Problem Set 2 (SOLUTIONS)
This problem set will take you through some Stata commands to estimate simple regression equations with dummy variables. You will learn how to interpret the estimated coefficients and test some linear hypotheses. Interpretation of these coefficients will be useful when we do treatment evaluation models later in term 1.
The hypothesis tests discussed in this problem set include standard T-tests and F-tests, which is assumed undergraduate knowledge for this module.
You will need to download the dataset problemset2.dta, which is available on Moodle.
Conditional Expectation Function
Consider the Conditional Expectation Function (CEF), \(E[Y_i|X_i]\). If \(X\) takes on discrete values: \(X_i\in\{x_1,x_2,...,x_m\}\), then
\[ E[Y_i|X_i] = E[Y_i|X_i=x_1]\cdot\mathbf{1}\{X_i = x_1\}+...+E[Y_i|X_i=x_m]\cdot\mathbf{1}\{X_i = x_m\} \] where \(\mathbf{1}\{X_i = x_m\}\) is a dummy variable, \(=1\) when \(X_i=x_m\). Since the values of \(X_i\) are mutually exclusive there is no overlap of these dummy variables.
Note, we do not need to assume that \(X\) is a single random variable. It can be a vector of random variables that takes on discrete values.
We can re-arrange this expression using anyone of the values of \(X\). The natural option is to choose the first, but this is arbitrary.
\[ \begin{aligned} E[Y_i|X_i] =& E[Y_i|X_i=x_1]+ (E[Y_i|X_i=x_2]-E[Y_i|X_i=x_1])\cdot\mathbf{1}\{X_i = x_2\}+... \\ &+(E[Y_i|X_i=x_m]-E[Y_i|X_i=x_1])\cdot\mathbf{1}\{X_i = x_m\} \end{aligned} \]
Since, \(E[Y_i|X_i = x_m]\) is a constant (\(X_i\) is set to a specific value), we can express the CEF as a function that is linear in parameters.
\[ E[Y_i|X_i] = \beta_1 + \beta_2D_{i2} + ... + \beta_m D_{im} \]
where \(D_{im}=\mathbf{1}\{X_i = x_m\}\).
Preamble
Create a do-file for this problem set and include a preamble that sets the directory and opens the data. For example,
clear
//or, to remove all stored values (including macros, matrices, scalars, etc.)
*clear all
* Replace $rootdir with the relevant path to on your local harddrive.
cd "$rootdir/problem-sets/ps-2"
cap log close
log using problem-set-2-log.txt, replace
use problem-set-2-data.dtaQuestions
1. Consider the \(E[ln(Wage_i)|Gender_i]\), where \(Gender_i\in\{1 ``Male'', 2 ``Female''\}\). Show that this CEF implies a linear model,
\[ ln(Wage_i) = \beta_1 + \beta_2 D_{i2} + \varepsilon_i \]
What do the parameters \(\beta_1\) and \(\beta_2\) imply?
\[ E[ln(Wage_i)|G_i] = E[ln(Wage_i)|G_i=1]+ (E[ln(Wage_i)|G_i=2]-E[ln(Wage_i)|G_i=1])\cdot\mathbf{1}\{G_i = 2\} \] \[ = \beta_1+ \beta_2D_{i2} \] \(\Rightarrow E[\varepsilon_i|D_{i2}]=0\).
2. Regress lwage (log wage) on just a set of binary indicators that will enable you to test the hypothesis that males and females are on average, paid the same wage, ceteris paribus. Test this hypothesis.
reg lwage female
* or
gen male=1-female
reg lwage male
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(1, 4163) = 491.72
Model | 93.6914807 1 93.6914807 Prob > F = 0.0000
Residual | 793.213421 4,163 .190538895 R-squared = 0.1056
-------------+---------------------------------- Adj R-squared = 0.1054
Total | 886.904902 4,164 .212993492 Root MSE = .43651
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4744661 .0213967 -22.17 0.000 -.516415 -.4325171
_cons | 6.729774 .00718 937.29 0.000 6.715697 6.74385
------------------------------------------------------------------------------
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(1, 4163) = 491.72
Model | 93.6914807 1 93.6914807 Prob > F = 0.0000
Residual | 793.213421 4,163 .190538895 R-squared = 0.1056
-------------+---------------------------------- Adj R-squared = 0.1054
Total | 886.904902 4,164 .212993492 Root MSE = .43651
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
male | .4744661 .0213967 22.17 0.000 .4325171 .516415
_cons | 6.255308 .020156 310.34 0.000 6.215791 6.294824
------------------------------------------------------------------------------Alternatively, you could use Stata’s factor notation:
reg lwage i.female
//note: defaults to smallest value as base category. This can be changed as follows.
