Student Solutions Manual to Accompany an Introduction to Econometrics: a Self-Contained Approach [1 ed.] 9780262317184, 9780262525404

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Student Solutions Manual to Accompany An Introduction to Econometrics: A Self-Contained Approach

Student Solutions Manual to Accompany An Introduction to Econometrics: A Self-Contained Approach

Frank Westhoff

The MIT Press Cambridge, Massachusetts London, England

© 2013 Massachusetts Institute of Technology All rights reserved. No part of this book may be reproduced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from the publisher. This book was printed and bound in the United States of America. ISBN: 978-0-262-52540-4

Contents

Chapter 1: Descriptive Statistics

1

Chapter 2: Essentials of Probability and Estimation Procedures

7

Chapter 3: Interval Estimates and the Central Limit Theorem

13

Chapter 4: Estimation Procedures, Estimates, and Hypothesis Testing

21

Chapter 5: Ordinary Least Squares Estimation Procedure—The Mechanics

25

Chapter 6: Ordinary Least Squares Estimation Procedure—The Properties

31

Chapter 7: Estimating the Variance of an Estimate’s Probability Distribution

37

Chapter 8: Interval Estimates and Hypothesis Testing

41

Chapter 9: One-Tailed Tests, Two-Tailed Tests, and Logarithms

49

Chapter 10: Multiple Regression Analysis—Introduction

59

Chapter 11: Hypothesis Testing and the Wald Test

63

Chapter 12: Model Specification and Development

67

Chapter 13: Dummy and Interaction Variables

73

Chapter 14: Omitted Explanatory Variables, Multicollinearity, and Irrelevant Explanatory Variables

77

Chapter 15: Other Regression Statistics and Pitfalls

83

Chapter 16: Heteroskedasticity

89

Chapter 17: Autocorrelation (Serial Correlation)

95

Chapter 18: Explanatory Variable/Error Term Independence Premise, Consistency, and Instrumental Variables

99

Chapter 19: Measurement Error and the Instrumental Variables Estimation Procedure

101

Chapter 20: Omitted Variables and the Instrumental Variable Estimation Procedure

105

Chapter 21: Panel Data and Omitted Variables

109

Chapter 22: Simultaneous Equations Models—Introduction

113

Chapter 23: Simultaneous Equations Models—Identification

117

Chapter 24: Binary and Truncated Dependent Variables

123

Chapter 25: Descriptive Statistics, Probability, and Random Variables— A Closer Look

129

Chapter 26: Estimating the Mean of a Population

133

Chapter 1: Descriptive Statistics Solutions to Chapter 1 Prep Questions 1. One reasonable way would be to calculate the average precipitation for each month and then choose the month with the highest average. Solutions to Chapter 1 Review Questions 1. a. The mean is the average of the values. It describes the center of the distribution. T

¦x

t

b.

Mean[ x ]

t 1

T

3. A histogram illustrates how the values of a single data variable are distributed. 5. a. Yes. b. No. 7. a. Cov[x, y] > 0 b. Cov[x, y] < 0 c. Cov[x, y] | 0

0 < CorrCoef[x, y] d 1 1 d CorrCoef[x, y] < 0 CorrCoef[x, y] | 0

Solutions to Chapter 1 Exercises 1. a. Inches of Precipitation 0–1 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11

1964 2 3 3 3 0 1 0 0 0 0 0

1975 0 0 2 3 3 1 1 0 1 0 1

1

b.

3. a. 1975 Precip TwoPlusPrecip b.

