# Heteroskedasticity-robust Standard Errors

The program below calculates the heteroskedasticity-robust standard errors.

In order to use the built-in heteroskedasticity tests in R, you’ll have to install the lmtest package.

install.packages("lmtest")


Then, assuming you saved regression results to myfit0, to run the Breusch-Pagan test,

library(lmtest)
bptest(myfit0)


## References

Problems, examples, and data from Wooldridge, Jeffrey M. Introductory Econometrics: A Modern Approach, 4th ed. South-Western Pub. 2009.

## Listing of the R program.

#
#  31 Oct 2012 D.S.Dixon
#
# Reproduces Example 8.2 in
#
#  Wooldridge, Jeffrey M. Introductory Econometrics: A Modern Approach. 4th ed. South-Western Pub. 2009.
#

library(car)

mydata <- read.table("GPA3.csv", sep=",",  header = TRUE, na.strings = ".")
cleandata <- na.omit(mydata[mydata$spring == 1,c("cumgpa","sat","hsperc","tothrs","female","black","white")]) myfit0 <- lm(cumgpa~sat + hsperc + tothrs + female + black + white, data=cleandata) output0 <- summary(myfit0) print(output0) # print the heteroskedasticity-robust standard errors print(sqrt(diag(hccm(myfit0, type = "hc0"))))  # Cost Function The first part of the program below is an example of regressing data, plotting the data points and the fit, and showing a few interesting points on the curve. The second part of the program is an example of computing the ATC, AVC and MC curves and plotting those. Note that MC could be computed discretely if the first derivative is not known. You would do this by calculating the differences and plotting them against the midpoints. mc <- TC[2:length(TC)] - TC[1:length(TC)-1] dQ <- 0.5 + Q[1:length(Q) - 1] plot(dQ, mc)  ## Listing of the R program. # # 28 Oct 2012 D.S.Dixon # library(car) source("RegReportLibrary.R") Q <- c(1:10) TC <- c(193,226,240,244,257,260,274,297,350,420) myfit0 <- lm(TC~Q+I(Q^2)+I(Q^3)) output0 <- summary(myfit0) print(output0) B0 <- myfit0$coefficients
B1 <- myfit0$coefficients B2 <- myfit0$coefficients
B3 <- myfit0$coefficients inflex.x <- - B2 / (3 * B3) inflex.y <- predict(myfit0, newdata=data.frame(Q=inflex.x)) inflex.slope = B1 + 2 * B2 * inflex.x - 3 * B3 * inflex.x^2 cat("inflection at Q =",inflex.x,", TC =", inflex.y,"\n") delta <- 0.01 * inflex.y QA = - (B2 + delta) / (3 * B3) QB = - (B2 - delta) / (3 * B3) cat("QA = ",QA,", QB = ",QB,"\n") png("ScatterPlusFit.png") plot(Q, TC) lines(Q,predict(myfit0)) points(c(inflex.x,inflex.x),c(inflex.y - 10,inflex.y + 10), pch=c(2,6)) lines(c(inflex.x,inflex.x),c(inflex.y - 10,inflex.y + 10)) points(c(QA,QB),predict(myfit0, data.frame(Q=c(QA,QB))), pch=4) legend(1,410,c("data","inflection","constant MC"),pch=c(1,2,4)) dev.off() PQ <- predict(myfit0) atc <- PQ / Q avc <- (PQ - B0) / Q # mc <- PQ[2:length(PQ)] - PQ[1:(length(PQ) - 1)] # dQ <- .5 + PQ[1:(length(PQ) - 1)] mc <- B1 + 2 * B2 * Q + 3 * B3 * Q^2 png("CostFunctions.png") plot(Q,atc,type="l",ylab="$",ylim=c(0,max(atc)),lty=1)
lines(Q,avc,lty=2)
# lines(dQ,mc,lty=3)
lines(Q,mc,lty=3)
legend(8,180,c("ATC","AVC","MC"),lty=c(1,2,3))
dev.off()


# Econometric equations in WordPress

Here’s a great way to get econometrics equations into your blog posts – you create a latex array environment and put each coefficient at the left of a new column. That way, you can get the standard errors to line up in the row below. It ends up looking like this: $\begin{array}{llllll} \widehat{log(wage)} = & 5.65\ + & 0.47\ educ\ + & .00078\ educ * pareduc\ + & .019\ exper\ + & .010\ tenure \\ & (.13) & (.010) & (.00021) & (.004) & (.003) \\ \multicolumn{3}{l}{n = 722, R^{2} = .169} \end{array}$

# More on creating interaction dummies

Here’s the dummies created all in one line (plus a line to add the column labels)

dummies <- 1*outer(apply(data.matrix(cleandata[,c("female","athlete")]) * c(2,1),1,sum),c(3:0),"==")
colnames(dummies)<-c("female_athlete","female_nonathlete","male_athlete","male_nonathlete")


# Interaction dummies

Here’s a couple of quick and clean ways to add new interaction dummy variables to your data
matrix so you can regress them the same way as other columns in the matrix

# make a copy of your data, in case you make a mistake

newdata <- cleandata
# create new columns with the new dummy variables
newdata[,"female_athlete"]    <- 1*(cleandata$female == 1 & cleandata$athlete == 1)
newdata[,"female_nonathlete"] <- 1*(cleandata$female == 1 & cleandata$athlete == 0)
newdata[,"male_athlete"]      <- 1*(cleandata$female == 0 & cleandata$athlete == 1)
newdata[,"male_nonathlete"]   <- 1*(cleandata$female == 0 & cleandata$athlete == 0)