reg lwage ib1.female
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(1, 4163) = 491.72
Model | 93.6914807 1 93.6914807 Prob > F = 0.0000
Residual | 793.213421 4,163 .190538895 R-squared = 0.1056
-------------+---------------------------------- Adj R-squared = 0.1054
Total | 886.904902 4,164 .212993492 Root MSE = .43651
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
1.female | -.4744661 .0213967 -22.17 0.000 -.516415 -.4325171
_cons | 6.729774 .00718 937.29 0.000 6.715697 6.74385
------------------------------------------------------------------------------
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(1, 4163) = 491.72
Model | 93.6914807 1 93.6914807 Prob > F = 0.0000
Residual | 793.213421 4,163 .190538895 R-squared = 0.1056
-------------+---------------------------------- Adj R-squared = 0.1054
Total | 886.904902 4,164 .212993492 Root MSE = .43651
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
0.female | .4744661 .0213967 22.17 0.000 .4325171 .516415
_cons | 6.255308 .020156 310.34 0.000 6.215791 6.294824
------------------------------------------------------------------------------It is evident from the test p-value that the difference is statistically significantly. Note, the standard reg command assume homoskedastic SEs. If we believe that the variance varies of (log of) wages varies with gender, we should estimate heteroskedastic SEs. However, in this instance it will not make a difference to the conclusion.
reg lwage female, r
Linear regression Number of obs = 4,165
F(1, 4163) = 520.64
Prob > F = 0.0000
R-squared = 0.1056
Root MSE = .43651
------------------------------------------------------------------------------
| Robust
lwage | Coefficient std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4744661 .0207939 -22.82 0.000 -.5152332 -.4336989
_cons | 6.729774 .0072089 933.53 0.000 6.71564 6.743907
------------------------------------------------------------------------------3. Extend the specification in (2) that will enable you to test the hypothesis that there is no difference in the wages between the following gender-ethnicity groups. Begin by defining the following dummy variables:
female_black=female\(\times\)blackmale_black= (1-female)\(\times\)blackfemale_nonblack=female\(\times\)(1-black)male_nonblack= (1-female)\(\times\)(1-black)
gen female_black=female*black
gen female_nonblack=female*(1-black)
gen male_black=(1-female)*black
gen male_nonblack=(1-female)*(1-black)Then estimate the following regressions:
lwageonfemale_black,female_nonblack,male_black,male_nonblack(without a constant: optionnocons)lwageonfemale,black,female_blacklwageonfemale_black,female_nonblack,male_black
For some of these exercises you may be able to use Stata’s factor notation. However, in some instances you will need to manually create the above dummy-variable interactions.
In each case, identify the base category and write down the parameters of the (implied) model in terms of conditional expectations.
* (a)
reg lwage female_black female_nonblack male_black male_nonblack
// note, Stata has dropped one variable due to perfect collinearity
reg lwage female_black female_nonblack male_black male_nonblack, noconsnote: male_black omitted because of collinearity.
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | -.4434409 .0523433 -8.47 0.000 -.5460617 -.34082
female_non~k | -.2101553 .0383456 -5.48 0.000 -.2853332 -.1349774
male_black | 0 (omitted)
male_nonbl~k | .2239593 .0317691 7.05 0.000 .1616749 .2862436
_cons | 6.517691 .0309152 210.82 0.000 6.457081 6.578301
------------------------------------------------------------------------------
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(4, 4161) > 99999.00
Model | 185756.485 4 46439.1213 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.9958
-------------+---------------------------------- Adj R-squared = 0.9958
Total | 186535.954 4,165 44.7865436 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | 6.07425 .0422382 143.81 0.000 5.991441 6.15706
female_non~k | 6.307536 .0226856 278.04 0.000 6.26306 6.352012
male_black | 6.517691 .0309152 210.82 0.000 6.457081 6.578301
male_nonbl~k | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------This model returns four parameter-estimates, each corresponding to the four gender-ethnicity groups. These are essentially conditional mean estimates.
