qnorm(c(0.005, 0.025, 0.05, 0.95, 0.975, 0.995))[1] -2.575829 -1.959964 -1.644854 1.644854 1.959964 2.575829qchisq(c(0.9, 0.95, 0.99), 1)[1] 2.705543 3.841459 6.634897qchisq(c(0.9, 0.95, 0.99), 2)[1] 4.605170 5.991465 9.210340qchisq(c(0.9, 0.95, 0.99), 3)[1] 6.251389 7.814728 11.344867qchisq(c(0.9, 0.95, 0.99), 11)[1] 17.27501 19.67514 24.72497qchisq(c(0.9, 0.95, 0.99), 12)[1] 18.54935 21.02607 26.21697qchisq(c(0.9, 0.95, 0.99), 13)[1] 19.81193 22.36203 27.68825A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable \(X\), with \(\mathbb{E}\left(X\right)=5\) and \(\mathsf{Var}\left(X\right)=0.04\). The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 5000 cm. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free.
What is the probability that for a chain of at least 5000 cm more than 1002 links are needed? For what percentage of the chains does the factory have to reimburse clients and provide free chains?
The sales department of the factory notices that it has to hand out a lot of free chains and asks the research lab what is wrong. After further investigations the research lab reports to the sales department that the expectation value 5 is incorrect, and that the correct value is 4.99 (cm). Do you think that it was necessary to report such a minor change of this value?
What commands would you have used if you answer the previous two questions using R?
Suppose you are provided two estimators \(S\) and \(T\) for a parameter \(\theta\). Let \(n\) be the sample size. Suppose \(\mathbb{E}\left(S\right)=\theta\) and \(\mathbb{E}\left(T\right)=\theta+3/\sqrt{n}\). Furthermore, \(\mathsf{Var}\left(S\right)=4\) and \(\mathsf{Var}\left(T\right)=40/n\).
Are \(S\) and \(T\) unbiased estimators of \(\theta\)? No explanation is needed.
Are \(S\) and \(T\) consistent estimators of \(\theta\)? Show why or why not.
Using your results, shed light on whether unbiasedness implies consistency and whether consistency implies unbiasedness.
Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During the past 44 weekdays, daily fees collected averaged $1264 with a standard deviation of $150.
What assumptions and conditions would you need to make in order to use these statistics for inference?
Describe the parameter of interest to the city. Be specific.
Compute a 95% confidence interval for the parameter in item 2.
The consultant who advised the city on this project predicted that parking revenues would average $1300 per day. On the basis of your confidence interval, do you think the consultant was correct? Why or why not?
Do you think the true value of the parameter in item 2 is in the confidence interval you computed in item 3?
Read the following R code and answer the following questions. In the code, you will see two digits preceding every line. These two digits represent the line number. Whenever possible, be very specific with your answers.
Let \(W_{1},\ldots,W_{15}\overset{\mathsf{IID}}{\sim}Exp\left(4\right)\). Note that \(\mathbb{E}\left(W_t\right)=1/4\) and \(\mathsf{Var}\left(W_t\right)=1/16\).
reps <- 10^4
wbar <- numeric(reps)
temp <- matrix(NA, nrow = reps, ncol = 2)
for (i in 1:reps)
{
w <- rexp(15, 4)
wbar[i] <- mean(w)
temp[i, ] <- c(mean(w)-1.96*sd(w)/sqrt(15), mean(w)+1.96*sd(w)/sqrt(15))
}
hist(wbar)
mean(wbar)
sd(wbar)
mean(temp[, 1] <= 1/4 & temp[, 2] >= 1/4)
mean(wbar-1/4 >= 0.1)Describe what specific statistical object line 8 is supposed to represent.
Describe what specific statistical object line 10 is supposed to represent.
What output do you expect line 11 to produce?
What output do you expect line 12 to produce?
What output do you expect line 13 to produce?
What output do you expect line 14 to produce?
With a perfectly balanced roulette wheel, in the long run red numbers should turn up 18 times in 38. To test its wheel, one casino records the results of 3800 plays. Here you will set up the statistical model and conduct a test of hypothesis. Let \(X_{t}=1\) when a red number turns up and \(X_t=0\) when a red number does not turn up.
Based on what is described, \(X_{1},\ldots,X_{3800}\) are IID draws from what distribution?
Under the assumption that in the long run red numbers should turn up 18 times in 38, what would be the mean \(\mu\) and variance \(\sigma^{2}\) of the distribution in the previous item?
If there are really too many or too few reds, what can you say about the mean \(\mu\) of the distribution in item 1?