# outer is a little more exotic: it gets the outer equals ("==") product of female and the vector(1,1,0)
#  and ANDs (&) it with the outer equals product of athlete and the vector(1,0,1)
mat <- 1*(outer(cleandata$female, c(1,1,0), "==") & outer(cleandata$athlete, c(1,0,1), "=="))
colnames(mat) <- c("female_athlete", "female_nonathlete", "male_athlete")
newdata <- cbind(cleandata, mat)

# this does the same thing, but for all four interaction dummies
mat <- 1*(outer(cleandata$female, c(1,1,0,0), "==") & outer(cleandata$athlete, c(1,0,1,0), "=="))
colnames(mat) <- c("female_athlete", "female_nonathlete", "male_athlete", "male_nonathlete")
newdata <- cbind(cleandata, mat)


# C5.2

Use the data in GPS2.RAW for this exercise.

(i) Using all 4,137 observationsl, estimate the equation $colgpa = \beta_{0} + \beta_{1} hsperc + \beta_{2} sat + u$

and report theresults in the standard form.

The regression results with the full data set are $\begin{array}{llll} \widehat{colgpa} = & 1.392 & -0.01352\ hsperc & +0.001476\ sat\\ & (0.0715) & (0.000549) & (6.531) \\ \multicolumn{4}{l}{n = 4137, R^{2} = 0.273, Adjusted R^{2} = 0.273}\\ \end{array}$

(ii) Reestimate the equation in part (i), using the first 2,070 observations.

The regression results with the first 2070 points is $\begin{array}{llll} \widehat{colgpa} = & 1.436 & -0.01275\ hsperc & +0.001468\ sat\\ & (0.0978) & (0.000719) & (8.858) \\ \multicolumn{4}{l}{n = 2070, R^{2} = 0.283, Adjusted R^{2} = 0.282}\\ \end{array}$

(iii) Find the ratio of the standard errors on hsperc from parts (i) and (ii). Compare this with the results from (5.10).

The ratio of the standard errors is $0.000549/0.000719 = 0.764$

Equation (5.10) $Se(\widehat{\beta_{j}}) \approx c_{j}/\sqrt{n}$

where $c_{j}$ is a positive constant that should be approximately the same for any n. That is, $Se(\widehat{\beta_{0}}) \sqrt{n_{0}} \approx Se(\widehat{\beta_{1}}) \sqrt{n_{1}}$

so that $Se(\widehat{\beta_{0}}) / Se(\widehat{\beta_{1}}) \approx \sqrt{n_{1}} / \sqrt{n_{0}}$

For parts (i) and (ii) $\sqrt{2070} / \sqrt{4137} = 0.707$

This is within about 7.5% of the ratio of the standard errors.

For the full dataset, the constant $c_{j}$ is 0.0353.

To examine convergence, compare this value for $c_{j}$ with values from subsets of different sizes. That is,
regress this model on two sets of 2068 (half the dataset), then on three sets of 1379, four sets of 1034, five sets of a 827,
and ten sets of 413. From each regression, multiply the standard error for $\beta_{1}$ by the square root of the number
of samples and plot these against the number of samples. The plot is shown here, and the sample R code is shown below. ## References

Problem and data from Wooldridge Introductory Econometrics: A Modern Approach, 4e.

## Listing of the R program.

#
#  10 Oct 2012 D.S.Dixon
#

# GPA2.DES
#
# sat       tothrs    colgpa    athlete   verbmath  hsize     hsrank    hsperc
# female    white     black     hsizesq
#
#   Obs:  4137
#
#   1. sat                      combined SAT score
#   2. tothrs                   total hours through fall semest
#   3. colgpa                   GPA after fall semester
#   4. athlete                  =1 if athlete
#   5. verbmath                 verbal/math SAT score
#   6. hsize                    size graduating class, 100s
#   7. hsrank                   rank in graduating class
#   8. hsperc                   high school percentile, from top
#   9. female                   =1 if female
#  10. white                    =1 if white
#  11. black                    =1 if black
#  12. hsizesq                  hsize^2
#

source("RegReportLibrary.R")

mydata <- read.table("gpa2.csv", sep=",",  header = TRUE, na.strings = ".")

myfit0 <- lm(colgpa~hsperc + sat, data=mydata)
output0 <- summary(myfit0)
print(output0)
wordpressFormat(myfit0)

nfull <- length(output0$residuals) sefull <- output0$coefficients[2,2]

myfit1 <- lm(colgpa~hsperc + sat, data=mydata[1:2070,])
output1 <- summary(myfit1)
print(output1)
wordpressFormat(myfit1)

separt <- output1$coefficients[2,2] npart <- length(output1$residuals)

cat("ratio of standard errors: ",(sefull/separt),"\n")
cat("ratio of square root observations ",((npart/nfull)^0.5),"\n")

