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Homework #5 Solutions
6.4 The following model allows the return to education to depend upon the total amount of both parents’ education, called pareduc :
log( wage )= β 0 + β 1 educ + β 2 educ * pareduc + β 3 exp er + β 4 tenure + u
(i) Show that, in decimal form, the return to another year of education in this
model is: Δ log( wage )/Δ educ = β 1 + β 2 pareduc What sign do you expect for
Holding other factors constant in this equation, we get the following expression when we solve for the effect of education on log(wages).
wage educ pareduc
wage pareduc educ
wage educ pareduc educ
1 2
1 2
1 2
log( )/
log( ) ( )*
log( )
The sign of β 2 is not obvious. One argument is that it is positive. For instance, a child
might get more out of another year of education if their parent has more education because then their parents are able to help with homework.
(ii) Using the data in WAGE2.RAW, the estimated equation is:
n 722 ,. 169
log( ) 5. 65. 047. 00078 *. 019 exp. 010
2
^
R
wage educ educ pareduc er tenure
(Only 722 observations contain full information on parents’ education.) Interpret the coefficient on the interaction term. It might help to choose two specific values for pareduc-for example pareduc=32 if both parents have a college education, or pareduc=24 if both parents have a high school education-and to compare the estimated returns to educ)
The effect of education on wages depends on the value of parent education, and so we use the values above for pareduc=32 and 24, and estimate the difference in the estimated return to education. Further, we assume that we are looking at the return to 1 more year of education.
Wage is predicted to be .047+.00078*(32-24)=.0062 higher if a child’s parents both have a college education relative to if they had high school educations. Since we have a log-level equation, the .0062 point increase can be interpreted as:
.0062*100%=.62% increase as a result of having parents with college degrees versus high school degrees.
(iii) When pareduc is added as a separate variable to the equation, we get:
n 722 ,. 174
log( ) 4. 94. 097. 033. 0016 *. 020 exp. 010
2
^
R
wage educ pareduc educ pareduc er tenure
Does the estimated return to education now depend positively on parent education? Test the null hypothesis that the return to education does not depend on parent education.
When we add pareduc by itself, the coefficient on the interaction of educ*pareduc is now negative, indicating return to another year of education is lower if your parents have higher levels of education.
To determine whether the return to education depends or doesn’t depend on parent education, we set out the following null hypothesis and alternative (assuming two sided test), and calculated the t-statistic.
1
0
t
H
H
educpareduc
educpareduc
The critical value c, for alpha=.05 and 722-5-1=716 degrees of freedom is _____ As a result, since |t|<c, we fail to reject the null hypothesis. That is, the return to education does not depend on parent education.
Note that the coefficient on pareduc=.033 has a t-statistic of .033/.017 which is significantly different under a two-sided hypothesis at alpha=.05. This suggests if we leave out the level term of an interaction from a regression, we may be omitting an important factor.
T-statistic: Reject null if |t|>c….|-6.16|>2.576 (The critical value=2.576 at alpha=.01, and n-k-1=526-3-1=522 degrees of freedom). P-value: Reject null if p-value<alpha……0.000<0.
This suggests that exper 2 is statistically different from zero at the 1% level.
(iii)Using the approximation % wage 100 ( 2 exp er ) exp er
^ 3
^
Δ ≈ β 2 + β Δ find the
approximate return to the fifth year of experience. What is the approximate return to the twentieth year of experience?
Following the formula, and the regression output in (i) we can calculate the return to the fifth and twentieth year of experience as the predicted percentage change in wage that occurs from moving from 4 years of experience to 5 years, and 19 years to 20 years.
Fifth year: % Δ wage ≈ 100 (. 0410 + 2 −. 007 * 4 )( 5 − 4 )=3.54%
Twentieth year: % Δ wage ≈ 100 (. 0410 + 2 −. 007 * 19 )( 20 − 19 )=1.44%
(iv) At what value of exper does additional experience actually lower predicted
log(wage)? How many people have more experience in this sample?
We know that at some point more experience (exper) lowers predicted wage since the coefficient on exper is positive and the coefficient on exper2 is negative. To calculate the point at which exper lowers predicted wage, we need to calculate the turn around point, given by the following formula:
exp | /( 2 * )| |.0410/(2*-.0007)| | 29. 28 | 29. 28
^ 3
^ 2
er *^ =β β = =− =
The turn around point occurs at 29.28 years of experience. Since the data reports experience in integers, we check how many people have experience greater than or equal to 29 years. See attached STATA output. There are 121 observations with 29 or more years of experience.
Wednesday October 21 17:21:17 2009 Page 1 ___ ____ ____ ____ ____tm /__ / ____/ / ____/ / / // / /___/ Statistics/Data Analysis
- reg colgpa hsize hsizesq hsperc sat female athlete satfemale log: C:\Documents and Settings\user\My Documents\Teaching\Fall 2009\Econ 8740\Homework Solut log type: smcl opened on: 21 Oct 2009, 16:56:
- do "C:\DOCUME~1\user\LOCALS~1\Temp\STD0d000000.tmp"
- /C 6.2/
- use "C:\Documents and Settings\user\My Documents\Teaching\Fall 2009\Econ 8740\Homework Assignments\Data\
- /(i)/
- reg lwage educ exper expersq Source SS df MS Number of obs = 526 F( 3, 522) = 74. Model 44.5393713 3 14.8464571 Prob > F = 0. Residual 103.79038 522 .198832146 R-squared = 0. Adj R-squared = 0. Total 148.329751 525 .28253286 Root MSE =.
lwage Coef. Std. Err. t P>|t| [95% Conf. Interval] educ .0903658 .007468 12.10 0.000 .0756948. exper .0410089 .0051965 7.89 0.000 .0308002. expersq -.0007136 .0001158 -6.16 0.000 -.000941 -. _cons .1279975 .1059323 1.21 0.227 -.0801085.
- /(iv)/
- sum exper if exper>=
Variable Obs Mean Std. Dev. Min Max exper 121 37.8595 5.997646 29 51
end of do-file
clear
do "C:\DOCUME~1\user\LOCALS~1\Temp\STD0d000000.tmp"
/*C 7.4 */
use "C:\Documents and Settings\user\My Documents\Teaching\Fall 2009\Econ 8740\Homework Assignments\Data\
/(ii)/
reg colgpa hsize hsizesq hsperc sat female athlete
Source SS df MS Number of obs = 4137 F( 6, 4130) = 284. Model 524.819305 6 87.4698842 Prob > F = 0. Residual 1269.37637 4130 .307355053 R-squared = 0. Adj R-squared = 0. Total 1794.19567 4136 .433799728 Root MSE =.
