QBA ASSIGNMENT PROJECT
Semester 1, 2011.
Lecturer: Dr Otto Konstandatos
This assignment is in two parts of unequal value.
Part 1 is technical, and is designed to help you understand the connection between Mathematics
(specifically multivariate optimisation using calculus), and econometric regression analysis, by
guiding you through the derivation of the least squares estimators which are used in multiple
regression.
This part may be typed or neatly hand-written in the space provided in the printed document.
Part 2 is applied, using multiple regression analysis to explore a real-world problem, namely the
relationship between a car’s ‘size’, and its fuel efficiency. Part 2 is designed to test your econometric
modelling skills using multiple regression and Eviews.
This part requires you to cut and paste your eviews output in the spaces below before printing your
solution.
This assignment may be done in groups of at most to five students who either are in the same formal
tutorial group as you, or who have the same tutor but are from another tutorial group. When you
hand in your assignment you must include this cover sheet. No names can be added onto the group
lists apart from the names that appear below.
Due Time/Date: 1:00pm Monday 30-05-2011
It must be deposited into your assigned tutor’s collection box on Level 3, Building 5, Discipline
of Finance. Ku-ring-gai students may submit their scripts after the lecture.
Name
Johanan Ottensooser
Jeremy Raymond
Fariba Razi
Kristina Coffey
Tutor’s Name:
Tutorial Day and Time:
Date stamp or tutor’s signature and date
Student Number
10873305
10596854
10449977
10837944
Quantitative Business Analysis
Part 1:
Question 1:
1. The sum of the average value of x is equal to the sum of all x values: that
is,
n
n
n
n
i=1
i=1
i=1
i=1
å xi = å x , since å xi - å x = 0 .
2. We are required to prove that
n
n
i=1
i=1
n
n
i=1
i=1
å(xi - x)(yi - y) = å xi (yi - y).
3. LHS = å(xi - x )(yi - y) = å(xi yi - xi y - xyi - xy)
4. Applying the rule of summation:
n
n
n
n
i=1
i=1
i=1
i=1
n
n
n
n
i=1
i=1
i=1
i=1
n
n
n
i=1
i=1
i=1
n
n
n
n
n
i=1
i=1
i=1
i=1
i=1
n
n
i=1
i=1
LHS = å xi yi - å xi y - å xyi + å xy
5. A constant may be moved outside of the summation, thus:
LHS = å xi yi - å yxi - xå yi + å xy
6. The sum of a constant is n multiplied by the constant, thus:
LHS = å xi yi - å yxi - xå yi + nxy
7. Applying the rule from point 1:
LHS = å xi yi - å yxi - xå y + nxy = å xi yi - å yxi - nxy + nxy
8. Thus, collecting like terms:
LHS = å xi yi - å yxi
9. Applying the rules of summation:
n
LHS = å(xi yi - yxi )
i=1
10. Factorizing:
n
LHS = å xi (yi - y) = RHS
i=1
Question 2:
n
1. Since
n
å(x - x)(y - y) = å x (y - y).
i
i
i
i=1
i
i=1
2. We are required to prove that
n
n
å(xi - x)2 = å xi (xi - x)
i=1
i=1
n
n
n
3. LHS = å(xi - x)2 = å(xi - x)(xi - x) = å(xi2 - xi x - xi x + x )
i=1
n
i=1
i=1
n
n
i=1
i=1
4. LHS = å xi2 - 2xå xi + å x
i=1
n
5. Since
2
n
åx = åx
i
i=1
n
i=1
n
n
i=1
i=1
LHS = å x - 2xå x + å x
2
i
i=1
n
6. Since
2
2
å k = nk ,
i=1
n
n
n
LHS = å xi2 - 2x ·nx + nx = å xi2 - 2nx + nx = å xi2 - nx
2
i=1
2
2
i=1
i=1
2
7. Applying the rules of summation, and the rule from Q1 point 1:
n
n
n
n
n
n
LHS = å x - å x = å x - å x · x = å x - xå xi
2
i
i=1
i=1
2
2
i
i=1
2
i
i=1
8. Applying the rules of summation:
n
LHS = å xi2 - xxi
i=1
9. Factorizing:
n
LHS = å xi (xi - x) = RHS
i=1
i=1
i=1
Question 3:
Part 1:
1. Show that
n
¶
f (b0 , b1 ) = -2å(yi - b0 - b1 xi )
¶b0
i=1
n
2.