N <- length(mydata$colgpa) N1 <- as.integer(N/10) N2 <- as.integer(N/5) N3 <- as.integer(N/4) N4 <- as.integer(N/3) N5 <- as.integer(N/2) mat <- matrix(nrow=25,ncol=2) row <- 1 sqrtn <- sqrt(N1) for(i in 0:9){ start <- N1 * i end <- N1 * (i + 1) - 1 output1 <- summary(lm(colgpa~hsperc + sat, data=mydata[start:end,])) mat[row,] <- c(N1,output1$coefficients[2,2]*sqrtn)
row <- row + 1
}

sqrtn <- sqrt(N2)
for(i in 0:4){
start <- N2 * i
end <- N2 * (i + 1) - 1
output1 <- summary(lm(colgpa~hsperc + sat, data=mydata[start:end,]))
mat[row,] <- c(N2,output1$coefficients[2,2]*sqrtn) row <- row + 1 } sqrtn <- sqrt(N3) for(i in 0:3){ start <- N3 * i end <- N3 * (i + 1) - 1 output1 <- summary(lm(colgpa~hsperc + sat, data=mydata[start:end,])) mat[row,] <- c(N3,output1$coefficients[2,2]*sqrtn)
row <- row + 1
}

sqrtn <- sqrt(N4)
for(i in 0:2){
start <- N4 * i
end <- N4 * (i + 1) - 1
output1 <- summary(lm(colgpa~hsperc + sat, data=mydata[start:end,]))
mat[row,] <- c(N4,output1$coefficients[2,2]*sqrtn) row <- row + 1 } sqrtn <- sqrt(N5) for(i in 0:1){ start <- N5 * i end <- N5 * (i + 1) - 1 output1 <- summary(lm(colgpa~hsperc + sat, data=mydata[start:end,])) mat[row,] <- c(N5,output1$coefficients[2,2]*sqrtn)
row <- row + 1
}

sqrtn <- sqrt(N)
mat[row,] <- c(N,output0$coefficients[2,2]*sqrtn) ## make a plot of constants png("f0d62b075365c15a9efcad7a9c046938.png") plot(mat, xlab="number of points per dataset", ylab="stderr") dev.off() cat("c = ",output0$coefficients[2,2]*sqrtn,"\n")


# Models with Interaction Terms

﻿#Interaction Terms#

Suppose you have an econometric model of the form

price = β0 + β1 sqrft + β2 bdrms + u

where

price = house price in 1000s of dollars
sqrft = size of house in square feet
bdrms = number of bedrooms

The estimator β2 is marginal home price per bedroom. Or is it? Ceteris paribus, a very small house with two small bedrooms may be more attractive than the same size house with three tiny bedrooms. In that case, the marginal price would be negative. The regression results are

Residuals:
Min       1Q   Median       3Q      Max
-127.627  -42.876   -7.051   32.589  229.003

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -19.31500   31.04662  -0.622    0.536
sqrft         0.12844    0.01382   9.291 1.39e-14 ***
bdrms        15.19819    9.48352   1.603    0.113
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 63.04 on 85 degrees of freedom
Multiple R-squared: 0.6319,     Adjusted R-squared: 0.6233
F-statistic: 72.96 on 2 and 85 DF,  p-value: < 2.2e-16


To take into account that increasing numbers of bedrooms may only be a good thing for a larger house, introduce a term that is the product of house size and number of bedrooms. Now the econometric model is

price = β0 + β1 sqrft + β2 bdrms + β3 sqrft * bdrms + u

The regression of this model is

Residuals:
Min       1Q   Median       3Q      Max
-146.858  -33.911   -8.586   27.900  212.400

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 181.69081   92.18885   1.971   0.0520 .
sqrft         0.03262    0.04364   0.747   0.4569
bdrms       -35.95534   24.01237  -1.497   0.1380
sqrft*bdrms   0.02345    0.01016   2.308   0.0234 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 61.5 on 84 degrees of freedom
Multiple R-squared: 0.6539,     Adjusted R-squared: 0.6415
F-statistic:  52.9 on 3 and 84 DF,  p-value: < 2.2e-16


Now the marginal price per bedroom is

marginal price per bedroom = -35.95534 + 0.02344804 * sqrft
(24.01237)     (0.01016)

For the values of sqrft in this dataset, the marginal price per bedroom ranges from -$8,500 to$55,020, with a mean of $11,260. Here’s a comparison of the two regressions Model 1 Model 2 Adj R-squared 0.62326 0.64152 F-statistic 72.964 52.896 (Intercept) -19.315 181.69 ….stderr ( 31.047 ) ( 92.189 )  sqrft 0.12844 0.032621 ….stderr. ( 0.013824 )  ( 0.043642 ) bdrms 15.198 -35.955 ….stderr . ( 9.4835 ) ( 24.012 ) sqrft*bdrms 0.023448 ….stderr.. ( 0.010157 )   10% significance  5% significance  1% significance ## Listing of the R programs. ### Program to print the regression results. # # 10 Oct 2012 D.S.Dixon # # HPRICE1.DES # # price assess bdrms lotsize sqrft colonial lprice lassess # llotsize lsqrft # # Obs: 88 # # 1. price house price,$1000s
#   2. assess                   assessed value, $1000s # 3. bdrms number of bedrooms # 4. lotsize size of lot in square feet # 5. sqrft size of house in square feet # 6. colonial =1 if home is colonial style # 7. lprice log(price) # 8. lassess log(assess # 9. llotsize log(lotsize) # 10. lsqrft log(sqrft) # mydata <- read.table("hprice1.csv", sep=",", header = TRUE, na.strings = ".") print(summary(mydata)) myfit0 <- lm(price~sqrft + bdrms, data=mydata) output0 <- summary(myfit0) print(output0) sqftrooms <- mydata$sqrft * mydata$bdrms myfit1 <- lm(price~sqrft + bdrms + sqftrooms, data=mydata) names(myfit1$coefficients) <- "sqrft*bdrms"
output1 <- summary(myfit1)
print(output1)

