 03/20/98 Time: 22:43
Sample: 1936 1987
Included observations: 52
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 11.71659 1.257868 9.314645 0.0000
LOILCON 0.564075 0.362768 1.554919 0.1267
LPCI -0.586528 0.290325 -2.020249 0.0491
LPRICEOK 0.707562 0.159539 4.435029 0.0001
LVEHICLE -0.258323 0.396029 -0.652283 0.5174
============================================================
R-squared 0.552816 Mean dependent var 10.62026
Adjusted R-squared 0.514758 S.D. dependent var 0.348383
S.E. of regression 0.242681 Akaike info criter-2.740802
Sum squared resid 2.768024 Schwarz criterion -2.553183
Log likelihood 2.476056 F-statistic 14.52556
Durbin-Watson stat 0.336451 Prob(F-statistic) 0.000000
============================================================
In this model there have been several changes in the significance level
of the coefficients. The variable LPCI and LOILCON are now individually more
significant then they were before. The coefficient of vehicle changed sign
between the two models, but it was not significantly different from 0 in
either case. Thus it does not make sense to worry about this coefficient
too much. This model seems to indicate that both an increase in price and an increase in
oil consumption cause increases in the number of wells drilled in the United
States each year. It also indicates that increases in per capita income cause
decreases in the number of wells drilled.
These results are not intuitive. The fact that the coefficient of PCI is
negative could perhaps be explained by the fact that if per capita income
increases consumers don't care as much about how expensive the oil is and are
more willing to buy from foreign competition. On the other hand, by this
reasoning, if oil consumption increases, it is hard to see why this would
cause an increase in the number of wells drilled. Perhaps both these
variables would be better used in a model which predicts the price of oil
(PRICEOK). The fit of the data to the model is actually better in part a,
(higher adjusted R2 ) where the only variable that has a significant
coefficient is PRICEOK.
Presidential vote appears to be related to the performance of the
economy. Note that the sign on GNP is positive.
============================================================
LS // Dependent Variable is PRSVOTE
Date: 03/20/98 Time: 23:01
Sample: 1 19
Included observations: 19
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 50.18956 1.802018 27.85186 0.0000
GNP 0.830757 0.286533 2.899338 0.0100
============================================================
R-squared 0.330871 Mean dependent var 53.05306
Adjusted R-squared 0.291510 S.D. dependent var 7.805461
S.E. of regression 6.569999 Akaike info criter 3.864328
Sum squared resid 733.8031 Schwarz criterion 3.963743
Log likelihood -61.67095 F-statistic 8.406161
Durbin-Watson stat 2.336690 Prob(F-statistic) 0.009977
============================================================
Adding military mobilization to the model does not seem to add much to the
model. This is not surprising given the results of the first homework where
we found that -- excluding the big outlier year of 1946 --
GNP and MILMOB
were significantly related.
============================================================
LS // Dependent Variable is PRSVOTE
Date: 03/20/98 Time: 23:04
Sample: 1 19
Included observations: 19
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 50.20606 1.862087 26.96225 0.0000
GNP 0.820986 0.306912 2.674985 0.0166
MILMOB 0.438611 3.765610 0.116478 0.9087
============================================================
R-squared 0.331438 Mean dependent var 53.05306
Adjusted R-squared 0.247868 S.D. dependent var 7.805461
S.E. of regression 6.769331 Akaike info criter 3.968744
Sum squared resid 733.1814 Schwarz criterion 4.117866
Log likelihood -61.66290 F-statistic 3.965978
Durbin-Watson stat 2.339763 Prob(F-statistic) 0.039915
============================================================
Now, introducing the lag of GNP into the model produces a rather surprising
result. Note that the sign is negative. This does not make sense even though
it is marginally significant. In addition, the last period here is 4 years
earlier (these are presidential elections) and that is probably too far back
to expect the voters to remember.
============================================================
LS // Dependent Variable is PRSVOTE
Date: 03/20/98 Time: 23:06
Sample(adjusted): 2 19
Included observations: 18 after adjusting endpoints
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 52.04742 2.224612 23.39617 0.0000
GNP 0.762056 0.316581 2.407146 0.0304
GNP(-1) -0.414956 0.300494 -1.380913 0.1890
MILMOB 1.031935 3.769765 0.273740 0.7883
============================================================
R-squared 0.428678 Mean dependent var 53.13232
Adjusted R-squared 0.306251 S.D. dependent var 8.023883
S.E. of regression 6.683218 Akaike info criter 3.992329
Sum squared resid 625.3156 Schwarz criterion 4.190190
Log likelihood -57.47186 F-statistic 3.501518
Durbin-Watson stat 2.532604 Prob(F-statistic) 0.044066
============================================================
Hence the model we should pick to predict the presidential votes is the
simplest one using only GNP.