f (b0 , b1 ) = å(y1 - b0 - b1 xi )2
i=1
3. Applying the chain rule
¶
( f (x))n = n( f (x))n-1 · f '(x) :
¶x
n
¶
f (b0 , b1 ) = 2å(-1)(yi - b0 - b1 xi )
¶b0
i=1
4. And simplifying
n
¶
f (b0 , b1 ) = -2å(yi - b0 - b1 xi ) , as required.
¶b0
i=1
Part 2:
¶
f (b0 , b1 ) = 0 when b0 º B̂0 = y - b1 x
¶b0
2. Substituting for b0 :
1. Then show that
n
n
¶
f (b0 , b1 ) = -2å(yi - (y - b1 x) - b1 xi ) = -2å(yi - y + b1 x - b1 xi )
¶b0
i=1
i=1
3. Applying the rules of summation:
n
n
n
n
n
n
én
ù
én
ù
= -2 êå yi - å y + å b1 x - å b1 xi ú = -2 êå yi - å y + b1 (å x - å xi )ú
ë i=1
û
ë i=1
û
i=1
i=1
i=1
i=1
i=1
i=1
n
4. Since
n
åx = åx :
i
i=1
n
i=1
n
n
n
é
ù
= -2 êå y - å y + b1 (å x - å x)ú
ë i=1
û
i=1
i=1
i=1
5. Collecting like terms:
= -2 [ 0 + b1 (0)]
¶
6. Thus, when b0 º B̂0 = y - b1 x ,
f (b0 , b1 ) = 0 , as required.
¶b0
Question 4:
Define
¶
f (b0 , b1 ) :
¶b1
n
1.
f (b0 , b1 ) = å(yi - b0 - b1 xi )2
i=1
2. Applying the chain rule:
¶
( f (x))n = n( f (x))n-1 · f '(x) :
¶x
n
¶
f (b0 , b1 ) = 2å(-xi )(yi - b0 - b1 xi )
¶b1
i=1
Question 5:
n
¶
Setting
f (b0 , b1 ) = 0 , show that b1 º B̂1 =
¶b1
å(x - x)(y - y)
i
i
i=1
n
å(x - x)
2
i
i=1
1.
n
¶
f (b0 , b1 ) = 0 = 2å(-xi )(yi - b0 - b1 xi )
¶b1
i=1
n
n
n
n
i=1
i=1
i=1
2. 0 = 2å(-xi yi + b0 xi + b1 x i2) = -å xi yi + b0 å xi + b1 å x i2
i=1
n
3.
n
n
å(x y ) - b å x = b å x
i i
i
0
i=1
1
i=1
i=1
n
n
å(x y ) - b å x
i i
4. b1 =
2
i
i
0
i=1
i=1
n
åx
2
i
i=1
5. Now, substitute b0 with y - b1 x :
n
b1 =
n
å(xi yi ) - (y - b1 x)å xi
i=1
i=1
n
åx
2
i
i=1
n
n
n
n
i=1
i=1
n
6. b1 å x = å(xi yi ) - yå xi + b1 xå xi
2
i
i=1
n
i=1
n
n
i=1
i=1
n
n
i=1
i=1
7. b1 (å x i2 + xå xi ) = å(xi yi ) - yå xi
i=1
n
n
i=1
8. b1 (å x i2 + å xi x ) = å(xi yi ) - å(xi y)
i=1
n
i=1
n
9. b1 (å x i2 + xi x) = å(xi yi - xi y)
i=1
n
i=1
n
10. b1 (å xi (xi + x) = å xi (yi - y)
i=1
n
11.