cat("marginal price per bedroom = ", output1$coefficients["bdrms",1], " + ", output1$coefficients["sqrft*bdrms",1], " * sqrft\n")
minmargin <- output1$coefficients["bdrms",1] + output1$coefficients["sqrft*bdrms",1] * min(mydata$sqrft) maxmargin <- output1$coefficients["bdrms",1]  + output1$coefficients["sqrft*bdrms",1] * max(mydata$sqrft)
meanmargin <- output1$coefficients["bdrms",1] + output1$coefficients["sqrft*bdrms",1] * mean(mydata$sqrft) cat("range = ", minmargin, " to ", maxmargin, ", with a mean of ", meanmargin,"\n")  ### Program to print the comparison table. # # 10 Oct 2012 D.S.Dixon # # This outputs a table in html form # # HPRICE1.DES # # price assess bdrms lotsize sqrft colonial lprice lassess # llotsize lsqrft # # Obs: 88 # # 1. price house price,$1000s
#   2. assess                   assessed value, $1000s # 3. bdrms number of bedrooms # 4. lotsize size of lot in square feet # 5. sqrft size of house in square feet # 6. colonial =1 if home is colonial style # 7. lprice log(price) # 8. lassess log(assess # 9. llotsize log(lotsize) # 10. lsqrft log(sqrft) # # # In order to use this, you have to install the xtable package. You do that by starting R and typing # # install.packages("xtable") # # or use the package tool in your GUI if you have one # library("xtable") mydata <- read.table("hprice1.csv", sep=",", header = TRUE, na.strings = ".") myfit0 <- lm(price~sqrft + bdrms, data=mydata) output0 <- summary(myfit0) sqftrooms <- mydata$sqrft * mydata$bdrms myfit1 <- lm(price~sqrft + bdrms + sqftrooms, data=mydata) names(myfit1$coefficients) <- "sqrft*bdrms"
output1 <- summary(myfit1)

mat <- matrix(nrow=10,ncol=2)
mat[1,1] <- format(output0$adj.r.squared , digits=5, width=0, justify="right") mat[2,1] <- format(output0$fstatistic     , digits=5, width=0, justify="right")

mat[3,1] <- format(output0$coefficients[1,1] , digits=5, width=0, justify="right") mat[5,1] <- format(output0$coefficients[2,1] , digits=5, width=0, justify="right")
mat[7,1] <- format(output0$coefficients[3,1] , digits=5, width=0, justify="right") sigstr <- c("","","","")[1 + (output0$coefficients[,4] <= 0.1) + (output0$coefficients[,4] <= 0.05) + (output0$coefficients[,4] <= 0.01)]

mat[4,1] <- paste("(", format(output0$coefficients[1,2], digits=5, width=0, justify="right"), ") ", sigstr) mat[6,1] <- paste("(", format(output0$coefficients[2,2], digits=5, width=0, justify="right"), ") ", sigstr)
mat[8,1] <- paste("(", format(output0$coefficients[3,2], digits=5, width=0, justify="right"), ") ", sigstr) mat[1,2] <- format(output1$adj.r.squared     , digits=5, width=0, justify="right")
mat[2,2] <- format(output1$fstatistic , digits=5, width=0, justify="right") mat[3,2] <- format(output1$coefficients[1,1] , digits=5, width=0, justify="right")
mat[5,2] <- format(output1$coefficients[2,1] , digits=5, width=0, justify="right") mat[7,2] <- format(output1$coefficients[3,1] , digits=5, width=0, justify="right")
mat[9,2] <- format(output1$coefficients[4,1] , digits=5, width=0, justify="right") sigstr <- c("","","","")[1 + (output1$coefficients[,4] <= 0.1) + (output1$coefficients[,4] <= 0.05) + (output1$coefficients[,4] <= 0.01)]

mat[4 ,2] <- paste("(", format(output1$coefficients[1,2], digits=5, width=0, justify="right"), ") ", sigstr) mat[6 ,2] <- paste("(", format(output1$coefficients[2,2], digits=5, width=0, justify="right"), ") ", sigstr)
mat[8 ,2] <- paste("(", format(output1$coefficients[3,2], digits=5, width=0, justify="right"), ") ", sigstr) mat[10,2] <- paste("(", format(output1$coefficients[4,2], digits=5, width=0, justify="right"), ") ", sigstr)

colnames(mat) <- c("Model 1","Model 2")

rnames <- c("Adj R-squared", "F-statistic",
rownames(output1$coefficients), "....stderr", rownames(output1$coefficients), "....stderr.",
rownames(output1$coefficients), "....stderr .", rownames(output1$coefficients), "....stderr..")

rownames(mat) <- rnames

newobject<-xtable(mat)
print.xtable(newobject, type="html")
# print(mat)


# Question 1

### a) Create a new error function based on u for which the expected value of the error function is zero.

Define a new error function v

v = u – α0

so that

E(v) = 0

### b) Express u in terms of your new error function and substitute it into the regression model. Show how this transformed regression model has a transformed constant term.

Rearranging

u = α0 + v

Now

y = (α0 + β0) + β1 x + β2 z + v

which has intercept term α0 + β0.

### c) Discuss the transformed constant term in terms of the original constant term and the transformed error function.

Considering the transformation as a function of v

U(v) = v + α0

then the intercept is merely transformed

U(β0) = β0 + α0

### d) Show how the transformed regression model has the same estimators for x and z as the untransformed regression model.

The estimators β1 and β2 are the same in both the untransformed and the transformed models. Thus, the estimators remain unbiased under a linear transform of the regression model

U(y) = y + α0

# Question 2

### a) Use this fact to show how the variance of your transformed error function in Question 1 relates to the variance of the orignal error term, Var(u).

