Ashenfelter, Levine, Zimmerman Statistics and Econometrics Methods and Applications ,Solution Manual

Ashenfelter, Levine, Zimmerman Statistics and Econometrics Methods and Applications By Orley Ashenfelte, David J. Zimmerman, Phillip B. Levine

Email: Richard@qwconsultancy.com

SOLUTIONS MANUAL

To

Statistics and Econometrics: Theory and Applications

by Orley Ashenfelter, Phillip B. Levine, and David J. Zimmerman

CONTENTS INSTRUCTOR’S NOTES SOLUTIONS TO CHAPTER PROBLEMS CHAPTER 2 BASIC PROBABILITY THEORY CHAPTER 3 RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS CHAPTER 4 MATHEMATICS OF EXPECTATIONS CHAPTER 5 MULTIVARIATE DISTRIBUTIONS CHAPTER 6 SAMPLING AND SAMPLING DISTRIBUTIONS CHAPTER 7 ESTIMATION CHAPTER 8 INTERVAL ESTIMATION AND HYPOTHESIS TESTING CHAPTER 9 SIMPLE LINEAR REGRESSION CHAPTER 10 INFERENCES IN SIMPLE LINEAR REGRESSION CHAPTER 11 MULTIPLE REGRESSION CHAPTER 12 EXTENDING THE MULTIPLE REGRESSION MODEL CHAPTER 13 SPECIFICATION ERROR, MULTICOLLINEARITY, AND MEASUREMENT ERROR CHAPTER 14 HETEROSKEDASTICITY AND SERIAL CORRELATION CHAPTER 15 SIMULTANEOUS EQUATIONS CHAPTER 16 QUALITATIVE CHOICE AND LIMITED DEPENDENT VARIABLE MODELS CHAPTER 17 LINEAR TIME SERIES MODELS CHAPTER 18 PANEL DATA MODELS

INSTRUCTOR’S NOTES

Although the market for introductory econometrics books is rather well-developed, we had never been completely satisfied with the choices available for our classes. As empirical labor economists, we found that we were unable to locate books that were able to stimulate in students our interest in conducting econometric exercises. We hope to further that goal in this book by providing boxes in each chapters that contain summaries of published research that address important real world questions.

In our own research, we have relied heavily on econometric tools to address such questions, which are unable to be answered by other means. For instance, economic analysis is often capable of providing theoretical predictions regarding the direction of a particular relationship but it cannot provide a magnitude of that relationship. Is the demand curve steep or flat? To what extent do wages rise with education? For other questions, economics cannot even tell us the direction of the relationship. Does labor supply rise with an increase in the wage?

Other questions outside the purview of economic analysis still require estimates of the impact of one variable on another. Does student performance improve with smaller class sizes? It is the desire to answer questions like these that have always motivated us to conduct econometric analyses and we have written this book in an attempt to motivate a future generation.

But this focus on real world questions does not diminish the need to understand the technical intricacies of a sophisticated econometric analysis. The credibility of econometric research exclusively relies upon whether or not it was conducted in an appropriate manner.

Research that addresses an important topic but that does not display a mastery of the technical components of the analysis is of little value. Therefore, we also seek to present a rigorous introduction to econometric techniques.

In the development of this book, there were two fundamental decisions that we made regarding the coverage of material. First, we had to make a choice between a book that covered all potentially relevant material or a book that focused just on the topics we considered most important. Some books on the market treat many greater topics than we do and serve a useful purpose as a reference. We have chosen to limit the topics covered to those that we address in our courses. From our experiences in teaching econometrics, we have found that those books which provide encyclopedic coverage of the field typically overwhelm students. When they go to read the book, they find so many topics that were not covered in lecture, that they tend not to use the book at all. On the other hand, those books may be invaluable for the relatively few students who pursue further study in econometrics. The book that we have written has largely been developed straight from our lecture notes, so we teach virtually all the material in it. We sought to present the material in such a way that faculty can cover a particular topic and simply assign to students to “read all of chapter 12.” Similarly, problems at the end of chapters are designed more like problem sets rather than an exhaustive coverage of all the topics in the chapter.

The second important decision we had to make was where to begin and end our coverage of material. Some schools teach a one-semester course in econometrics that begins with basic

probability theory and ends with regression analysis. The curriculum at other institutions includes a two-semester sequence. Probability and statistics is covered in the first semester and regression analysis, including more sophisticated topics like time series and panel data methods, are covered in the second semester. We have chosen to write a book that is flexible enough to cover both types of instruction. For those who teach all the material in one semester, coverage could begin from the beginning of the book and end wherever the instructor makes it (perhaps through multiple regression in chapter 12). Alternatively, in a two-semester sequence, the first half of the book (through, say, chapters 9 or 10 on simple linear regression) could be covered in the first semester and the remainder in the second semester. Of course, the second semester would have to review the basics of probability and statistics, so we have provided Appendix A1 to facilitate that process. In some institutions, that first semester course would be taught by, say, the math department, which would not be likely to adopt this book. Nevertheless, instructors could assign Appendix A1 followed by the material in Chapters 9 through 18 for a self-contained treatment of the second semester’s material. For those who choose to use this book for both semesters of the two-semester sequence, students will appreciate the ability to buy a single book.

SOLUTIONS TO CHAPTER 2

1) Yes. P( A  B / B) =

P( A  B  B) P( A  B) = = P( A / B) P( B) P( B)

2) a) S={Heads, Tails}. The probability for each is .5 b) S={all of the cards in the deck}. The odds of even A occurring is ¼. c) The cast of two unbiased coins, where event A occurs if the roll is less than 5.

S={(1,1),(1,2),(1,3),...,(6,5),(6,6)}, i.e. list of all 36 possible combinations Pr(A)=Pr{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6/36 = 1/6

3) S{all of the cards in the deck} a) P(A)=12/52 b) P(A and B) = 2/52 c) P(A or B)=16/52 d) P(C or D)=1 e) P(C and D)=0 4)

Prob(Win/Three Point attempt) = .3 Prob(Win/Two Point attemp)=P(make two pointer and win in overtime) = .5*.5 = .25 (assuming independence) So, go for it! Try the three pointer.

5) P(good shot)=P( good / correct ) P(correct ) + P( good / correct ') P(correct ') = (1/3)*(1/4) +(1/5)(3/4) = 7/30=.2333 and P(correct ' good ) P( good ) P( good  correct ') P ( good / correct ') P (correct ') = = P( good ) P ( good )

P(correct '/ good ) =

=(1/5)(3/4)/.233 = .643

6) Let: H=hired I=Interviewed Given: P(H)=.04 so P(~H)=.96 P(I/H)=.98 P(I/~H)=.01 We want to know P(H/I).

P( H  I ) P( I  H ) = P( I ) P( I ) P( I / H ) P( H ) = P( I / H ) P( H ) + P ( I / H ') P( H ')

P( H / I ) =

= (.98)*(.04) / ( (.98*.04) + .01*.96)) = .803. Thus better than an 80% chance.

7) The possibilities are (let A, B, C indicate acceptance at that school): Outcome Prob note: (P(~A)=1-P(A)) Also, we assume independence.

ABC

.2*.4*.6=.048

i.e. got in to all three

ABC’ AB’C AB’C’ A’BC A’BC’ A’B’C A’B’C’

.2*.4*.4=.032 .2*.6*.6=.072 .2*.6*.4=.048 .8*.4*.6=.192 .8*.4*.4=.128 .8*.6*.6=.288 .8*.6*.4

i.e. got in to A and B but not C

=.192

i.e. didn’t get in to any

This is the whole distribution.

This is a simple way to solve the problem. The odds you get rejected at all three is .192 or 19.2%. The odds you get accepted to at least one is simply 1 minus the odds you don’t get into any – which is 1-.192=.808.

8)

Q1 = Use marijuana on the job ?

Q 2 = Wish you win the lottery ?

P( yes ) = .44 P(Q1) = 0.7 P(Q 2) = 0.3 Want : P( yes / Q1) P( yes ) = P ( yes  Q1) + P ( yes  Q 2) = P( yes / Q1) P (Q1) + P ( yes / Q 2) P (Q 2) we assume that P ( yes / Q 2) = 1, and solving for P ( yes / Q1), we get : P( yes ) − P ( yes / Q 2) P (Q 2) P( yes / Q1) = P(Q1) (.44) − (1)(0.3) = (0.7) = 0.2

9) Fred should switch doors after a booby prize has been shown behind one of the other doors. There are a few different ways to reason the answer. Suppose that Fred initially choose door #1. Then Fred has a 1/3 probabilit of being correct. This also means there is a 2/3 probability the car is behind either door #2 or door #3. Now suppose Monty opens

door #2 to reveal a booby prize. The question is what happens to the 1/3 probability previously associated with that door? The answer is that this probability is added to the probability that the car is behind door #3. The revealing of the booby prize adds no information about the likelihood of door#1 winning, essentially because Monty was at liberty to reveal only what was behind #2 or #3.

Alternatively, you may wish to think of it like this: because Fred has a 1/3 chance of winning when he first chooses door #1, if he never changes how can his chances improve? So, at the ‘final decision’ stage if there is only one other door, it must have a 2/3 chance of concealing the car. Thus if Fred changes doors he doubles his chance of winning.

(If you’re still not convinced, suppose there are 100 doors concealing a lone car. You choose a door, Monty opens 98 of the remaining 99 doors, and offers you the chance to change. Would you now?)

10) a) Male applied Arts Sciences Total

500 1500 2000

admitte d 50 450 500

Female applied 1500 500 2000

admitte d 180 200 380

b) Yes, once you break the admittance rates down by discipline type, women appear to be admitted at higher rates than men in each discipline type. c) The appearance of discrimination occurs because women disproportionately apply to

the disciplines with lower overall acceptance rates (the arts), while men apply to disciplines with higher overall acceptance rates (the sciences).

Given that women are more likely to be admitted than men conditional on their choice of discipline, it does not appear that there is sex discrimination in the way the university decides who to admit. However, the data could suggest discrimination through some other mechanism. For instance, we could imagine that the university encourages men to apply to the sciences and women to apply to the arts, which could lead to the disproportionate application numbers. We might also think that the university favors men by maintaining large departments for the male-dominated fields and small departments for female-dominated fields. This could lead to the overall lower acceptance rates in the arts than in the sciences.

Overall, then, we can not say if any form of discrimination is occurring, but there certainly does not appear to be evidence of discrimination in the admittance process.

SOLUTIONS TO CHAPTER 3 1) a) This is a proper pdf. All probabilities are positive and the the sum of all probabilities is one. b) This is not a proper pdf. f(2)<0.

2) f ( x) = cx3 =0

for x = 1, 2,3, 4 otherwise

a) What must c be in order for f(x) to be a proper probability function?

To be a proper p.d.f., f(x) must satisfy two criteria:

(1)

f ( x)  0  x

(2)

 f ( x)dx = 1 x

(1) If c  0 , then f ( x)  0 4

 cx = c(1) + c(2) + c(3) + c(4) = 1 3

3

3

3

3

i =1

c + 8c + 27c + 64c = 1 100c = 1 1 c= 100

(2)

b) What is the probability that 1  x  2 ?

2 1 P(1  x  2) =  ( )xi i =1 100 1 1 = (1)3 + (2)3 100 100 9 = 100

3) Suppose X is a continuous random variable having the p.d.f: f ( x) = cx3 =0

for 1  x  3 otherwise

a) What must c be in order for f(x) to be a proper p.d.f.

(1) c  0 for f ( x)  0 (see 3-2)

3

 f ( x)dx = 1 1

3

3 3 c c  1 =  cx 3 dx =  x 4  = 34 − 14  1 1  4 1 4 4 1 c= 4 4 = 20 3 −1

(2)

b) What is the probability that 1  x  2 ?

2

1 3 x dx 20 1

P(1  x  2) = 

2 1 x4| 20* 4 1 1  24 − 14  = 20* 4  = 0.1875

=

4) Suppose X is a continuous random variable having the p.d.f.: f ( x ) = x, 0  x 1 = 2 − x, 1  x  2 =0 otherwise

a) Graph the probability density function. f ( x) 1

0.5

0 0

1

2

b) Graph the cumulative probability distribution. F ( x) =  f ( x)dx 1

0.5

0 0

0.5

1

1.5

2

c) Calculate P(0.5  x  1.5) .

1.5

P(0.5  x  1.5) =  f ( x)dx 0.5 1

1.5

0.5

1

=  xdx +  (2 − x)dx 3 3 = + 8 8 3 = 4

5) .5

a) P( X  1/ 2) =  6 x − 6 x 2 dx = .5 0

x

b) F ( X ) =  6t − 6t 2 dt = 3x 2 − 2 x3 0

c) F (.5) = 3(.5)2 − 2(.5)3 = .5

6) a) Find the value of m such that P( X  m) = 0.5 (that is, find the median, m) m

 f ( x)dx = 0.5, 0  x  4 0

m

1 0.5 =  xdx 8 0 =

1 2m 1 m2 x | =  m2 − 02  = 16 0 16 16

m = 2 2  2.828

b) Find the value of m such that P( X  m) = 0.25 4

1 =  xdx 8 m

0.25 =

1 24 1 2 16 − m 2 x | =  4 − m 2  = 16 m 16 16

m = 2 3  3.464

7) P(insured ) = 0.5 and

P(no more than one lacks insurance) = P(exactly zero uninsured)+P(exactly one uninsured) P(exactly zero)= ( 300 ) (0.95)30 (1 − 0.95) 0 =0.2146 P(exactly one)= ( 301 ) (0.95) 29 (1 − 0.95)1=0.3389 P(no more than one lacks insurance)  0.2146+0.3389=0.5535

8) The distribution becomes approximately normal as n increases.

SOLUTIONS TO CHAPTER 4 1) a) E(attendance)=.2*500 + .6*1000 + .2*2000 = 1100 b) E(profits)=5*E(attendance) – 2000 = 3500. c) 5000 2) This is possible if the transferring student has an IQ score that is below the Williams average, but above the Amherst average.

3) Payoff $50 $x

P(Payoff) 2/6 4/6

2 4 = x* 6 6 100 = 4 x 25 = x 50*

4) The probability density function for the random variable X is given by f ( x) = 6 x(1 − x) =0

0  x 1 otherwise

a) Find the mean, E(X): v

E( X )

=  x  f ( x)dx u

1

1

=  x  6 x(1 − x)dx =  (6 x 2 − 6 x 3 )dx 0

0

1

3   =  2 x3 − x 4  2 0  1 = 2

b) Find the median, m:

m

1

 f ( x)dx = 2 0

m

=  (6 x 2 − 6 x 3 )dx 0

= 3 x 2 − 2 x 3 

m 0

= 3m 2 − 2m3 = m=

1 2

1 2

This answer makes sense, because if we notice that f(x) is symmetric (it is a parabola), then it follows that the mean equals the median. c) var( X ) =   X − E ( X )   f ( x)dx 2

= E ( X 2 ) − E ( X )2 1 note : E ( X ) = (from part a) 2 E ( X 2 ) =  x 2  f ( x)dx 1

=  (6 x3 − 6 x 4 )dx 0

1

6  3 =  x 4 − x5  5 0 2 3 6 3 = − = 2 5 10 2

var( X ) =

3 1 1 −  = 10  2  20

5) a) c

 f ( x)dx = 1 0

c

→ 0

2 2

xdx = 1 c

 2   2 2 →  x2  = c =1  4 0 4 → c = 2 /

b)

2 /

E ( X ) =  xf ( x)dx 0

2 /

  =  x3  6  0 2

=

4 3

var( X ) = E ( X 2 ) − E ( X ) 2 2 /

 4  =  x 2 f ( x)dx −    3  0

2

2  2  16 = 2 − 2 = 2 9    9

c) 3 4 1 3 P(  X  ) =  f ( x)dx 2 4 1 2

= 1/ 64

d) This represents a deviation from the mean of Thus, the upper bound is

3 which is 1.06 standard deviations.