b1 (å xi (xi + x)
i=1
n
å x (x + x)
i
i=1
n
i
=
i
i=1
n
å x (y - y)
i
12. b1 =
i
i=1
n
å x (x + x)
i
i=1
i
å x (y - y)
i
i=1
n
å x (x + x)
i
i=1
i
, when b0 = y - b1 x
n
13. Substitute (as proved above)
n
å x (y - y) = å(x - x)(y - y), and
i
i
i=1
n
n
i=1
i=1
å xi (xi - x) = å(xi - x)2
n
å(x - x)(y - y)
i
14. Thus, b1 =
i
i=1
n
å(x - x)
i
i=1
2
, as required
i
i=1
i
Question 6
Use the appropriate second-order test from Mathematics to show that the above
choices for (b0 , b1 ) give a minimum for f (b0, b1 ).
By deriving this two factor function twice, in all its forms, and using the formula
D = AC - B2 , it is possible to prove, if D > 0 , that the outcome is a local minimum.
The function is:
n
1.
f (b0 , b1 ) = å(y1 - b0 - b1 xi )2
i=1
The first derivatives are:
n
n
n
¶
2.
f (b0 , b1 ) = -2å(yi - b0 - b1 xi ) = -2å yi + 2nb0 - 2b1 å xi
¶b0
i=1
i=1
i=1
n
n
n
n
¶
f (b0 , b1 ) = 2å(-xi )(yi - b0 - b1 xi ) = -2å xi yi + 2b0 å xi + 2b1 å xi2
¶b1
i=1
i=1
i=1
i=1
The second derivatives are:
¶2
4.
f (b0 , b1 ) = 2n = A
¶b02
3.
5.
n
¶2
f (b0 , b1 ) = 2å xi = B
¶b0 b1
i=1
n
¶2
f (b0 , b1 ) = 2å xi2 = C
2
¶b1
i=1
Applying the test D = AC - B2
6.
n
n
7. D = (2n)(2å x ) - (2å xi )2
2
i
i=1
é n 2 n
ù
8. = 4 ênå x i - (å xi )2 ú
ë i=1
û
i
9. since
n
n
i=1
i=1
i=1
å xi = å x
n
é n 2 n 2ù é n 2 n
ù é n 2
ù é n 2 2 2ù
én 2
= 4 ênå x i - (å x ) ú = 4ênå x i - (å x )(å x )ú = 4 ênå x i - (nx)(nx)ú = 4 ênå x i - n x ú = 4n êå x i
ë i=1
û ë i=1
û ë i=1
û ë i=1
û
ë i=1
i
i
i
é n 2 n 2ù
10. = 4n êå x i - å x ú
ë i=1
û
i=1
Thus, D is always positive, and a local minimum exists. (b0 , b1 ) , thus, gives a
minimum for f (b0, b1 ).
Part 2: Econometrics
Question 1
(1 mark) Read the given data file cars.xls into EVIEWS, and run a regression of
KMLIT against the rest of the variables assuming homoskedastic errors. Copy and
paste the EVIEWS output into the space below, and report the estimated
equation with the standard errors below the coefficients.
Regression output
Dependent Variable: KMLIT
Method: Least Squares
Date: 05/23/11 Time: 12:35
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
CYL
ENGCM3
HP
WTKG
C
-0.160554
2.72E-05
-0.015504
-0.004089
16.23109
0.146039
0.000196
0.004587
0.000561
0.540492
-1.099391
0.139041
-3.379960
-7.286719
30.03019
0.2723
0.8895
0.0008
0.0000
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
F-statistic
Prob(F-statistic)
0.704063
0.701004
1.511960
884.6908
-715.7633
230.1772
0.000000
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
Durbin-Watson stat
8.297429
2.765078
3.677364
3.728017
3.697439
0.887708
Equation
Y
= B1(X1)
KMLIT
= -0.160554(CYL)
Std. Errors
(0.146039)
+B2(X2)
+2.72E-05(ENGCM3)
(0.000196)
+B3(X3)
+B4(X4)
+U
-0.015504(HP) -0.004089(WTKG) 16.23109
(0.004587)
(0.000561)
0.540492
Question 2
(1 mark) Comment on the sign of each estimated coefficient in turn, and state
whether this is what you expect. Ignore significance at this stage.
