1
Chapter 12
Solutions to Exercises
Solutions to Exercises in Chapter 12
12.1
(a) The least-squares estimated equation is given by
I!t = 6.22 + 0.770 Yt − 0.184 Rt
(2.51) (0.072) (0.126)
R2 = 0.816
Both b2 and b3 have the expected signs; income is expected to have a positive effect on
investment whereas an increase in the interest rate should reduce investment. The
standard errors for b1 and b2 are relatively small suggesting that the corresponding
coefficients are significantly different from zero. However, the standard error of b3 is
large, yielding a t ratio that is less than two. Based on this standard error, we can question
whether we should include Rt in the equation, although economic theory suggests Rt
should have a strong influence on It.
(b) The plot of the least squares residuals in Figure 12.1 reveals a few long runs of negative
and positive residuals, suggesting the existence of autocorrelation.
e
8
4
0
0
5
10
15
20
25
30
35
t
-4
-8
Figure 12.1 Residuals for Investment Equation
(c) In this context, the Durbin-Watson test is a test for H0: ρ = 0 against H1: ρ > 0 in the firstorder autoregressive model et = ρet−1 + vt. We find the computed value for the DurbinWatson statistic is d = 0.852 and the p-value of the test if P(d < 0.852) = 0.000114.
Because the p-value is less than 0.05, we reject H0 and conclude that autocorrelation does
exist.
(d) The estimator for ρ introduced on p. 248 of the text is not the one utilized by all software.
Different programs often employ slightly different estimators that yield different
estimates. Consequently, for this question we report three estimates for ρ and the three
corresponding generalized least-squares estimated equations. The results from using
SHAZAM, EViews and SAS are:
2
Chapter 12
Solutions to Exercises
I!t = 8.43 + 0.742 Yt − 0.285 Rt
(2.86) (0.114) (0.079)
ρ̂ = 0.5677
(SHAZAM)
Iˆt = 7.33 + 0.785Yt − 0.296 Rt
(3.73) (0.147) (0.080)
ρˆ = 0.6146
(EViews)
I!t = 8.41 + 0.742 Yt − 0.285 Rt
(2.90) (0.115) (0.081)
ρ! = 0.5616
(SAS)
Comparing these results with those form part (a), we find that there has been little change
in the coefficient estimates, but a considerable change in the standard errors. The
standard error on the coefficient of Yt has increased, suggesting that, if we did not correct
for autocorrelation, our confidence interval for β 2 would be too narrow, giving us a false
sense of the reliability of our estimate. The opposite has occurred for β 3. Here the
standard error has dropped after correcting for autocorrelation. From the results in part
(a) we might be misled into thinking that interest rate has no impact on investment (the
estimated coefficient is not significant). After correcting for autocorrelation we have a
relatively narrow confidence interval that does not include zero.
(e) Given the next year values of Y and R are YT+1 = 36 and RT+1 = 14, the appropriate
forecast using the SHAZAM results is
I!T + 1 = β! 1 + β! 2 YT + 1 + β! 3RT + 1 + ρ! ~
eT = 8.43 + 0.742(36) − 0.285(14) + 0.5677(2.1513) = 32.356
With EViews and SAS, the result turns out to be 32.646 and 32.346, respectively.
If autocorrelation is ignored, our prediction is
I!T +1 = b1 + b2 YT +1 + b3 RT +1 = 6.22 + 0.77(36) − 0.184(14) = 31.363
There is not a large difference between the two predictions.