6

1 = .89 1.062

6) Notice that:

var( X ) = E ( X 2 ) − E ( X ) 2 3 = E( X 2 ) − 4 E( X 2 ) = 7 So E (Y ) = 2*7 + 5* 2 + 4 = 28 7)

a) Give an upper bound on the probability that the number of people is outside the range [75, 125].

Use Chebyshev’s Inequality:

P ( X − E ( X )  k X ) 

1 k2

var( X ) =  X = 10

k= 

(125 − 100 )

( 10 )

=

5 10 std . deviations 2

1 1 2 = = 2 2 125 k  5 10     2 

b) Give the range for X for which the probability is at most 1/100 that the number of people who show up is outside that range.

1 1  k 2 100 100  k 2 k  10  k X = 10 10 Range = 100 − 10 10,100 + 10 10 

8) N=number of cars sold (a given) D=# of safety devices B=# of cars that blow up P=Probability any given car blows up  =Profit

Notice, each car is a Bernoulli random variable; i.e., the car either blows up or it doesn’t.

B is the sum of these independent random variables. The B follows a binomial distribution. N

Thus, P( B) =   P B (1 − P) N − B and E ( B) = NP, var( B) = NP(1 − P) B 



Next, notice that P depends on D; i.e., P(device 1 fails  device 2 fails  device 3 fails  ) Since the devices are independent, this implies that P= ( .1)  ( .1)  = ( .1) D Then, total profits are:  =(10000-5000)N - 100DN - 1,000,000B where the first part of the expression is the net

revenue excluding blow-ups and devices, the second part is the cost of D devices, and the final part is the cost of each blowup. a)

E(  ) = (10000-5000)N - 100DN-1,000,000E(B) = 5000N-100DN-1,000,000N(.1)D

To maximize our expected profit w.r.t. number of devices, we set the first derivative to zero, and solve for D: d = −100 N − N (.1D ) ln(.1) 1, 000, 000 dD

f ( x)  df ( x)  f ( x)  note : da =a ln(a)  dx d ( x)   

 D* = 4.36

Thus, we set D=4. (You can prove this by showing that E ( ) |D = 4  E ( ) |D =5 ) b) Since we now know that D=4 in order to maximize expected profit: E ( ) = 5000 N -100 ( 4 ) N - N (.1)

D

= 4500 N or $4500 per car var( ) = -1, 000, 0002  var( B) = -1, 000, 0002  NP(1- P) = -1, 000, 0002  N ( (0.1) 4 )(1- (0.1) 4 ) = 99,990, 000 N  20  4 0 4 20  (0.1 ) (1 − 0.1 ) = .002 = .2% 0  

c) P( B  1) = 1 − P(0) = 1 − 

d) I’ll leave the ethics up to you. We took the insurance company’s valuation into our calculations. We needn’t do it in that fashion.

SOLUTIONS TO CHAPTER 5 1) a) X 10 20

f(X) .4 .6

Y 10 20 30

f(Y) .5 .3 .2

b) E(X)=16, E(Y)=17 var(X)=24, var(Y)=61 c) No. f(X,Y) does not equal f(X)f(Y) for all values of X and Y. 2) a) Fill in the missing elements in the table.

Y

X

0

1

P(X)

-10

0.18

0

0.18

0

0

0.3

0.3

10

0.07

0.45

0.52

P(Y)

0.25

0.75

b) Compute E(X) and E(Y). Should you take the bet? Why? E( X ) E (Y )

= −10(0.18) + 0(0.3) + 10(0.52) = 3.4 = 0(0.25) + 1(0.75) = 0.75

E(X) > 0, so take the bet!

c) Your friend’s roommate is willing to tell you if she studied for the exam or not. But, he wants you to pay $1 for the information. Should you pay for the information and how should you use it?

−10(0.18) + 0(0) + 10(0.07) 0.25 = −4.4 −10(0) + 0(0.3) + 10(0.45) E ( X | Y = 1) = 0.75 =6

E ( X | Y = 0) =

This means your strategy is if your friend studied (Y=1), take the bet; if she didn’t (Y=0), don’t take the bet.

If Y = 0 , then your earnings are -1 (0 because you don’t take the bet, but –1 for the cost of the information).

If Y = 1 , then your earnings are 5 (6-1 for cost of information).

Then your expected earnings are (−1)(0.25) + (6)(0.75) = 4.25 , which is greater than E(X)— your expected winning if you hadn’t bought information.

So pay for the information. On average, you’ll be better off.

3) a) E[3X+2Y]=3*2+2*1=8 b) Var[X-Y]=4+2=6 c) Var[3X+2Y]=9*4+4*2=44 d) The answer in part a) doesn’t change. In part b) you’d subtract the covariance (i.e., 6-3=3) and in part d you’d add 3*2*3=18 (i.e. 44+18=62) (see question 5 below).

4) Let X, Y, and Z be random variables. Show Cov(X+Y,Z)=Cov(X,Z)+Cov(Y,Z).

Cov( X + Y , Z ) = E  ( X + Y ) Z  − E ( X + Y ) E ( Z ) = E  XZ + YZ  − E ( X + Y ) E ( Z ) = E ( XZ ) + E (YZ ) − E ( X + Y ) E ( Z ) = E ( XZ ) + E (YZ ) −  E ( X ) + E (Y )  E ( Z ) = E ( XZ ) + E (YZ ) − E ( X ) E ( Z ) − E (Y ) E ( Z ) =  E ( XZ ) − E ( X ) E ( Z )  +  E (YZ ) − E (Y ) E ( Z )  = cov( X , Z ) + cov(Y , Z )

5)

Cov(aX + b, cY + d ) = E  (aX + b − aE ( X ) − b)(cY + d − cE (Y ) − d  = E[ac( X − E ( X )(Y − E (Y )] = acCov( X , Y )

6) a) Show f ( x, y ) is a proper joint probability density function.

(1) By inspection, f ( x, y)  0  x, y

(2) 1 =   f ( x, y )dydx x y

6  4 =    x + y dydx 5 5  0 0 1 1

1

6  4 =   xy + y 2  dx 5 10  0 0 1

 4 6  =   x +  − ( 0 + 0 ) dx 5 10    0  1

1

6  4 =  x2 + x 10 10  0 =1

b) Find P ( x  .5 and y  .5) 0.5 0.5

4

6 

   5 x + 5 y dydx = ?

0 0

... =

1 8

c) Find the marginal distribution for the random variable X.

 f ( x, y)dy y

6  4 =   x + y dy 5 5   0 1

1

3  4 =  xy + y 2  5 0 5 4 3 = x + = f ( x) 5 5

d) Find the mean of X. 1

4

3 

 x  f ( x)dx =   5 x + 5 x dx x

2

0

1

3  4 =  x3 + x 2  10  0 15 4 3 = + 15 10 17 = 30

e) What is the probability that y  .5 given that x  .5 ?

P( y  0.5  x  0.5) P( x  0.5) 1 note : P( y  0.5  x  0.5) = (from part b) 8 0.5 3 4 P( x  0.5) =   x +  dx 5 5 0 

P ( y  0.5 | x  0.5) =

0.5

3  4 =  x2 + x 5 0 10 2 = 5 P ( y  0.5 | x  0.5) =

P( y  0.5  x  0.5) P( x  0.5)

1   5 8 = =  2  16   5

f) Are X and Y independent? Explain.

If X and Y were independent, then f ( x, y) = g ( x)  h( y) 4 5

In part c), we found g ( x) = x +

3 5

h( y ) =  f ( x, y )dx x

6  4 =   x + y dx 5 5  0 1

1

6  4 =  x 2 + xy  5 0 10 2 6 = + y 5 5

Multiplying g(x) and h(y), we have: 3  2 6  4 g ( x )  h( y ) =  x +    + y  5  5 5  5 8 24 6 18 = x+ xy + + y 25 25 25 25 24 9 12 3 =  x + y + xy +  55 5 5 5 2  ( 2 x + 3 y ) = f ( x, y ) 5

independent.

7)

Since f(x,y)  g(x)  h(y) , X and Y are not

1

f ( x) =  2 xdy = 2 xy 0 = 2 x 1

0

1

1

0

0

f ( y ) =  2 xdx = x 2 = 1 1

2 E ( X ) =  2 x dx = x 3 = 2 / 3 0 3 0 1

2

1

a,b)

E ( X 2 ) =  2 x3dx = 0

1

1 4 x = 1/ 2 2 0 1

y2 E (Y ) =  ydy = = 1/ 2 0 2 0 1

1

y3 E (Y ) =  y dy = = 1/ 3 0 3 0 1

2

2

Var ( X ) = E ( X 2 ) − E ( X ) 2 = 1/ 2 − (2 / 3) 2 = .055 Var (Y ) = .083 cov( X , Y ) = E ( XY ) − E ( X ) E (Y ) 1 1

c) E ( XY ) =   2 x 2 dydx =1/ 3 0 0

cov( X , Y ) = 1/ 3 − 1/ 3 = 0 1

d) E ( X / Y ] =  x 2 xdx =2 / 3 0

8) From the table we can calcuate: P( Hit1, Hit2 ) =.743 P( Hit1, Miss2 ) =.101 P( Hit1, either ) =.843 P( Miss1, either ) =.157 P( Miss1, Hit2 ) =.142 P( Miss1, Miss2 ) =.015 P(either , Hit2 ) =.885 P(either , Miss2 ) =.115

Then, the conditional probabilities are: P( Hit2 / Hit1 ) =

P( Hit2 , Hit1 ) =.743/.843 =.88 P( Hit1 )

P( Hit2 / Miss1 ) =

P( Hit2 , Miss1 ) =.142/.157 =.91 P( Miss1 )

Apparently, Larry Bird suffers from the Cold Hand!

9) No. Two variables may move together but not be causally related. Shoe size and earnings are likely related, but only because shoe size is associated with age. There is no causal connection.

SOLUTIONS TO CHAPTER 6 1) a) Calculate the sample mean and the sample standard deviation.

1 n  Xi n i =1 100 + 120 + 210 + 175 + 155 + 110 + 160 + 185 + 200 + 220 = 10 = 163.5

X =

1 n  ( X i − X )2 n − 1 i =1 16152.5 = 10 − 1  1794.72

Sx = 2

S x = S x  42.36 lbs 2

b) Suppose the scales underreport weight by ten pounds. What is the mean and the standard deviation of the correct weight (in pounds)?

1 n  ( X i + 10) n i =1 1 n 1 n 1 n 1 =  X i + 10 =  X i + (10n ) n i =1 n i =1 n i =1 n

X'=

= X + 10 = 163.5 + 10 = 173.5 2 1 n ( X i + 10) − X ')   n − 1 i =1 2 1 n ( X i + 10 ) − ( X + 10 )  =    n − 1 i =1 2 1 n =   X i + 10 − X − 10 n − 1 i =1  2 1 n  X i − X  =  n − 1 i =1 = 1794.72 (note : variance did not change)

S x '2 =

so

S x'  42.36 lbs

c) What is the mean and standard deviation in kilograms? (Note: 1 kilogram=2.2046 pounds)

X kg = =

1 n  Xi   n i =1  2.2046  1 1 n  Xi    2.2046  n i =1 

1 X 2.2046 = 74.16 kg

=

1 n  Xi  − X kg   n − 1 i =1  2.2046  = 369.26 kg

2

S xkg = 2

so S xkg  19.22 kg

2)  4 4! = 6 different samples  = 2 2!(4 − 2)!  

3) The value of X can take on different values, depending on the specific sample chosen from the population.

4)  X = 68 P(58  X  78)

 X2 = 25

→

X = 5

78 − 68   58 − 68 = P Z  5   5 = P(−2  Z  2) = (0.9772) − (1 − 0.9772) = 0.9544

5)  X = 700

 X = 40

750 − 700   P( X  750) = P  Z   40   = P ( Z  1.25 ) = 1 − 0.8944 = 0.1056

6) 8−7  6−7 P ( 6  X  8) = P  Z  25   25 1  1 = P−  Z   5 5  = 0.5793 − (1 − 0.5793) = 0.1586    6−7  8 − 7  P ( 6  X  8) = P  Z  25 25    100   100  1 1  = P− Z 5   5 10   10 = P ( −2  Z  2 ) = 0.9772 − (1 − 0.9772 ) = 0.9544

7) a) Sketch the probability density function of X. On the same graph, sketch the probability density function of the sample mean X from a sample of 16 observations.

b) Calculate P (180  X  220 ) 220 − 200   180 − 200 P (180  X  220 ) = P  X  20 20   = P ( −1  Z  1) = 0.6826

c) Calculate P (160  X  240 )

240 − 200   160 − 200 P (160  X  240 ) = P  X  20 20   = P ( −2  Z  2 ) = 0.9544

d) Calculate P (195  X  205) for n = 4,10,20,50,100. Graph the calculated probabilities against n.

1.2 1 0.8

Prob

   195 − 200 205 − 200   f (n) = P (195  X  205 ) = P  Z 20 20     n n    n n f (n) = P  − Z  4   4 f (4) = 0.3830 f (10) = 0.5704 f (20) = 0.7372 f (50) = 0.9216 f (100) = 0.9876

0.6 0.4 0.2 0 0

50

100

150

n

e) Calculate P (195  X  205) for  = 2,4,12,16,20,30. Graph the calculated probabilities against  .

1.2 1 0.8

Prob

   195 − 200 205 − 200   f ( ) = P (195  X  205 ) = P  Z       16 16   20   20 f ( ) = P  − Z     f (2)  1 f (4)  1 f (12) = 0.9050 f (16) = 0.7888 f (20) = 0.6826 f (30) = 0.4972

0.6 0.4 0.2 0 0

10

20

sigma

30

40

f) Suppose we wanted to conduct a survey. It is desired that we produce an interval estimate of the mean that is 5 from  with 99% confidence. Assume a value of  = 15.

How big should the sample be?

We want P ( −5  X −   5 ) = 0.99, given  X = 15       In general P  − Z 1 −   X  X −   Z  1 −   X  = 1 −  2    2   Thus,1 −  = 0.99   = 0.01 



= 0.005  1 −

2 15 15   So, P  − Z.995  X −   Z.995  = 0.99 n n  15  Z.995 =5 n 15    n =  Z.995  5 

 2

= 0.995

2

2

15   =  2.576   5  = 59.7 (round up)  60 observations

8) How often would the sample mean deviate from the population by more than: a) One standard deviation?

The probability that the sample would not deviate from the mean by more than one standard deviation is: P ( −1  Z  1) = 0.6826 So the probability that the sample would deviate from the mean by more than one standard deviation would be: 1 − P ( −1  Z  1) = 1 − 0.6826 = 0.3174

b) Two standard deviations?

1 − P ( −2  Z  2 ) = 1 − 0.9544 = 0.0456

c) Three standard deviations?