2

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 4.39 3.04 3.97 2.87 2.10 4.68 10.56 6.13 8.63 4.90 5.08 3.90 6.39 5.04 5.97 4.87 4.10 6.68 12.56 8.13 10.63 6.90 7.08 5.90

The histograms have identical shape. The TwoPlusPrecip histogram is just “slided” 2 inches to the right. ii. The center has risen (shifted to the right) by 2 inches. iii. The spread is unchanged. c. i. Mean[TwoPlusPrecip] = Mean[2 + Precip] = 2 + Mean[Precip] = 2 + 5.02 = 7.02 ii. Var[TwoPlusPrecip] = Var[2 + Precip] = Var[Precip] = 5.41 d. i. 7.02 ii. 64.91 iii. 5.41 e. Yes, the answers are consistent. The histogram reveals that the x center of the TwoPlusPrecip distribution is farther to the right; this is consistent with its higher mean. x spread of the two distributions is the same; this is consistent with their equal variances. The appropriate equation and statistical software provide the same answers. i.

3

5. a.

b. Yes, SatMath and SatVerbal appear to be correlated. As SatVerbal increases, SatMath tends to increase also. c. SatMath SatVerbal Mean 737.0 719.0 Variance 2,774.3 3,402.3 Covariance 2,103.7 Correlation Coefficient .6847 d. i. Mean[SatSum]= Mean[SatMath] + Mean[SatVerbal] = 737.0 + 719.0 = 1,456.0 ii. Var[SatSum] = Var[SatMath] + 2Cov[SatMath, SatVerbal] + Var[SatVerbal] = 2,774.3 + 2u2,103.7 + 3,402.3 = 10,384.0 e. i. 1,456.0 ii. 311,520 iii. 10,384.0 f. Yes 7. a. Mean[SatMath] = = =

4

x1  x2  ...  x30 30 x1  x2  ...  x10  x11  x12  ...  x30 30 x1  x2  ...  x10 x  x  ...  x30 + 11 12 30 30

= = = b. WgtFemale + WgtMale =

10 x1  x2  ...  x10 20 x11  x12  ...  x30 u u + 30 10 30 20 1 2 u Mean[SatMathFemale] + u Mean[SatMathMale] 3 3 WgtFemale Mean[SatMathFemale] + WgtMaleMean[SatMathMale]

1 2 + =1 3 3

c. Mean[SatMath] = 737 Mean[SatMathFemale] = 725 Mean[SatMathMale] = 743 Yes, the results are consistent. Mean[SatMath] = WgtFemale Mean[SatMathFemale] = = = =

+

WgtMaleMean[SatMathMale]

10 20 u Mean[SatMathFemale] + u Mean[SatMathMale] 30 30 1 2 u Mean[SatMathFemale] + u Mean[SatMathMale] 3 3 1 2 u 725 + u 743 3 3 725  1, 486 2, 211 = = 737 3 3

5

Chapter 2: Essentials of Probability and Estimation Procedures Solutions to Chapter 2 Prep Questions 1. a. There are 13 chances out of 52 that the card drawn is a heart; that is, the probability that the card drawn will be a heart equals

13 52

1 . 4

b. There are 4 chances out of 52 that the card drawn is an ace; that is, the probability that the card drawn will be an ace equals

4 52

1 . 13

c. There are 26 chances out of 52 that the card drawn is a red card; that is, the probability that the card drawn will be a red card equals

3. a. b. c. d. e.

26 52

1 . 2

Mean[cx] = c Mean[x] Mean[x + y] = Mean[x] + Mean[y] Var[cx] = Var[cx] = c2Var[x] Var[x + y] = Var[x + y] = Var[x] + 2Cov[x, y] + Var[y] When x and y are independent: Var[x + y] = Var[x + y] = Var[x] + Var[y]

Solutions to Chapter 2 Review Questions 1. A random process is a process whose outcome is uncertain; that is, we cannot determine the outcome with certainty until the process is completed. 3. A random variable is a variable that is associated with a random process. The value of a random variable cannot be determined with certainty before the experiment is conducted. 5. a. Beforehand, we cannot determine the numerical value of the random variable with certainty. b. Beforehand, we can often describe the probability distribution of the random variable. 7. a. Center of the random variable’s probability distribution. b. Spread of the random variable’s probability distribution.