The variance of the transformed error function is

Var(v) = Var(u – α0) = Var(u)

That is, the transformed error function has the same variance as the untransformed error function.

### b) Discuss your results from Questions 1 and 2a in terms of OLS bias and efficiency.

The results in Question 1 show that a regression model is not biased by a linear transformation. The result in Question 2a shows that a linear transformation does not affect variance, and therefore will not impact OLS efficiency.

# Question 3

### a) What do you expect to be the signs of the estimators?

Expected signs:

 income positive (college is a normal good) nchild positive or negative (complex interaction of other factors) pareduc positive (propensity for education) pctsh negative (opposite to income) pctfood negative (opposite to income) pctclo negative (opposite to income) pctoth positive (opposite to shelter, food, and clothing)

### b) Which variables would you consider examining for censoring?

Natural censoring is likely with income and number of children. This is further exacerbated with selection bias if, for example, the data only include home-owners, families with children, or families on assistance.

### c) Do you see any potential problems with colinearity?

The four budget terms are perfectly colinear.

### d) What counsel will you give your colleague about this model?

I will suggest that pctoth be dropped to eliminate the perfect colinearity.

### d) What counsel will you give your colleague about the data?

I will recommend plotting histograms of all the variables to get a sense of their distributions and look for censoring, selection bias, and possible binning, particularly of income.

# Question 4

### Listed below is the summary of a regression of cash dividends on after-tax profit. These are national aggregate statistics from the U.S. Department of Commerce for 1974 through 1986.

    Call:
lm(formula = dividends ~ profit, data = mydata)

Residuals:
Min     1Q Median     3Q    Max
-9332  -4706  -2600   5203  11056

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  373.3014  9530.3787   0.039  0.96946
profit         0.4200     0.1154   3.641  0.00388 **
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 6956 on 11 degrees of freedom
Multiple R-squared: 0.5465, Adjusted R-squared: 0.5052
F-statistic: 13.25 on 1 and 11 DF,  p-value: 0.003884


### a) Describe the goodness of fit and signficance of the model. Make your statement in terms of dividends and profit and not in terms of variables, parameters, etc. Use R2 and the F-test to support your statement.

This model is a measure of what fraction of after-tax profits companies distribute to shareholders in the form of dividends. Although profit is the principle component of dividends, dividends are not the only way in which profit is spent. The regression results indicate that there is strong, pausibly causal, relationship between profit and dividends, but that this relationship only tells about half the story. The strength of the relationship is indicated by the F-statistics of 13.25, which is significant at the one percent level. The extent to which this only tells part of the story is reflected in an R2 of about 0.55, meaning that nearly half of the variation in dividends and profit is not captured in this relationship. Given that many firms do not pay dividends, this is not surprising.

### b) Discuss the meanings of the estimators and their significance. Again, make your statements in terms of the model, then use regression results to support them.

Many stocks do not pay dividends, with profit going into stock value, so the fraction will be less than one, but should be considerably more than zero. A constant term would reflect dividends paid irrespective of profit and would be, in principle, zero. The regression results show that, on the average, between 1974 and 1986, U.S. corporations distributed 42% of their profits to shareholders. This result is signicant at the one-percent level. The intercept is not significantly different from zero, as anticipated.

### c) With a 95% confidence interval, assess the null hypothesis that there is no relationship between profit and dividends.

The null hypothesis that there is no relationship between profit and dividends is rejected if the F-statistic indicates a probability greater than five percent that all parameters are zero. The F-statistic indicates that the probability of all parameters being zero is less than 0.4%, for a confidence level of 99.6%, which certainly falls within the 95% confidence interval.

# C4.2

### (i) Using the same model as Problem 3.4, state and test the null hypothesis that the rank of law schools has no ceteris paribus effect on median starting salary.

The median starting salary for new law school graduates is determined by

log(salary) = β0 + β1 LSAT + β2 GPA + β3 log(libvol) + β4 log(cost) + β5 rank + u

where LSAT is the median LSAT score for the graduating class, GPA is the median college GPA for the class, libvol is the number of volumes in the law school library, cost is the annual cost of atttending law school, and rank is a law school ranking (with rank = 1 being the best).

The null hypothesis that rank has no effect on log(salary) is

H0: β5 = 0

The alternative is

H1: β5 ≠ 0

which is a two-tailed test.

The regression results are:

    Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  8.3432330  0.5325192  15.667  < 2e-16 ***
LSAT         0.0046964  0.0040105   1.171  0.24373
GPA          0.2475247  0.0900370   2.749  0.00683 **
llibvol      0.0949926  0.0332544   2.857  0.00499 **
lcost        0.0375544  0.0321061   1.170  0.24426
rank        -0.0033246  0.0003485  -9.541  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1124 on 130 degrees of freedom
(20 observations deleted due to missingness)
Multiple R-squared: 0.8417, Adjusted R-squared: 0.8356
F-statistic: 138.2 on 5 and 130 DF,  p-value: < 2.2e-16


The estimated model is

log(salary) = 8.34 + 0.00470 LSAT + 0.248 GPA + 0.0950 log(libvol) +
(0.53)    (0.0040)                 (0.090)         (0.033)
0.0376 log(cost) – 0.00332 rank
(0.032)                 (0.00035)
n = 136, Adj. R2 = 0.8356

Note from the regression results that the t-value is -9.541 which is highly significant. The critical value at one percent significance for 120 degrees of freedom is 2.617, and clearly |t|>>2.617. Thus we reject the null hypothesis.