* (b)
reg lwage female black female_black
// or, using factor notation:
reg lwage i.female##i.black
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
female_black | -.0093263 .057515 -0.16 0.871 -.1220865 .1034339
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
1.female | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
1.black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
|
female#black |
1 1 | -.0093263 .057515 -0.16 0.871 -.1220865 .1034339
|
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------In this model, we have the following:
\(\beta_1 = E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_2 = E[ln(Wage_i)|F = 1,B = 0]-E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_3 = E[ln(Wage_i)|F = 0,B = 1]-E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_4 = (E[ln(Wage_i)|F = 1,B = 1]-E[ln(Wage_i)|F = 0,B = 1])-(E[ln(Wage_i)|F = 1,B = 0]-E[ln(Wage_i)|F = 0,B = 0])\)
* (c)
reg lwage female_black female_nonblack male_black
// note, this one is harder to replicate using factor notation.
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | -.6674001 .0428671 -15.57 0.000 -.7514426 -.5833576
female_non~k | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
male_black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------In this model, we have the following:
\(\beta_1 = E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_2 = E[ln(Wage_i)|F = 1,B = 1]-E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_3 = E[ln(Wage_i)|F = 1,B = 0]-E[ln(Wage_i)|F = 0,B = 0]\)
\(\beta_4 = E[ln(Wage_i)|F = 0,B = 1]-E[ln(Wage_i)|F = 0,B = 0]\)
4. Compare the estimated coefficients with the sample average values for the lwage for the four subgroups. What do you see?
table female black, stat(mean lwage)
------------------------------------------------
| black
| 0 1 Total
---------------+--------------------------------
female or male |
0 | 6.74165 6.517691 6.729774
1 | 6.307536 6.07425 6.255308
Total | 6.700755 6.363002 6.676346
------------------------------------------------You can compute the coefficients from each model simply using these averages.
5. In each of the above models, describe the null hypothesis you would test to evaluate whether there is a significant earnings difference between the earnings of black and non-black females.
\(H_0: \beta_1 = \beta_2\)
\(H_0: \beta_3 + \beta_4 = 0\)
\(H_0: \beta_2 - \beta_3 = 0\)
6. Verify your solution to 4 by performing a test using the three set of regression output. You can use the post-estimation test command.
reg lwage female_black female_nonblack male_black male_nonblack, nocons
test female_black = female_nonblack
reg lwage female black female_black
test female_black + black = 0
reg lwage female_black female_nonblack male_black
test female_black - female_nonblack = 0
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(4, 4161) > 99999.00
Model | 185756.485 4 46439.1213 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.9958
-------------+---------------------------------- Adj R-squared = 0.9958
Total | 186535.954 4,165 44.7865436 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | 6.07425 .0422382 143.81 0.000 5.991441 6.15706
female_non~k | 6.307536 .0226856 278.04 0.000 6.26306 6.352012
male_black | 6.517691 .0309152 210.82 0.000 6.457081 6.578301
male_nonbl~k | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) female_black - female_nonblack = 0
F( 1, 4161) = 23.68
Prob > F = 0.0000
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
female_black | -.0093263 .057515 -0.16 0.871 -.1220865 .1034339
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) black + female_black = 0
F( 1, 4161) = 23.68
Prob > F = 0.0000
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | -.6674001 .0428671 -15.57 0.000 -.7514426 -.5833576
female_non~k | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
male_black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) female_black - female_nonblack = 0
F( 1, 4161) = 23.68
Prob > F = 0.00007. In each case, test equality across all four gender-ethnicity groups. Again, you should get the same result.
reg lwage female_black female_nonblack male_black male_nonblack, nocons
test female_black = female_nonblack = male_black = male_nonblack
reg lwage female black female_black
test female_black = black = female = 0
reg lwage female_black female_nonblack male_black
test female_black = female_nonblack = male_black = 0
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(4, 4161) > 99999.00
Model | 185756.485 4 46439.1213 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.9958
-------------+---------------------------------- Adj R-squared = 0.9958
Total | 186535.954 4,165 44.7865436 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | 6.07425 .0422382 143.81 0.000 5.991441 6.15706
female_non~k | 6.307536 .0226856 278.04 0.000 6.26306 6.352012
male_black | 6.517691 .0309152 210.82 0.000 6.457081 6.578301
male_nonbl~k | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) female_black - female_nonblack = 0
( 2) female_black - male_black = 0
( 3) female_black - male_nonblack = 0
F( 3, 4161) = 191.17
Prob > F = 0.0000
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
female_black | -.0093263 .057515 -0.16 0.871 -.1220865 .1034339
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) - black + female_black = 0
( 2) - female + female_black = 0
( 3) female_black = 0
F( 3, 4161) = 191.17
Prob > F = 0.0000
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female_black | -.6674001 .0428671 -15.57 0.000 -.7514426 -.5833576
female_non~k | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
male_black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
( 1) female_black - female_nonblack = 0
( 2) female_black - male_black = 0
( 3) female_black = 0
F( 3, 4161) = 191.17
Prob > F = 0.00008. Try to replicate the F-statistic for one of the above models. Hint, the F-stat for these models is the same as that of the whole model.