7

Solutions to Chapter 2 Exercises 1. Suppose that you have a deck composed of the following 10 cards: 2k 2j 2i 2h 3k 3j 3i 4k 4j 5i Consider the following experiment: x Thoroughly shuffle the deck of 10 cards. x Draw one card. x Replace the card drawn. a. A random variable describes the outcome of a random process. The value of a random variable cannot be determined beforehand, before the random process is completed. v is a random variable because we cannot determine if it will equal 2, 3, 4, or 5 before the experiment is conducted. b. Since there are a total of 10 cards in the deck, the probability of drawing any one card is

1 : 10 Card 2k k 2j 2i 2h 3k

Probability

1 10 1 10 1 10 1 10 1 10

Card

8

2k 2j 2i 2h

3

3k 3j 3i

4

4k 4j

5

5i

1 10 1 10 1 10 1 10 1 10

3j 3i 4k 4j 5i

Now it is easy to calculate the probability distribution of v: v 2

Probability

Prob[v]

1 1 1 1 4 + + + = = .4 10 10 10 10 10 1 1 1 3 + + = = .3 10 10 10 10 1 1 2 + = = .2 10 10 10 1 = .1 10

v equals 2 whenever the 2k, 2j, 2i, or 2h is drawn. Therefore, the probability that v equals 2 is

1 1 1 1 1 + + + = = .4. Using analogous logic, the probability that v 10 10 10 10 10

equals 3 is .3, 4 is .2, and 5 is .1. c. Mean[v]= ¦ vProb[v] All v

Var[v]

= 2u.4 + 3u.3 + 4u.2 + 5u.1 = .8 + .9 + .8 + .5 = = ¦ (v  Mean[v])2 Prob[v]

3

All v

v 2 3 4 5

Mean[v] 3 3 3 3

v  Mean[v] 2  3 = 1 33=0 43=1 53=2

(v  Mean[v])2 1 0 1 4

Prob(v) .4 .3 .2 .1

Var[v]

= 1u.4 + 0u.3 + 1u.2 + 4u.1 = .4 + 0 + .2 + .4 = 1 d. v Portion of the repetitions 2 | 40% or .4 3 | 30% or .3 4 | 20% or .2 5 | 10% or .1 The simulation confirms the relative frequency interpretation of probability. After many, many repetitions, the distribution of the numerical values mirrors the probability distribution. e. Mean of the Numerical Values of v = 3 Variance of the Numerical Values of v = 1 The mean of the probability distribution equals the mean of the numerical values. The variance of the probability distribution equals the variance of the numerical values. 3. a. i.

If a 2 was drawn on the first draw, only three 2’s remain in the deck of 9 cards. Therefore, there are 3 chances in 9 of drawing a 2 on the second draw: Prob[v2 = 2 IF v1 = 2] =

3 9

1 = .3333… 3

ii. If a 2 was not drawn on the first draw, four 2’s remain in the deck of 9 cards. Therefore, there are 4 chances in 9 of drawing a 2 on the second draw: Prob[v2 = 2 IF v1 z 2] =

4 = .4444… 9

b. No. The random variables v1 and v2 are not independent. Knowing whether v1 equals 2 or not helps us predict the value of v2 because the value of v1 affects the probability of v2. c. i. If a 2 was drawn on the first draw, 3,999 2’s remain in the deck of 9,999 cards. Therefore, there are 3,999 chances in 9,999 of drawing a 2 on the second draw:

9

Prob[v2 = 2 IF v1 = 2] =

3,999 = .39993999… 9,999

ii. If a 2 was not drawn on the first draw, 4,000 2’s remain in the deck of 9,999 cards. Therefore, there are 4,000 chances in 9,999 of drawing a 2 on the second draw: Prob[v2 = 2 IF v1 z 2] = d.

