Problem set 2 - Statistics and econometrics - Msc Business Analytics - Imperial College London

1. Problem set 2
Jonathan Zimmermann
25 October 2015
Exercise 1
Use the data in discrim.RData to answer this question. These are ZIP code­level data on prices for
various items at fast­food restaurants, along with characteristics of the zip code population, in New
Jersey and Pennsylvania. The idea is to see whether fast­food restaurants charge higher prices in
areas with a larger concentration of blacks.
a. Consider a model to explain the price of soda in its log form, log(psoda), in terms of the proportion of
the population that is black and log of median income:
Estimate the model and report your results (including SRF, the sample size, and R­squared). If
prpblck increases by .20, what is the estimated percentage change in psoda?
>>
load("discrim.RData")
The regression model is:
The SRF is:
a = lm(log(psoda)~prpblck+log(income),data)
a
##
## Call:
## lm(formula = log(psoda) ~ prpblck + log(income), data = data)
##
## Coefficients:
## (Intercept) prpblck log(income)
## ‐0.79377 0.12158 0.07651
The sample size is:
summary(a)$df[2]+length(coef(a)) # = Degrees of freedom + number of coefficients. nrow(da
ta) would have counted rows with missing values even though they are not computed into th
e regression
log(psoda) = + prpblck + log(income) + uβ0
β1
β2
log(psoda) = + prpblck + log(income) + uβ0
β1
β2

2. ## [1] 401
The r­squared is:
summary(a)$r.squared
## [1] 0.06809228
If the proportion of black (prpblck) increases by “0.2” (i.e. 20 percentage points), the estimated change in
the price of the soda (psoda) is 2.4316053%.
b. Compare the estimate from part (a) with the simple regression estimate from log(psoda) on prpblck.
Is the discrimination effect larger or smaller when you control for income? Explain.
>>
The same regression without including “income” as a parameter (i.e. without “controlling for income”) would
look like:
b = lm(log(psoda)~prpblck,data)
b
##
## Call:
## lm(formula = log(psoda) ~ prpblck, data = data)
##
## Coefficients:
## (Intercept) prpblck
## 0.03312 0.06247
As we can see, became smaller (went from 0.1215803 to 0.0624662), i.e. the discrimination effect is
larger when we control for income. So compared to when we control for income, the beta of our last
regression is negatively biased (omitted variable bias). Since we know that income is positively correlated
with the price of soda, the only reason why beta1 could be negatively biased when we don’t control for
income is that income is negatively correlated with being black (which we know empirically to be realistic).
We can actually check and see that this is correct pretty easily by running the following regression:
lm(income~prpblck,data)
##
## Call:
## lm(formula = income ~ prpblck, data = data)
##
## Coefficients:
## (Intercept) prpblck
## 50608 ‐31322
β1

3. The fact that prices are positively correlated with the median income is also quite intuitive (and empirically
true in many other studies): if the purchasing power of individuals is higher, they will be less price sensitive,
making it more profitable to increase prices.
c. Now add the variable prppov to the regression in part (a). What happens to ?
>>
c = lm(log(psoda)~prpblck+log(income)+prppov,data)
c
##
## Call:
## lm(formula = log(psoda) ~ prpblck + log(income) + prppov, data = data)
##
## Coefficients:
## (Intercept) prpblck log(income) prppov
## ‐1.46333 0.07281 0.13696 0.38036
The estimated “beta prpblck” becomes smaller than what it was when we add “prppov” in the model,
i.e. the discrimination effect becomes smaller when we control for the proportion of poverty.
Note: At the first glance, it seems quite unintuitive that the rate of poverty could have a positive impact on
the prices in an area. But there are different explanations that could (at least partially) explain this result. A
possible explanation could be that stores are mostly interested by the price­sensitivity of non­poor
customers. Indeed, poor customers generally spend most of their budget on necessity goods, therefore
only having a small fraction of their income to dedicate to superior goods such as high priced soda.
Therefore, ceteris paribus (i.e. for an identical median income in the area), a higher fraction of poor people
generally means a more unequal distribution of wealth (only up to 50% of poverty rate, because after that
the effect is inversed in terms of wealth distribution), i.e. more “rich people” who are more likely to consume
superior goods. A strategy for a restaurant could even be to artificially increase prices just to deter poors
from coming.
d. Find the correlation between log(income) and prppov. Is it roughly what you expected?
>>
cor(data$prppov,log(data$income), use="complete")
## [1] ‐0.838467
There is a very strong negative correlation between the log of the median income and the poverty rate.
This is roughly what we could expect as poverty is a function of income (you are considered “poor” when
your income is below a certain treshold, given your family size and a few other variables). The sign of the
correlation (negative) was also expected since an area with many poor people will generally have a lower
median income.
e. Evaluate the following statement: ``Because log(income) and prppov are so highly correlated, they
β
^
prpblck