12.2
From the equation for the AR(1) error model we have
var(et ) = ρ2 var(et −1 ) + var(vt ) + 2ρ cov(et − 1 , vt )
(
)
from which we get σ 2e = ρ2 σ 2e + σ 2v + 0 , or σ 2e 1 − ρ2 = σ 2v , and hence
σ 2e =
σ 2v
1 − ρ2
To find E (et et −1 ) we note that et et −1 = ρet2−1 + et − 1vt . Taking expectations,
( )
E (et et − 1 ) = ρE et2− 1 + 0 = ρσ 2e
Similarly, et et − 2 = ρet − 1et − 2 + et − 2 vt , and
E (et et − 2 ) = ρE(et − 1et − 2 ) + 0 = ρE (et et −1 ) = ρ2 σ 2e
3
Chapter 12
12.3
Solutions to Exercises
(a) The least-squares estimated equation is
∧
ln( JVt ) = 3.5027 − 1.6116 ln(Ut)
(0.2829) (0.1555)
R2 = 0.8299
Using the value tc = 2.074, a 95% confidence interval for β 2 is
b2 ± tcse(b2) = (−1.9342, −1.2890)
(b) The value of the Durbin-Watson statistic is d = 1.09. In terms of its p-value, we find that
P(d < 1.09) = 0.0088. Since this p-value is less than 0.05, we reject H0 and conclude that
positive autocorrelation exists. The existence of autocorrelation means the original
assumption for et , that the et are independent, is not correct. This problem also causes the
confidence interval for β 2 in part (a) to be incorrect, meaning we will have a false sense
of the reliability of the coefficient estimate.
(c) After correcting for autocorrelation, the estimated equation is
∧
ln( JVt ) = 3.5137 − 1.616 ln(Ut)
(0.2377) (0.123)
^
ln( JVt ) = 3.5034 − 1.600 ln(U t )
(0.2487) (0.132)
∧
ln( JVt ) = 3.5138 − 1.616 ln(Ut)
(0.2437) (0.127)
ρ! = 0.4472
(SHAZAM)
ρˆ = 0.4486
(EViews)
ρ! = 0.4318
(SAS)
The results for SAS, EViews and SHAZAM differ slightly because they use different
estimators for ρ. The 95% confidence intervals for β 2 from SHAZAM, EViews and SAS
are, respectively, (−1.872, −1.360), (–1.875, –1.325) and (−1.879, −1.353). These
confidence intervals are slightly narrower than that given in part (a). A direct comparison
with the interval in part (a) is difficult because the least squares standard errors are
incorrect in the presence of AR(1) errors. However, given the change in standard errors is
not great, and that we know least squares is less efficient, one could conjecture that the
least squares confidence interval is narrower than it should be, implying unjustified
reliability.
12.4
(a) The marginal functions are obtained by differentiating the total functions. That is,
mr =
d ( tr )
= β1 + 2β 2 q
dq
mc =
d ( tc)
= α 2 + 2α 3q
dq
(b) If mc = mr, then β1 + 2β 2 q ∗ = α 2 + 2α 3q ∗ , and 2 q ∗ (β 2 − α 3 ) = α 2 − β1 , leading to the
solution
q∗ =
α 2 − β1
2(β 2 − α 3 )
4
Chapter 12
Solutions to Exercises
(c) The least squares estimated models (with standard errors in parentheses) are
∧
tri = 174.28 qi − 0.5024 qi2
(4.54) (0.0235)
∧
tri = 2066.1 − 1.5784 qi + 0.22768 qi2
(727.2) (9.452)
(0.02889)
The statistical models that are appropriate for the above estimation are
tri = β 1qi + β 2 qi2 + e1i
tci = α1 + α2qi + α3 qi2 + e2i
where e1i and e2i are both independent random variables with zero means and constant
variances σ12 and σ 22 , respectively. The profit maximizing level of output suggested
from the least squares estimates is
q∗ =
−1.5784 − 174. 28
= 120.43
2 ( −0. 5024 − 0. 2277 )
or q ∗ ≈ 120 .
(d) For q = 120, total cost and total revenue predictions in each of the next three months are
given by
∧
tr T + h = 174.28(120) − 0.5024(120)2 = 13679
h = 1,2,3
∧
tc T + h = 2066.1 − 1.5784(120) + 0.22768(120)2 = 5155
h = 1,2,3
Thus, a prediction for profit is π! = 13679 − 5155 = 8524.
(e) The Durbin-Watson statistics and p-values for each of the equations are
Equation
revenue
cost
DW-value
0.287
1.076
p-value
0.00000
0.00028
Both p-values are substantially below a nominal significance level of 0.05. We therefore
conclude that the errors in both equations are autocorrelated.
(f) The answers to the remaining parts of this question could depend on the computer
software package that is being used. Different software packages often use different
estimators for ρ and varying estimates of ρ flow through to the generalized least squares
estimates and their standard errors, as well as predictions of future values. From this part
on we report SHAZAM, EViews and SAS estimates. Generalized least squares estimates
of the coefficients appear in the table below. Standard errors are in parentheses.