Ans
The coefficient “CYL”, or number of cylinders, has a negative sign: as the number
of cylinders increases, the fuel efficiency decreases. This is to be expected, since,
cetirus paribus, increase in the number of cylinders with out an increase in
horsepower, weight or engine capacity would increase the friction on the motor,
decreasing efficiency and reducing the amount of kilometres that you are able to
drive per litre.
The coefficient “ENGCM3”, or the capacity of the engine in centimeters squared,
has a positive sign: increasing engine capacity increases the efficiency of the
engine. Whilst an engine with a higher capacity may run at lower revs, the
positive sign is still unexpected, since more petrol is used.
The coefficient “HP”, or the power of the engine, has a negative sign: as power
increases, the fuel efficiency decreases. Whilst this is to be expected at the
higher range (increasing power above a certain point would decrease efficiency)
this is not to be expected at the lower range: where there is not enough power,
you would need to use more throttle to maintain speed, using more petrol.
However, this is a relatively rare case, and in most cases, the negative
relationship is to be expected.
The coefficient “WTKG”, or weight in kilos has a negative sign: the heavier the
car, the less fuel efficient. This is to be expected since more power will need to
be used to move the increased weight.
Question 3
(1 mark) Interpret the estimated effect of the engine power (HP) on kilometers
travelled.
Answer
An increase of 1 hp will lead to a 0.0155 kilometres per litre decrease in
efficiency.
Further, this coefficient is statistically significant (with a 99.92% chance of being
non-zero).
Question 4
(1 mark) Test whether the data supports the hypothesis that Engine size does
affect a car’s mileage (i.e. how far it can travel per litre). Formulate and carry out
an appropriate hypothesis test using the t-statistic approach at the 5%
significance level.
Answer
To prove engine size (ENGCM3) affects mileage (KMLIT), ENGCM3≠0.
1. H0 :KMLIT=0
2. H1 :KMLIT≠0
3. The level of significance is 5%; this is a two tailed test, so 2.5% per tail.
a. tcrit = −1.96
4. tact =
̂1 −𝛽1,0
𝛽
̂1
SE𝛽
(2.72×10−05 )−0
a. =
0.000196
b. ≈ 0.1388 (4.d.p)
5. Since tact < |tcrit |, we do not reject the null hypothesis: thus, engine size
does not statistically significantly affect mileage at the 5% level.
Question 5
(1 mark) Test whether the number of cylinders affects a car’s mileage.
Formulate and carry out an appropriate hypothesis test using the p-values
approach, at the 5% level.
Answer
1. H0 :CYL=0
2. H1 :CYL≠0
3. The level of significance is 5%; this is a two tailed test.
a. Thus, if P-val<0.05, accept at the 5% level.
4. tact =
̂1 −𝛽1,0
𝛽
̂1
SE𝛽
(−0.160554)−0
a. = 0.146039
b. -1.099391≈ 1.10
5. P-Val=2Φ(-|tact |)
a. 2 Φ-1.10
b. 2(0.1357)
c. 0.2714= 27.14%
6. Since P-Val>5%, we reject H0 , Cylinder size is not a statistically significant
variable in determining mileage.
Question 6
(2 marks) Test the following hypotheses about the coefficients on CYL (B1) and
ENGCM3 (B2). Clearly specify the rejection region if you are using critical values,
and clearly state your conclusions. When using p-values, calculate and compare
your p-values to the test size then state your conclusion.
(Hint, assume the Central Limit Theorem holds)
(a) H0: B1  0 , H1: B1  0 , with α=0.05 using the critical-value approach.
(b) H0: B1  0 , H1: B1  0 , with α=0.05 using the critical-value approach.
(c) H0: B2  0 , H1: B2  0 , with α=0.05 using the p-value approach.
(d) H0: B2  0 , H1: B2  0 , with α=0.05 using the p-value approach.
Answer
Part A
To prove that B>0, we must reject H0: B1=0, in the right tail of the distribution.