1 − P ( −3  Z  3) = 1 − 0.9974 = 0.0026

9) The total amount poured is: 36

a) E (Y ) =  xi = 36* 2 = 72 where xi is the amount in the ith cup. i =1

36

var(Y ) =  var( xi ) = 36*1/ 4 = 9 i =1

Using Chebyshev’s inequality, we have: P(Y  63) = P(Y  72 − 3*3)  P(Y  Y − 3 Y ) + P(Y  Y + 3 Y ) 

b) 1 1 = 4*36 144 63 1.75 − 2 P (Y  63) = P( X  ) = P( Z  ) = .0013 1 36 144

Notice that  X = 2 and varX =

1 = 1/ 9 32

SOLUTIONS TO CHAPTER 7 1) a) Find the maximum likelihood estimator for .

L( ) = f ( xi |  ) = f ( x1 |  )  f ( x2 |  )  f ( x3 |  )  ...  f ( xn |  ) =  e − x1  e − x2  ...  e − xn n

=  e − xi i =1

 n  ln L( ) = ln   e − xi   i =1 

(

n

=  ln  e − xi i =1 n

)

(

=  ln  + ln e − xi i =1

)

n

=  ( ln  −  xi ) i =1

max w.r.t.  : L( ) 

n 1  =   − xi  = 0   i =1  n

n

1

 −  x = 0 i =1

i =1

i

n

n =  xi i =1

ˆ = n

n

x i =1

i

b) Suppose a random sample of 5 observations on x yields data: 1, .8, 1.3, 1.6, 2. What is the maximum likelihood estimate for ? ˆ = n

n

 xi

;n = 5

i =1

n

 x = 1 + 0.8 + 1.3 + 1.6 + 2 = 6.7 i =1

so

i

ˆ =

( 5)

( 6.7 )

 0.746

c) Is the maximum likelihood estimator always the best choice for an estimator?

Not always—this is why we have (and learn!) many different methods of estimation. We must choose our best estimator based upon important criteria—unbiasedness and efficiency—and the potential tradeoff we may be forced to make between the two. d) Outline how you would find a method of moments estimator.

1) Choose a moment:e.g.,the sample mean, E ( X ) 2) Define E ( X ) as a function of your parameter that you are trying to estimate: E ( X ) = f ( ) 3) Solve for your parameter:  = f -1 ( E ( X ) ) 4) Substitute the population statistic with the sample statistic: ˆ = f -1 ( X )

2) Suppose f ( x) = ( + 1) x for 0  x  1 a) Find a method of moments estimator for .

1

E ( X ) =   x  ( + 1) x  dx 0

1

=  ( + 1) x +1  dx 0

( + 1)   + 2 1 x ( + 2 )   0 ( + 1)   + 2  + 2  ( + 1) = 1 −0  = ( + 2 )  ( + 2 )

=

E ( X )  ( + 2 ) = ( + 1)

 E( X ) + 2E( X ) =  + 1  E( X ) −  = 1 − 2E( X )   E ( X ) − 1 = 1 − 2 E ( X ) =

1 − 2E( X ) E( X ) −1



ˆ =

1− 2X X −1

b) Suppose a random sample of 5 observations on x yields data: 1,4,2,7,3. What is the method of moments estimate for ?

1− 2X ; n=5 X −1 1 n 1+ 4 + 2 + 7 + 3 X =  Xi = = 3.4 n i =1 5

ˆ =

so

ˆ =

1 − 2 ( 3.4 )

( 3.4 ) − 1

 2.4167

3) The MLE estimator will be the sample mean.

4) a) Find a method of moments estimator for b.  1 =   x   dx b 0 b

E( X )

b

1  =  x2   2b  0 1 b = b2 = = E ( X ) 2b 2 b = 2  E( X ) bˆ = 2 X

b) Find a maximum likelihood estimator for b.

L(b) = f ( x1 | b)  f ( x2 | b)  ...  f ( xn | b) n

1 i =1 b 1 = n b

=

In order to maximize L(b) =

1 , pick b to be as small as possible, subject to our constraint, b  x. bn

 bˆ = max{all values of x}

MSE = E[(ˆ −  ) 2 ] = E[(ˆ −  + E (ˆ) − E (ˆ)) 2 ]

5)

(

)

= E  ((ˆ − E (ˆ)) + ( E (ˆ) −  )  = var(ˆ) + Bias 2 + 2 E ((ˆ − E (ˆ))( E (ˆ) −  ))   and, notice that the last part of the equation 2 E ((ˆ − E (ˆ))( E (ˆ) −  )) is zero.

2

6) Not always. Sometimes we may make tradeoffs between bias and efficiency.

7) A more efficient estimator has a smaller variance. Thus, between a) and b), the one with a smaller variance will be more efficient.

1  4 x + 12 x2 + 3x3  var  1 = 2 var ( 4 x1 + 12 x2 + 3 x3 )  19   19 1 =  var ( 4 x1 ) + var (12 x2 ) + var ( 3 x3 )  361  1  42 var ( x1 ) + 122 var ( x2 ) + 32 var ( x3 )  = 361  1 = (16 x2 + 144 x2 + 9 x2 ) 361 169 2 = x 361 x +x +x  1 var  1 2 3  = 2 var ( x1 + x2 + x3 ) 3   3 1 =  var ( x1 ) + var ( x2 ) + var ( x3 )  9 1 = ( x2 +  x2 +  x2 ) 9 3 1 =  x2 =  x2 9 3

Since

169 2 1 2  x   x  b) is more efficient than a) 361 3

8) a) E (Y ) =

1 n 1 n E ( Y ) =  =  i n n i =1 i =1

b) Yes, ̂ is a better estimator than Y since for the first two options the estimator is always correct and for other values of Y it is no worse. I.e. ̂ has a smaller sampling variance.

SOLUTIONS TO CHAPTER 8 1) False. The significance level gives the probability of committing a type I error. That is, the probability of concluding the alternative when the null is true.

2) If we drew repeated samples and formed such intervals the population mean would be found within the bounds 95% of the time.

3) a) With a sample of 10 draws from the distribution, what is the cutoff for a test with probability of Type I error of at most 0.01?

P (Type I error ) = P ( choose H1 | H 0 is true distribution )     c − 1  = 0.01 = PZ   3    n     3 = P  2.326 + 1  c  = 0.01 10   So c  2.274

b) What is the probability of a Type II error?

P (Type II error ) = P ( choose H 0 | H1 is true distribution )     c − 2 =? = PZ   3    n      (2.274) − 2  = PZ    3   10   = P ( Z  0.500 ) = 0.6915

c) Repeat parts a) and b) using a sample size of 100 draws. What does this tell you about keeping the probability of a Type I error constant as the sample size becomes larger?   3 P (Type I error ) = P  2.326 + 1  c  = 0.01 100   So c  1.403     (1.403) − 2  P (Type II error ) = P  Z    3   100   = P ( Z  −0.29 ) = 0.6141

So, while keeping the probability of a Type I error constant, if we increase the sample size, we can decrease the probability of a Type II error.

4)

Examiner's Decision

Innocent Guilty

Suspect's True Status Innocent Guilty 131 15 9 125

a) What would be the numerical value of the Type I and Type II errors in this context?

Since we “normally” think of suspects as innocent until proven guilty, “Innocent” becomes the null hypothesis, while “Guilty” becomes the alternative hypothesis.

Type I error = P ( decision = guilty | true status = innocent ) 9  0.0643 140 Type II error = P ( decision = innocent | true status = guilty ) =

=

15  0.1071 140

b) In a judicial setting do we typically think of Type I or Type II errors carrying equal weight?

In a judicial setting, we are probably more likely concerned of convicting a innocent suspect than releasing a guilty one, thus, we place more weight on Type I errors.

5) a) What is the 95% confidence interval for the (true) mean per student expenditure?       P  − Z 1 −   X + X    Z 1 −   X + X  = 1 −   2   2  where,  = 0.05, X = 200,  = 150, and n = 225 150 150   P  − Z 0.975  + 200    Z 0.975  + 200  = 0.95 225 225   P  −1.96 10 + 200    1.96 10 + 200  = 0.95 P 180.4    219.6 = 0.95

b) Form a lower bound for the (true) mean per student expenditure. Use a 99% confidence interval.

P  − Z (1 −  )  X + X    = 1 −  where  = 0.01, X = 200,  = 150, and n = 225 150   P  −2.326  + 200    = 0.99 225   P 176.74    = 0.99

c) Use a decision rule to test the hypothesis that the (true) mean per student expenditure is less than $225. Use a 1% significance level.

H 0 :  = 225 H1 :   225

Conclude H1 when :

X −



 − Z (1 −  )

n X −

 n

=

( 200 ) − ( 225)

 150     225  = −2.5  − Z 0.99 = −2.326

So, conclude H1

d) Suppose that somehow you knew that the true mean per student expenditure was $215.

What (approximately) is the probability that you would see a sample mean of $200 or lower if this was the case?    200 − ( 215 )   P ( X  200 ) = PZ  150     225   = P ( Z  −1.5 ) = 0.0668

6) a) If the manufacturer’s claim is true, what is the probability of a sample mean less than or equal to the one you observed?

1 n 29 + 29.5 + 28.8 + 28.7 + 31 + 26 + 28 + 30.5 + 30.2 + 28.3 Xi =  n i =1 10 = 29

X =

 X2 =

2 n

=

32 9 = = 0.9 10 10

 X =  X2 = 0.9  0.9487  29 −   P ( X  29 ) = P  Z   X   29 − 30   = PZ   0.9487   = P ( Z  −1.0540 ) = 0.1469

b) After you turn in your report, you regret not having been able to make more confident statements about the validity of the manufacturer’s claim. You decide that, in future test of other cars, if the sample mean is at least one mile per gallon less than the manufacturer’s claim, you would like to reject the claim. How many tankfuls must you sample in a future test to make the probability less than 0.05 that you reject a claim when in fact it is true? Continue to assume that miles per gallon on a tank of gas is independently, normally distributed with a standard deviation of 3.

We want P ( X −   −1)  0.05 

 

 X −  −1  P    0.05 X   X     1  n  = PZ  − PZ  −   0.05          n  n  −1.645 3 n  25

−

7) a)     ˆ (1 − ˆ )    ˆ (1 − ˆ )  For proportions: P ˆ − Z 1 −     ˆ + Z 1 −   = 1− n n  2  2   Thus,1 −  = 0.90   = 0.10 

 2

= 0.05  1 −

 2

= 0.90

 .45(1 − .45) .45(1 − .45)  So, P  .45 − 1.645    .45 + 1.645  = 0.90 1000 1000    [.42,.47]

b)   .45(1 − .45) P  .45 − 1.645    .  = 0.95 1000    [.42, ]

8) a) The papers should be the same, on average, in every aspect except for information on the identity of the author. Otherwise, the estimated “treatment effects” would not be credible. b)

10.60 − 14.1 1.12 + 1.42

= 1.96 suggesting we can reject the hypothesis that the acceptance

rates are the same at the 5% significance level. c) 5%

10.0 − 11.2

d)

= .26 so we cannot reject the null that acceptance rates are the same

3.02 + 3.52

for women.

11 − 15 1.22 + 1.52

= 2.08 so we can reject the null that acceptance rates are the same for men.

e) The results do not suggest discrimination against female authors.

9)

We want P ( −5  X −   5 ) = 0.99, given  X = 15       In general P  − Z 1 −   X  X −   Z  1 −   X  = 1 −  2   2    Thus,1 −  = 0.99   = 0.01 



= 0.005  1 −

2 15 15   So, P  − Z.995  X −   Z.995  = 0.99 n n  15  Z.995 =5 n 15    n =  Z.995  5 

 2

= 0.995

2

2

15   =  2.576   5  = 59.7 (round up)  60 observations

10) a) Let C0 =7384 and E0 =6889. Also, var(C0 ) = Test: H 0 : C0 −  E0 = 0

32762 31462 and var( E0 ) = 1095 939

var(C0 − E0 ) = var(C0 ) + var( E0 ) Notice that

32762 31462 →  C0 − E0 = + = 143.1 939 1095 C − E0 7384 − 6889 = = 3.45 And t = 0  C0 − E0 143.1 Thus, we can reject the null. This suggests famililies were not randomly assigned to each group. b) Let C1 =8541 and E1 =7604

32492 36962 + = 165.7 1 1 939 1095 C − E1 8541 − 7604 t= 1 = = 5.66  C1 − E1 165.7

 C −E =

Again, we reject the null that the means are the same. The probability of getting a tstatistic this large if the null was true is essentially zero. Ideally, we’d probably want to use a one-tailed test because our alternative hypothesis is that the controls earn more. c) C1 − E1 is likely to be a biased estimate of the program effect since the controls earned more even in the pre-expermental period. We might compare the difference in earnings before and after. The is called a difference-in-difference estimator.

t=

(C0 − E0 ) − (C1 − E1 )

 (C − E ) −(C − E ) 0

0

1

and

1

 (C − E )−(C − E ) = 0

0

1

1

var(C0 ) + var(C1 ) + var( E0 ) + var( E1 ) − 2 cov(C0 , C1 ) − 2 cov( E0 , E1 ) Thus, t =

442 = 3.03 145.4

Again, we can reject the null.

11) a) Let there be n1 trainees at cost a+b each. Let there be n2 controls at cost a each.

Let P =

n1 n where n = n1 + n2 . 1 − P = 2 n n

var( X 1 − X 2 ) =

2 n1

+

2 n2

  D2

2 Thus,

2

 D2 = n + n

1 − P which is the sampling variance of earnings difference.

P

=



2

NP(1 − P)

b) Total costs are n1 (a + b) + n2 a . If b=0 then the experiment simply costs na and we see costs is not a function of P. To minimize sampling variance,

2

NP (1 − P ) given n, you should maximize over P. i.e. make P(1-P) as large as possible. 

for a

dP(1 − P) = 1 − 2 P = 0  P = 1/ 2 . dP

c) If b does not equal zero we must minimize the sampling variance subject to the budget constraint, R, where R  n1 (a + b) + n2 a . The can be written as R n= . Substituting this into the sampling variance and minimizing with a + Pb respect to P gives us: (1 − 2 P)(a + bP) − b( P − P 2 ) =0  (a + Pb) which yields (a + bP) 2 Min Rp(1 − P) P  a − 2 Pa − P 2b = 0 Applying the quadratic formula gives us: b b b P=− + 1+ a a a 2

Optimal proportion trainees

Cost ratio b/a

SOLUTIONS TO CHAPTER 9

1) In the derivation of the OLS estimator, we found that: n

ˆ1 =

n

n

n Yi X i − Yi  X i i =1

i =1

i =1

  n X −   X i  i =1  i =1  n

n

2

2 i

Substituting in from the values given yields:

ˆ1 =

500 18, 000 − 1400 3000 4,800, 000 = = 0.2 500*66, 000 − 30002 24, 000, 000

The intercept is obtained from the formula  0 = Y −  1 X . To get the sample means, we have: X=

1 500 1 Xi = 3000 = 6  500 i =1 500

and

Y =

1 500 1 Yi = 1400 = 2.8  500 i =1 500

Therefore, ˆ0 = Y − ˆ1 X = 2.8 − 0.2 6 = 1.6

2A)

14

suicide

13

12

11

10 3

4

5

6 unemrate

7

8

9

B) year 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 means sums

Y X Y - Y(bar) suicide unemrate 10.7 3.6 -1.31 11.1 3.5 -0.91 11.5 4.9 -0.51 11.7 5.9 -0.31 12 5.6 -0.01 12 4.9 -0.01 12.1 5.6 0.09 12.7 8.5 0.69 12.5 7.7 0.49 13.3 7 1.29 12.5 6 0.49 12.01

X - X(bar)(X - X(bar))^2 (Y - Y(bar)) * (y - y(bar))^2 (X - X(bar)) TSS -2.15 4.62 2.82 1.72 -2.25 5.06 2.05 0.83 -0.85 0.72 0.43 0.26 0.15 0.02 -0.05 0.10 -0.15 0.02 0.00 0.00 -0.85 0.72 0.01 0.00 -0.15 0.02 -0.01 0.01 2.75 7.56 1.90 0.48 1.95 3.80 0.96 0.24 1.25 1.56 1.61 1.66 0.25 0.06 0.12 0.24

Y(hat) Y - Y(hat) 11.14 11.10 11.66 12.07 11.95 11.66 11.95 13.13 12.80 12.52 12.11

-0.44 0.00 -0.16 -0.37 0.05 0.34 0.15 -0.43 -0.30 0.78 0.39

5.75 24.19

9.84

Beta1(hat) = Beta0(hat) =

0.407 9.672

5.53

0.01

. regress suicide unemrate Source | SS df MS ---------+-----------------------------Model | 3.99946502 1 3.99946502 Residual | 1.52962611 9 .169958457 ---------+-----------------------------Total | 5.52909113 10 .552909113

Number of obs = F( 1, 9) = Prob > F = R-squared = Adj R-squared = Root MSE =

11 23.53 0.0009 0.7233 0.6926 .41226

-----------------------------------------------------------------------------suicide | Coef.

Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------unemrate | .4066376 .0838258 4.851 0.001 .2170103 .5962648 _cons | 9.672773 .4973995 19.447 0.000 8.547577 10.79797 ------------------------------------------------------------------------------

C) If suicide rates were reported 1,000 in the population rather than per 100,000 in the population, the coefficient would decrease by a multiple of 100.

D) see the above table.

E) The average estimated error is about zero because the positive and negative values cancel each other out. This is a result of the fact that the regression line goes right through the middle of the scatter plot.

F) Given an unemployment rate of 5.8 percent in 1979, this model would predict a suicide rate of 9.673 + 0.407*5.8 = 12.03 per 100,000 population.

3A and B) Your answers to this question will depend specifically on the time period used in constructing your dataset.

4A) . regress unemp inflat if year <= 1967 Source | SS df MS -------------+-----------------------------Model | .774129073 1 .774129073 Residual | 11.5551564 12 .962929701 -------------+-----------------------------Total | 12.3292855 13 .948406575

Number of obs = 14 F( 1, 12) = 0.80 Prob > F = 0.3876 R-squared = 0.0628 Adj R-squared = -0.0153 Root MSE = .98129

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------inflation | -.2303902 25.69535 -0.90 0.388 -79.02438 32.94634 _cons | 5.465446 .4913862 11.12 0.000 4.394808 6.536085 ------------------------------------------------------------------------------

B) The intercept indicates that if there was no inflation, the unemployment rate would be 5.5 percent. The slope indicates that if the inflation rate increased by 1 percentage point, the unemployment rate would fall by -0.23 percentage points.

C) . regress unemp inflat Source | SS df MS -------------+-----------------------------Model | 8.3835527 1 8.3835527 Residual | 76.2986689 34 2.2440785 -------------+-----------------------------Total | 84.6822216 35 2.41949205

Number of obs = F( 1, 34) = Prob > F = R-squared = Adj R-squared = Root MSE =

36 3.74 0.0616 0.0990 0.0725 1.498

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------inflation | .1462325 .075657 1.93 0.062 -.0075209 .299986 _cons | 5.277482 .4166605 12.67 0.000 4.430726 6.124238 ------------------------------------------------------------------------------

The intercept is similar to what it was in part A, but the coefficient is now positive and indicates that a 1 percentage point increase in inflation now is estimated to INCREASE unemployment by 0.15 percentage points.

D) 10 1982

1983

civilian unemployment

1975

8 1976

1981

1984 1985

1977

1986 1961

1980

1958 1987

6

1978 1971

1963 1959 1954 1962 1960

1972

1979 1974

1988 1989

1964

19701973 1955

1965 1957 1956

4

1966 1967 1968

0

1969

5 10 proportionate change in CPI

15

The observations for the 1970s and early 1980s are all in the northeast portion of the graph. If we only used these data, the relationship looks negative, but when we add them to the earlier years, we observe a positive relationship.

E) The assumption of the regression model that is not satisfied by these data is that the parameters of the model do not appear to have been constant over the sample period.

SOLUTIONS TO CHAPTER 10 1) TSS (Y-Y(bar))^2

Y(hat)

3934.71

0.282

3.324

0.306

0.094

0.051

16198.35

0.755

2.642

-0.412

0.170

0.209

6843.80

0.053

3.396

-0.066

0.004

0.088

-1.61

52.89

0.049

3.073

0.247

0.061

0.001

62.73 -247.27

10.72 180.28

3934.71 61143.80

0.029 0.532

3.324 2.212

-0.054 0.158

0.003 0.025

0.051 0.787

0.51

122.73

62.70

15061.98

0.261

3.539

0.071

0.005

0.194

0.13

32.73

4.28

1071.07

0.017

3.217

0.013

0.000

0.014

1300

-0.51

-127.27

64.79

16198.35

0.259

2.642

-0.052

0.003

0.209

1450

0.20

22.73

4.57

516.53

0.040

3.181

0.119

0.014

0.007

3.21 1550 mean 3.0991 1427.27

0.11

122.73

13.61

15061.98

0.012

3.539

-0.329

0.109

0.194

3.9968E-15 2.2737E-13

502.37

140018.18

2.29

34.09

0.00

0.49

1.80

X

3.63

1490

0.53

62.73

33.30

2.23

1300

-0.87

-127.27

110.61

3.33

1510

0.23

82.73

19.10

3.32

1420

0.22

-7.27

3.27 2.37

1490 1180

0.17 -0.73

3.61

1550

3.23

1460

2.59 3.3

sum

Y - Y(bar)

X - X(bar)

(Y - Y(bar))* (X - X(bar)) (X - X(bar))^2

Y

Beta1

0.00358791

Beta0

-2.0218361

R-squared

0.78707289

ei USS Y - Y(hat) (Y - Y(hat))^2

ESS Y(hat) - Y(bar)^2

2A) food

mean sum

(Y - Y(bar))*

TSS

share

ln(inc)

Y - Y(bar)

X - X(bar)

(X - X(bar))

(X - X(bar))^2

0.54

9.2

0.04

0.03

0.00

0.00

0.001

0.67

8.9

0.17

-0.27

-0.05

0.07

0.36

9.4

-0.14

0.23

-0.03

0.05

0.55

9.2

0.05

0.03

0.00

0.61

8.8

0.11

-0.37

0.39

9.4

-0.11

0.47

9.2

0.29 0.62

USS

(Y-Y(bar))^2 (Y - Y(hat))^2

ei

ei^2

Y(hat)

Y - Y(hat)

0.0028

0.487

0.053

0.003

0.028

0.0010

0.639

0.031

0.001

0.020

0.0007

0.386

-0.026

0.001

0.00

0.002

0.0040

0.487

0.063

0.004

-0.04

0.14

0.012

0.0063

0.689

-0.079

0.006

0.23

-0.03

0.05

0.013

0.0000

0.386

0.004

0.000

-0.03

0.03

0.00

0.00

0.001

0.0003

0.487

-0.017

0.000

9.5

-0.21

0.33

-0.07

0.11

0.045

0.0020

0.335

-0.045

0.002

9

0.12

-0.17

-0.02

0.03

0.014

0.0010

0.588

0.032

0.001

0.52

9.1

0.02

-0.07

0.00

0.00

0.000

0.0003

0.537

-0.017

0.000

0.502

9.17 5.551E-17 1.7764E-14

-0.23

0.46

0.1366

0.0184

5.02

0.00 sigma^2 e =

Beta1

-0.5062907

Beta0

5.14468547

R-squared

0.86532105

s.d. (beta1)

0.07061812

0.0023

2B) Engel’s law states that the share of the household budget devoted to food should decline as income rises. Based on the empirical specification of the Engel curve, this law can be interpreted as telling us that a β1 is negative. Therefore, we have the null and alternative hypotheses: H0: β1 < 0 H1: β1 not < 0 2C) See chart for computation of R2. The fact that the R2 = 0.865 indicates that differences in log income can explain 86.5 percent of the variance in food shares.

3) The formulas for the two estimated regression coefficients would be:

n

 (Y − Y )( X − X )

ˆ = i =1

i

i

n

( X − X )

n

 (Y − Y )( X − X )

and ˆ = i =1

2

i

i =1

i

i

n

 (Y − Y )

2

i

i =1

One is the reciprocal of the other if ˆ1  ˆ1 = 1 , or if: n

n

 (Y − Y )( X − X )  (Y − Y )( X − X ) i =1

i

i

n

( X − X ) i =1

i

 i =1

2

i

i

n

 (Y − Y ) i =1

=1

2

i

This does not hold in general. One can show (for extra credit!) that this is true, however, if the R2 from a regression of Y on X is equal to 1.

4A) 7

lnprice

6.5

6

5.5 1960

1965

1970 year

1975

1980

B) You should probably have “eye-balled a slope of between, say, -0.4 and -0.5.

C) . regress lnprice vintage Source | SS df MS -------------+-----------------------------Model | .868465144 1 .868465144 Residual | 2.06913692 13 .159164379 -------------+-----------------------------Total | 2.93760207 14 .209828719

Number of obs = F( 1, 13) = Prob > F = R-squared = Adj R-squared = Root MSE =

15 5.46 0.0362 0.2956 0.2415 .39895

-----------------------------------------------------------------------------lnprice | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------vintage | -.0456397 .0195384 -2.34 0.036 -.0878499 -.0034295 _cons | 5.418722 .3652171 14.84 0.000 4.629718 6.207725 ------------------------------------------------------------------------------

The estimated return from holding a wine for an additional year is 4.6 percent.

D) A 95 percent confidence interval for holding a wine for an additional year spans from 0.3 percent to 8.8 percent.

E) year 1960 1962 1963 1964 1966 1967 1968 1969 1970 1971 1973 1974 1975 1976 1978

vintage -27 -25 -24 -23 -21 -20 -19 -18 -17 -16 -14 -13 -12 -11 -9

lnprice 6.4085 6.5793 6.0521 6.7007 6.7056 6.1633 7.0130 6.0776 6.7957 5.5984 5.8230 6.3750 5.5215 5.8861 5.8579

lnpricehat 6.6510 6.5597 6.5141 6.4684 6.3772 6.3315 6.2859 6.2402 6.1946 6.1490 6.0577 6.0120 5.9664 5.9208 5.8295

resid -0.2425 0.0195 -0.4620 0.2323 0.3285 -0.1682 0.7271 -0.1626 0.6011 -0.5505 -0.2346 0.3630 -0.4449 -0.0347 0.0285

The average of these errors is about zero.

F) Good vintages are those with positive errors and bad vintages are those with negative errors.

It is a good vintage if the price is higher than what one would have expected solely based upon its age. The opposite is true for bad vintages. The best vintages by this criterion are 1968, 1970, and 1974.

G) The R2 for this model is 0.296. This means that vintage explains about 30 percent of the variation in the log price of wine.

SOLUTIONS TO CHAPTER 11 1A) Holding constant a student’s math score, every additional point on his/her verbal score will lead to a 0.002 increase in his/her GPA. Holding constant a student’s verbal score, every additional point on his/her math score will increase his/her GPA by 0.0015.

B) The t-statistic for the verbal score is t = 0.002/0.0011 = 1.82 and the t-statistic for the math score is t = 0.0015/0.0011 = 1.36. The critical value for a significance test at the 5% level is 1.96 and at the 10% level is 1.645. Therefore, the coefficient on the math score is not statistically significant at either level, but the coefficient on the verbal score is statistically significant at the 10% level. The p-value for this coefficient is 6.9%.

C) The value of Z from the standard normal distribution for which 99% of the distribution falls between it and –Z is 2.576. Therefore, a 99% confidence interval for the coefficient on the math SAT score is (0.0015 – (2.576∙0.0011) < β < (0.0015 + (2.576∙0.0011)) -0.0013 < β < 0.0043 D) If verbal and math scores were divided by 100, then the estimated equation would be: GPA = 0.65 + 0.20V + 0.15M (0.11) (0.11) E) A student who scored 500 on both parts of the SAT would be predicted to have a GPA of 0.65 + 500∙0.0020 + 500∙0.0015 = 2.4.

2A) Every additional phone line reduces the wait time by 1.2 minutes, holding staff constant. If an additional staff member is working, the wait time is reduced by 7 minutes, holding phone lines constant.

B) A 95% confidence interval for waiting times is based upon the statistics that the mean waiting time is 15 minutes and the standard deviation is 5 minutes. Therefore, the confidence interval is: (15 – (5∙1.96) < waiting time < (15 + (5∙1.96)) 5.2 < waiting time < 24.8 C) The p-value for the phone line coefficient is based upon an estimate of 1.2 and a standard deviation of 0.8, so the t-statistic is t = 1.2/0.8 = 1.5. From the standard normal distribution, we know that 0.0668 percent of the distribution is below -1.5 and the same would be above 1.5, so the p-value is 13.36%.

3A) The regression equation is:

ln( wage)i = 0.721 + 0.102Si + 0.011Ei +  i This equation tells us that an individual with no education or experience would earn, in logs, 0.721 per hour. Holding constant experience, for every additional year of schooling an individual would increase his/her wage by about 10 percent, on average. Holding constant education, every additional year of experience increases wages by about 1.1 percent, on average. . regress lnhrwage educ exp Source | SS df MS -------------+-----------------------------Model | 14238.8902 2 7119.44509 Residual | 45111.3137179557 .251236731 -------------+-----------------------------Total | 59350.2038179559 .330533161

Number of obs = 179560 F( 2,179557) =28337.60 Prob > F = 0.0000 R-squared = 0.2399 Adj R-squared = 0.2399 Root MSE = .50124

-----------------------------------------------------------------------------lnhrwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------educ | .1015499 .0004471 227.14 0.000 .1006737 .1024262 exp | .0106627 .0000932 114.44 0.000 .01048 .0108453 _cons | .7211546 .0065248 110.53 0.000 .7083662 .7339431 ------------------------------------------------------------------------------

B) Education and experience can explain 24 percent of the variation in log wages.

C) To conduct this test, we construct the relevant t-statistic: t=

0.1015 − 0.1 0.0015 = = 3.33 .00045 .00045

This t-statistic is greater than the critical value of 1.96 for a test at the 5% significance level. Therefore, we can reject this null hypothesis that the return to a year of education is 10 percent.

D) We may choose to include experience squared in the regression because it may be the case that wages increase with experience faster for younger workers than for older workers. Such a hypothesis would be consistent with the predictions of the Human Capital Model, which suggests that younger individuals should invest more in human capital than older workers since they have a longer period to recoup the returns on the investment.

