Exam Guide Solution
Question 1
1.
dat <- mroz
colSums(is.na(dat[, c("hours", "wage")]))
## hours
##
0
wage
325
The missing values are not necessarily due to data errors. In particular, women who are not participating
in the labour force may not have an observed wage because they are not working.
2.
ggplot(dat, aes(x = hours)) +
geom_histogram() +
labs(title = "Distribution of Hours Worked",
x = "Hours worked",
y = "Count")
## ‘stat_bin()‘ using ‘bins = 30‘. Pick better value ‘binwidth‘.
1
Distribution of Hours Worked
300
Count
200
100
0
0
1000
2000
3000
4000
5000
Hours worked
The distribution is highly skewed and contains a large mass at zero because many women do not participate
in the labour force.
The majority are around a standard full-time work schedule (40*50=2000). Some very large values appear
a little unusual, but they could plausibly reflect individuals working extremely long hours. Therefore, it is
difficult to justify removing observations solely based on visual inspection.
3.
ggplot(dat, aes(x = huswage, y = wage)) +
geom_point() +
labs(title = "Scatter Plot of Husband's Wage and Wage",
x = "Husband's Wage",
y = "Wage")
## Warning: Removed 325 rows containing missing values or values outside the scale range
## (‘geom_point()‘).
2
Scatter Plot of Husband's Wage and Wage
25
20
Wage
15
10
5
0
0
10
20
30
Husband's Wage
The raw scatter plot is difficult to interpret because there are many overlapping observations.
binsreg(y = dat$wage,
x = dat$huswage)
3
40
dat$wage
5
4
3
0
10
20
dat$huswage
## Call: binsreg
##
## Binscatter Plot
## Bin/Degree selection method (binsmethod) = IMSE direct plug-in (select # of bins)
## Placement (binspos)
= Quantile-spaced
## Derivative (deriv)
= 0
##
## Group (by)
= Full Sample
## Sample size (n)
= 428
## # of distinct values (Ndist)
= 401
## # of clusters (Nclust)
= NA
## dots, degree (p)
= 0
## dots, smoothness (s)
= 0
## # of bins (nbins)
= 10
From the binscatter, it is easier to see a positive relationship.
Part 4
reg_huswage <- feols(data = dat,
hours ~ huswage,
vcov = "hetero")
summary(reg_huswage)
4
## OLS estimation, Dep. Var.: hours
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error t value
Pr(>|t|)
## (Intercept) 892.5182
58.43901 15.27264 < 2.2e-16 ***
## huswage
-20.3072
5.89784 -3.44315 0.00060684 ***
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 866.5
Adj. R2: 0.008403
The intercept is 893, representing the predicted number of annual hours worked when the husband’s hourly
wage is zero (i.e. they do not work).
The coefficient on huswage is -20. This means that a one dollar increase in the husband’s hourly wage is
associated with wives working about 20 fewer hours per year on average.
The relationship is negative: married women tend to work fewer hours when their husbands earn higher
wages. This is consistent with the idea that higher household income may reduce the wife’s labour supply,
but this simple regression should not be interpreted causally.
Part 5
The estimate from Part 4 may suffer from omitted variable bias if educ is related both to the husband’s
wage and to the wife’s hours worked. For example, more educated women may tend to have higher-earning
husbands. At the same time, education may increase women’s own labour market opportunities and therefore
increase their labour supply. If education is omitted, the coefficient on huswage may partly capture differences
in education rather than only the relationship between husbands’ wages and wives’ hours worked. Therefore,
not controlling for education could positively bias the estimated relationship between huswage and hours.
Part 6
reg_huswage_educ <- feols(data = dat,
hours ~ huswage + educ,
vcov = "hetero")
summary(reg_huswage_educ)
## OLS estimation, Dep. Var.: hours
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error t value
Pr(>|t|)
## (Intercept) 271.5826 159.72247 1.70034 8.9481e-02 .
## huswage
-28.8693
6.22317 -4.63900 4.1307e-06 ***
## educ
55.7506
13.52526 4.12196 4.1758e-05 ***
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 857.9
Adj. R2: 0.026692
After controlling for education, the coefficient on huswage is -29. This means that, holding education fixed,
a one dollar increase in the husband’s hourly wage is associated with wives working 29 fewer hours per
year. The interpretation has changed relative to Part 4. The coefficient is no longer just the unconditional
relationship between husbands’ wages and wives’ hours worked. It is now the relationship conditional on
wives having the same level of education.