4. have no business being in the same regression."
>>
As long as they are at least partially linearly independant, two highly correlated variables can be used in
the same regression. And in our case, both variables actually have very different meanings: as discussed
in c), income only represents the median wealth of an area, while poverty rate is more representative of
low income side of the distribution of wealth in that area. Theoretically, an area could have for example a
poverty rate of 1% and yet have the same median income as an area with a poverty rate of 45%. Actually,
as we saw in c), despite having such a high negative corelation, beta_prppov and beta_log(income) are
both positively with soda prices, which proves they carry very different information.
Exercise 2
Using the data set ceosal1.RData to answer the following questions. Consider an equation to
explain salaries of CEOs in terms of annual firm sales, return on equity (roe, in percentage form),
and return on the firm’s stock (ros, in percentage form):
a) In terms of the model parameters, state the null hypothesis that, after controlling for sales and roe, ros
has no effect on CEO salary. State the alternative that better stock market performance increases a CEO’s
salary.
>>
load("ceosal1.RData")
In the model, our null hypothesis and one­sided alternative are:
b. Estimate the model and report your results (including SRF, the sample size, and R­squared). By
what percentage is salary predicted to increase if ros increases by 50 basis points? Does ros have a
practically large effect on salary?
>>
The regression model is:
The SRF is:
a = lm(log(salary)~log(sales)+roe+ros,data)
a
log(salary) = + log(sales) + roe + ros + uβ0
β1
β2
β3
: = 0H0
β3
: > 0H1
β3
log(salary) = + log(sales) + roe + ros + uβ0
β1
β2
β3

5. ##
## Call:
## lm(formula = log(salary) ~ log(sales) + roe + ros, data = data)
##
## Coefficients:
## (Intercept) log(sales) roe ros
## 4.3117124 0.2803149 0.0174167 0.0002417
The sample size is:
summary(a)$df[2]+length(coef(a)) # = Degrees of freedom + number of coefficients. nrow(da
ta) would have counted rows with missing values even though they are not computed into th
e regression
## [1] 209
The r­squared is:
summary(a)$r.squared
## [1] 0.2826848
If “ros” increases by 50 basis points (i.e. “+0.5” since ros is stored in a percentage format already), salary is
predicted to increase by 0.0120828%. This effect is practically and economically unsignificant.
c. Test the null hypothesis that ros has no effect on salary against the alternative that ros has a
positive effect. Carry out the test at the 10% significance level (please show clearly the test statistic
and the critical value used in your testing).
>>
The t­statistic of of the model defined in b) with the null hypothesis defined in a) is:
a.beta_ros.t_statistic = coef(a)[["ros"]]/summary(a)$coefficients["ros","Std. Error"]
a.beta_ros.t_statistic
## [1] 0.4460218
At the 10% significance level, the one­sided t­test critical value is:
a.beta_ros.critical_value = qt(0.9,summary(a)$df[2]+length(coef(a))) #P(T > critical_valu
e) = 10%
a.beta_ros.critical_value
## [1] 1.285615
βros

6. Do we reject H0?
#Reject H0 if a.beta_ros.t_statistic > a.beta_ros.critical_value.
a.beta_ros.t_statistic > a.beta_ros.critical_value
## [1] FALSE
We cannot reject H0, so we cannot say that is statistically significant at the 10% significance level
(i.e. if the true was zero, we would have more than 10% chance of getting an estimated with the
value calculated in b)).
d. Would you include ros in a final model explaining CEO compensation in terms of firm performance?
Explain.
>>
Probably not. Despite having very conservatively set the significance level at 10%, the t­statistic was very
far from the critical value, making it highly unlikely for “ros” to have a significant effect on the salary. The
main reason why we would want to avoid adding control variables with non­significant betas is that they
might (and probably are) biased, therefore reducing the quality and reliability of the model. But even if the
estimated was unbiased, we would still prefer not adding it into the model to avoid other undesirable
but more complex effects such as a higher variance.
βros
βros
βros
βros