### (ii) Are features of the incoming class of students — namely, LSAT and GPA — individually or jointly significant for explaining salary? (Be sure to account for missing data on LSAT and GPA.)

Based on the t-value in the regression results, the coefficient on LSAT is not significant but GPA is significant at the 1% level, which would dominate any joint significance.

unrestricted SSR = 1.6427, df = 130
restricted SSR = 1.8942, df = 132
F(2,130) = 9.95, 1% critical value for F(2,inf) = 4.61


Thus, at the 1% level, we reject the null hypothesis that LCAT and GPA are jointly insignificant. From the R linearHypothesis test,

    Linear hypothesis test

Hypothesis:
LSAT = 0
GPA = 0

Model 1: restricted model
Model 2: lsalary ~ LSAT + GPA + llibvol + lcost + rank

Res.Df    RSS Df Sum of Sq      F    Pr(>F)
1    132 1.8942
2    130 1.6427  2   0.25151 9.9517 9.518e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


Here, at the 0.1% level, we reject the null hypothesis that LCAT and GPA are jointly insignificant.

### (iii) Test whether incoming class size (clsize) or the size of the faculty (faculty) needs to be added to this equation; carry out a single test. (Be careful to account for missing data on clsize and faculty.)

To test the joint significance of these variables, first create a model that includes them. Then, do the test manually with an unrestricted regression of this model, then a restricted model that omits these two variables.

For the unrestricted model

unrestricted SSR = 1.5732, df = 123
restricted SSR = 1.5974, df = 125
F(2,123) = 0.9484, 10% critical value for F(2,120) = 2.35


Thus, at the 10% level, we fail to reject the null hypothesis that clsize and faculty are jointly insignificant. From the R linearHypothesis test,

    Linear hypothesis test

Hypothesis:
clsize = 0
faculty = 0

Model 1: restricted model
Model 2: lsalary ~ LSAT + GPA + llibvol + lcost + rank + clsize + faculty

Res.Df    RSS Df Sum of Sq      F Pr(>F)
1    125 1.5974
2    123 1.5732  2  0.024259 0.9484 0.3902


Here, at the 39% level, we fail to reject the null hypothesis that clsize and faculty are jointly insignificant.

### (iv) What factors might influence the rank of the law school that are not included in the salary regression?

There are salary differences based on gender and race/ethnicity across the labor market, and these may have some correlation with rank for a few law schools, but probably not over the entire data set. Individual programs are frequently ranked by the frequency and quality of publications by their faculty, so that is very likely to correlate with rank.

## References

Problem and data from Wooldridge Introductory Econometrics: A Modern Approach, 4e.

## Listing of the R program.

#
#  1 Oct 2012 D.S.Dixon
#
#
# LAWSCH85.DES
#
# rank      salary    cost      LSAT      GPA       libvol    faculty   age
# clsize    north     south     east      west      lsalary   studfac   top10
# r11_25    r26_40    r41_60    llibvol   lcost
#
#   Obs:   156
#
#   1. rank                     law school ranking
#   2. salary                   median starting salary
#   3. cost                     law school cost
#   4. LSAT                     median LSAT score
#   5. GPA                      median college GPA
#   6. libvol                   no. volumes in lib., 1000s
#   7. faculty                  no. of faculty
#   8. age                      age of law sch., years
#   9. clsize                   size of entering class
#  10. north                    =1 if law sch in north
#  11. south                    =1 if law sch in south
#  12. east                     =1 if law sch in east
#  13. west                     =1 if law sch in west
#  14. lsalary                  log(salary)
#  15. studfac                  student-faculty ratio
#  16. top10                    =1 if ranked in top 10
#  17. r11_25                   =1 if ranked 11-25
#  18. r26_40                   =1 if ranked 26-40
#  19. r41_60                   =1 if ranked 41-60
#  20. llibvol                  log(libvol)
#  21. lcost                    log(cost)
#

mydata <- read.table("LAWSCH85.csv", sep=",",  header = TRUE, na.strings = ".", )

print(summary(mydata))

# eliminate missing data in LSAT and GPA
cleandata<-mydata[!is.na(mydata$LSAT) & !is.na(mydata$GPA),]

# unrestricted model
myfit0<-lm(lsalary~LSAT+GPA+llibvol+lcost+rank, data=cleandata)
fitsum0 <- summary(myfit0)

SSR0 <- deviance(myfit0)
df0 <- df.residual(myfit0)

print(fitsum0)
print(paste("n = ",length(myfit0$residuals))) # restricted model (LSAT=0, GPA=0) myfit0Rest<-lm(lsalary~llibvol+lcost+rank, data=cleandata) fitsum0Rest <- summary(myfit0Rest) SSR0Rest <- deviance(myfit0Rest) df0Rest <- df.residual(myfit0Rest) print(fitsum0Rest) print(paste("n = ",length(myfit0Rest$residuals)))

print(paste("SSR0 = ",SSR0))
print(paste("SSR0Rest = ",SSR0Rest))
q <- df0Rest - df0
nk1 <- df0
print(paste("df0 = ",df0))
print(paste("df0Rest = ",df0Rest))
print(paste("q = ",q))
print(paste("nk1 = ",nk1))
F0 <- ((SSR0Rest - SSR0)/q)/(SSR0/nk1)
print("old school")
print(paste("F = ",F0))

library(car)

# joint significance of LSAT and GPA
hypmatrix <- rbind(c(0,1,0,0,0,0),c(0,0,1,0,0,0))
rhs <- c(0,0)

myfit<-lm(lsalary~LSAT+GPA+llibvol+lcost+rank, data=mydata)
hyp <- linearHypothesis(myfit, hypmatrix, rhs)

print("new school")
print(hyp)