5
Chapter 12
Solutions to Exercises
Equation
tr
Variable
SHAZAM
tc
EViews
SAS
SHAZAM
EViews
SAS
2347.7
(608.4)
2354.7
(616.0)
2318.6
(633.2)
constant
qi
172.92
(5.90)
171.58
(8.01)
175.28
(4.60)
−5.500
(7.920)
-5.752
(8.020)
−5.119
(8.263)
qi2
−0.5123
(0.0188)
-0.5085
(0.0248)
−0.5190
(0.0155)
0.2410
(0.0244)
0.2415
(0.0247)
0.2396
(0.0254)
ρ
0.8882
0.9495
0.7965
0.4439
0.4595
0.3629
(g) The profit maximizing level of output suggested by the SHAZAM results in part (f) is
q∗ =
−5.4999 − 172.92
= 118.42
2 ( −0.51233 − 0.24101)
or, q ∗ ≈ 118 . Using the EViews and SAS results in part (f) we obtain q ∗ ≈ 118 and
q∗ ≈ 119 , respectively.
(h) In this case, because the errors are assumed autocorrelated, the total revenue and total
cost errors for month 48 have a bearing on the predictions, and the predictions will be
different in each of the future three months. For the total revenue function the SHAZAM
estimated error for month 48 is
~
e = 8435 − [172.92(83) − 0.51233(83)2] = −2387.98
1,48
Therefore, for q = 118, total revenue predictions for the next 3 months (using the
SHAZAM results) are given by
∧
∧
∧
∧
∧
∧
tr T +1 = tr 49 = 172.92(118) − 0.51233(118)2 + (0.88817)1(−2387.98) = 11150
tr T + 2 = tr 50 = 172.92(118) − 0.51233(118)2 + (0.88817)2(−2387.98) = 11387
tr T + 3 = tr 51 = 172.92(118) − 0.51233(118)2 + (0.88817)3(−2387.98) = 11598
Using the EViews results the corresponding predictions are
∧
∧
∧
tr 49 = 10979
tr 50 = 11090
tr 51 = 11195
Using the SAS results the corresponding predictions are
∧
∧
∧
tr 49 = 11487
tr 50 = 11899
tr 51 = 12226
For the total cost function, the SHAZAM error for the last sample month is
~
e
= 4829 − 2347.7 + 5.4999(83) − 0.24101(83)2 = 1277.45
2 , 48
6
Chapter 12
Solutions to Exercises
Therefore, using the SHAZAM results, total cost predictions for q = 118 are
∧
∧
tc T +1 = tc 49 = 2347.7 − 5.4999(118) + 0.24101(118)2 + (0.44394)1(1277.45) = 5622
∧
∧
∧
∧
tc T + 2 = tc 50 = 2347.7 − 5.4999(118) + 0.24101(118)2 + (0.44394)2(1277.45) = 5306
tc T + 3 = tc 51 = 2347.7 − 5.4999(118) + 0.24101(118)2 + (0.44394)3(1277.45) = 5166
The corresponding predictions from EViews are
∧
∧
∧
tc 49 = 5630
tc 50 = 5311
tc 51 = 5164
The corresponding predictions from SAS are
∧
∧
∧
tc 49 = 5569
tc 50 = 5272
tc 51 = 5164
Profits for the months of 49, 50 and 51 (obtained by subtracting total cost from total
revenue) are
(SHAZAM)
π! 49 = 5528
π! 50 = 6081
π! 51 = 6432
(EViews)
π! 49 = 5349
π! 50 = 5779
π! 51 = 6031
(SAS)
π! 49 = 5918
π! 50 = 6627
π! 51 = 7062
Because ~
e1T is negative, and its impact declines as we predict further into the future, the
total revenue predictions become larger the further into the future we predict. The
e2 T is positive. Combining these
opposite happens with total cost; it declines because ~
two influences means that the predictions for profit increase over time. These predictions
are, however, much lower than 8524, the prediction for profit that was obtained when
autocorrelation was ignored. Thus, even although autocorrelation has little impact on the
optimal setting for q, it has considerable impact on the predictions of profit. This impact
is caused by a relatively large negative residual for revenue, and a relatively large
positive residual for cost, in month 48.