5
In this case, the coefficient is even more negative, so the basic conclusion is still that higher husbands’ wages
are associated with lower wives’ labour supply but even more so now.
Part 7
dat <- dat %>%
mutate(youngkids = ifelse(kidslt6 > 0, 1, 0))
reg_interaction <- feols(data = dat,
hours ~ huswage + educ + youngkids + huswage:youngkids,
vcov = "hetero")
summary(reg_interaction)
## OLS estimation, Dep. Var.: hours
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error
t value
Pr(>|t|)
## (Intercept)
258.81267 158.90578 1.628718 1.0379e-01
## huswage
-26.90084
6.76561 -3.976114 7.6848e-05 ***
## educ
63.70396
13.12041 4.855334 1.4638e-06 ***
## youngkids
-442.33909 129.42863 -3.417629 6.6586e-04 ***
## huswage:youngkids
-8.79580
11.52198 -0.763393 4.4547e-01
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 833.8
Adj. R2: 0.078103
For women without young children, the effect of huswage on hours is given by the coefficient on huswage,
-26.9.
For women with young children, the effect is the sum of the coefficient on huswage and the interaction
coefficient, (-26.9 - 8.8 = -35.7). Thus, for women with young children, a one dollar increase in the husband’s
hourly wage is associated with about 35.7 additional annual hours worked. For women without young
children, the corresponding effect is about 26.9 hours.
Part 8
youngkids would be a mediator if husbands’ wages affected whether the household has young children, and
this in turn affected wives’ hours worked. In that case, part of the effect of huswage on hours would operate
through the presence of young children.
reg_mediator <- feols(data = dat,
hours ~ educ + huswage + youngkids,
vcov = "hetero")
summary(reg_mediator)
## OLS estimation, Dep. Var.: hours
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error t value
Pr(>|t|)
## (Intercept) 274.8705 154.96937 1.77371 7.6518e-02 .
6
## educ
63.3471
13.08888 4.83976 1.5791e-06 ***
## huswage
-28.4799
5.93349 -4.79986 1.9171e-06 ***
## youngkids
-509.8797
66.78329 -7.63484 6.8998e-14 ***
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 833.9
Adj. R2: 0.079058
In the model without youngkids, the coefficient on huswage is -28.9. After adding youngkids, the coefficient
is -28.5. The coefficient changes only modestly. Therefore, if youngkids were considered a potential mediator,
it does not appear to be a very important mechanism for the relationship between husbands’ wages and wives’
hours worked.
Part 9
dat <- dat %>%
mutate(lhuswage = log(huswage))
reg_log_huswage <- feols(data = dat,
hours ~ lhuswage + educ + youngkids,
vcov = "hetero")
summary(reg_log_huswage)
## OLS estimation, Dep. Var.: hours
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error t value
Pr(>|t|)
## (Intercept) 358.8141
163.1034 2.19992 2.8117e-02 *
## lhuswage
-108.6494
51.7974 -2.09758 3.6277e-02 *
## educ
55.6455
13.1756 4.22337 2.7030e-05 ***
## youngkids
-510.6521
67.3640 -7.58049 1.0193e-13 ***
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 839.7
Adj. R2: 0.06636
The coefficient on lhuswage is -108. Since the independent variable is in logs and the dependent variable is in
levels, a 1 percent increase in the husband’s wage is associated with approximately -108/100=1.08 additional
annual hours worked.
r2(reg_mediator, type = "r2")
##
r2
## 0.08273216
r2(reg_log_huswage, type = "r2")
##
r2
## 0.07008427
The model with the larger R2 , the level-level regression, fits the sample slightly better. Yes, it is possible to
compare R2 values here because the dependent variable is the same in both models.