# now eliminate missing data from clsize and faculty
cleanerdata<-cleandata[!is.na(cleandata$clsize) & !is.na(cleandata$faculty),]

print(summary(cleanerdata))

# unrestricted model
myfit1<-lm(lsalary~LSAT+GPA+llibvol+lcost+rank+clsize+faculty, data=cleanerdata)
fitsum1 <- summary(myfit1)
SSR1 <- deviance(myfit1)
df1 <- df.residual(myfit1)

# restricted model (clsize=0, faculty=0)
myfit1Rest<-lm(lsalary~LSAT+GPA+llibvol+lcost+rank, data=cleanerdata)
fitsum1Rest <- summary(myfit1Rest)
SSR1Rest <- deviance(myfit1Rest)
df1Rest <- df.residual(myfit1Rest)

print("Testing clsize and faculty")

print(paste("n = ",length(myfit1Rest\$residuals)))

print(paste("SSR1 = ",SSR1))
print(paste("SSR1Rest = ",SSR1Rest))
q <- df1Rest - df1
nk1 <- df1
print(paste("df1 = ",df1))
print(paste("df1Rest = ",df1Rest))
print(paste("q = ",q))
print(paste("nk1 = ",nk1))
F1 <- ((SSR1Rest - SSR1)/q)/(SSR1/nk1)
print("old school")
print(paste("F = ",F1))

library(car)

# joint significance of LSAT and GPA
hypmatrix <- rbind(c(0,0,0,0,0,0,1,0),c(0,0,0,0,0,0,0,1))
rhs <- c(0,0)

myfit<-lm(lsalary~LSAT+GPA+llibvol+lcost+rank+clsize+faculty, data=mydata)
hyp <- linearHypothesis(myfit, hypmatrix, rhs)

print("new school")
print(hyp)


# An Example of Hypothesis Testing Using R

## Installing the car library

If you haven’t done it already, you’ll have to install the car library, which includes, among many other useful things, the linearHypothesis command.

Start R and type

install.packages('car',.Library)


It asks you to select a mirror (server) from which to download. Doesn’t much matter, I suppose, but I picked Tennessee.

Once you do this, you never have to do it again, on the same computer, anyway.

## The example econometric model

The model is from the textbook, from section 4.4, starting on page 140.

log(wage) = β0 + β1 jc + β2 univ + β3 exper + u

where

jc = number of years attending a tyo-year college.
univ = number of years at a four-year college.
exper = months in the workforce.


The null hypothesis is that, in terms of impact on wages, a year at a two-year college is equivalent to a year at a four-year college. That is

H0: β1 = β2

The alternative is that a year at a two-year college earns a lower wage than a year at a four-year college:

H1: β1 < β2

## Setting up the hypothesis test

### Getting to know the data

Here is a summary of the data

> print(summary(mydata))
female          phsrank            BA               AA
Min.   :0.0000   Min.   : 0.00   Min.   :0.0000   Min.   :0.00000
1st Qu.:0.0000   1st Qu.:44.00   1st Qu.:0.0000   1st Qu.:0.00000
Median :1.0000   Median :50.00   Median :0.0000   Median :0.00000
Mean   :0.5196   Mean   :56.16   Mean   :0.3065   Mean   :0.04406
3rd Qu.:1.0000   3rd Qu.:76.00   3rd Qu.:1.0000   3rd Qu.:0.00000
Max.   :1.0000   Max.   :99.00   Max.   :1.0000   Max.   :1.00000
black            hispanic             id            exper
Min.   :0.00000   Min.   :0.00000   Min.   :   19   Min.   :  3.0
1st Qu.:0.00000   1st Qu.:0.00000   1st Qu.:19372   1st Qu.:104.0
Median :0.00000   Median :0.00000   Median :39301   Median :129.0
Mean   :0.09508   Mean   :0.04687   Mean   :40616   Mean   :122.4
3rd Qu.:0.00000   3rd Qu.:0.00000   3rd Qu.:58842   3rd Qu.:149.0
Max.   :1.00000   Max.   :1.00000   Max.   :89958   Max.   :166.0
jc              univ           lwage            stotal
Min.   :0.0000   Min.   :0.000   Min.   :0.5555   Min.   :-3.32480
1st Qu.:0.0000   1st Qu.:0.000   1st Qu.:1.9253   1st Qu.:-0.32734
Median :0.0000   Median :0.200   Median :2.2763   Median : 0.00000
Mean   :0.3389   Mean   :1.926   Mean   :2.2481   Mean   : 0.04748
3rd Qu.:0.0000   3rd Qu.:4.200   3rd Qu.:2.5969   3rd Qu.: 0.61079
Max.   :3.8333   Max.   :7.500   Max.   :3.9120   Max.   : 2.23537
smcity          medcity           submed            lgcity
Min.   :0.0000   Min.   :0.0000   Min.   :0.00000   Min.   :0.00000
1st Qu.:0.0000   1st Qu.:0.0000   1st Qu.:0.00000   1st Qu.:0.00000
Median :0.0000   Median :0.0000   Median :0.00000   Median :0.00000
Mean   :0.2854   Mean   :0.1174   Mean   :0.06861   Mean   :0.09448
3rd Qu.:1.0000   3rd Qu.:0.0000   3rd Qu.:0.00000   3rd Qu.:0.00000
Max.   :1.0000   Max.   :1.0000   Max.   :1.00000   Max.   :1.00000
sublg            vlgcity            subvlg              ne
Min.   :0.00000   Min.   :0.00000   Min.   :0.00000   Min.   :0.0000
1st Qu.:0.00000   1st Qu.:0.00000   1st Qu.:0.00000   1st Qu.:0.0000
Median :0.00000   Median :0.00000   Median :0.00000   Median :0.0000
Mean   :0.08709   Mean   :0.05855   Mean   :0.06358   Mean   :0.2107
3rd Qu.:0.00000   3rd Qu.:0.00000   3rd Qu.:0.00000   3rd Qu.:0.0000
Max.   :1.00000   Max.   :1.00000   Max.   :1.00000   Max.   :1.0000
nc             south           totcoll
Min.   :0.0000   Min.   :0.0000   Min.   : 0.000
1st Qu.:0.0000   1st Qu.:0.0000   1st Qu.: 0.000
Median :0.0000   Median :0.0000   Median : 1.507
Mean   :0.2988   Mean   :0.3271   Mean   : 2.265
3rd Qu.:1.0000   3rd Qu.:1.0000   3rd Qu.: 4.367
Max.   :1.0000   Max.   :1.0000   Max.   :10.067