reg lwage female black female_black
ereturn list
scalar fstat = (e(r2)*e(df_r))/((1-e(r2))*e(df_m))
scalar list fstat
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(3, 4161) = 191.17
Model | 107.436063 3 35.8120209 Prob > F = 0.0000
Residual | 779.468839 4,161 .187327287 R-squared = 0.1211
-------------+---------------------------------- Adj R-squared = 0.1205
Total | 886.904902 4,164 .212993492 Root MSE = .43281
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
female | -.4341146 .0238361 -18.21 0.000 -.480846 -.3873832
black | -.2239593 .0317691 -7.05 0.000 -.2862436 -.1616749
female_black | -.0093263 .057515 -0.16 0.871 -.1220865 .1034339
_cons | 6.74165 .0073159 921.51 0.000 6.727307 6.755993
------------------------------------------------------------------------------
scalars:
e(N) = 4165
e(df_m) = 3
e(df_r) = 4161
e(F) = 191.1735422330181
e(r2) = .1211359442526956
e(rmse) = .4328132235793649
e(mss) = 107.4360627542734
e(rss) = 779.4688391479763
e(r2_a) = .1205023003768864
e(ll) = -2419.902951629166
e(ll_0) = -2688.805870567022
e(rank) = 4
macros:
e(cmdline) : "regress lwage female black female_black"
e(title) : "Linear regression"
e(marginsok) : "XB default"
e(vce) : "ols"
e(depvar) : "lwage"
e(cmd) : "regress"
e(properties) : "b V"
e(predict) : "regres_p"
e(model) : "ols"
e(estat_cmd) : "regress_estat"
matrices:
e(b) : 1 x 4
e(V) : 4 x 4
e(beta) : 1 x 3
functions:
e(sample)
fstat = 191.17354In the case where the F-test corresponds to the test of the entire model, you can write the F-statistic in terms of \(R^2\).
9. Estimate the following model:
\[ lwage = \beta_1 + \beta_2F + \beta_3B + \beta_4F\times B + \beta_5exp + \beta_6exp^2 + \beta_7educ + \varepsilon \]
Interpret the estimated coeffiecients \(\hat{\beta}_7\).
Interpret the effect of experience variable
exp. Use the median level of experience to make your calculation.
sum educ, det
reg lwage i.female##i.black exper expsq educ
di (exp(_b[educ])-1)*100
years of education
-------------------------------------------------------------
Percentiles Smallest
1% 6 4
5% 8 4
10% 9 4 Obs 4,165
25% 12 4 Sum of wgt. 4,165
50% 12 Mean 12.84538
Largest Std. dev. 2.787995
75% 16 17
90% 17 17 Variance 7.772916
95% 17 17 Skewness -.2581161
99% 17 17 Kurtosis 2.71273
Source | SS df MS Number of obs = 4,165
-------------+---------------------------------- F(6, 4158) = 411.81
Model | 330.586 6 55.0976667 Prob > F = 0.0000
Residual | 556.318902 4,158 .13379483 R-squared = 0.3727
-------------+---------------------------------- Adj R-squared = 0.3718
Total | 886.904902 4,164 .212993492 Root MSE = .36578
------------------------------------------------------------------------------
lwage | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
1.female | -.4032047 .0203208 -19.84 0.000 -.4430444 -.363365
1.black | -.1551546 .0269249 -5.76 0.000 -.2079417 -.1023674
|
female#black |
1 1 | -.002071 .0488405 -0.04 0.966 -.0978246 .0936826
|
exper | .0427346 .0022404 19.07 0.000 .0383422 .047127
expsq | -.0006982 .0000494 -14.14 0.000 -.0007951 -.0006014
educ | .0731837 .0020983 34.88 0.000 .0690698 .0772975
_cons | 5.303667 .0362462 146.32 0.000 5.232606 5.374729
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7.5928119- A one unit increase in years of educ is associated with an increase of 7.59% in expected wages, holding other regressors fixed.