7
Question 2
Part 1
dat <- dat %>%
mutate(age2 = (age - mean(age))ˆ2)
reg_lpm <- feols(data = dat,
inlf ~ age + age2 + educ + exper + kidslt6 + huswage,
vcov = "hetero")
summary(reg_lpm)
## OLS estimation, Dep. Var.: inlf
## Observations: 753
## Standard-errors: Heteroskedasticity-robust
##
Estimate Std. Error t value
Pr(>|t|)
## (Intercept) 0.776447
0.133505 5.81587 8.9454e-09 ***
## age
-0.018300
0.002169 -8.43819 < 2.2e-16 ***
## age2
-0.000322
0.000249 -1.29551 1.9554e-01
## educ
0.038206
0.007156 5.33890 1.2423e-07 ***
## exper
0.022856
0.002008 11.38292 < 2.2e-16 ***
## kidslt6
-0.267166
0.031576 -8.46096 < 2.2e-16 ***
## huswage
-0.007691
0.004059 -1.89470 5.8519e-02 .
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 0.428015
Adj. R2: 0.247235
It may be appropriate to include a squared effect of age because the relationship between age and labour
force participation may be nonlinear. For example, participation may rise with age at younger ages but fall
later in life.
The coefficient on age is negative and indicates that for someone who is average age, getting one year older
is associated with decreasing the probability of being in the labour force by 1.8 percentage points.
Part 2
If huswage contains classical measurement error, the estimated effect of huswage on inlf will usually be
biased toward zero. This is called attenuation bias. Intuitively, classical measurement error adds noise to the
regressor. This makes it harder for the regression to detect the true relationship, so the estimated coefficient
is smaller in absolute value than it would be if huswage were measured without error.
Part 3
dat <- dat %>%
mutate(phat_lpm = predict(reg_lpm, newdata = dat))
min(dat$phat_lpm)
## [1] -0.2603922
8
max(dat$phat_lpm)
## [1] 1.121379
The minimum predicted probability from the LPM is -0.26, and the maximum predicted probability is 1.12.
These values are outside the interval from zero to one. This does not make sense if we interpret them literally
as probabilities, but it is a known limitation of the linear probability model. The LPM is linear and therefore
does not force fitted values to lie between zero and one.
Part 4
reg_probit <- feglm(data = dat,
inlf ~ age + age2 + educ + exper + kidslt6 + huswage,
family = binomial(link = "probit"))
summary(reg_probit)
## GLM estimation, family = binomial(link = "probit"), Dep. Var.: inlf
## Observations: 753
## Standard-errors: IID
##
Estimate Std. Error z value
Pr(>|z|)
## (Intercept) 0.792408
0.438530 1.80696 7.0768e-02 .
## age
-0.057456
0.007882 -7.28919 3.1183e-13 ***
## age2
-0.001053
0.000852 -1.23585 2.1651e-01
## educ
0.125949
0.024831 5.07223 3.9317e-07 ***
## exper
0.071720
0.007532 9.52143 < 2.2e-16 ***
## kidslt6
-0.845201
0.118972 -7.10419 1.2103e-12 ***
## huswage
-0.023772
0.012644 -1.88017 6.0085e-02 .
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## Log-Likelihood: -406.9
Adj. Pseudo R2: 0.198092
##
BIC: 860.1
Squared Cor.: 0.258922
avg_slopes(reg_probit)
##
##
Term Estimate Std. Error
z Pr(>|z|)
S
2.5 %
97.5 %
## age
-0.017572
0.00216 -8.15
<0.001 51.3 -0.021798 -0.013345
## age2
-0.000322
0.00026 -1.24
0.2152
2.2 -0.000831 0.000187
## educ
0.038519
0.00723 5.33
<0.001 23.3 0.024354 0.052685
## exper
0.021934
0.00188 11.67
<0.001 102.1 0.018250 0.025619
## huswage -0.007270
0.00384 -1.89
0.0584
4.1 -0.014799 0.000258
## kidslt6 -0.258492
0.03272 -7.90
<0.001 48.4 -0.322619 -0.194365
##
## Type: response
## Comparison: dY/dX
The average marginal effect of educ is 0.04. This means that one additional year of education is associated
with an increase in the probability of labour force participation of 4 percentage points, holding the other
variables fixed.
