Problem set 4 - Statistics and Econometrics - Msc Business Analytics - Imperial College London

1. Problem set 4
Jonathan Zimmermann
7 November 2015
Exercise 1
Use the data in meap00_01.RData to answer this question. The original source of this data
set is Michigan Department of Education (www.michigan.gov/mde).
a) Estimate the model
math4 = β0 + β1lunch + β2log(enroll) + β3log(exppp) + u
by OLS and obtain the usual standard errors and the robust standard errors. How do they generally compare?
>>
library(dplyr)
library(ggvis)
library(sandwich)
library(lmtest)
library(car)
load("meap00_01.RData")
Standard OLS:
a = lm(math4 ~ lunch + log(enroll) + log(exppp), data) %>% print()
##
## Call:
## lm(formula = math4 ~ lunch + log(enroll) + log(exppp), data = data)
##
## Coefficients:
## (Intercept) lunch log(enroll) log(exppp)
## 91.9324 -0.4487 -5.3992 3.5248
summary(a)
##
## Call:
## lm(formula = math4 ~ lunch + log(enroll) + log(exppp), data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -56.326 -8.758 0.914 9.164 51.416
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
1

2. ## (Intercept) 91.93240 19.96170 4.605 4.42e-06 ***
## lunch -0.44874 0.01464 -30.648 < 2e-16 ***
## log(enroll) -5.39915 0.94041 -5.741 1.11e-08 ***
## log(exppp) 3.52475 2.09785 1.680 0.0931 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 15.3 on 1688 degrees of freedom
## Multiple R-squared: 0.3729, Adjusted R-squared: 0.3718
## F-statistic: 334.6 on 3 and 1688 DF, p-value: < 2.2e-16
Robust OLS:
b = coeftest(a, vcov = vcovHC(a, "HC1")) %>% print()
##
## t test of coefficients:
##
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 91.932404 23.087103 3.9820 7.125e-05 ***
## lunch -0.448743 0.016584 -27.0583 < 2.2e-16 ***
## log(enroll) -5.399152 1.131175 -4.7730 1.971e-06 ***
## log(exppp) 3.524751 2.353737 1.4975 0.1344
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The sample size is:
summary(a)$df[2] + length(coef(a)) # = Degrees of freedom + number of coefficients. nrow(data) would ha
## [1] 1692
The r-squared is:
summary(a)$r.squared
## [1] 0.3728875
Regular OLS regression results:
math4 = 91.93240
(19.96170)
− 0.44874
(0.01464)
lunch − 5.39915
(0.94041)
log(enroll) + 3.52475
(2.09785)
log(exppp)
Robust OLS regression results:
math4 = 91.93240
(23.087103)
− 0.44874
(0.016584)
lunch − 5.39915
(1.131175)
log(enroll) + 3.52475
(2.353737)
log(exppp)
As we can see, the two models give the same estimates for the coefficients, but the standard errors are larger
in the robust version. This is a general result of heteroskedasticity-robust regressions: the standard errors are
usually larger to compensate for the presence (or suspicion) of heteroskedasticity.
Note: When variances tends to increase as X is far from its mean, we might have a robust standard error
smaller than the regular standard error. But this case is quite rare in practice, and in almost any data set,
the robust standard errors are typically larger or equal.
2