12.5
(a) For the AR(1) error model et = ρet −1 + vt , we are testing H0: ρ = 0 against H1: ρ > 0. The
computed value of the Durbin-Watson statistic is 0.6634 and its p-value is 0.000182.
Since 0.000182 < 0.01, we reject H0 and conclude that the errors are autocorrelated.
(b) The least squares results yield b2 = −0.3857 and se(b2) = 0.0360. The generalized least
squares results from SHAZAM are β! = −0.3752 and se(β! ) = 0.0527. Using EViews,
2
2
we obtain βˆ 2 = −0.5286 and se(βˆ 2 ) = 0.1129 . Using SAS, we obtain β! 2 = −0.3780 and
se(β! 2 ) = 0.0497. The different results arise from using different estimators for ρ, and
because EViews automatic command drops the first observation; the other two software
packages do not. In small samples dropping the first observation can make a substantial
difference. This is one of those occasions. The different estimates for ρ are ρ! = 0.5583
7
Chapter 12
Solutions to Exercises
(SHAZAM), ρˆ = 0.6128 (EViews) and ρ! = 0.4691 (SAS). Using tc = 2.145 (14 degrees
of freedom), the two confidence intervals for β 2 are:
LS = (−0.463, −0.308)
GLS = (−0.488, −0.262)
GLS = (−0.771, −0.286)
GLS = (−0.485, −0.271)
(SHAZAM)
(EViews)
(SAS)
The wider interval obtained using GLS estimates suggests that incorrect standard errors
have made the least squares interval too narrow. The least squares interval suggests our
information about β 2 is more reliable than it actually is.
(c) To answer this question we need to predict unit cost for a cumulative production value of
3800. Our prediction will be different depending on whether or not we assume the
existence of autocorrelated errors.
(i) Without autocorrelated errors our prediction is computed from the least squares
results
ln(u!1971 ) = b1 + b2ln(3800) = 6.0191 − 0.3857(8.24276) = 2.83989
u!1971 = exp(2.83989) = 17.1139
(ii) When the existence of autocorrelated errors is recognized, the relevant prediction
using the SHAZAM estimated value for ρ is
ln(u!1971 ) = β! 1 + β! 2 ln(3800) + ρ! e~1970
= 5.9281 − 0.37524(8.24276) + 0.55826(−0.068952) = 2.79659
u!1971 = exp(2.79659) = 16.3886
Using the EViews estimated value for ρ, the results are ln(u!1971 ) = 2.7643 and u!1971 =
15.8684. Using the SAS estimated value for ρ, the results are ln(u!1971 ) = 2.80335 and
u!1971 = 16.4999. Since unit cost is 16.4163 in 1970, the least squares prediction suggests
cost will be greater in 1971, whereas the generalized least squares prediction suggests
cost will be less. Intuitively, we would expect cost to decline with the increase in
cumulative production. The least squares results do not predict a decline because the
residual for 1970 is negative; the actual value for 1970 is below the value that would be
predicted for that year. The generalized least squares results recognize the negative
residual in 1970 and make allowance for the fact that another negative residual is likely
for 1971.
12.6
(a) The least squares estimated equation is
Q! t = 0.1973 − 1.044 Pt + 0.0033 It + 0.0035 Ft
(0.2702) (0.834)
(0.0012) (0.0004)
R2 = 0.719
(b) The calculated t-value for the hypothesis H0: β 2 = 0 is −1.252, indicating that b2 is not
significantly different from zero. From this result one might be tempted to conclude that
price is not a relevant variable for explaining the demand. However, a close look at the
data shows that there has been little variation in price. Thus, it is more likely that the
8
Chapter 12
Solutions to Exercises
variation in price is too small to get an accurate estimate of its effect; not that price is
unimportant.
(c) The Durbin-Watson statistic is 1.0212. With k = 4 and T = 30 we have d L = 1.214 and
dU = 1.650 . Since d = 1.0212 < d L , on the basis of this test we conclude that
autocorrelation exists. This decision is confirmed by a p-value of 0.0003. The Lagrange
multiplier test gives t = 2.0277 with a p-value of 0.0534. Thus, at a 5% significance level,
the Lagrange multiplier test does not reject a null hypothesis of no autocorrelation.