9
dat <- dat %>%
mutate(phat_probit = predict(reg_probit, newdata = dat, type = "response"))
Part 5
dat <- dat %>%
mutate(pred_lpm = ifelse(phat_lpm >= 0.5, 1, 0),
pred_probit = ifelse(phat_probit >= 0.5, 1, 0))
mean(dat$pred_lpm == dat$inlf)
## [1] 0.7343958
mean(dat$pred_probit == dat$inlf)
## [1] 0.7343958
mean(dat$inlf)
## [1] 0.5683931
The LPM hit ratio is and the probit hit ratio are 0.73, and the naive hit ratio is 0.57.
Both the LPM and probit models improve on the naive model. The LPM and probit have the same hit
ratios so neither dominates the other, but we can also check the Brier score:
mean((dat$phat_lpm - dat$inlf)ˆ2)
## [1] 0.1831966
mean((dat$phat_probit - dat$inlf)ˆ2)
## [1] 0.1818039
Now we see that the probit has a slightly lower Brier score, so it is the preferred method.
Question 3
Part 1
ggplot(dat, aes(x = RainMM, y = Visits)) +
geom_point(alpha = 0.5) +
geom_smooth(method = "lm", se = FALSE, colour = "firebrick") +
labs(x = "Daily rainfall (mm)", y = "Daily gym check-ins",
title = "Gym visits vs. rainfall") +
theme_minimal()
10
## ‘geom_smooth()‘ using formula = ’y ~ x’
Daily gym check−ins
Gym visits vs. rainfall
1000
500
0
10
20
Daily rainfall (mm)
cor(dat$RainMM, dat$Visits)
## [1] 0.05844299
The scatter shows essentially no clear relationship — the cloud of points is roughly flat and the correlation
is tiny (~0.06). At most one might argue for a weak positive association.
Crucially, no causal conclusion can be drawn from this plot. It is a purely bivariate correlation and rainfall is
not randomly assigned. Confounders can drive both variables simultaneously — for example, day of week and
seasonality affect both attendance and weather patterns. The omitted-variable problem and the descriptive
nature of a scatter mean we cannot infer causality.
Part 2
dat <- dat %>%
mutate(L1_Visits = lag(Visits, n = 1),
L1_RainMM = lag(RainMM, n = 1),
DoW
= as.factor(wday(Date)))
datTrain <- dat[1:500, ]
datTest <- dat[501:nrow(dat), ]
nrow(datTrain); nrow(datTest)
11
## [1] 500
## [1] 231
The model forecasts gym visits one day ahead. A forecast for day t is made at the end of day t − 1, so every
regressor must be observable at that point.
• RainMM: Today’s rainfall is not known the day before. We therefore use the one-period lag L1_RainMM
(yesterday’s rainfall), which is already observed.
• Promo: Promotions are decided and announced in advance by the gym. Whether a promotion runs
on day t is already known the day before, so contemporaneous Promo can be used directly without
lagging.
Note that row 1 of the training set has NA lags and is dropped automatically by feols, so models trained
on datTrain use 499 observations.
Part 3
m3 <- feols(Visits ~ L1_Visits + L1_RainMM + Promo,
data = datTrain, vcov = vcov_NW(time = "Date"))
## NOTE: 1 observation removed because of NA values (RHS: 1).
summary(m3)
## OLS estimation, Dep. Var.: Visits
## Observations: 499
## Standard-errors: Newey-West (L=20)
##
Estimate Std. Error
t value
Pr(>|t|)
## (Intercept) 8.253125
5.290187
1.56008 1.1938e-01
## L1_Visits
0.982026
0.007944 123.61694 < 2.2e-16 ***
## L1_RainMM
-1.856102
0.397867 -4.66514 3.9684e-06 ***
## Promo
51.017902
4.739265 10.76494 < 2.2e-16 ***
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 39.4
Adj. R2: 0.974604
pred3 <- predict(m3, newdata = datTest)
mse3 <- mean((datTest$Visits - pred3)ˆ2)
mse3
## [1] 1542.539
The estimated coefficient on L1_Visits is approximately 0.98, indicating strong day-to-day persistence in
gym attendance. L1_RainMM is approximately -1.9 (more rain yesterday is associated with slightly fewer
visits today), and Promo is approximately +51 (a promotion adds around 51 check-ins).