In this model, the first derivative of the log wage with respect to experience is:

 ln( wage) =  2 + 2  3 E = 0.0389-2  0.00064  E E

This expression is maximized (since the second derivative is negative) at the value of E for which the first derivative equals zero:  ln( wage) 0.0389 = 0 when E = =30.4 E 0.00128

A worker hits his/her maximum wage after 30 years of experience. . regress lnhrwage educ exp exp2 Source | SS df MS -------------+-----------------------------Model | 17111.9161 3 5703.97204 Residual | 42238.2877179556 .235237406 -------------+-----------------------------Total | 59350.2038179559 .330533161

Number of obs = 179560 F( 3,179556) =24247.73 Prob > F = 0.0000 R-squared = 0.2883 Adj R-squared = 0.2883 Root MSE = .48501

-----------------------------------------------------------------------------lnhrwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------educ | .0929638 .0004395 211.51 0.000 .0921023 .0938252 exp | .0388713 .0002707 143.59 0.000 .0383407 .0394019 exp2 | -.0006351 5.75e-06 -110.51 0.000 -.0006464 -.0006238 _cons | .635635 .0063609 99.93 0.000 .6231679 .6481022 ------------------------------------------------------------------------------

4A) Holding constant temperature and harvest rainfall, every additional millimeter of rain in the winter increases the price of wine by 0.13 percent. Holding constant rain at harvest time and in the winter, an additional degree (Celsius) of temperature during the growing season increases wine prices by 71 percent. Holding constant winter rain and temperature, a 1 millimeter increase in rainfall during the harvest lowers the price of wine by .36 percent. . regress lnprice wintrain avetemp harvrain Source | SS df MS -------------+-----------------------------Model | 7.75557784 3 2.58519261 Residual | 2.71468685 23 .118029863 -------------+-----------------------------Total | 10.4702647 26 .402702488

Number of obs = F( 3, 23) = Prob > F = R-squared = Adj R-squared = Root MSE =

27 21.90 0.0000 0.7407 0.7069 .34355

-----------------------------------------------------------------------------lnprice | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------wintrain | .001282 .0005765 2.22 0.036 .0000894 .0024747 avetemp | .7123178 .1087674 6.55 0.000 .4873154 .9373202 harvrain | -.0036242 .0009646 -3.76 0.001 -.0056196 -.0016288 _cons | -13.44433 1.969396 -6.83 0.000 -17.51834 -9.370326

------------------------------------------------------------------------------

B) The R2 for this regression is 0.74, meaning that 74 percent of the variation in wine prices can be explained by these 3 factors.

C) Many years before we can determine whether a vintage is high or low quality, we will have data on the growing conditions in the year the grapes were grown. We can use these factors and the coefficients from our regression model to predict what the subsequent price of the wine from that vintage will be. To do this, all we need to do is to multiply those conditions by the coefficient estimates and add on the intercept to get the predicted price.

5A) Ceteris paribus: 1) Every additional yard of driving distance lowers your score by a fraction of a shot.

2) For every additional green you hit in regulation, your score falls by -0.17 shots.

3) For every additional putt taken on the green, on average, your score rises by 14 shots.

It is somewhat surprising that the further you drive the ball has such a small impact on your score. The reason for this may be that what matters is whether or not you make the green in regulation and longer drives are more likely to lead to accomplishing this. But if you don’t get to the green in fewer shots, the length of the drive is probably irrelevant. . regress score drivedis greenreg puttave Source | SS df MS -------------+-----------------------------Model | 94.5087415 3 31.5029138 Residual | 21.2374555 187 .113569281 -------------+-----------------------------Total | 115.746197 190 .609190511

Number of obs = F( 3, 187) = Prob > F = R-squared = Adj R-squared = Root MSE =

191 277.39 0.0000 0.8165 0.8136 .337

-----------------------------------------------------------------------------score | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------drivedis | -.000937 .00338 -0.28 0.782 -.0076048 .0057307 greenreg | -.1669289 .0083485 -20.00 0.000 -.1833982 -.1504596 puttave | 14.28765 .879934 16.24 0.000 12.55178 16.02353 _cons | 56.88066 1.938216 29.35 0.000 53.05708 60.70424 ------------------------------------------------------------------------------

B) The 5 players with the largest negative residuals (which is a good thing in golf where the lowest score wins!), along with the values of their residuals, are: SHIGEKI MARUYAMA PETER JACOBSEN CHRIS RILEY TOM BYRUM LEN MATTIACE

-.7817349 -.7161538 -.7003437 -.6840691 -.620189

Luck may not be the appropriate interpretation of these results. The residual in this regression captures all factors related to a player’s score that aren’t incorporated into the 3 variables included in the model. Some players “do the little things right” and lower their scores in ways that are not captured by these 3 variables. These things may lower their scores beyond that which we’d predict based on driving distance, hitting greens in regulation, and average number of putts.

C) Using this model, it is hard to completely explain Corey Pavin’s fall over the period.

The only observable weakness in his game is that he only 10.5 greens in regulation in 2000, but 11.9 greens in regulation in 1995. The 1.4 decrease in greens hit in regulation would raise his score by 0.234 shots. But his putts per round fell by 0.48, or .027 when averaged over 18 holes. This reduction should reduce his score by 0.38 shots per round.

The small decline in driving distance would have a negligible effect on his score given the small coefficient on this variable. In the end, it looks like he just may have been unlucky in 2000!

SOLUTIONS TO CHAPTER 12

1A) The tables below present the fraction of houses that are in different locations and of different styles. . tab locat location of house | Freq.

Percent Cum. -------------------------------+----------------------------------East Meadow (Nassau County) | 74 20.44 20.44 Farmingdale (Nassau County) | 60 16.57 37.02 Levittown (Nassau County) | 99 27.35 64.36 Islip (Suffolk County) | 84 23.20 87.57 Islip Terrace (Suffolk County) | 45 12.43 100.00 -------------------------------+----------------------------------Total | 362 100.00 . tab style architectural | style of house | Freq.

Percent Cum. ---------------+----------------------------------cape | 112 30.94 30.94 expanded ranch | 38 10.50 41.44 colonial | 47 12.98 54.42 ranch | 119 32.87 87.29 split level | 46 12.71 100.00 ---------------+----------------------------------Total | 362 100.00

B) The variables that are significantly different from zero at the 5% level are: Positive effects Negative Effects Number of bathrooms Number of bedrooms Number of rooms Age 2 car garage modern bath central air conditioning ranch (relative to expanded ranch) fireplace modern kitchen town of East Meadow (relative to Islip Terrace) town of Farmingdale (relative to Islip Terrace) town of Levittown (relative to Islip Terrace) colonial (relative to expanded ranch) Most of these results make sense (without any specific knowledge of these towns!) except the negative effect of bedrooms and having a modern bath. It is important to keep in mind, however, that these are ceteris paribus effects. Therefore, the estimated effect of an additional bedroom is conditional upon the house being of the same size overall. For a house of a fixed size, it may be less desirable to have more bedrooms if they are all small.

. regress value lotsize bed bath rooms age garage1 garage2 eik cac firepl sewer ba > se modkit modbath eastmead farming levit islip cape colonial ranch splevel oilhe > at hotwater forcedair agpool igpool Source | SS df MS -------------+-----------------------------Model | 349858.75 27 12957.7315 Residual | 128493.607 334 384.711398 -------------+-----------------------------Total | 478352.357 361 1325.07578

Number of obs = F( 27, 334) = Prob > F = R-squared = Adj R-squared = Root MSE =

362 33.68 0.0000 0.7314 0.7097 19.614

-----------------------------------------------------------------------------value | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lotsize | .4516799 .2530251 1.79 0.075 -.0460438 .9494036 bed | -4.873818 1.867904 -2.61 0.009 -8.548156 -1.19948 bath | 14.1439 2.363729 5.98 0.000 9.49423 18.79357 rooms | 4.994374 1.177994 4.24 0.000 2.677152 7.311595 age | -.3614255 .1028868 -3.51 0.001 -.5638132 -.1590378 garage1 | 2.539624 2.515014 1.01 0.313 -2.407639 7.486886 garage2 | 12.28309 3.84924 3.19 0.002 4.711283 19.8549 eik | 6.07231 4.562753 1.33 0.184 -2.903043 15.04766 cac | 15.86984 4.261128 3.72 0.000 7.487806 24.25187 firepl | 9.4127 2.464543 3.82 0.000 4.564718 14.26068 sewer | 2.847816 2.796176 1.02 0.309 -2.652519 8.34815 base | 4.515098 2.956653 1.53 0.128 -1.300909 10.33111 modkit | 8.144916 3.990237 2.04 0.042 .2957544 15.99408 modbath | -9.777903 4.228515 -2.31 0.021 -18.09578 -1.460026 eastmead | 62.20131 4.837042 12.86 0.000 52.6864 71.71621 farming | 39.54235 4.715008 8.39 0.000 30.2675 48.81721 levit | 33.30781 5.163984 6.45 0.000 23.14977 43.46584 islip | 3.182055 3.853396 0.83 0.410 -4.39793 10.76204 cape | -7.263188 4.409009 -1.65 0.100 -15.93611 1.409738 colonial | 10.02934 4.740438 2.12 0.035 .7044588 19.35421 ranch | -9.765846 4.23149 -2.31 0.022 -18.08957 -1.442116 splevel | 6.331974 5.132867 1.23 0.218 -3.764848 16.4288 oilheat | -5.628714 3.833556 -1.47 0.143 -13.16967 1.912242 hotwater | 4.349174 6.054125 0.72 0.473 -7.559847 16.25819 forcedair | -4.499314 6.554489 -0.69 0.493 -17.3926 8.393968 agpool | 1.853501 3.662928 0.51 0.613 -5.351815 9.058816 igpool | 4.420653 5.177338 0.85 0.394 -5.763647 14.60495 _cons | 103.8126 13.13156 7.91 0.000 77.98156 129.6436 ------------------------------------------------------------------------------

C) You could use this model to find “bargains” by looking at the residuals. A bargain would be a house with a large negative residual.

2A) The mean percentage of women in male- and female-dominated jobs is 88.8 and 8.8 percent, respectively.

B) Male- and female-dominated jobs pay $17,188 and $20,586 on average, respectively. The difference does not necessarily reflect discrimination because differences in wages can be due to differences in compensable factor scores. In fact, male jobs due have higher point totals on these scores than female jobs. For instance, total points averaged over all female jobs is 178, but it is 215 for male jobs. So it is perhaps not surprising that women’s jobs pay less than men’s jobs.

. sum if femjob == 1 Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------wage | 34 17188.24 4991.736 11300 26200 pctfem | 34 88.83823 8.584095 71.7 99.4 number | 34 777.5 1096.219 15 4361 skills | 34 117.0784 44.31473 63.5 212 effort | 34 28.41176 21.11163 8 92 respon | 34 30.70098 20.73347 10 92 wc | 34 1.803922 3.52611 0 12.66667 points | 34 177.9951 86.61959 86 407 points2 | 34 176.1912 85.5496 81.5 396 femjob | 34 1 0 1 1 malejob | 34 0 0 0 0 . sum if malejob == 1 Variable | Obs Mean Std. Dev. Min Max ---------+----------------------------------------------------wage | 50 20586 7775.603 11800 48400 pctfem | 50 8.726 9.779942 0 29.4 number | 50 390.58 1351.249 10 9496 skills | 50 132.0667 47.20489 65.66666 304 effort | 50 33.81 22.92465 3 115 respon | 50 39.37333 25.71269 11 132 wc | 50 9.406667 6.267937 0 22 points | 50 214.6567 91.25041 103.5 551 points2 | 50 205.25 94.7365 86.33333 551 femjob | 50 0 0 0 0 malejob | 50 1 0 1 1

C) The results indicate that every additional point awarded for skill increases the wages of a job by $158, holding all other factors constant. Every additional point awarded for effort increases wages by $99, holding all other factors constant. Every additional point awarded for responsibility actually lowers wages by $166, holding all other factors constant. Every additional point awarded for working conditions increases wages by $39, holding all other factors constant. One might have expected that all of these compensable factors would be positively related to wages. In this case, three of them have the expected sign, but responsibility is negatively related to wages after controlling for the other factors. This result may not have been what you expected. The only two factors whose coefficients are statistically significant different from zero are skills and responsibility. These four factors explain 72.15 percent of the variation in wages across jobs.

. regress wage skill effort respon wc Source | SS df MS ---------+-----------------------------Model | 3.8187e+09 4 954676714 Residual | 1.4739e+09 111 13278661.5 ---------+-----------------------------Total | 5.2926e+09 115 46022941.5

Number of obs = F( 4, 111) = Prob > F = R-squared = Adj R-squared = Root MSE =

116 71.90 0.0000 0.7215 0.7115 3644.0

-----------------------------------------------------------------------------wage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------skills | 158.1902 32.96567 4.799 0.000 92.86651 223.5139 effort | 99.01591 68.38141 1.448 0.150 -36.48641 234.5182 respon | -165.7459 69.33838 -2.390 0.019 -303.1445 -28.34727 wc | 39.12169 58.88546 0.664 0.508 -77.56378 155.8072 _cons | 1872.702 2118.375 0.884 0.379 -2324.998 6070.403 ------------------------------------------------------------------------------

D) The formula for the f-statistic is:

R c F = 1− R c

2 unrestricted

2 − Rrestricted

2 unrestricted

 (k ) h

 (n − k ) h

2

The R2 restricted model is 0.6883 and the R2 for the unrestricted model is 0.7215. With a sample size of 116, 3 slope coefficients we are testing (k2) and 5 variables overall (k), the Fstatistic is computed as

F=

(0.7215 − 0.6883) / 3 O L M N(1 − 0.7215) / 111 P Q= 4.41

With (3,111) degrees of freedom, the critical value of the 5 distribution with at the five percent level is roughly2.68. Therefore, we can reject the null hypothesis that these three coefficients are all equal to zero. . regress wage points Source | SS df MS ---------+-----------------------------Model | 3.6429e+09 1 3.6429e+09 Residual | 1.6498e+09 114 14471656.0 ---------+-----------------------------Total | 5.2926e+09 115 46022941.5

Number of obs = F( 1, 114) = Prob > F = R-squared = Adj R-squared = Root MSE =

116 251.72 0.0000 0.6883 0.6856 3804.2

-----------------------------------------------------------------------------wage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------points | 61.06354 3.848746 15.866 0.000 53.4392 68.68788 _cons | 7135.366 879.7495 8.111 0.000 5392.589 8878.143 ------------------------------------------------------------------------------

. regress wage points skills effort wc Source | SS df MS ---------+-----------------------------Model | 3.8187e+09 4 954676717 Residual | 1.4739e+09 111 13278661.3 ---------+-----------------------------Total | 5.2926e+09 115 46022941.5

Number of obs = F( 4, 111) = Prob > F = R-squared = Adj R-squared = Root MSE =

116 71.90 0.0000 0.7215 0.7115 3644.0

-----------------------------------------------------------------------------wage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------points | -165.7459 69.33838 -2.390 0.019 -303.1445 -28.34728 skills | 323.9361 91.18049 3.553 0.001 143.2559 504.6163 effort | 264.7618 120.8532 2.191 0.031 25.28316 504.2405 wc | 204.8676 97.03843 2.111 0.037 12.57947 397.1557 _cons | 1872.702 2118.375 0.884 0.379 -2324.998 6070.403 -----------------------------------------------------------------------------. testparm skills effort wc ( 1) ( 2) ( 3)

skills = 0.0 effort = 0.0 wc = 0.0 F(

3, 111) = Prob > F =

4.41 0.0057

E) Holding constant differences in compensable factors, female-dominated jobs are paid almost $2,000 less than other jobs. This difference is statistically significant at the five percent level. Since the compensable factors are supposed to control for differences in the value of the job to the firm, one could interpret this finding as indicating that the discrimination exists in salary determination. In this case, the discrimination is against women’s dominated jobs, not women directly. . regress wage respon skill effort wc femjob Source | SS df MS ---------+-----------------------------Model | 3.8855e+09 5 777106708 Residual | 1.4071e+09 110 12791861.2 ---------+-----------------------------Total | 5.2926e+09 115 46022941.5

Number of obs = F( 5, 110) = Prob > F = R-squared = Adj R-squared = Root MSE =

116 60.75 0.0000 0.7341 0.7221 3576.6

-----------------------------------------------------------------------------wage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------respon | -168.8315 68.06892 -2.480 0.015 -303.7281 -33.93485 skills | 156.1875 32.36763 4.825 0.000 92.04242 220.3325 effort | 93.02009 67.16751 1.385 0.169 -40.09014 226.1303 wc | -36.36607 66.56694 -0.546 0.586 -168.2861 95.55397 femjob | -1964.794 859.6249 -2.286 0.024 -3668.369 -261.2195 _cons | 3472.875 2193.886 1.583 0.116 -874.891 7820.642 ------------------------------------------------------------------------------

3A) We can test the hypothesis that β1 = 0 by conducting a t-test.