However, it is “close” to rejection. Gathering further evidence from the plot of least
squares residuals in Figure 12.2, we see that these residuals tend to exhibit runs of
positive and negative values. Thus, overall, there is evidence of autocorrelated errors.
e
1E-01
8E-02
4E-02
0E+00
0
5
10
15
20
25
30
-4E-02
35
t
-8E-02
Figure 12.2 Residuals for Ice Cream Equation
(d) Using SHAZAM for the autocorrelation correction we find that ρ! = 0.40063 and the
estimated equation is
Q! t = 0.3374 − 1.176 Pt + 0.0022 It + 0.0033 Ft
(0.2867) (0.835) (0.0015) (0.0006)
With EViews, ρ! = 0.4007, and the estimated equation is
Qˆ t = 0.1572 − 0.8924 Pt + 0.0032 I t + 0.0036 Ft
(0.2998) (0.8295) (0.0016) (0.0006)
With SAS, ρ! = 0.32977, and the estimated equation is
Q! t = 0.3075 − 1.171 Pt + 0.0025 It + 0.0034 Ft
(0.2895) (0.856) (0.0015) (0.0005)
The reason that the estimated equations from EViews and SHAZAM are substantially
different, despite the fact that they use virtually identical estimates of ρ , is that EViews
drops the first observation when estimating the β s; SHAZAM does not.
9
Chapter 12
Solutions to Exercises
(e) In this case we find that, in addition to β! 2 , β! 3 is also not significantly different from
zero. Thus, some doubt is cast on the relevance of income in the demand for ice cream.
Again, it may be that the variation in income is inadequate to capture its effect.
12.7
(a) The Durbin-Watson p-value computed from the residuals for the transformed model
(from SHAZAM) is 0.0295. Thus, if we were testing H0: θ = 0 against H0: θ > 0 at a 5%
level of significance, where θ is the autocorrelation parameter in the model
vt = θvt −1 + ut , we would reject H0. The correction for autocorrelation in Exercise 12.6
does not seem to have cured the problem. A similar result is obtained with SAS. For
EViews the Durbin-Watson statistic of 1.55 falls in the inconclusive region, suggesting
there is still a potential problem.
(b) If we combine the model in part (a) with the traditional AR(1) error specification, we
have
et = ρet −1 + vt
vt = θvt −1 + ut
where the ut are independent random errors with ut ~ ( 0, σ 2u ) . Note that
et − 1 = ρet − 2 + vt − 1
and hence
θvt − 1 = θet − 1 − θρet − 2
from which we obtain
et = ρet − 1 + θet − 1 − θρet − 2 + ut = (ρ + θ)et − 1 − θρet − 2 + ut
We call this model an AR(2) error process. Under this assumed process our SHAZAMestimated model is
Q! t = 0.6381 − 0.9437 Pt − 0.0016 It + 0.0028 Ft
(0.2925) (0.7516)
(0.0021) (0.0007)
The EViews estimated model is
Qˆ t = 0.1106 − 0.8709 Pt + 0.0036 I t + 0.0037 Ft
(0.304) (0.852) (0.0016) (0.0007)
The SAS-estimated model is
Q! t = 0.2887 − 1.2043 Pt + 0.0028 It + 0.0034 Ft
(0.2905) (0.8637)
(0.0014) (0.0006)
The estimated model appears very sensitive to the way in which the AR(2) parameters
are estimated. We again find that β! 2 and β! 3 are not significantly different from zero.
Also, the SHAZAM results give β! < 0; a negative effect of income on the demand for
3
ice cream does not seem plausible.
10
Chapter 12
12.8
Solutions to Exercises
(a) The estimate for the AR(1) parameter ρ is
T
ρˆ =
∑ eˆ eˆ
t =2
T
t t −1
∑ eˆ
t =1
=
2
t
55453
= 0.8758
63316
The approximate Durbin-Watson statistic is d * = 2(1 − ρˆ ) = 0.2484 .
Based on T = 90 and K = 5 , d L = 1.566 and dU = 1.751 . Since d * is less than d L we
conclude that positive autocorrelation is present.