3. b) Apply the White test for heteroskedasticity. What is the value of the F test? What do you conclude?
>>
We can perform the White test in R using a modified Breusch-Pagan function:
fitted.a <- a$fitted.values
bptest(a, ~fitted.a + I(fitted.a^2))
##
## studentized Breusch-Pagan test
##
## data: a
## BP = 229.78, df = 2, p-value < 2.2e-16
The resulting p-value is extremely low. We can therefore safely reject the null hypothesis that the model
has homoskedastic variance, i.e. Gauss-Markov’s assumption 5 doesn’t hold and OLS is not the best linear
estimator.
Alternatively, we could have computed the White test manually:
data$fittedValues <- fitted.a
data$residual <- a$residuals
data$fittedValuesSq <- data$fittedValues^2
data$residualMathSq <- data$residual^2
whiteRegression <- lm(residualMathSq ~ fittedValues + fittedValuesSq,
data)
summary(whiteRegression)
##
## Call:
## lm(formula = residualMathSq ~ fittedValues + fittedValuesSq,
## data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -593.52 -163.92 -72.42 52.12 2735.87
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1664.12800 315.21436 5.279 1.46e-07 ***
## fittedValues -28.29903 9.21685 -3.070 0.00217 **
## fittedValuesSq 0.11553 0.06591 1.753 0.07982 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 365.1 on 1689 degrees of freedom
## Multiple R-squared: 0.1358, Adjusted R-squared: 0.1348
## F-statistic: 132.7 on 2 and 1689 DF, p-value: < 2.2e-16
whiteRsquared <- summary(whiteRegression)$r.squared
# Calculating the F-statistic
fStatistic <- (whiteRsquared/2)/((1 - whiteRsquared)/(1692 -
3)) %>% print
3

4. ## [1] 0.0005116607
# Critical value for alpha = 5%
qf(0.95, df1 = 2, df2 = 1692 - 3)
## [1] 3.001052
# Critical value for alpha = 1%
qf(0.99, df1 = 2, df2 = 1692 - 3)
## [1] 4.617749
c) Obtain ˆgi as the fitted values from the regression log(ˆu2
i ) on math4i, math4
2
i where math4i are the
OLS fitted values and the ˆui are the OLS residuals. Let ˆhi = exp(ˆgi). Use the ˆhi to obtain WLS
estimates. Are there big differences with the OLS coefficients?
>>
fitted_square = fitted.a^2
c = lm(log(a$residuals^2) ~ fitted.a + fitted_square) %>% print()
##
## Call:
## lm(formula = log(a$residuals^2) ~ fitted.a + fitted_square)
##
## Coefficients:
## (Intercept) fitted.a fitted_square
## 4.2075441 0.0596369 -0.0008415
g = c$fitted.values
h = exp(g)
WLS_model = lm(math4 ~ lunch + log(enroll) + log(exppp), data,
weight = 1/h) %>% print()
##
## Call:
## lm(formula = math4 ~ lunch + log(enroll) + log(exppp), data = data,
## weights = 1/h)
##
## Coefficients:
## (Intercept) lunch log(enroll) log(exppp)
## 50.4782 -0.4486 -2.6473 6.4742
summary(WLS_model)
##
## Call:
## lm(formula = math4 ~ lunch + log(enroll) + log(exppp), data = data,
## weights = 1/h)
4

5. ##
## Weighted Residuals:
## Min 1Q Median 3Q Max
## -10.7122 -1.1914 0.1403 1.3741 3.9151
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 50.47815 16.51032 3.057 0.002268 **
## lunch -0.44859 0.01461 -30.697 < 2e-16 ***
## log(enroll) -2.64728 0.83594 -3.167 0.001569 **
## log(exppp) 6.47420 1.68588 3.840 0.000127 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.9 on 1688 degrees of freedom
## Multiple R-squared: 0.36, Adjusted R-squared: 0.3588
## F-statistic: 316.5 on 3 and 1688 DF, p-value: < 2.2e-16
Regular OLS regression results:
math4 = 91.93240
(19.96170)
− 0.44874
(0.01464)
lunch − 5.39915
(0.94041)
log(enroll) + 3.52475
(2.09785)
log(exppp)
Regular WLS regression results:
math4 = 50.47815
(16.51032)
− 0.44859
(0.01461)
lunch − 2.64728
(0.83594)
log(enroll) + 6.47420
(1.68588)
log(exppp)
Yes, there are significant differences with WLS compared to the OLS regression. None of the coefficients
changed sign and they all remained statistically significant, but their values changed a lot. In particular:
• The intercept decreased from 91.93240 to 50.47815
• The coefficient log(enroll) increased from -5.39915 to -2.64728 (i.e. increasing school enrollment is now
estimated to have less impact on 4th grade math)
• The coefficient log(exppp) increased from 3.52475 to 6.47420 (i.e. spending more on pupils is now
expected to have a bigger impact on 4th grade math)
The coefficient “lunch” almost didn’t change.
The significance of some coefficients also changed. In particular, log(exppp) is now highly statistically
significant.
Reminder: the interpretation of the coefficients doesn’t change when WLS is used instead of OLS. And in
both models, they are unbiased (as long as assumptions 1 to 4 held in the original model).
d) Obtain the standard errors for WLS that allow misspecification of the variance function. Do these differ
much from the usual WLS standard errors?
>>
If we are not sure of the specification of the variance function, we can couple WLS with the use of robust
standard errors:
5