B) We can test the hypothesis that β1 = 0 and β4 = β5 by conducting an F-test. To do this, we would estimate a regression including all 5 X variables (the unrestricted model) and then a second regression imposing the restrictions (the restricted model). We would then calculate the F-statistic for this test and compare its value to the critical values obtained from a table of the F distribution.

C) We can test the hypothesis that β1 = 0, β3 = 2, and β4 = β5 analogously to the procedure described in part B.

SOLUTIONS TO CHAPTER 13

1A) Adults make more money than children and adults have larger feet!

B) People who smoke tend to have less education, and less-educated people earn less.

C) Those who go to college would be more productive workers than those who graduate from high school but do not go on to college regardless of the difference in educational attainment.

D) Welfare families are, by definition, poor families and children from poor families are more likely to grow up and be poor themselves. Therefore, they are more likely to be eligible for welfare and then to receive it.

2) The X variable, profits would be perfectly correlated with the intercept and we could not estimate this model.

3A) On average, men make $12.66 per hour and women make $9.66 per hour. There is a $3 per hour wage differential between men and women. It may not represent evidence of discrimination because men and women may have different characteristics that would affect their productivity and could lead to a wage differential.

B) These results indicate that women make 25 percent less than men. From part A, since women earn $3 less than men, from a base male wage of $12.66, this represents a 23.7 percent differential. This regression result approximates the wage gap implicit in the results from part A. . regress lnhrwage female Source | SS df MS -------------+-----------------------------Model | 2875.86117 1 2875.86117 Residual | 56474.3427179558 .314518666 -------------+-----------------------------Total | 59350.2038179559 .330533161

Number of obs = 179560 F( 1,179558) = 9143.69 Prob > F = 0.0000 R-squared = 0.0485 Adj R-squared = 0.0485 Root MSE = .56082

-----------------------------------------------------------------------------lnhrwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------female | -.2531924 .0026478 -95.62 0.000 -.258382 -.2480027 _cons | 2.371012 .0018483 1282.84 0.000 2.367389 2.374634 ------------------------------------------------------------------------------

C) When union status, education, and experience are added to the model, the female coefficient falls just slightly. Since excluding these variables had little impact on the female coefficient, it must be the case that each of these variables is not strongly related to female. In other words, men and women must have similar levels of educational attainment, labor market experience, and the differential in union status must not affect

their wages that much. In reality, one shortcoming with this methodology is that we do not really have a good measure of labor market experience. It has been estimated using the algorithm that experience equals age - education - 6. If women are more likely to take time out of the labor force because of family responsibilities, this could affect their wage, but is not incorporated into our measure of experience.

These results go beyond just a raw wage differential between men and women since they control for some factors, but it would still be premature to conclude that discrimination exists against women. To draw such a conclusion, we would want to know whether women make less than men after holding constant any characteristics that may be attributable to productivity differences. For instance, we would certainly want to include a better measure of labor market experience. Other factors that matter may be even more difficult to measure, including things like level of motivation, commitment to the job, etc.

This is why it is so hard to prove discrimination! . regress lnhrwage female educ exp union Source | SS df MS -------------+-----------------------------Model | 18540.0294 4 4635.00735 Residual | 40810.1745179555 .227285091 -------------+-----------------------------Total | 59350.2038179559 .330533161

Number of obs = 179560 F( 4,179555) =20392.92 Prob > F = 0.0000 R-squared = 0.3124 Adj R-squared = 0.3124 Root MSE = .47674

-----------------------------------------------------------------------------lnhrwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------female | -.2478511 .0022606 -109.64 0.000 -.2522818 -.2434205 educ | .1011611 .0004257 237.64 0.000 .1003268 .1019954 exp | .0099321 .0000892 111.34 0.000 .0097572 .0101069 union | .2248583 .0030883 72.81 0.000 .2188053 .2309112 _cons | .8241264 .0062873 131.08 0.000 .8118034 .8364494 ------------------------------------------------------------------------------

D) If we included a series of dummy variables indicating a worker’s occupation, the coefficient on the female dummy variable is likely to fall. One reason why we may want to include such variables is that women may have preferences for different sorts of jobs than men, perhaps because of one’s ability to integrate family and worklife, and these jobs pay less than jobs men are more likely to take. If so, any wage gap that exists without controlling for this effect would overstate the extent of the wage differential. On the other hand, women may be more likely to be employed in these jobs because they are shut out of typically male-dominated jobs, indicating these variables would capture part of the discrimination that exists against women.

4) The parameter, α, provides a measure of the productivity of labor. We could estimate this parameter by taking the natural log of both sides of this equation:

log Yt = log A +  log Lt +  log Kt + log  t

The coefficient on log L is the productivity parameter we are interested in. This specification has imposed the restriction that labor productivity is constant over the sample period, but we could test whether or not this restriction is true. One way to do this is to create a dummy variable indicating whether a particular observation came from after, say, 1983. Let’s call this variable DV83. Then we could estimate the model:

log Yt = log A + 1 log Lt +  2 log Lt  DV 83t +  log Kt +  DV 83t + log  t A hypothesis test of α2 = 0 provides an indication of whether or not labor productivity changed in the period following the late 1980s.

This specification may have the problem that it still restricts the productivity of capital to be the same before and after the early 1980s. This may be inappropriate since firms that are using “too much” labor may also be using “too little” capital. Hence, we could also include an interaction for capital with DV83 to test this proposition as well.

SOLUTIONS TO CHAPTER 14 1A) Serial correlation in likely to be present here. It is plausible that there may be “random” (i.e. otherwise unaccounted for by the model) shocks to stock prices that are likely to persist over multiple periods. This persistence represents serial correlation.

B) Heteroskedasticity is likely to be a problem in this regression. Although some lowincome households spend more on housing than others, the range of their housing options are relatively limited. On the other hand high income households could vary a great deal regarding the housing expenditures depending upon their preferences for housing. Therefore, the variance of the residuals will likely be much greater for higher income compared to lower income households.

C) Heteroskedasticity may be a problem in this regression as well for a similar reason. Those cities with relatively low median income do not have the option to spend a great deal of money on their schools. Those cities with relatively high median income have more choices how to allocate their money and may spend more on schools, but may choose to spend it on other things instead. Therefore, the variance of the residuals is likely to be much greater for the high median income cities.

2A) This regression is likely to be affected by heteroskedasticity. A unit of observation represents an average within a location. The Central Limit Theorem tells us that the variance of a sample average is equal to  2 / n , where σ2 is the variance of the underlying population.

Now consider the residual from this regression, which represents the mean residual in each location. Even if the underlying variance of this residual from micro data were constant across individuals, we know from the Central Limit Theorem that the mean residuals across locations will have different variances, depending upon the size of the sample used to construct the mean. If we estimate this model using OLS, the estimated parameters of the model will be unbiased, but the standard errors will be calculated incorrectly and will bias our hypothesis tests.

B) This model can be estimated using weighted least squares, applying a function of the sample sizes used to construct the mean within each village as weights. This approach is described in detail in the part of section 14.2.5 related to “grouped data.”

3) With cross-sectional data we have observations that represent different people at a point in time. Suppose, for instance, we were estimating a regression model using data from a representative sample of the population. Given the size of the country and the relatively small fraction of people contained in the sample, the chances that the residual in any one respondent’s outcome could be affected by another respondent seems remote.

On the other hand, it is not impossible for heteroskedasticity to creep into an analysis using cross-sectional data. Suppose a local satellite dish provider is estimating a model of the

determinants of using their services (rather than cable or broadcast TV) as a function of household characteristics. Suppose that one development of houses is located in a heavily wooded area with very tall trees, so that a dish on any of these houses cannot communicate with the satellite. This would amount to a negative residual that is likely to be correlated among all of the houses in this area.

4A) No, we cannot reject the null hypothesis that b = 0 at the 10% level. The t-statistic for this coefficient is -1.68 and its p-value is 0.110, just less than the 10 percent cut-off we had set. The point estimate, however, is negative, which is consistent with the negative relationship between inflation and unemployment as stated in the Philips Curve.

. regress unemp infl if year <= 1970 Source | SS df MS -------------+-----------------------------Model | 3.41554144 1 3.41554144 Residual | 21.7419581 18 1.20788656 -------------+-----------------------------Total | 25.1574995 19 1.32407892

Number of obs = F( 1, 18) = Prob > F = R-squared = Adj R-squared = Root MSE =

20 2.83 0.1099 0.1358 0.0878 1.099

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------infl | -.206935 .1230601 -1.68 0.110 -.4654746 .0516047 _cons | 5.127604 .3869477 13.25 0.000 4.314657 5.94055 ------------------------------------------------------------------------------

When we conduct a Durbin-Watson test, however, we see that first order serial correlation is likely to be a problem in the model. The Durbin-Watson statistic equals 1.19, which is just below the lower bound of 1.20 at the 5 percent level of significance when n = 20 and k = 1 (excluding the constant). . dwstat Durbin-Watson d-statistic(

2,

20) =

1.188325

When we re-estimate the model using the Cochrane-Orcutt procedure, our new estimate is slightly positive, not negative as the Philips Curve indicates, but is not close to being statistically significantly different from zero. . prais unemp infl if year <= 1970, corc Iteration 0: rho = 0.0000 Iteration 1: rho = 0.4008 Iteration 2: rho = 0.5215 Iteration 3: rho = 0.5409 Iteration 4: rho = 0.5433 Iteration 5: rho = 0.5436 Iteration 6: rho = 0.5437 Iteration 7: rho = 0.5437 Iteration 8: rho = 0.5437 Cochrane-Orcutt AR(1) regression -- iterated estimates Source |

SS

df

MS

Number of obs =

19

-------------+-----------------------------Model | .097964755 1 .097964755 Residual | 16.3677846 17 .962810862 -------------+-----------------------------Total | 16.4657494 18 .914763856

F( 1, 17) = 0.10 Prob > F = 0.7536 R-squared = 0.0059 Adj R-squared = -0.0525 Root MSE = .98123

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------infl | .0525499 .1647431 0.32 0.754 -.2950277 .4001275 _cons | 4.689642 .5937807 7.90 0.000 3.436874 5.942409 -------------+---------------------------------------------------------------rho | .5436678 -----------------------------------------------------------------------------Durbin-Watson statistic (original) 1.188325 Durbin-Watson statistic (transformed) 2.134001

B) In no case do we estimate a negative and significant relationship between inflation and unemployment. In fact, adding the 1970s led to a considerably more positive relationship, although it is not statistically significant. . prais unemp infl if year <= 1980, corc Iteration 0: rho = 0.0000 Iteration 1: rho = 0.5479 Iteration 2: rho = 0.5679 Iteration 3: rho = 0.5716 Iteration 4: rho = 0.5723 Iteration 5: rho = 0.5725 Iteration 6: rho = 0.5725 Iteration 7: rho = 0.5725 Iteration 8: rho = 0.5725 Iteration 9: rho = 0.5725 Cochrane-Orcutt AR(1) regression -- iterated estimates Source | SS df MS -------------+-----------------------------Model | 1.92327239 1 1.92327239 Residual | 28.6014298 27 1.05931222 -------------+-----------------------------Total | 30.5247022 28 1.09016794

Number of obs = F( 1, 27) = Prob > F = R-squared = Adj R-squared = Root MSE =

29 1.82 0.1890 0.0630 0.0283 1.0292

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------infl | .1108722 .0822837 1.35 0.189 -.0579602 .2797045 _cons | 4.983472 .5740396 8.68 0.000 3.80564 6.161304 -------------+---------------------------------------------------------------rho | .5725233 -----------------------------------------------------------------------------Durbin-Watson statistic (original) 0.783530 Durbin-Watson statistic (transformed) 1.912052

. prais unemp infl , corc Iteration 0: rho = 0.0000 Iteration 1: rho = 0.6798 Iteration 2: rho = 0.7621 Iteration 3: rho = 0.7790

Iteration 4: Iteration 5: Iteration 6: Iteration 7: Iteration 8: Iteration 9:

rho = 0.7820 rho = 0.7825 rho = 0.7826 rho = 0.7826 rho = 0.7826 rho = 0.7826

Cochrane-Orcutt AR(1) regression -- iterated estimates Source | SS df MS -------------+-----------------------------Model | .219439672 1 .219439672 Residual | 43.9715999 46 .955904346 -------------+-----------------------------Total | 44.1910396 47 .940234885

Number of obs = 48 F( 1, 46) = 0.23 Prob > F = 0.6341 R-squared = 0.0050 Adj R-squared = -0.0167 Root MSE = .9777

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------infl | -.0374634 .0781911 -0.48 0.634 -.1948539 .119927 _cons | 5.955129 .706415 8.43 0.000 4.533189 7.377069 -------------+---------------------------------------------------------------rho | .7826191 -----------------------------------------------------------------------------Durbin-Watson statistic (original) 0.560631 Durbin-Watson statistic (transformed) 1.833323

C) Friedman would expect a negative coefficient on the change in inflation and, in fact, this is what we observe. This would be based on the assumption that the change in inflation is unanticipated or, in other words, that the expectation of inflation next year is today’s value of inflation. We estimate this model using the Cochrane-Orcutt method because a DurbinWatson test based on the OLS specification indicates the presence of serial correlation. . prais unemp chnginfl, corc Iteration 0: rho = 0.0000 Iteration 1: rho = 0.7230 Iteration 2: rho = 0.7153 Iteration 3: rho = 0.7146 Iteration 4: rho = 0.7146 Iteration 5: rho = 0.7146 Iteration 6: rho = 0.7146 Cochrane-Orcutt AR(1) regression -- iterated estimates Source | SS df MS -------------+-----------------------------Model | 11.7722082 1 11.7722082 Residual | 31.6128652 45 .702508116 -------------+-----------------------------Total | 43.3850734 46 .94315377

Number of obs = F( 1, 45) = Prob > F = R-squared = Adj R-squared = Root MSE =

47 16.76 0.0002 0.2713 0.2551 .83816

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------chnginfl | -.2635225 .0643746 -4.09 0.000 -.3931796 -.1338654 _cons | 5.971091 .4289468 13.92 0.000 5.107148 6.835034 -------------+---------------------------------------------------------------rho | .7145685 ------------------------------------------------------------------------------

Durbin-Watson statistic (original) 0.391181 Durbin-Watson statistic (transformed) 2.110510

D) Model 2 represents a restricted form of model 3, where the restriction from model 3 is that b = -c. Below, we report estimates from the unrestricted model and the results of an F-test on this proposition. The p-value for this test is 0.0001, indicating that we can safely reject the null hypothesis. This would suggest that the model represented by equation 3 may be the best to describe the data. . prais unemp infl laginf, corc Iteration 0: rho = 0.0000 Iteration 1: rho = 0.5922 Iteration 2: rho = 0.6068 Iteration 3: rho = 0.6114 Iteration 4: rho = 0.6128 Iteration 5: rho = 0.6132 Iteration 6: rho = 0.6134 Iteration 7: rho = 0.6134 Iteration 8: rho = 0.6134 Iteration 9: rho = 0.6134 Iteration 10: rho = 0.6134 Iteration 11: rho = 0.6134 Cochrane-Orcutt AR(1) regression -- iterated estimates Source | SS df MS -------------+-----------------------------Model | 15.6348052 2 7.81740262 Residual | 29.5047652 44 .670562845 -------------+-----------------------------Total | 45.1395704 46 .981295009

Number of obs = F( 2, 44) = Prob > F = R-squared = Adj R-squared = Root MSE =

47 11.66 0.0001 0.3464 0.3167 .81888

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------infl | -.1958499 .0780701 -2.51 0.016 -.3531899 -.0385099 laginfl | .3546264 .0734421 4.83 0.000 .2066137 .5026391 _cons | 5.292884 .4263725 12.41 0.000 4.433586 6.152181 -------------+---------------------------------------------------------------rho | .613423 -----------------------------------------------------------------------------Durbin-Watson statistic (original) 0.571898 Durbin-Watson statistic (transformed) 1.953961 . test infl=laginfl ( 1)

infl - laginfl = 0.0 F(

1, 44) = Prob > F =

17.38 0.0001

SOLUTIONS TO CHAPTER 15

1) In a supply and demand model of price-determination, we have two equations that reflect the relationship between price and quantity, one representing supply and one representing demand. All we observe in the data is the equilibrium price and quantity, which represents the interaction of supply and demand and does not tell us anything about what the supply curve or demand curve look like. To identify supply and demand separately, we would need at least one factor that is related to the demand for the product, but not the supply, and at least one factor that is related to supply, but not demand.