Newey-West standard errors are used because time-series residuals are typically serially correlated (and
possibly heteroskedastic). Newey-West gives HAC-valid (heteroskedasticity and autocorrelation consistent)
inference in this setting.
The out-of-sample MSE for this model is approximately 1543.
12
Part 4
m4 <- feols(Visits ~ L1_Visits + L1_RainMM + Promo + DoW,
data = datTrain, vcov = vcov_NW(time = "Date"))
## NOTE: 1 observation removed because of NA values (RHS: 1).
summary(m4)
## OLS estimation, Dep. Var.: Visits
## Observations: 499
## Standard-errors: Newey-West (L=9)
##
Estimate Std. Error
t value
Pr(>|t|)
## (Intercept) -33.595934
6.122883 -5.48695 6.5197e-08 ***
## L1_Visits
0.987225
0.007054 139.94345 < 2.2e-16 ***
## L1_RainMM
-1.802309
0.276616 -6.51556 1.7757e-10 ***
## Promo
49.350781
2.876018 17.15941 < 2.2e-16 ***
## DoW2
83.299655
5.455722 15.26831 < 2.2e-16 ***
## DoW3
61.760564
4.461940 13.84164 < 2.2e-16 ***
## DoW4
50.488806
4.496109 11.22945 < 2.2e-16 ***
## DoW5
37.408751
4.625969
8.08668 4.7004e-15 ***
## DoW6
21.878252
4.346724
5.03327 6.7479e-07 ***
## DoW7
11.659714
4.582552
2.54437 1.1248e-02 *
## --## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
## RMSE: 28.5
Adj. R2: 0.986526
pred4 <- predict(m4, newdata = datTest)
mse4 <- mean((datTest$Visits - pred4)ˆ2)
c(MSE_no_DoW = mse3, MSE_with_DoW = mse4,
improvement_pct = 100 * (mse3 - mse4) / mse3)
##
##
MSE_no_DoW
1542.53942
MSE_with_DoW improvement_pct
852.46227
44.73644
Adding the day-of-week factor cuts the out-of-sample MSE from approximately 1543 to approximately 853,
an improvement of around 45%. The preferred model is the one with the day-of-week dummies: gym
attendance has a strong weekly cycle that the lagged visits variable alone cannot fully capture. The model
is judged on out-of-sample performance, so the gain is genuine and not merely in-sample overfitting. It is
also worth noting that the coefficients on the other regressors are largely unchanged by the inclusion of the
day-of-week factor.
Part 5
A time series has a unit root if it is non-stationary of the random-walk type. In an AR(1) model, Visitst =
ρ · Visitst−1 + ut , a unit root means ρ = 1: shocks are permanent (they never die out), the series has
no tendency to revert to a fixed mean, and its variance grows without bound over time. The estimated
persistence of approximately 0.98 is close to 1, which may suggest a unit root.
This is problematic for several reasons:
13
• Spurious regression: two unrelated unit-root series can appear highly significantly related (high R2 ,
large t-statistics) purely by chance.
• Invalid inference: OLS t- and F -statistics no longer follow their standard distributions under a unit
root, so p-values are invalid.
• Non-constant mean and variance: these violate the stationarity assumptions underlying our forecasting and inference tools.
The standard correction is to take first differences. If Visits non-stationary, then ∆Visitst = Visitst −
Visitst−1 is typically stationary (I(0)), and we model the differenced series instead.
dat <- dat %>%
mutate(Visits_diff = Visits - lag(Visits, n = 1))
head(dat[, c("Date", "Visits", "Visits_diff")])
##
Date Visits Visits_diff
## 1 2023-01-01
1000
NA
## 2 2023-01-02
1096
96
## 3 2023-01-03
1152
56
## 4 2023-01-04
1167
15
## 5 2023-01-05
1159
-8
## 6 2023-01-06
1152
-7
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