(b) The estimate for the AR(1) parameter of ρ is
T
ρˆ =
∑ eˆ eˆ
t =2
T
t t −1
∑ eˆ
t =1
2
t
=
621
= 0.0505
12292
The approximate Durbin-Watson statistic is d * = 2(1 − ρˆ ) = 1.8990 .
Based on T = 90 and K = 6 , d L = 1.542 and dU = 1.776 . Since d * > dU , we cannot
reject a null hypothesis of no positive autocorrelation.
12.9
(a) From the residual plots the residuals tend to exhibit runs of positive and negative values,
suggesting autocorrelated errors. The Durbin-Watson statistic is 1.124. With T = 26 and
K = 2 we obtain d L = 1.302 and dU = 1.461 . Since the value of the Durbin-Watson
statistic is less than d L , we conclude that there is evidence of positive autocorrelation.
(b) The estimates, their standard errors and the confidence intervals obtained from least
squares and generalised least squares (GLS) are presented in the table below. For the
least squares method T = 26, K = 2 and the critical t value is t0.025 = 2.064 . For GLS, T =
25 and K = 2, so t0.025 = 2.069 . The estimates obtained from least squares and GLS are
very similar. However, the standard errors from GLS are much higher than from least
squares, resulting in GLS confidence intervals which are much wider than those obtained
from least squares. Hence, ignoring autocorrelation means the estimates are less reliable
than they appear.
Least squares
GLS
Estimate
(se)
Confidence interval
Estimate
(se)
Confidence interval
β1
-387.97
(112.66)
(-620.49, -155.45)
-343.85
(192.17)
(-741.45, 53.75)
β2
24.7646
(0.9715)
(22.759, 26.770)
24.3882
(1.5741)
(21.131, 27.645)
11
Chapter 12
Solutions to Exercises
(c) Because of the evidence of autocorrelation the forecasts are based on the GLS results.
DISP86 = βˆ 1 + βˆ 2 DUR86 + ρˆ e"85
= −343.85 + 24.3882(190) + 0.4186( −277.2)
= 4173.87
DISP87 = βˆ 1 + βˆ 2 DUR87 + ρˆ 2 e"85
= −343.85 + 24.3882(195) + 0.41862 ( −277.2)
= 4363.28
DISP88 = βˆ 1 + βˆ 2 DUR88 + ρˆ 3 e"85
= −343.85 + 24.3882(192) + 0.41863 ( −277.2)
= 4318.35
12.10 (a) The LS estimated equation is
ln( POWt ) = −0.1708 + 0.0082 t − 0.000037 t 2 + 0.9365ln( PROt )
(0.4147) (0.0005) (0.000004) (0.0899)
The positive sign for b2 and the negative sign for b3 , and their relative magnitudes,
suggest that the trend for ln(POW) is increasing at a decreasing rate. A positive b4
implies the elasticity of power use with respect to productivity is positive. The value of
the Durbin-Watson statistic is 0.3924 with a very small corresponding p-value of 0.0000.
This suggests a problem of autocorrelation in the errors.
(b) The table below compares the least squares results with those generalised least squares
ones obtained using SHAZAM, EViews and SAS. Apart from β1 , the least squares and
GLS estimates are similar, although the EViews estimator that drops the first observation
has led to some discrepancies. The most noticeable change is the increase in standard
errors when GLS is used. Ignoring autocorrelation conveys a false sense of reliability.
GLS
LS
EViews
SAS
SHAZAM
β1
-0.1708
(0.4147)
-0.2551
(0.4904)
-0.3013
(0.4819)
-0.3035
(0.4805)
β2
0.0082
(0.0005)
0.0070
(0.0015)
0.0083
(0.0011)
0.0083
(0.0011)
β3
-0.000037
(0.000004)
0.000027
(0.000012)
-0.000036
(0.000010)
-0.000036
(0.000010)
β4
0.9365
(0.0899)
0.9612
(0.1055)
0.9629
(0.1044)
0.9633
(0.1041)
12
Chapter 12
Solutions to Exercises
(c) The p-values (from SHAZAM) for testing the hypothesis H 0 : β 4 = 1 are 0.4817 and
0.7249 before and after correcting for autocorrelation, respectively. We do not reject H 0
in both cases. Correcting for autocorrelation has led to a large change in the p-value, but
the test decision is still the same.