2A) The identification problem indicates that we cannot estimate all the parameters of this system of equations.

B) Consider, for instance, the coefficient, β1 in equation 1. This coefficient is supposed to tell us the impact of the variable Y2 on the variable Y1. But in this model, Y1 and Y2 are jointly determined. If Y1 goes up, then Y2 is affected and the change in Y2 has an impact on Y1 and so on. If we estimate β1 using OLS, our estimate will capture all of these effects and will not tell us the independent effect of Y2 on the Y1.

C) For each equation in this model to be identified, we would need at least one exogenous variable (i.e. the Xs) to be excluded from one equation, but included in the other. i) ii) iii)

Neither equation is identified Neither equation is identified since X1 drops out of both equations Again, neither equation would be identified

D) If β3 = β4 = 0, then equation 1 is identified and we can use two-stage least squares to estimate the model. To do this, we would use equation 2 to obtain predicted values for Y2 and then use these predicted values rather than the observed values in estimating equation 1. Equation 2 cannot be estimated in the same way since it still is not identified.

3A) Labor and the wage are endogenous. The remainder of the variables are exogenous.

B) The reduced for equation for ln L is:

ln Li =

a b  a b +b  ab 1 b0 + 0 1 +  2 1 5  ln Schi + 3 1 ln Othwag i 1 − a1 1 − a1  1 − a1  1 − a1

+

b b2 b be +e ln landi + 3 ln REi + 4 ln Oth expi + 1 2 1 1 − a1 1 − a1 1 − a1 1 − a1

The reduced for equation lnW is:

ln Li =

ab ab +a  a 1 a0 + 1 0 +  1 5 2  ln Schi + 3 ln Othwagi 1 − b1 1 − b1  1 − b1  1 − b1

+

ab a a1b2 ae +e ln landi + 1 3 ln REi + 3 ln Oth expi + 1 1 2 1 − b1 1 − b1 1 − b1 1 − b1

C) Equation 1 is exactly identified because there is only 1 exogenous variable in equation 2 that is not also contained in equation 1. Equation 2 is overidentified because there are 3 exogenous variables contained in equation 1 that are not contained in equation 2.

D) The OLS estimates of these equations are: . regress lnlabor lnwage lnland lnre lnothexp lnsch Source | SS df MS -------------+-----------------------------Model | 11.0167377 5 2.20334753 Residual | 1.61159521 33 .048836218 -------------+-----------------------------Total | 12.6283329 38 .332324549

Number of obs = F( 5, 33) = Prob > F = R-squared = Adj R-squared = Root MSE =

39 45.12 0.0000 0.8724 0.8530 .22099

-----------------------------------------------------------------------------lnlabor | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lnwage | -1.105774 .2893065 -3.82 0.001 -1.694373 -.517176 lnland | .3118018 .0837263 3.72 0.001 .1414594 .4821441 lnre | .4446036 .1140206 3.90 0.000 .2126269 .6765803 lnothexp | .2645385 .1127825 2.35 0.025 .0350809 .4939961 lnsch | -.2623222 .6082036 -0.43 0.669 -1.499722 .9750773 _cons | .8757252 1.943497 0.45 0.655 -3.078349 4.8298 -----------------------------------------------------------------------------. regress lnwage lnlabor lnsch lnothwag Source | SS df MS -------------+-----------------------------Model | 1.871368 3 .623789334 Residual | .465284236 35 .013293835 -------------+-----------------------------Total | 2.33665224 38 .061490848

Number of obs = F( 3, 35) = Prob > F = R-squared = Adj R-squared = Root MSE =

39 46.92 0.0000 0.8009 0.7838 .1153

-----------------------------------------------------------------------------lnwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lnlabor | -.0109471 .0340323 -0.32 0.750 -.0800364 .0581422 lnsch | 1.164415 .2384458 4.88 0.000 .6803442 1.648486 lnothwage | .4682868 .1323288 3.54 0.001 .1996451 .7369286 _cons | -2.972985 .6356697 -4.68 0.000 -4.263463 -1.682507 ------------------------------------------------------------------------------

Two stage least squares estimates are: . regress lnlabor lnwage lnland lnre lnothexp lnsch (lnland lnre lnothexp lnsch ln > othwag) Instrumental variables (2SLS) regression Source | SS df MS -------------+-----------------------------Model | 10.9478902 5 2.18957803 Residual | 1.68044271 33 .050922506 -------------+-----------------------------Total | 12.6283329 38 .332324549

Number of obs = F( 5, 33) = Prob > F = R-squared = Adj R-squared = Root MSE =

39 41.63 0.0000 0.8669 0.8468 .22566

-----------------------------------------------------------------------------lnlabor | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lnwage | -1.449278 .6012275 -2.41 0.022 -2.672485 -.2260715 lnland | .3286769 .0892823 3.68 0.001 .1470308 .5103231 lnre | .4687022 .1220887 3.84 0.001 .2203109 .7170935 lnothexp | .2497877 .117341 2.13 0.041 .0110556 .4885198 lnsch | .3420969 1.111155 0.31 0.760 -1.918564 2.602758 _cons | -.8591309 3.306457 -0.26 0.797 -7.586169 5.867907 -----------------------------------------------------------------------------. regress lnwage lnlabor lnsch lnothwag (lnsch lnothwag lnland lnre lnothexp) Instrumental variables (2SLS) regression Source | SS df MS -------------+-----------------------------Model | 1.84881748 3 .616272493 Residual | .487834759 35 .013938136 -------------+-----------------------------Total | 2.33665224 38 .061490848

Number of obs = F( 3, 35) = Prob > F = R-squared = Adj R-squared = Root MSE =

39 44.97 0.0000 0.7912 0.7733 .11806

-----------------------------------------------------------------------------lnwage | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lnlabor | .0333775 .0384072 0.87 0.391 -.0445932 .1113482 lnsch | 1.234335 .245481 5.03 0.000 .7359817 1.732687 lnothwage | .4643323 .1355053 3.43 0.002 .189242 .7394226 _cons | -3.576787 .6870601 -5.21 0.000 -4.971593 -2.181981 ------------------------------------------------------------------------------

OLS and TSLS estimates are not that much different, particularly when you take into consideration the increase in the size of the standard errors associated with the TSLS estimates. Although technically OLS is an inappropriate technique to use for this problem because of the endogeneity problem, in practice correcting the problem by estimating the model using TSLS does not change the results much.

4A) In model I: Equation 1 is underidentified.

Equation 2 is underidentified.

Equation 3 is just identified.

In model II: Equation 1 is overidentified.

Equation 2 does not contain any endogenous right hand side variables.

B) The parameters of the identified equations can be estimated using two-stage least squares.

5) Note that we can re-write this as a system of two equations (one with q as the dependent variable and one with p as the dependent variable) as follows: q = 0 + 1p + 2z + u1 (eq1) p = -0/1 + (1/1)q - (2/1)z - u2/1 (eq2) This is useful because it allows us to analyze the system as a pair of simultaneous equations. (In the initial specification, we did not have an explanatory equation for p.)

Written in this form, we can also use the order condition to tell whether or not each equation is identified. (a) i. As written, both equations are under-identified. ii. If 2 = 0, then the supply equation is just identified because z does not belong in that equation (giving us an additional exogenous variable for the one endogenous variable). The demand equation is still under-identified. iii. If 2 = 2 = 0, we are back where we started, with both equations under-identified. (b) Obtain the reduced forms: q = 0 + 1[-0/1 + (1/1)q - (2/1)z - u2/1]+ 2z + u1  q(1-1/1) = 0 + ....

After some algebra, we get q* = 11 + 21z + vq

with

11 = (01 - 10)/(1 - 1) 21 = (21 - 12)/(1 - 1) vq = (u2 - u1)/(1 - 1)

p = -0/1 + (1/1)[0 + 1p + 2z + u1] - (2/1)z - u2/1  p(1-1/1) = -0/1 + ....

After some algebra, we get p* = 12 + 22z + vp

12 = (10- 01)/(1- 1) 22 = (12- 21)/(1- 1) vp = (1u2- 1u1)/(1- 1) These "star" equations are our reduced-form equations. with

To see what restrictions are implied by each set of assumptions, we need to plug these reduced form equations into each of the original explanatory equations for q and p. I'll do this first for eq1: 11 + 21z + vq = 0 + 1[12 + 22z + vp] + 2z + u1

By taking the "expectation" of both sides to eliminate the error terms and collecting like terms we get: 11 - 0 - 112 + (21 - 122 - 2)z = 0

(eq3)

This has to hold for any value of z. Therefore, we can choose some values of z that allow us to determine the relationship between the structural and reduced form parameters: Setting z=0 implies 11 - 0 - 112 = 0 Setting z=1 implies 21 - 122 - 2 = 0 Note that additional values of 'z' will not give us any more equations! We can also see that, as written, we have only two equations for three unknowns (0, 1 and 2), so we can not solve for the structural parameters from the reduced-form parameters.

Following the same procedure for eq2, I get the following equation: 112 + 0 - 11 + (122 - 21 + 2)z = 0 Setting z=0 implies 112 + 0 - 11 = 0 Setting z=1 implies 122 - 21 + 2 = 0 Again we see that, as written, we have only two equations for three unknowns (0, 1 and 2), so we can not solve for the structural parameters from the reduced-form parameters.

Under condition (ii), 2 = 0, we see that we now have two equations for two unknown supply parameters. Therefore, we can solve for these, yielding the following results: 122 - 21 = 0 112 + 0 - 11 = 0 =>

=> 1 = 21/22 0 = 11 - 112 = 11 - (21/22)12

Note that condition (ii) does not help us to solve for 0, 1 and 2 so the demand equation is still under-identified.

Under condition (iii), 2 = 2 = 0, note that 21 and 22 = 0. This leaves us with only one relationship for each set of structural parameters: 11 - 0 - 112 = 0 for the demand parameters, and 112 + 0 - 11 = 0 for the supply parameters.

As the order condition indicated, we are not able to solve for the structural parameters in terms of the reduced form parameters under this condition.

SOLUTIONS TO CHAPTER 16

1) Any example with a dummy dependent variable would be appropriate here.

2) OLS has two main limitations in terms of estimating a model with a dummy dependent variable. The first is that the errors are heteroskedastic, although we could use whitecorrected standard errors to address this problem. The second is that OLS may lead to predicted probabilities that are greater than unity or less than zero and this problem cannot be corrected.

3) The reason why OLS and probit/logit models often provide similar estimates of the derivative is that is common that the slope for most of the range of predicted probabilities is similar. Only if we had an outcome that is typically very likely or very unlikely does the curved part of the logit/probit line lead to different derivatives than what you would get from OLS.

P(Y=1)

1

Non-linear model (Logit or Probit)

.5

Linear Probability Model

0 X

4A) Overall, 9.2% of the daughters received welfare. When we split the sample according to mother’s welfare receipt, we see that daughters whose mothers received welfare are much more likely to receive welfare themselves compared to daughters whose mothers did not receive welfare (23% compared to 6.9%). . sum recwelfd Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------recwelfd | 3263 .0922464 .2894178 0 1

. sum recwelfd if recwelfm == 0 Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------recwelfd | 2708 .0694239 .2542206 0 1 . sum recwelfd if recwelfm == 1 Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------recwelfd | 456 .2302632 .4214636 0 1

B) Estimates from a linear probability model indicate that a daughter of a mother who received welfare is 16 percentage points more likely to receive welfare herself compared to a daughter of a mother who did not receive welfare. . regress recwelfd recwelfm Source | SS df MS -------------+-----------------------------Model | 10.0962708 1 10.0962708 Residual | 255.77067 3162 .08088889 -------------+-----------------------------Total | 265.866941 3163 .084055308

Number of obs = F( 1, 3162) = Prob > F = R-squared = Adj R-squared = Root MSE =

3164 124.82 0.0000 0.0380 0.0377 .28441

-----------------------------------------------------------------------------recwelfd | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------recwelfm | .1608392 .0143965 11.17 0.000 .1326119 .1890666 _cons | .0694239 .0054654 12.70 0.000 .0587079 .08014 ------------------------------------------------------------------------------

C) In this case, the point estimate from the probit model is identical to that from the linear probability model. . dprobit recwelfd recwelfm Iteration 0: Iteration 1: Iteration 2: Iteration 3:

log likelihood = -976.16414 log likelihood = -929.76919 log likelihood = -928.86722 log likelihood = -928.86703

Probit estimates Log likelihood = -928.86703

Number of obs = 3164 LR chi2(1) = 94.59 Prob > chi2 = 0.0000 Pseudo R2 = 0.0485

-----------------------------------------------------------------------------recwelfd | dF/dx Std. Err. z P>|z| x-bar [ 95% C.I. ] ---------+-------------------------------------------------------------------recwelfm*| .1608392 .0203112 9.96 0.000 .144121 .12103 .200648 ---------+-------------------------------------------------------------------obs. P | .0926043 pred. P | .0848543 (at x-bar) -----------------------------------------------------------------------------(*) dF/dx is for discrete change of dummy variable from 0 to 1 z and P>|z| are the test of the underlying coefficient being 0

D) When we include income in the model, the coefficient on mother’s welfare receipt falls by about one-half. This is a perfect example of omitted variable bias. Daughters whose mothers received welfare grew up in poor families (by the eligibility rules for welfare). Children who grow up in poor families are more likely to be poor regardless of the welfare system and, therefore, eligible for welfare themselves. Therefore, without controlling for differences in income while growing up, we would attribute part of the daughter’s welfare receipt to welfare dependency when a “poverty trap” may be partly to blame. . dprobit recwelfd recwelfm lnfaminc Iteration 0: Iteration 1: Iteration 2: Iteration 3:

log likelihood = -854.09485 log likelihood = -789.66528 log likelihood = -788.50743 log likelihood = -788.50705

Probit estimates Log likelihood = -788.50705

Number of obs = 2664 LR chi2(2) = 131.18 Prob > chi2 = 0.0000 Pseudo R2 = 0.0768

-----------------------------------------------------------------------------recwelfd | dF/dx Std. Err. z P>|z| x-bar [ 95% C.I. ] ---------+-------------------------------------------------------------------recwelfm*| .0876142 .020068 5.23 0.000 .155781 .048282 .126947 lnfaminc | -.0486469 .0070004 -6.80 0.000 9.48903 -.062367 -.034926 ---------+-------------------------------------------------------------------obs. P | .097973 pred. P | .0837966 (at x-bar) -----------------------------------------------------------------------------(*) dF/dx is for discrete change of dummy variable from 0 to 1 z and P>|z| are the test of the underlying coefficient being 0

5) The factors that lead to a statistically significant (at the 10% level of lower) impact on the likelihood that the death penalty is imposed are: (A) white victim, (B) aggravating circumstances, (C) victim was a stranger, and (D) there were multiple stab wounds. Each of these factors increased the likelihood that the death penalty was imposed. The magnitude of the effect for factors A, B, and C are quite large. In each case falling into that category increases the likelihood that the death penalty is imposed by roughly 30 percent.