This brings us to our preferred model:
============================================================
LS // Dependent Variable is PRSVOTE
Date: 03/20/98 Time: 23:08
Sample: 1 19
Included observations: 19
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 45.55513 2.154329 21.14586 0.0000
GNP 1.088018 0.252156 4.314866 0.0005
REPUB 7.911778 2.656529 2.978239 0.0089
============================================================
R-squared 0.569517 Mean dependent var 53.05306
Adjusted R-squared 0.515707 S.D. dependent var 7.805461
S.E. of regression 5.431912 Akaike info criter 3.528521
Sum squared resid 472.0906 Schwarz criterion 3.677643
Log likelihood -57.48079 F-statistic 10.58379
Durbin-Watson stat 2.459840 Prob(F-statistic) 0.001179
============================================================
Note the large and significant coefficient on the REPUB dummy variable,
indicating that when the incumbent president is a Republican then they can
expect 7.91 percent more of the vote than when the incumbent president is a
Democrat. The coefficient of GNP is still positive and significant as well.
The dummy variable REPUB should stay in the model because it tells us to add
7.91 percent to any prediction about presidential vote when the incumbent is a
Republican and 0 when the incumbent president is a Democrat. Another way to
say this is that the coefficient of the REPUB dummy variable gives an estimate
of the difference that one would expect between the percentage vote for the
Republican party over and above the Democratic party when all other factors
(here only GNP) are held fixed.
The two models whose estimation was requested are shown below:
============================================================
LS // Dependent Variable is HOUSVOTE
Date: 03/20/98 Time: 23:10
Sample: 1 19
Included observations: 19
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 24.80678 7.923629 3.130735 0.0065
GNP 0.089298 0.225553 0.395908 0.6974
PRSVOTE 0.467304 0.156172 2.992236 0.0086
============================================================
R-squared 0.494500 Mean dependent var 49.90647
Adjusted R-squared 0.431312 S.D. dependent var 5.609913
S.E. of regression 4.230514 Akaike info criter 3.028586
Sum squared resid 286.3560 Schwarz criterion 3.177708
Log likelihood -52.73140 F-statistic 7.825901
Durbin-Watson stat 1.845674 Prob(F-statistic) 0.004264
============================================================
============================================================
LS // Dependent Variable is HOUSVOTE
Date: 03/20/98 Time: 23:10
Sample: 1 19
Included observations: 19
============================================================
Variable CoefficienStd. Errort-Statistic Prob.
============================================================
C 11.51443 5.967311 1.929584 0.0728
GNP -0.465363 0.190953 -2.437049 0.0277
PRSVOTE 0.824670 0.128708 6.407271 0.0000
REPUB -7.927619 1.705135 -4.649262 0.0003
============================================================
R-squared 0.792916 Mean dependent var 49.90647
Adjusted R-squared 0.751499 S.D. dependent var 5.609913
S.E. of regression 2.796532 Akaike info criter 2.241424
Sum squared resid 117.3089 Schwarz criterion 2.440253
Log likelihood -44.25336 F-statistic 19.14481
Durbin-Watson stat 2.238649 Prob(F-statistic) 0.000022
============================================================
The estimated models are:
HOUSVOTE = 24.80678 + .089298*GNP + .467304*PRSVOTE
and
HOUSVOTE = 11.51443 - .465363*GNP + .82467*PRSVOTE - 7.927619*REPUB
The results of these two models are hard to follow. In the first model the
coefficient of GNP is not significantly different from 0 and it is positive.
In the second equation the coefficient of the variable GNP is negative and it
is significantly different from 0. An interpretation of the first model
would say that the House of Representatives vote for the incumbent's party
goes up by a factor of .467304 for each percent increase in the presidential
vote for the incumbent party's presidential candidate.
The increase in the House vote due to a unit change in presidential vote is
even greater in the second model. This is because we are controlling for two
factors which negatively influence the House vote as they increase. The fact
that the incumbent is a Republican negatively influences the House vote by 7.9
percent. For each unit increase in GNP growth, there is a decrease in House
vote for the incumbent's party by .465363. This last statement does not make
sense if one believes that a higher GNP implies that the voters are more
likely to vote for the incumbent. One way that this might be explained is
that the predictive power of the variable PRSVOTE on HOUSVOTE
could be much
greater than that of GNP, and in this model GNP is in
essence a correction term.
Note that in the first model GNP does not influence the House vote (large
p-value) whereas it does in the second model. The r-square is much higher in
the second model with the REPUB dummy variable indicating that it provides
much of the punch of the model. Because GNP was not significant in the first
model and it was in the second, the relationship between GNP and REPUB should
be investigated.
Below is the correlation matrix for the variables used in the two models.
Correlation Matrix
============================================================
HOUSVOTE GNP PRSVOTE REPUB
============================================================
HOUSVOTE 1.000000 0.460028 0.699677 -0.270831
GNP 0.460028 1.000000 0.575214 -0.342567
PRSVOTE 0.699677 0.575214 1.000000 0.261906
REPUB -0.270831 -0.342567 0.261906 1.000000
============================================================
The variables REPUB and GNP are negatively correlated.
Thus when REPUB is
added to the model, the variable GNP becomes a better predictor.
However, we
should be very suspicious about the role GNP plays in this model.
The coefficient on GNP is negative -- which is counter-intuitive
-- and it is not
significant when used by itself in the absence of REPUB.