You are asked to determine if the roulette wheel is working fine or the roulette wheel has a mechanical problem (too many or too few reds). Based on your previous answers, write down the null hypothesis (in mathematical form). Write down the alternative hypothesis (in mathematical form).
Suppose that you observe 1890 red numbers in 3800 plays. What would you conclude?
What R commands would you use to give a \(p\)-value?
Recall our example on I SEE THE MOUSE and the definition of \(Y\) and \(X_1\) for that example. You already know that the CEF of \(Y\) given \(X_1\) is different from the BLP of \(Y\) given \(X_1\).
lm(). After that, we test the null hypothesis that \(\beta_1=1\) against the alternative that \(\beta_1\neq 1\) at the 5% level. Below you find the R code to evaluate the properties of the procedure previously described. Discuss the Monte Carlo findings evaluating the procedure used to implement the test. [,1] [,2]
[1,] 1 0
[2,] 3 2
[3,] 3 1
[4,] 5 1get.test.stat <- function(n)
{
# Follow the joint distribution of I SEE THE MOUSE
data <- source[sample(nrow(source),size=n,replace=TRUE),]
temp <- lm(data[,1]~data[,2])
beta1hat <- coef(temp)[[2]]
se.beta1hat <- sqrt(diag(vcov(temp)))[[2]]
teststat <- (beta1hat-1)/se.beta1hat
return(teststat)
}
results <- replicate(10^4, get.test.stat(50))
mean(results > 1.96 | results < -1.96)[1] 0.0293[1] 0.0169Recall that the CEF can be equal to the BLP when you have a saturated model. This is the case where your \(Y\) could be continuous or discrete and the \(X\)’s all have to be binary or dummy variables.
Consider the workplace smoking ban dataset described here. You will try to make sense of what a linear probability model is all about, how to interpret the coefficients of a linear probability model, how they relate to summaries of the data, and what a saturated model really means. The latter is usually used to justify the use of linear regression in settings where \(Y\) is discrete.
Let \(Y\) be the variable smoker, \(X_1\) be the variable smkban, and \(X_2\) be the variable female. All variables are dummies.
Below you will find R code related to this exercise.
library(foreign)
smoking <- read.dta("https://www.princeton.edu/~mwatson/Stock-Watson_3u/Students/EE_Datasets/Smoking.dta")
lm(smoker ~ 1, data = smoking)
mean(smoking$smoker)
lm(smoker ~ smkban, data = smoking)
tapply(smoking$smoker, smoking$smkban, mean)
lm(smoker ~ female, data = smoking)
tapply(smoking$smoker, smoking$female, mean)
lm(smoker ~ smkban + female, data = smoking)
lm(smoker ~ smkban + female + smkban:female, data = smoking)
tapply(smoking$smoker, list(smoking$smkban, smoking$female), mean)Here is the R output:
Call:
lm(formula = smoker ~ 1, data = smoking)
Coefficients:
(Intercept)
0.2423 [1] 0.2423
Call:
lm(formula = smoker ~ smkban, data = smoking)
Coefficients:
(Intercept) smkban
0.28960 -0.07756 0 1
0.2895951 0.2120367
Call:
lm(formula = smoker ~ female, data = smoking)
Coefficients:
(Intercept) female
0.25762 -0.02718 0 1
0.2576209 0.2304417
Call:
lm(formula = smoker ~ smkban + female, data = smoking)
Coefficients:
(Intercept) smkban female
0.29877 -0.07538 -0.01864
Call:
lm(formula = smoker ~ smkban + female + smkban:female, data = smoking)
Coefficients:
(Intercept) smkban female smkban:female
0.31045 -0.09676 -0.04236 0.03965 0 1
0 0.3104493 0.2680895
1 0.2136860 0.2109795lm() command?black? How would you apply least squares so that you can recover \(\mathbb{E}\left(Y|X_1, X_2, X_3\right)\)?Below you will find R code estimating the model \[\mathrm{sat} = \beta_0+\beta_1 \mathrm{hsize}+\beta_2\mathrm{hsize}^2+\beta_3 \mathrm{female}+\beta_4\mathrm{black}+\beta_5\mathrm{female}\times\mathrm{black}+\varepsilon \] The description1 of the dataset could be found here.
Suppose that the conditions of Version 2 of the correctly specified linear regression model hold.
library(foreign)
gpa2 <- read.dta("http://fmwww.bc.edu/ec-p/data/wooldridge/gpa2.dta")
summary(gpa2$hsize) Min. 1st Qu. Median Mean 3rd Qu. Max.