In order to prove that B1 is greater than 0, we must show that Tact(B)>Tcrit(B).
Thus, Tact(B) must be in the rejection region, in the right 5% of the distribution,
to the right of Tcrit(B)=2.57.
Cyl(B1)
TactCyl(B1)=-1.10 (from the E-views output). This is to the left of TcritCyl(B1),
thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is statistically
significantly greater than 0.
ENGCM3(B2)
TactENGCM3(B2)=0.14 (from the E-views output). This is to the left of
TcritCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that
ENGCM3(B2) is statistically significantly greater than 0.
Part B
To prove that B<0, we must reject H0: B1=0, in the left tail of the distribution.
In order to prove that B1 is less than 0, we must show that Tact(B)<Tcrit(B).
Thus, Tact(B) must be in the rejection region, in the left 2.5% of the distribution,
to the left of Tcrit(B)=-2.57.
Cyl(B1)
TactCyl(B1)=-1.10 (from the E-views output). This is to the right of TcritCyl(B1),
thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is statistically
significantly less than 0.
ENGCM3(B2)
TactENGCM3(B2)=0.14 (from the E-views output). This is to the right of
TcritCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that
ENGCM3(B2) is statistically significantly greater than 0.
Part C
To prove that B>0, we must reject H0: B1=0, in the right tail of the distribution.
In order to prove that B1 is greater than 0, we must show that P-Val(B)>PCrit(B). Thus, P-Val(B) must be in the rejection region, in the right 5% of the
distribution (P-Val(B) must be equal to or greater than 95%).
Cyl(B1)
P-Val Cyl(B1)=.27=27% (from the E-views output). This is to the left of PcritCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is
statistically significantly greater than 0.
ENGCM3(B2)
P-Val ENGCM3(B2) =.89= 89% (from the E-views output). This is to the left of PcritCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is
statistically significantly greater than 0.
Part C
To prove that B<0, we must reject H0: B1=0, in the left tail of the distribution.
In order to prove that B1 is less than 0, we must show that P-Val(B)<P-Crit(B).
Thus, P-Val(B) must be in the rejection region, in the left 5% of the distribution
(P-Val(B) must be equal to or greater than 5%).
Cyl(B1)
P-Val Cyl(B1)=.27=27% (from the E-views output). This is to the right of PcritCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is
statistically significantly less than 0.
ENGCM3(B2)
P-Val ENGCM3(B2) =.89= 89% (from the E-views output). This is to the right of
P-critCyl(B1), thus, we cannot reject H0, and, thus, we cannot show that Cyl(B1) is
statistically significantly less than 0.
Question 7
(1 mark) Formulate a hypothesis test to test whether a unit increase in a car’s
weight (mass) has a greater detrimental effect on fuel efficiency than a unit
increase in the power of the car’s engine, rather than the same effect. Use reparameterization to convert the model to allow you to test this hypothesis using
a simple t-test.
Answer
We want to test that B4>B3 (that weight has a greater effect on efficiency than
engine power).
The original regression is of the form
Yi = 𝛽0 + 𝛽1 (Cyl)+𝛽2 (ENGCM3) + 𝛽3 (HP) + 𝐵4 (WTKG)
However, it is not possible to do the appropriate hypothesis test on this: thus, we
use re-parameterization: 𝛽4 − 𝛽3 = 𝜃. Thus:
Yi = 𝛽0 + 𝛽1 (Cyl)+𝛽2 (ENGCM3) + 𝛽3 (HP) + (𝜃 + 𝛽3 )(WTKG)
Yi = 𝛽0 + 𝛽1 (Cyl)+𝛽2 (ENGCM3) + 𝛽3 ((HP) + (WTKG)) + (𝜃)(WTKG)
Now, we substitute (HP) + (WTKG) = 𝑋̅
Yi = 𝛽0 + 𝛽1 (Cyl)+𝛽2 (ENGCM3) + 𝛽3 (𝑋̅) + (𝜃)(WTKG)
Now, we can do a hypothesis test on θ to solve the problem.
H0: θ=0; H1: θ>0.