6. WLS_model_robust = coeftest(WLS_model, vcov = vcovHC(WLS_model,
"HC1")) %>% print()
##
## t test of coefficients:
##
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 50.478152 18.925179 2.6672 0.0077206 **
## lunch -0.448592 0.014254 -31.4711 < 2.2e-16 ***
## log(enroll) -2.647282 1.054608 -2.5102 0.0121590 *
## log(exppp) 6.474200 1.812868 3.5712 0.0003652 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Regular WLS regression results:
math4 = 50.47815
(16.51032)
− 0.44859
(0.01461)
lunch − 2.64728
(0.83594)
log(enroll) + 6.47420
(1.68588)
log(exppp)
Robust WLS regression results:
math4 = 50.47815
(18.925179
− 0.44859
(0.014254)
lunch − 2.64728
(1.054608)
log(enroll) + 6.47420
(1.812868)
log(exppp)
As expected, the robust standard errors are usually larger than the regular WLS. But in this case, the
differences are small. The only relatively important impact this has is that, now, the coefficient log(enroll) is
statistically significant at the 1.2159% significance level only, compared to 0.1569% in the regular model.
e) For estimating the effect of spending on math4, does OLS or WLS appear to be more precise?
>>
WLS appears to be considerably more precise for this coefficient.
In the regular OLS model, log(exppp) is statistically significant at the 9.31% level whereas in the regular
WLS model it is statistically significant at the 0.0127% level. The same can be observed if we compare the
robust OLS model to the robust WLS model, in which the significance level for log(exppp) are respectively
13.44% and 0.03652%.
Exercise 2
Use the data from jtrain.RData for this exercise.
a) Consider the simple regression model
log(scrap) = β0 + β1grant + u
where scrap is the firm scrap rate and grant is a dummy variable indicating whether a firm received a job
training grant. Can you think of some reasons why the unobserved factors in u might be correlated with
grant?
>>
6

7. load("jtrain.RData")
There is an infinity of potential reasons. We are interested by factors that have an impact on both grant and
scrap and are not already included in the regression (i.e. unobserved factors).
For example, the attribution of a grant could be linked to the attributes of the workforce, i.e. a firm is more
likely to receive a job training grant if its workers have specific attributes (e.g. age, previous qualifications,
etc.). Those attributes, in turn, are very likely to affect the scrap rate, for example if the age of a worker has
a causal impact on his efficiency.
To give a comprehensive answer to this question, it would be useful to understand the process of attribution
of grants in more detail. If being eligible to receive such a grant requires to meet some specific criteria at the
firm level, such as being based in a specific geographical area, then these criteria might also, in turn, have an
impact on the scrap rate (for example, some geographical areas may increase the likelihood of defect because
of weather conditions).
b) Estimate the simple regression model using the data for 1988. (You should have 54 observations.) Does
receiving a job training grant significantly lower a firm’s scrap rate?
>>
The SRF is:
data_88 = filter(data, year == "1988")
model_88 = lm(log(scrap) ~ grant, data_88) %>% print
##
## Call:
## lm(formula = log(scrap) ~ grant, data = data_88)
##
## Coefficients:
## (Intercept) grant
## 0.4085 0.0566
summary(model_88)
##
## Call:
## lm(formula = log(scrap) ~ grant, data = data_88)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.4043 -0.9536 -0.0465 0.9636 2.8103
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.4085 0.2406 1.698 0.0954 .
## grant 0.0566 0.4056 0.140 0.8895
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.423 on 52 degrees of freedom
## (103 observations deleted due to missingness)
## Multiple R-squared: 0.0003744, Adjusted R-squared: -0.01885
## F-statistic: 0.01948 on 1 and 52 DF, p-value: 0.8895
7