sum exper, det
return list
di (exp(_b[exper]+2*r(p50)*_b[expsq])-1)*100
years of full-time work experience
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Percentiles Smallest
1% 3 1
5% 5 1
10% 7 1 Obs 4,165
25% 11 1 Sum of wgt. 4,165
50% 18 Mean 19.85378
Largest Std. dev. 10.96637
75% 29 50
90% 36 50 Variance 120.2613
95% 39 50 Skewness .4000014
99% 44 51 Kurtosis 2.072064
scalars:
r(N) = 4165
r(sum_w) = 4165
r(mean) = 19.85378151260504
r(Var) = 120.2612759224727
r(sd) = 10.96637022548814
r(skewness) = .4000013893781186
r(kurtosis) = 2.072064120792274
r(sum) = 82691
r(min) = 1
r(max) = 51
r(p1) = 3
r(p5) = 5
r(p10) = 7
r(p25) = 11
r(p50) = 18
r(p75) = 29
r(p90) = 36
r(p95) = 39
r(p99) = 44
1.7754427- A one unit incease in years of experience is associated with an increase of 1.78% in expected wages, holding other regressors fixed.
10. Theoretically, how would you test the following restrictions for the model below?
\(\beta_2 = \beta_3\)
\(\beta_4 + \beta_5 = 1\)
\(\beta_2 = \beta_3\) and \(\beta_4 + \beta_5 = 1\)
\[ Y = \beta_1 + \beta_2X_2 + \beta_3X_3 + \beta_4X_4 + \beta_5X_5 + \varepsilon \]
- Restrictions 1: \(H_0: \beta_2 = \beta_3\Rightarrow \beta_2-\beta_3 = 0\). This can be written as a simple T-test (or F-tests),
\[ \text{T-stat} = \frac{\hat{\beta}_2-\hat{\beta}_3}{\sqrt{\hat{Var}(\hat{\beta}_2-\hat{\beta}_3)}} \]
where,
\[ Var(\hat{\beta}_2-\hat{\beta}_3) = Var(\hat{\beta}_2) + Var(\hat{\beta}_3)-2\cdot Cov(\hat{\beta}_2,\hat{\beta}_3) \]
Alternatively, rewrite the model, adding and subtracting \(\beta_3X_2\) (or \(\beta_2X_3\)):
\[ Y = \beta_1 + (\beta_2-\beta_3)X_2 + \beta_3(X_2+X_3) + \beta_4X_4 + \beta_5X_5 + \varepsilon \]
Then test the hypothese that the coefficient on \(X_2\) is \(=0\).
- Restrictions 1: \(H_0: \beta_4 + \beta_5 = 1\). This can be written as a simple T-test,
\[ \text{T-stat} = \frac{\hat{\beta}_4+\hat{\beta}_5-1}{\sqrt{\hat{Var}(\hat{\beta}_4+\hat{\beta}_5})} \]
where,
\[ Var(\hat{\beta}_4+\hat{\beta}_5) = Var(\hat{\beta}_4) + Var(\hat{\beta}_5)+2\cdot Cov(\hat{\beta}_4,\hat{\beta}_5) \]
Alternatively, rewrite the model, adding and subtracting \(\beta_5X_4-X_5\) (or \(\beta_2X_3\)):
\[ Y-X_4 = \beta_1 + \beta_2X_2 + \beta_3X_3 + (\beta_4+\beta_5-1)X_4 + (\beta_5)(X_5-X_4) + \varepsilon \]
Then test the hypothese that the coefficient on \(X_4\) is \(=0\).
- To test both of the linear restrictions simultaneously, we would use an F-test.
Step 1: estimate the unrestricted model
\[ Y = \beta_1 + \beta_2X_2 + \beta_3X_3 + \beta_4X_4 + \beta_5X_5 + \varepsilon \]
Store the \(SSR_U\)
Step 2: estimate the restricted model
\[ (Y-X_4) = \gamma_1 + \gamma_2(X_2+X_3) + \gamma_5(X_5-X_4) + \varepsilon \]
Store the \(SSR_R\).
Step 3: Compute the F-statistic
\[ \text{F-stat} = \frac{(SSR_R-SSR_U)/(df_R-df_U)}{SSR_U/df_U} \]
Postamble
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