If we can reject H0, then θ>0, and B4>B3, (weight is statistically significantly
greater than hp in effect).
Dependent Variable: KMLIT
Method: Least Squares
Date: 05/26/11 Time: 14:59
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
C
CYL
ENGCM3
HP+WTKG
WTKG
16.23109
-0.160554
2.72E-05
-0.015504
0.011415
0.540492
0.146039
0.000196
0.004587
0.004721
30.03019
-1.099391
0.139041
-3.379960
2.418110
0.0000
0.2723
0.8895
0.0008
0.0161
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
F-statistic
Prob(F-statistic)
0.704063
0.701004
1.511960
884.6908
-715.7633
230.1772
0.000000
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
Durbin-Watson stat
8.297429
2.765078
3.677364
3.728017
3.697439
0.887708
1. H0: θ=0; H1: θ>0.
2. Rejection region is the right tail of the distribution, equal to or above Pcrit (0.05)
3. P-val (θ)=0.016 (θ is positive, and significant)
4. Thus, we can reject H0 and accept H1: θ>0.
Since θ is statistically significantly greater than 0, WTKG-HP>0, thus weight has
a greater per-unit effect on mileage than hp.
Question 8
(1 mark) Verify that the “OLS Wonder Equation” gives a standard error for the
CYL coefficient close to 0.145. You will need to run a regression of CYL on all the
other independent variables, and you must include this regression output below.
(Remember, the OLS Wonder Equation gives an estimate of the homoskedasticity
consistent standard error).
Dependent Variable: CYL
Method: Least Squares
Date: 05/26/11 Time: 13:33
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
ENGCM3
HP
WTKG
C
0.000915
-0.002908
0.000426
2.285539
4.97E-05
0.001588
0.000194
0.147783
18.42238
-1.831399
2.195681
15.46555
0.0000
0.0678
0.0287
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
F-statistic
Prob(F-statistic)
1. SE(Cyl)≈
0.905786
0.905057
0.525600
107.1870
-302.0733
1243.421
0.000000
𝑆u
̂
𝑆xi
×
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
Durbin-Watson stat
1
2
√𝑛(1−𝑅𝑥𝑖
𝑜𝑛 𝑥 )
1.511
2. SE(Cyl)≈ 1.706 ×
3. SE(Cyl)≈0.1459
1
√392(1−.906)
5.471939
1.705783
1.561598
1.602121
1.577659
1.479510
Question 9
(1 mark) Test the following joint hypothesis about the coefficients on CYL (B1)
and ENGCM3 (B2): H0: B1  0 and B2  0 , H1: B1  0 or B2  0 , with α=0.05.
Along with the previous results, what do you conclude about B1 and B2? Is this
consistent with your intuition?
Answer
The Regression is as follows:
Yi = 𝛽0 + 𝛽1 (Cyl)+𝛽2 (ENGCM3) + 𝛽3 (HP) + 𝐵4 (WTKG) + 𝐶
To create the restricted regression, we assume the null hypothesis (H0:
B1=B2=0) is true. Thus, the following regression is formed:
Yi = 𝛽0 + 𝛽3 (HP) + 𝐵4 (WTKG) + 𝐶
This has two restrictions and 392 observations, thus, the F-crit value is (at 5%
significance) 3.00.
Dependent Variable: KMLIT
Method: Least Squares
Date: 05/26/11 Time: 14:21
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
HP
WTKG
C
-0.016962
-0.004488
16.13025
0.003947
0.000394
0.282533
-4.297243
-11.38779
57.09155
0.0000
0.0000
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
F-statistic
Prob(F-statistic)
0.702602
0.701073
1.511786
889.0585
-716.7285
459.5047
0.000000
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
Durbin-Watson stat
8.297429
2.765078
3.672084
3.702477
3.684130
0.882036
(𝑅 2 −𝑅 2 )/𝑞
𝑅
1. 𝐹 = (1−𝑅2𝑈𝑅)/(𝑛−𝑘−1)
𝑈𝑅
2.