8. Regression results:
log(scrap) = 0.4085
(0.2406)
+ 0.0566
(0.4056)
grant
The sample size is:
summary(model_88)$df[2] + length(coef(model_88)) # = Degrees of freedom + number of coefficients. nrow(
## [1] 54
The r-squared is:
summary(model_88)$r.squared
## [1] 0.0003744378
No, there is no evidence that receiving a job grant significantly lowers a firm’s scrap rate since the coefficient
is positive, at least based on 1988’s data.
c) Now, add as an explanatory variable log(scrap87). How does this change the estimated effect of grant?
Interpret the coefficient on grant. Is it statistically significant at the 5% level against the one-sided
alternative H1 : βgrant < 0?
>>
new_data_88 = filter(data, year == "1987") %>% select(scrap87 = scrap) %>%
cbind(data_88)
new_model_88 = lm(log(scrap) ~ grant + log(scrap87), new_data_88) %>%
print
##
## Call:
## lm(formula = log(scrap) ~ grant + log(scrap87), data = new_data_88)
##
## Coefficients:
## (Intercept) grant log(scrap87)
## 0.02124 -0.25397 0.83116
summary(new_model_88)
##
## Call:
## lm(formula = log(scrap) ~ grant + log(scrap87), data = new_data_88)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9146 -0.1763 0.0057 0.2308 1.5991
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.02124 0.08910 0.238 0.8126
8

9. ## grant -0.25397 0.14703 -1.727 0.0902 .
## log(scrap87) 0.83116 0.04444 18.701 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5127 on 51 degrees of freedom
## (103 observations deleted due to missingness)
## Multiple R-squared: 0.8728, Adjusted R-squared: 0.8678
## F-statistic: 174.9 on 2 and 51 DF, p-value: < 2.2e-16
The sample size is:
summary(new_model_88)$df[2] + length(coef(new_model_88)) # = Degrees of freedom + number of coefficient
## [1] 54
The r-squared is:
summary(new_model_88)$r.squared
## [1] 0.8727805
Let’s compare the two models, (2) being the one with the lagged dependant variable.
Independant Variables (1) (2)
(Intercept) 0.4085 (0.2406) 0.02124 (0.08910)
grant 0.0566 (0.4056) -0.25397 (0.14703)
log(scrap87) – 0.83116 (0.04444)
Observations 54 54
R-Squared 0.0003744378 0.8727805
Adding the lagged independant variable changed the sign of grant and increasing its statistical significance.
In the initial model (1), the interpretation of Beta_grant was “If a firm receives a job training grant, its scrap
rate is expected to increase by approximately 5.66%”. In the new model (2), the interpretation of Beta_grant
is “If a firm receives a job training grant, its scrap rate is expected to decrease by approximately 25.397%”.
Thus, by controlling for the scrap rate of the previous year, the estimated effect of grant changed considerably.
We want to test the following hypothesis:
H0 : βgrant = 0H1 : βgrant < 0
The t-statistic of βgrant is:
beta_grant.t_statistic = (coef(new_model_88)[["grant"]]/summary(new_model_88)$coefficients["grant",
"Std. Error"]) %>% print
## [1] -1.727319
At the 5% significance level, the one-sided t-test critical value is:
9