12.11 The model with the restriction β 4 = 1 imposed is
ln( POWt ) = β1 + β2 t + β3 t 2 + ln( PROt ) + et
Or it can be written as
 POWt
ln 
 PROt

2
 = β1 + β2 t + β3 t + et

The estimated equation with the restriction β 4 = 1 imposed is
∧
 POW
t

ln
 PROt


 = −0.4728 + 0.0082 t − 0.000036 t 2

 (0.0271) (0.0011) (0.000010)
(SHAZAM)
 ∧
POWt
ln 
 PROt


 = −0.4339 + 0.0069 t − 0.000026 t 2

 (0.0403) (0.0015) (0.000012)
(EViews)
 ∧
POWt
ln 
 PROt


 = −0.4724 + 0.0082 t − 0.000036 t 2


(0.0263) (0.0011) (0.000009)
(SAS)
We can use the exponential of the within sample prediction from this equation to plot the
trend for ( POWt PROt ) as in the graph below. It appears that the trend is increasing at a
decreasing rate over time.
1.1
POW_PRO
1.0
0.9
0.8
0.7
0.6
20
40
60
T
80
100
13
Chapter 12
Solutions to Exercises
12.12 (a) The model where β1 = α1 + δ1Dt and β 4 = α4 + δ4 Dt can be written as
ln( POWt ) = α1 + δ1Dt + β2 t + β3 t 2 + α4 ln( PROt ) + δ4 Dt ln( PROt ) + et
The dummy variable Dt is introduced to capture the structural change in power use (if
any) after the year 1985.
The LS estimation result is
^
ln( POWt ) = 0.3287 + 0.1995Dt + 0.0091t − 0.000041t 2 + 0.8257ln( PROt )
(0.6524) (1.0510) (0.0007) (0.000005) (0.1423)
−0.051Dt ln( PROt )
(0.224)
Both of the dummy variable parameters estimates δˆ 1 and δˆ 4 are not significantly
different from zero. To test for autocorrelation we obtain the Durbin-Watson statistic as
0.4283 with a very small p-value of 0.0000, suggesting the presence of autocorrelation.
After allowing for autocorrelation the GLS estimated equation is
^
ln( POWt ) = −0.5576 + 1.0625Dt + 0.0083 t − 0.000037 t 2 + 1.0182ln( PROt )
(0.5761) (1.1130) (0.0012) (0.000009) (0.1253)
−0.2293Dt ln( PROt )
(SHAZAM)
(0.2364)
^
ln( POWt ) = −0.5727 + 1.2259 Dt + 0.0069 t − 0.000027 t 2 + 1.0302ln( PROt )
(0.5901) (1.1333) (0.0017) (0.000012) (0.1279)
−0.2626 Dt ln( PROt ) (EViews)
(0.2408)
^
ln( POWt ) = −0.5532 + 1.0549 Dt + 0.0083 t − 0.000037 t 2 + 1.01732ln( PROt )
(0.5793) (1.1164) (0.0012) (0.000009) (0.1260)
−0.2278Dt ln( PROt ) (SAS)
(0.2373)
The estimates for δ1 and δ 4 change substantially after allowing for autocorrelation. The
others have changed slightly.
(b) For testing H 0 : δ1 = δ4 = 0 the test outcomes appear in the table below.
χ2 Test
F Test
χ2 Value
p-value
F value
p-value
LS
6.0901
0.0476
3.0450
0.0518
GLS (EViews)
1.4013
0.4963
0.7007
0.4986
GLS (SAS)
1.3524
0.5086
0.7760
0.4629
GLS (Shazam)
1.5539
0.4598
0.7769
0.4624
14
Chapter 12
Solutions to Exercises
Allowing for autocorrelation changes the test values considerably. Before correcting for
autocorrelation, the null hypothesis is bordering on rejection at a 5% significance level.
After correcting for autocorrelation there is no support for the null hypothesis.
(c) If H 0 : δ1 = δ4 = 0 is rejected, we can conclude there is a structural change in the pattern
of power use from the year 1985.