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Exam October 24, 2013, Answers

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Econometrics (6012B0212Y)

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Solutions Final Examination Econometrics October 24th, 2013.

Question 1. 1). Model lnYtLK   01 2  ln 3 ln 4 D 80 ut

1a). 1 lnYYY/ tt







is the relative change of Y given a one-unit increase of t, so 100β 1 gives

the percentage change of output that results each year, when the other explanatory variables L, K and D80 remain constant. So β 1 gives an autonomous yearly increase of productivity. This can be attributed to technical progress, because over time new production possibilities become available. CI: 0±1×0 = (0, 0) 1b). There are constant returns to scale if  23   1. Given the results, this seems possible, because summing up the two CI’s gives (0, 1), which includes the value 1. ADDITION November 26th 2013, 13:00h: this is an incorrect way to calculate the CI since the covariance of the estimated parameters I ignored in this calculation. This is suggested by ‘this seems possible’but perhaps not clear to everybody. Correct procedures below. Formal test: H02 3:1  versus H12 3:1  . T test: write the model as lnYL    ln  01 t( 2 31) lnL 3 (lnKL Duln )  480 t,

define  231 , apply LS to this model and use

ˆ

~ ( 37) or (0,1) ()ˆ

ttdfN SE

.

F test: apply LS to the restricted model lnYLln    01 3t(lnKL Duln )  480 t, and use

min

()/

~( 1, 37)

/

ru numerator deno ator u

SSR SSR

F F df df SSR

, where SSRu = 0.

1c). (i). According to the p-value = 0 > 0 it is not significant (two-sided test). If you use the t-value -1 and the critical values ±1, then it is significant. There is a difference in results, because the p-value is based on t(df=37), while the critical values are based on N(0,1). (ii). lnYtLKˆ101 2  ˆˆ ˆ  ln ˆ 3 ln ˆ 4 when D80=

001 2 3 lnYtLKˆ   ˆˆ ˆln ˆln when D80= So the effect of the strike is: lnYYˆˆ 104 ln ˆ 0. This leads to ln( /YYˆˆ 10 )0, YY eˆˆ 10 /  0, and

101 0. 00

ˆˆ ˆ 100 ˆˆ 100 1 100 1 10%

YY Y e YY

   

.

The strike has had a considerable impact, so it is economically significant. There has been a notable decrease of productivity.

The additional regression in (b) would be

. regress lnYperL T lnKperL D Source | SS df MS Number of obs = 42 -------------+------------------------------ F( 3, 38) = 341. Model | 3 3 1 Prob > F = 0. Residual | .119472874 38 .003144023 R-squared = 0. -------------+------------------------------ Adj R-squared = 0.

Total | 3 41 .081515608 Root MSE =.

lnYperL | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+----------------------------------------------------------------

T | .0167981 .0018495 9 0 .013054. lnKperL | .1391929 .0397808 3 0 .0586609. D80 | - .0576849 -1 0 -. _cons | - .0209505 -0 0 -.

And give an F value of 3. . test ( lnLconstr+ lnKconstr=1) ( 1) lnLconstr + lnKconstr = 1 F( 1, 37) = 3. Prob > F = 0.

Question 2. a) The problem here is omitted variable bias: ability is not part of the equation. If ability in indeed relevant and if there is a correlation between schooling and ability, the OLS-estimate of the effect of schooling will be biased. Since we expect a positive effect of ability on wage and a positive correlation the bias will be positive. The estimated effect is larger than the actual effect of schooling:

ˆ cov( , ) var( ) SS

Su S

 , cov(S,u) is positive since ability is part of u and has a positive effect on ln(W)

and is positively correlated with S: u = θAbility + u* ,where θ > 0. NB: Some students argue there is a negative bias between schooling and ability. If the reasoning is correct, this is also a correct solution. b) To solve the problem in a). we can use IV- or 2SLS (page 360, Stock and Watson). We also could simply add GRADE to the Mincerian wage equation but since GRADE is probably not perfectly describing ability, so that we have measurement error (and a potentially resulting bias). The question also is whether the remaining variation is correlated with the other explanatory variables in the model. If yes, we need to apply IV or 2SLS. Two instruments are suggested: GRADE and AGE. Both are relevant (F-value combined > 10 and (t-values) 2 separate > 10), but GRADE has a direct effect on wages (t-value 3). AGE and GRADE are likely also exogenous (no test provided, but from theory we can expect this, esp. AGE). AGE seems to be the only appropriate instrument. To get the best estimation results (efficient) apply 2SLS.