10. beta_grant.critical_value = qt(0.05, summary(new_model_88)$df[2] +
length(coef(new_model_88))) %>% print #P(T < critical_value) = 5%
## [1] -1.673565
Do we reject H0?
# Reject H0 if beta_grant.t_statistic <
# -beta_grant.critical_value.
beta_grant.t_statistic < -beta_grant.critical_value
## [1] TRUE
We can reject H0 at the 5% significance level against the one-sided alternative that βgrant is lower than 0.
This means that, in this new model, βgrant is statistically significant and smaller than 0.
d) Test the null hypothesis that the parameter on log(scrap87) is one against the two-sided alternative.
Report the p-value for the test.
We want to test the following hypothesis:
H0 : βlog(scrap87) = 1H1 : βlog(scrap87) = 1
The t-statistic of βlog(scrap87) is:
beta_logScrap87.t_statistic = ((coef(new_model_88)[["log(scrap87)"]] -
1)/summary(new_model_88)$coefficients["log(scrap87)", "Std. Error"]) %>%
print
## [1] -3.798889
At the 5% significance level, the two-sided t-test critical value is:
beta_logScrap87.critical_value = qt(0.05/2, summary(new_model_88)$df[2] +
length(coef(new_model_88))) %>% print #P(T < critical_value) = 5%
## [1] -2.004879
Do we reject H0?
# Reject H0 if abs(beta_logScrap87.t_statistic) >
# beta_logScrap87.critical_value.
abs(beta_logScrap87.t_statistic) > beta_logScrap87.critical_value
## [1] TRUE
We can reject H0 at the 5% significance level against the two-sided alternative.
We could also have performed that test using the “car” package:
10

11. linearHypothesis(new_model_88, "log(scrap87)=1")
## Linear hypothesis test
##
## Hypothesis:
## log(scrap87) = 1
##
## Model 1: restricted model
## Model 2: log(scrap) ~ grant + log(scrap87)
##
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 52 17.197
## 2 51 13.404 1 3.793 14.432 0.0003885 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The p-value for the test is 0.0003885.
e) Repeat parts (c) and (d), using heteroskedasticity-robust standard errors, and briefly discuss any notable
differences.
We still want to test the following hypothesis, except that now it is with robust standard errors:
H0 : βgrant = 0H1 : βgrant < 0
linearHypothesis(new_model_88, "grant=0", white.adjust = "hc1")
## Linear hypothesis test
##
## Hypothesis:
## grant = 0
##
## Model 1: restricted model
## Model 2: log(scrap) ~ grant + log(scrap87)
##
## Note: Coefficient covariance matrix supplied.
##
## Res.Df Df F Pr(>F)
## 1 52
## 2 51 1 3.0105 0.08876 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The one-sided p-value is 0.04438 (two-sided p-value divided by two), so we can still reject the null hypothesis
at the 5% significance level (the p-value is actually lower with the robust standard errors in this case).
Regarding log(scrap87), we still want to test the following hypothesis, but this time with robust standard
errors:
H0 : βlog(scrap87) = 1H1 : βlog(scrap87) = 1
11

12. linearHypothesis(new_model_88, "log(scrap87)=1", white.adjust = "hc1")
## Linear hypothesis test
##
## Hypothesis:
## log(scrap87) = 1
##
## Model 1: restricted model
## Model 2: log(scrap) ~ grant + log(scrap87)
##
## Note: Coefficient covariance matrix supplied.
##
## Res.Df Df F Pr(>F)
## 1 52
## 2 51 1 5.271 0.02583 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The p-value is now 0.02583, compared to 0.0003885 with the non-robust standard errors. So we can still
reject the null hypothesis at the 2.583% significance level, but including robustness in the hypothesis testing
considerably reduced our confidence.
12