0.03 1.65 2.51 2.80 3.68 9.40 result <- lm(sat ~ hsize + hsizesq + female + black + female:black, data=gpa2)
result
Call:
lm(formula = sat ~ hsize + hsizesq + female + black + female:black,
data = gpa2)
Coefficients:
(Intercept) hsize hsizesq female black
1028.097 19.297 -2.195 -45.091 -169.813
female:black
62.306 library(sandwich)
vcovHC(result) (Intercept) hsize hsizesq female black
(Intercept) 40.560861 -20.31005096 2.3502359605 -9.052500e+00 -11.59581004
hsize -20.310051 14.62999282 -1.9129657400 8.208576e-02 1.28851117
hsizesq 2.350236 -1.91296574 0.2764387415 -6.278926e-04 -0.08077265
female -9.052500 0.08208576 -0.0006278926 1.789903e+01 8.83660577
black -11.595810 1.28851117 -0.0807726456 8.836606e+00 242.04401938
female:black 8.248203 0.05581902 0.0326601173 -1.785279e+01 -241.54439938
female:black
(Intercept) 8.24820279
hsize 0.05581902
hsizesq 0.03266012
female -17.85279148
black -241.54439938
female:black 379.20953866library(car)Loading required package: carDatalinearHypothesis(result, c("hsize", "hsizesq"), rhs = 0, vcov = vcovHC(result), test = "Chisq")Linear hypothesis test
Hypothesis:
hsize = 0
hsizesq = 0
Model 1: restricted model
Model 2: sat ~ hsize + hsizesq + female + black + female:black
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 4133
2 4131 2 29.552 3.826e-07 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1linearHypothesis(result, c("hsize + 3 * hsizesq"), rhs = 0, vcov = vcovHC(result), test = "Chisq")Linear hypothesis test
Hypothesis:
hsize + 3 hsizesq = 0
Model 1: restricted model
Model 2: sat ~ hsize + hsizesq + female + black + female:black
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 4132
2 4131 1 28.653 8.656e-08 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Consider the following equation \(Y=\beta_0+\beta_1X+\varepsilon\). You have encountered different types of assumptions on \(\varepsilon\).
A \(\varepsilon\) independent of \(X\) and \(\mathbb{E}\left(\varepsilon\right)=0\)
B \(\mathbb{E}\left(\varepsilon|X\right)=0\)
C \(\mathbb{E}\left(\varepsilon\right)=0\) and \(\mathbb{E}\left(X\varepsilon\right)=0\)
So far, you have encountered multiple applications where the target parameter is an elasticity, or at least you have encountered applications where the variables may be logarithmically transformed. Another potential issue with logarithmic transformation is explored in this exercise. Recall that the elasticity of \(Y\) with respect to \(X\) is the ratio of the percentage difference of \(Y\) to the percentage difference in \(X\), i.e. \[\frac{\partial \log Y}{\partial \log X}=\frac{\dfrac{1}{Y}\partial Y}{\dfrac{1}{X}\partial X}=\frac{X}{Y}\cdot\frac{\partial Y}{\partial X}\]
Consider a stylized situation where \(Z=\alpha_{1}W^{\alpha_{2}}V^{\alpha_{3}}\eta\) and \(\mathbb{E}\left(\eta|W, V\right)=1\). The target parameter of interest is \(\alpha_{2}\).
A model inspired by the theory of demand and supply could be written as: \[\begin{eqnarray}q_t^d &=&\alpha_0+\alpha_1 p_t+u_t \\ q_t^s &=&\beta_0+\beta_1 p_t+v_t\\ q_t^d &=& q_t^s=q_t\end{eqnarray}\] Assume IID sampling over \(t\), \(\mathbb{E}\left(u_t\right)=0\), \(\mathbb{E}\left(v_t\right)=0\), \(\mathsf{Var}\left(u_t\right)=\sigma^2_u\), \(\mathsf{Var}\left(v_t\right)=\sigma^2_v\), and \(\mathsf{Cov}\left(u_t,v_t\right)=0\). The variables \((q_t, p_t)\) for all \(t\) are observable, but \((u_t,v_t)\) are unobservable.
In the previous exercise, you encountered a situation where we are unable to use an instrumental variables strategy. Consider a modification of the model as follows: \[\begin{eqnarray}q_t^d &=&\alpha_0+\alpha_1 p_t+u_t \\ q_t^s &=&\beta_0+\beta_1 p_t+\beta_2 s_t + v_t\\ q_t^d &=& q_t^s=q_t\end{eqnarray}\] Assume IID sampling over \(t\), \(\mathbb{E}\left(u_t\right)=0\), \(\mathbb{E}\left(v_t\right)=0\), \(\mathsf{Var}\left(u_t\right)=\sigma^2_u\), \(\mathsf{Var}\left(v_t\right)=\sigma^2_v\), and \(\mathsf{Cov}\left(u_t,v_t\right)=0\). The variables \((q_t, p_t, s_t)\) for all \(t\) are observable, but \((u_t,v_t)\) are unobservable. Finally, assume that \(\mathbb{E}\left(s_tu_t\right)=0\) and \(\mathbb{E}\left(s_tv_t\right)=0\).