. regress death blkdef-yngvict Source | SS df MS -------------+-----------------------------Model | 5.81563261 8 .726954076 Residual | 19.1743674 91 .210707334 -------------+-----------------------------Total | 24.99 99 .252424242

Number of obs = F( 8, 91) = Prob > F = R-squared = Adj R-squared = Root MSE =

100 3.45 0.0016 0.2327 0.1653 .45903

-----------------------------------------------------------------------------death | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------blkdef | -.0869168 .1102374 -0.79 0.432 -.3058898 .1320562 whtvict | .3052246 .1207463 2.53 0.013 .065377 .5450722 aggcirc | .0678655 .0371417 1.83 0.071 -.0059119 .1416429 fevict | .0790273 .1061283 0.74 0.458 -.1317836 .2898382 stranger | .3563889 .101464 3.51 0.001 .1548431 .5579348 multvic | .0499414 .1394044 0.36 0.721 -.2269684 .3268511 multstab | .2836468 .1517671 1.87 0.065 -.0178198 .5851134 yngvict | .050356 .1773002 0.28 0.777 -.3018291 .4025411 _cons | -.1867882 .2003356 -0.93 0.354 -.5847302 .2111538 ------------------------------------------------------------------------------

B) Estimates from the probit model are very similar to that from the linear probability model. The probit model has the advantage that we can correctly estimate standard errors, which are flawed in OLS because of the inherent heteroskedasticity in the model. In addition, the probit model limits our predicted probabilities to be above 0 and below 1, but the linear probability model does not. . dprobit death blkdef-yngvict Iteration 0: Iteration 1: Iteration 2: Iteration 3: Iteration 4:

log likelihood = -69.294717 log likelihood = -56.636345 log likelihood = -56.288085 log likelihood = -56.286721 log likelihood = -56.286721

Probit estimates Log likelihood = -56.286721

Number of obs = 100 LR chi2(8) = 26.02 Prob > chi2 = 0.0010 Pseudo R2 = 0.1877

-----------------------------------------------------------------------------death | dF/dx Std. Err. z P>|z| x-bar [ 95% C.I. ] ---------+-------------------------------------------------------------------blkdef*| -.1102602 .1281565 -0.86 0.393 .53 -.361442 .140922 whtvict*| .3375192 .1190266 2.51 0.012 .74 .104231 .570807 aggcirc | .0890258 .0454779 1.96 0.050 3.62 -.000109 .178161 fevict*| .094722 .1268127 0.74 0.457 .31 -.153826 .34327 stranger*| .4018255 .1072414 3.41 0.001 .51 .191636 .612015 multvic*| .0654536 .1619635 0.40 0.687 .14 -.251989 .382896 multstab*| .3191978 .1537351 1.79 0.074 .12 .017883 .620513 yngvict*| .0193519 .2305202 0.08 0.933 .08 -.432459 .471163 ---------+-------------------------------------------------------------------obs. P | .49 pred. P | .4807727 (at x-bar) -----------------------------------------------------------------------------(*) dF/dx is for discrete change of dummy variable from 0 to 1 z and P>|z| are the test of the underlying coefficient being 0

C) The predicted probability for the first set of circumstances is 0.92. If the only thing that changes is that the defendant is not black, then the predicted probability increases to 1.01. This is clearly impossible and points out one of the weaknesses in the linear probability model.

SOLUTIONS TO CHAPTER 17

1) random walk: A model that exhibits a random walk can be described as an AR(1) of the form: yt = 1 yt −1 +  t where  = 1. In this model, today’s value is equal to yesterday’s value plus a random error term.

ARMA: An ARMA model is a model that has both an autoregressive and a moving average component. Formally, an ARMA model is specified as:

yt = 0 + 1 yt −1 + 2 yt −2 + ... + m yt −m +  t +  1 t −1 +  2 t −2 + ... +  k t −k where this general version includes an autoregressive component of order m, AR(m), and a moving average component of order k, MA(k) and, hence, we say that this model is an ARMA(m,k). stationarity: If time series observations are drawn from the same underlying distribution, we say that the time series process is stationary. unit root: A unit root exists in the context of an AR(1) model of the form yt = 1 yt −1 +  t where  = 1 (i.e. a random walk). The unit root represents the fact that  = 1.

2) An MA(2) takes the form:

yt =  0 +  t + 1 t −1 +  2 t −2 Suppose that a shock occurs in period t, but there had been no shocks in the periods preceding that and none in the following periods. We would then have:

yt −1 =  0 yt =  0 +  t yt +1 =  0 +  1 t yt + 2 =  0 +  2 t yt +3 =  0

3) 0.25 0.20 0.15 0.10 0.05 0.00 1

2

3

4

5

6

7

8

9

10

4A) The upper bound for a Durbin-Watson test with 1 explanatory variable and 45 observations at the 5% level is 1.566. The Durbin-Watson statistic for this model 1.83, which is well above this cut-off, providing little evidence of first order serial correlation.

Based on this model the predictions that we get for 1997 through 1999 are considerably higher than what we actually observed. . regress unemp lagun1 if year <= 1996 Source | SS df MS -------------+-----------------------------Model | 63.2154238 1 63.2154238 Residual | 42.9476863 43 .998783403 -------------+-----------------------------Total | 106.16311 44 2.41279796

Number of obs = F( 1, 43) = Prob > F = R-squared = Adj R-squared = Root MSE =

45 63.29 0.0000 0.5955 0.5860 .99939

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lagun1 | .7502622 .0943055 7.96 0.000 .5600769 .9404474 _cons | 1.492371 .5658874 2.64 0.012 .3511505 2.633592 -----------------------------------------------------------------------------. dwstat Durbin-Watson d-statistic(

2,

45) =

1.831658

Based on this model the predictions that we get for 1997 through 1999 are considerably higher than what we actually observed. year 1997 1998 1999

unemp 4.9 4.5 4.2

unemphat 5.543787 5.168656 4.868551

B) In the AR(2) model, the 2nd lag of the unemployment rate is not statistically significant. In addition, the predictions for subsequent unemployment are similar, but slightly worse than what we obtained in the AR(1) model. As a result, one would probably prefer the AR(1). . regress unemp lagun1 lagun2 if year <= 1996 Source | SS df MS -------------+-----------------------------Model | 56.6636686 2 28.3318343 Residual | 41.2763304 41 1.00673977 -------------+-----------------------------Total | 97.9399991 43 2.2776744

Number of obs = F( 2, 41) = Prob > F = R-squared = Adj R-squared = Root MSE =

44 28.14 0.0000 0.5786 0.5580 1.0034

-----------------------------------------------------------------------------unemp | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lagun1 | .8224242 .1532625 5.37 0.000 .5129044 1.131944 lagun2 | -.1201272 .148901 -0.81 0.424 -.4208388 .1805844 _cons | 1.788476 .6138583 2.91 0.006 .5487635 3.028188 -----------------------------------------------------------------------------. list year unemp unemphat if year >= 1997 47.

48.

49.

year 1997 1998 1999

unemp 4.9 4.5 4.2

unemphat 5.556854 5.169667 4.900761

SOLUTIONS TO CHAPTER 18

1) In the estimated model, omitting the individual fixed effects could have an effect on both the coefficient estimates and their standard errors. The coefficient estimates would be affected by a standard omitted variable bias problem. If the individual fixed effect is correlated with both the outcome measure and any of the included explanatory variables, the coefficient estimates we would obtain by applying OLS to this model would be biased.

The standard errors that we estimate when we apply OLS will also be incorrect, because the formula we use to calculate these standard errors are based upon the assumption that the residuals between different observations are uncorrelated. If different observations in the sample reflect repeated observations for the same person, it seems like this assumption is unlikely to be satisfied. An individual who tends to be “luckier” than others in one period is likely to have a similar experience in a different period.

2A) The main problem with identifying a causal relationship in this case is that you would worry about the endogeneity of the imposition of the law. For instance, it is quite conceivable that the reason California instituted this law is because it has high crime rates. What we want to know is did the change in the law reduce the level even if it started out being high. But a comparison of the trend in crime rates in California before and after the law change is problematic as well because other things (economic growth?) could have been going on at the same time that would have altered the crime rate.

B) One approach to overcoming these problems is to employ a quasi-experimental design. California (and any other state instituting such a law) would represent the quasitreatment group since that is the location where the policy change took place. States with no such law changes would represent a quasi-control group. We could then compare the change in treatment group crime rates after the law was implemented with the change in control state crime rates at the same time. If crime rates fell in California by more than than they did in the other states, we would attribute credit to the law change.

C) The model we would estimate is:

CRst =  0 +  s +  t + 1lawst + 1 X1st +  2 X 2 st +  3 X 3st + ... +  k X kst +  st where CR represents the crime rate, law represents the presence of a “three strikes” law, the Xs represent other state characteristics that may change over time, and γs and γt represent state and year fixed effects.

D) One could use state level data aggregated by state to estimate this model. One may consider weighting this regression by the state population in each year, so this variable would be needed as well.

E) In this specification, we would examine the value of ˆ1 . We would consider the policy to be effective if this estimate was negative and statistically significant.

3A) The results of this regression show that public policies have have mixed effects in terms of reducing vehicle fatalities. Increasing the beer tax and the minimum legal drinking age along with instituting a per se law show no statistically significant effect on vehicle fatalities. Dram shop laws are found to reduce fatalities, but implied consent laws actually lead to a statistically significant INCREASE in fatalities. regress vfr beertax mlda perse dramlaw impcond unrate yngdrv perinc vmiles yr2yr7 Source | SS df MS -------------+-----------------------------Model | 45.2646089 15 3.01764059 Residual | 63.6321327 319 .19947377 -------------+-----------------------------Total | 108.896742 334 .326038148

Number of obs = F( 15, 319) = Prob > F = R-squared = Adj R-squared = Root MSE =

335 15.13 0.0000 0.4157 0.3882 .44662

-----------------------------------------------------------------------------vfrall | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------beertax | .0593581 .0594128 1.00 0.319 -.0575324 .1762486 mlda | -.0166876 .030208 -0.55 0.581 -.0761197 .0427446 perse | -.0165753 .0566646 -0.29 0.770 -.1280589 .0949084 dramlaw | -.1237995 .0545775 -2.27 0.024 -.2311769 -.0164222 impcond | .1121463 .0560813 2.00 0.046 .0018105 .2224822 unrate | .0189943 .0140215 1.35 0.176 -.008592 .0465807 yngdrv | 2.156517 1.388299 1.55 0.121 -.5748615 4.887896 perinc | -.0001007 .0000162 -6.21 0.000 -.0001327 -.0000688 vmiles | .0001281 .0000187 6.85 0.000 .0000913 .0001649 yr2 | -.071154 .0922751 -0.77 0.441 -.2526986 .1103906 yr3 | -.0300197 .098297 -0.31 0.760 -.223412 .1633725 yr4 | .0024444 .1006729 0.02 0.981 -.1956223 .2005112 yr5 | .1260969 .1061374 1.19 0.236 -.0827208 .3349145 yr6 | .1602504 .1134054 1.41 0.159 -.0628666 .3833674 yr7 | .198747 .123598 1.61 0.109 -.0444232 .4419172 _cons | 2.150031 .8172604 2.63 0.009 .5421299 3.757932 ------------------------------------------------------------------------------

B) The regression results are shown below. An f-test examining the null hypothesis that the state fixed effects jointly have no influence on the model is strongly rejected. The fstatistic for this test is 49.9, which is well above the critical value for a test with 47 and 272 degrees of freedom (the p-value for this test approaches zero). . areg vfr beertax mlda perse dramlaw impcond unrate yngdrv perinc vmiles yr2yr7, absorb(state) Number of obs = F( 15, 272) = Prob > F = R-squared = Adj R-squared = Root MSE =

335 11.45 0.0000 0.9393 0.9255 .1559

-----------------------------------------------------------------------------vfrall | Coef.

Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------beertax | -.468338 .1675023 -2.80 0.006 -.7981039 -.1385722 mlda | .0086224 .0182162 0.47 0.636 -.0272403 .0444851 perse | -.0695977 .0426074 -1.63 0.104 -.1534799 .0142846 dramlaw | -.1695569 .0527664 -3.21 0.001 -.2734394 -.0656744 impcond | .0675874 .0828815 0.82 0.416 -.0955833 .2307582 unrate | -.0715429 .0107478 -6.66 0.000 -.0927022 -.0503835 yngdrv | .3587643 .8776127 0.41 0.683 -1.369013 2.086541 perinc | .000061 .0000221 2.75 0.006 .0000174 .0001045 vmiles | .0000117 8.83e-06 1.33 0.185 -5.63e-06 .0000291 yr2 | -.0693518 .0337095 -2.06 0.041 -.1357164 -.0029872 yr3 | -.2315689 .0441478 -5.25 0.000 -.3184838 -.1446541 yr4 | -.3043802 .0490644 -6.20 0.000 -.4009744 -.2077861 yr5 | -.2415773 .0575327 -4.20 0.000 -.3548434 -.1283113 yr6 | -.3196 .0676172 -4.73 0.000 -.4527195 -.1864805 yr7 | -.3889697 .079887 -4.87 0.000 -.5462451 -.2316942 _cons | 1.937521 .5076522 3.82 0.000 .9380936 2.936947 -------------+---------------------------------------------------------------state | F(47, 272) = 49.915 0.000 (48 categories)

C) In the model with state fixed effects, public policies appear to have a greater impact on reducing vehicle fatalities. An increase in the beer tax and dram shop laws statistically significantly reduce deaths associated with drunk drivers. The other policies do not have a statistically significant effect. One explanation for the difference in results between the two models is that states differ in both their level of alcohol related driving deaths and the types of laws they have in place. For instance, states that have strict drunk driving laws may have them because they have a high rate of alcohol-related automobile deaths. This relationship does not address the question of whether or not these deaths declined when the law was instituted. Include state fixed effects allows us to answer questions such as this.

D) Because of the problem that omitted state characteristics may impose bias in the regression results, a model that includes state fixed effects is probably appropriate.

E) The policy recommendations that make sense based on these results is that states can benefit from raising the beer tax and imposing dram shop laws.