(0.704−0.703)/2
(1−0.704)/(392−2−1)
3. 0.657
Since the F statistic for this is smaller than the F critical value, we cannot reject
the null hypothesis: thus, the restriction is not void, on CYL (B1) and ENGCM3
(B2) are jointly statistically insignificant.
This is consistent with the tests conducted above, and our expectations for the
values.
Question 10
(2 marks) Can you explain any conflict between the implications of the results
obtained about B1 and B2 and your expectations? (Hint: Run auxiliary regressions
for the explanatory variables in question against the others, and compute the
correlations between all the explanatory variables. What do you notice?).
Answer
The results above suggest that CYL and ENGCM3 are insignificant variables.
Whilst the R squared coefficient of the regression on KMLIT including all the
variables is slightly higher than the restricted regression (KMLIT against HP and
WTKG), this is to be expected by adding variables. Whilst this improves the
predictive power on KMLIT, it does not allow us to further understand causation,
since the standard errors would be higher.
Further, it seems that a regression on KMLIT against the variables HP and WTKG
gives a slightly better adjusted R sqared coefficient: as such, the variables are
more explanatory in such a model.
Thus, we believe that the coefficients HP and WTKG correllate with CYL and
ENGCM3. Whilst leaving the in the regression formula would cause Ommited
Variable Bias, this would be slight compared to the inaccuracy on each variable
caused by the correlation.
This correlation is proved below, and shows that our expectations about CYL and
ENGCM3 are not invalidated, but, rather, that their effect is absorbed by the
effect of WTKG and HP on KMLIT.
CYL is correlated with WTKG and HP
CYL is correlated with both WTKG and HP, as shown in the following scatter
plots (an increase in CYL, generally, is correlated with an increase in HP and
WTKG):
2,400
200
2,000
160
1,600
HP
WTKG
240
120
1,200
80
800
400
40
2
3
4
5
6
7
8
9
2
3
4
5
6
7
8
9
CYL
CYL
Further, an auxillary regression shows that WTKG and HP are excellent
explanators of CYL (with an R squared coefficient of greater than 0.8)
Dependent Variable: CYL
Method: Least Squares
Date: 05/27/11 Time: 13:41
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
HP
WTKG
0.011212
0.003171
0.001856
0.000147
6.042161
21.55655
0.0000
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
Durbin-Watson stat
0.821785
0.821328
0.721029
202.7542
-427.0075
1.584778
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
5.471939
1.705783
2.188814
2.209075
2.196844
ENGCM3 is correlated with WTKG and HP
ENGCM3 is correlated with both WTKG and HP, as shown in the following scatter
plots (an increase in ENGCM3, generally, is correlated with an increase in HP and
WTKG):
2,400
200
2,000
160
1,600
HP
WTKG
240
120
1,200
80
800
40
1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000
400
1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000
ENGCM3
ENGCM3
Further, an auxillary regression shows that WTKG and HP are excellent
explanators of ENGCM3 (with an R squared coefficient of greater than 0.75).
Dependent Variable: ENGCM3
Method: Least Squares
Date: 05/27/11 Time: 13:47
Sample: 1 392
Included observations: 392
Variable
Coefficient
Std. Error
t-Statistic
Prob.
HP
WTKG
21.30413
0.831055
2.074520
0.164440
10.26942
5.053839
0.0000
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
Durbin-Watson stat
0.779611
0.779046
806.0712
2.53E+08
-3178.553
0.812699
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
Hannan-Quinn criter.
Correlations computed
Using Eviews, we computed the following correlations:
CYL
ENGCM3
HP
CYL
1.00
0.95
0.84
ENGCM3
0.95
1.00
0.90
HP
0.84
0.90
1.00
WTKG
0.90
0.93
0.86
3185.650
1714.836
16.22731
16.24757
16.23534
WTKG
0.90
0.93
0.86
1.00
Conclusion
Both CYL and ENGCM3 are correlated closely (and explained well) by HP and
WTKG. Thus, the effects of CYL and ENGCM3 on KMLIT is explained via HP and
WTKG, and their effect ong KMLIT outside of this is too small to be significant.