During the 1880s, a cartel known as the Joint Executive Committee (JEC) controlled the rail transport of grain from the Midwest to eastern cities of the United States. The cartel preceded the Sherman Antiturst Act of 1890, and it legally operated to increase the price of grain above what would have been the competitive price. From time to time, cheating by members of the cartel brought about a temporary collapse of the collusive price-setting agreement. In This exercise, you will use variations in supply associated with the cartel’s collapses to estimate the elasticity of demand for rail transport of grain.
Data on weekly observations on the rail shipping price and other factors from 1880 to 1886 are loaded in R as the object JEC. The description of the variables may be found here.
A model for the demand for rail transport was specified as \[\log Q_t = \beta_0+\beta_1\log P_t + \beta_2 \mathrm{ice}_t + \sum_{j=1}^{12} \beta_{2+j} \mathrm{seas}_{jt}+\varepsilon_t\] and empirical results based on the estimation of such a model are presented below along with the R code and a summary of the output.
library(foreign)
JEC <- read.dta("https://www.princeton.edu/~mwatson/Stock-Watson_3u/Students/EE_Datasets/JEC.dta")
hist(JEC$price)
JEC$logQ <- log(JEC$quantity)
JEC$logP <- log(JEC$price)
result.1 <- lm(logQ ~ logP + ice, data = JEC)
result.2 <- lm(logQ ~ logP + ice + seas1 + seas2 + seas3 + seas4 + seas5 + seas6 + seas7 + seas8 + seas9 + seas10 + seas11 + seas12, data = JEC)
library(AER)
result.3 <- ivreg(logQ ~ logP + ice + seas1 + seas2 + seas3 + seas4 + seas5 + seas6 + seas7 + seas8 + seas9 + seas10 + seas11 + seas12 | cartel + ice + seas1 + seas2 + seas3 + seas4 + seas5 + seas6 + seas7 + seas8 + seas9 + seas10 + seas11 + seas12, data = JEC)
result.4 <- lm(logP ~ cartel + ice + seas1 + seas2 + seas3 + seas4 + seas5 + seas6 + seas7 + seas8 + seas9 + seas10 + seas11 + seas12, data = JEC)
library(tibble)
rows <- tribble(~label, ~result.1, ~result.2, ~result.3, ~result.4,
'Seasonal dummies', 'No', 'Yes', 'Yes', 'Yes',
'Sum of squared residuals', format(sum(residuals(result.1)^2), digits = 4), format(sum(residuals(result.2)^2), digits = 4), format(sum(residuals(result.3)^2), digits = 4), format(sum(residuals(result.4)^2), digits = 4))
attr(rows, 'position') <- c(7, 8)
library(modelsummary)
results <- list()
results[["result.1"]] <- result.1
results[["result.2"]] <- result.2
results[["result.3"]] <- result.3
results[["result.4"]] <- result.4
modelsummary(results, coef_map = c("logP", "ice", "cartel"), gof_map = c("nobs"), add_rows = rows)| result.1 | result.2 | result.3 | result.4 | |
|---|---|---|---|---|
| logP | −0.634 | −0.639 | −0.867 | |
| (0.082) | (0.082) | (0.132) | ||
| ice | 0.410 | 0.448 | 0.423 | 0.035 |
| (0.048) | (0.120) | (0.122) | (0.064) | |
| cartel | 0.358 | |||
| (0.025) | ||||
| Seasonal dummies | No | Yes | Yes | Yes |
| Sum of squared residuals | 54.54 | 49.4 | 50.6 | 13.99 |
| Num.Obs. | 328 | 328 | 328 | 328 |
result.2.cartel as part of an instrumental variables strategy. Consult the R code related to how the strategy was implemented.
cartel could be exogenous with respect to \(\varepsilon_t\).cartel would satisfy instrument relevance.cartel is a weak instrument? If you can, show your calculations and communicate your finding. If not, what information would you need to be able to determine whether cartel is a weak instrument.Of course, descriptions of the variables will be made available in the exam.↩︎
Note that this question could have been reframed in terms of a confidence interval.↩︎
Click the links to review price elasticity of demand and monopoly pricing.↩︎