Step S1:    01   AGE 2 LEX 3 LEX 2   4 FE 5 GRADE

and estimate this equation with OLS. Calculate the fitted value: Sˆfor every observation and then

estimate the original model (including GRADE) with OLS were S is replaced by Sˆor apply IV with Sˆas the instrument for S. 2 01 2 3 4 5 2 01 2 3 4 5

2: ln( ) ˆ ln( ) using as an instument of Sˆ

Step OLS on W S LEX LEX FE GRADE u or IV on W S LEX LEX FE GRADE u S

    

    

     

     

c). (i) to get a better instrument Sˆ(higher correlation with S, resulting in smaller estimated standard errors of the estimated coefficients of the equation of interest). (ii) Not relevant, correct specification if the S equation (STEP 1) is not a requirement we prefer a

high as possible correlation between S and Sˆ, without a correlation between Sˆand W. So the estimates remain consistent (but perhaps with larger standard errors of the estimated coefficients of the equation of interest).

Question 3.

where 122

1

() exp( ) 2

zz 

is the pdf of the standard normal distribution. Since ()z goes to

zero as ||z becomes large, we see that the marginal effect of PRATIO also goes to zero. Note that the non-linearity does not imply that the other explanatory variables influence the marginal effect (after they are chosen), i. the marginal effect is a partial derivative holding the other variables fixed! c) ML is more efficient (i. the ML estimators are expected to have smaller standard errors). Note that NLS is also consistent and asymptotically normal distribution (see p. 437 of Stock and Watson). Furthermore,  (bbX 011 ... bXkk) always lies between 0 and 1, so NLS also respects the boundaries of probabilities!

d1)

ˆ 1

[ 1] 0.

1 exp( (0 1 0 1))

PCoke    

d2) We see that PCokeˆ[  1] 0 0, so we predict COKE 1 which corresponds to the

prediction that the shopper chooses a Coke.

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Exam October 24, 2013, Answers

Vak: Econometrics (6012B0212Y)

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Studenten deelden 315 documenten in dit vak
Was dit document nuttig?
Solutions Final Examination Econometrics October 24th, 2013.
Question 1.
1). Model 01 2 3 4
ln ln ln 80 t
YtLKDu


1a). 1
ln /YYY
tt



is the relative change of Y given a one-unit increase of t, so 100β1 gives
the percentage change of output that results each year, when the other explanatory variables L, K
and D80 remain constant. So β1 gives an autonomous yearly increase of productivity. This can be
attributed to technical progress, because over time new production possibilities become available.
CI: 0.0170208±1.96×0.0017926 = (0.0135, 0.0205)
1b). There are constant returns to scale if 23
1
. Given the results, this seems possible,
because summing up the two CI’s gives (0.9084, 1.4967), which includes the value 1. ADDITION
November 26th 2013, 13:00h: this is an incorrect way to calculate the CI since the covariance of the
estimated parameters I ignored in this calculation. This is suggested by ‘this seems possible’but
perhaps not clear to everybody. Correct procedures below.
Formal test: 02 3
:1H

versus 12 3
:1H
.
T test: write the model as 01 23 3 4
ln ln ( 1) ln (ln ln ) 80 t
YL t L KL Du


,
define 23
1


, apply LS to this model and use ˆ~ ( 37) or (0,1)
ˆ
()
ttdfN
SE
.
F test: apply LS to the restricted model 01 3 4
ln ln (ln ln ) 80 t
YL t KL Du


, and use
min
()/1
~( 1, 37)
/37
ru
numerator deno ator
u
SSR SSR
F F df df
SSR
, where SSRu = 0.10882408.
1c). (i). According to the p-value = 0.054 > 0.05 it is not significant (two-sided test). If you use the
t-value -1.99 and the critical values ±1.96, then it is significant. There is a difference in results,
because the p-value is based on t(df=37), while the critical values are based on N(0,1).
(ii). 101 2 3 4
ˆˆ ˆ ˆ ˆ
ˆ
ln ln lnYtLK


when D80=1
001 2 3
ˆˆ ˆ ˆ
ˆ
ln ln lnYtLK

 when D80=0
So the effect of the strike is: 104
ˆ
ˆˆ
ln ln 0.1108467YY
 .
This leads to 10
ˆˆ
ln( / ) 0.1108467YY , 0.1108467
10
ˆˆ
/YY e
, and

0.1108467
10 1
00
ˆˆ ˆ
100 100 1 100 1 10.4924%
ˆˆ
YY Ye
YY




.
The strike has had a considerable impact, so it is economically significant. There has been a notable
decrease of productivity.
The additional regression in (b) would be
. regress lnYperL T lnKperL D80
Source | SS df MS Number of obs = 42
-------------+------------------------------ F( 3, 38) = 341.67
Model | 3.22266706 3 1.07422235 Prob > F = 0.0000
Residual | .119472874 38 .003144023 R-squared = 0.9643
-------------+------------------------------ Adj R-squared = 0.9614
Total | 3.34213993 41 .081515608 Root MSE = .05607
------------------------------------------------------------------------------
lnYperL | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------