The diligent student will also do histograms of the variables of interest.

### Getting to know the model

Here is a the regression model

> myfit <- lm(lwage~jc+univ+exper,data=mydata)


Here is a summary of the results from the regression

> print(summary(myfit))

Call:
lm(formula = lwage ~ jc + univ + exper, data = mydata)

Residuals:
Min       1Q   Median       3Q      Max
-2.10362 -0.28132  0.00551  0.28518  1.78167

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.4723256  0.0210602  69.910   <2e-16 ***
jc          0.0666967  0.0068288   9.767   <2e-16 ***
univ        0.0768763  0.0023087  33.298   <2e-16 ***
exper       0.0049442  0.0001575  31.397   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4301 on 6759 degrees of freedom
Multiple R-squared: 0.2224,	Adjusted R-squared: 0.2221
F-statistic: 644.5 on 3 and 6759 DF,  p-value: < 2.2e-16


### Creating a new variable

The book gives an example of using a t-test on a created variable

θ = β1 – β2

so that the hypotheses are

H0: θ = 0
H1: θ < 0

This is an instructive exercise, and I encourage students to try it. I may include it in this blog post sometime, but for now, I’ll skip over it.

### Setting up linearHypothesis

The general matrix form for a hypothesis test is

M β = Y

where

M = n x k matrix, n = number of hypotheses, k = number of estimators
β = k-length vector of estimators
Y = n-length vector of constraint values

For the null hypothesis

H0: β1 – β2 = 0

the hypothesis matrices are

M = [0, 1, -1, 0, 0]
β = [β0, β1, β2, β3]
Y = 

> hypomatrix <- c(0,1,-1,0)
> hypotest <- linearHypothesis(myfit,hypomatrix,0)
> print(hypotest)
Linear hypothesis test

Hypothesis:
jc - univ = 0

Model 1: restricted model
Model 2: lwage ~ jc + univ + exper

Res.Df    RSS Df Sum of Sq     F Pr(>F)
1   6760 1250.9
2   6759 1250.5  1   0.39853 2.154 0.1422


With F(1,6759) = 2.154, we fail to reject the null hypothesis at the 15% level. That is, a year at a two-year college is equivalent to a year at a four-year college as far as wage is concerned.

### Listing of the R program used to answer these questions.

#
#  28 Aug 2012 D.S.Dixon
#
# This is the dataset for the first EC 460 homework assignment
#

# from TWOYEAR.DES
#
# female    phsrank   BA        AA        black     hispanic  id        exper
# jc        univ      lwage     stotal    smcity    medcity   submed    lgcity
# sublg     vlgcity   subvlg    ne        nc        south     totcoll
#
#   Obs:      6763
#
#  1. female                            =1 if female
#  2. phsrank                           % high school rank; 100 = best
#  3. BA                                =1 if Bachelor's degree
#  4. AA                                =1 if Associate's degree
#  5. black                             =1 if African-American
#  6. hispanic                          =1 if Hispanic
#  7. id                                ID Number
#  8. exper                             total (actual) work experience
#  9. jc                                total 2-year credits
# 10. univ                              total 4-year credits
# 11. lwage                             log hourly wage
# 12. stotal                            total standardized test score
# 13. smcity                            =1 if small city, 1972
# 14. medcity                           =1 if med. city, 1972
# 15. submed                            =1 if suburb med. city, 1972
# 16. lgcity                            =1 if large city, 1972
# 17. sublg                             =1 if suburb large city, 1972
# 18. vlgcity                           =1 if very large city, 1972
# 19. subvlg                            =1 if sub. very lge. city, 1972
# 20. ne                                =1 if northeast
# 21. nc                                =1 if north central
# 22. south                             =1 if south
# 23. totcoll                           jc + univ
#

# fetch the library with linearyHypothesis
library(car)

# read the data
mydata <- read.table("twoyear.csv", sep=",",  header = TRUE, na.strings = ".")

# summarize the data
### @export "data-summary"
print(summary(mydata))
### @end

# run the regression model
### @export "regression-model"
myfit <- lm(lwage~jc+univ+exper,data=mydata)
### @end

# summarize the regression
### @export "regression-summary"
print(summary(myfit))
### @end

# summarize the regression
### @export "hypothesis-test"
hypomatrix <- c(0,1,-1,0)
hypotest <- linearHypothesis(myfit,hypomatrix,0)
print(hypotest)
### @end