4, 000 = .40004000… 9,999

Size of Population: Number of Cards in Original Deck 10 10,000 .4444… .40004000… Prob[v2 = 2 IF v1 z 2] .3333… .39993999… Prob[v2 = 2 IF v1 = 2] Difference .1111… .00010001…

As the size of the population increases, the probability of drawing a 2 on the 2nd draw is affected less by the card drawn on the 1st draw. When there are 10,000 cards in the original deck, the probability of drawing a 2 on the 2nd draw is .400 to the nearest thousandths regardless of what card was drawn on the 1st draw. e. As the size of the population increases, the independence assumption becomes a better approximation of reality. f. A professional pollster need not be concerned. 5. a. Since Archie’s arrows always land within 20 centimeters of the center, the probability that v is less than or equal to 20 must equal 1. b. Prob[Points t 6]

=

1 u.10u20 2

= 1.0

c. Prob[Points = 10]

=

1 u (.10  .08) u 4 + .08 u 4 2

= .04 + .32 = .36

10

d. Prob[Points = 9]

=

1 u (.08.06) u 4 + .06 u 4 2

= .04 + .24 = .28

e. Prob[Points = 7 or Points = 9] =

1 u (.06.02) u 4 + .02 u 8 2

= .08 + .16 = .24

30 1 = 60 2 5 1 ii. = 60 12 25 5 = iii. 60 12 1 1 7 + = b. i. 2 12 12 5 ii. 12 5 5 5 5 5 3,125 c. u u u u = = .0126 12 12 12 12 12 248,832

7. a. i.

11

Chapter 3: Interval Estimates and the Central Limit Theorem Solutions to Chapter 3 Prep Questions 1. a. A random variable describes the outcome of a random process (a draw, a coin toss, etc.). The value of a random variable cannot be determined beforehand, before the random process is completed. b. i. The value of the random variable beforehand. ii. The probability distribution of the random variable. 3.

[T1  v  v

Var

1

2



]

 vT

Since Var[cx] = c2Var[x]

=

1 Var  v1  v2  T2

[

]

 vT

Since Var[x + y] = Var[x] + Var[y] when x and y are independent; hence, the covariances are all 0.

=

1 (Var[v1 ]  Var[v2 ]  T2

 Var[vT ]

)

Since Var[v1] = Var[v2] = … = Var[vT] = p(1  p)

=

1 ( p(1  p)  p(1  p)  T2

 p(1 (1  p)

)

How many p(1  p) terms are there? A total of T.

=

1 (T u p(1  p)) T2 Simplifying.

p(1  p) T 5. x x

The mean of the probability distribution describes the distribution’s center. The variance of the probability distribution describes the distribution’s spread.

Solutions to Chapter 3 Review Questions 1. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close to the actual value is small. 13

3. As the sample size increases, the probability distribution of the random variable approaches the normal distribution. Solutions to Chapter 3 Exercises 1. a. Mean[SatSum]

= Mean[SatMath + SatVerbal] = Mean[SatMath] + Mean[SatVerbal] = 515 + 460 = 975

= Mean[SatMath  SatVerbal] = Mean[SatMath + (1)SatVerbal)] = Mean[SatMath]  Mean[SatVerbal] = 515  460 = 55 b. CorrCoef[SatMath, SatVerbal] = .0 Since SatMath and SatVerbal are independent. Mean[SatDiff]

CorrCoef[SatMath, SatVerbal] =

Cov[ SatMath, SatVerbal ] Var[ SatMath] Var[ SatVerbal ]

.0 =

Cov[ SatMath, SatVerbal ] 100 u100

Cov[SatMath SatVerbal] = .0 Var[SatSum] = Var[SatMath + SatVerbal] = Var[SatMath] + 2Cov[SatMath, SatVerbal] + Var[SatVerbal] = 10,000 + 0 + 10,000 = 20,000 Var[SatDiff] = Var[SatMath  SatVerbal] = Var[SatMath + (1)SatVerbal)] = Var[SatMath] + 2Cov[SatMath, (1)SatVerbal] + Var[(1)SatVerbal] = Var[SatMath]  2Cov[SatMath, SatVerbal] + Var[(1)SatVerbal] = 10,000 + 0 + 10,000 = 20,000 c. CorrCoef[SatMath, SatVerbal] = 1.0 Perfect positive correlation. CorrCoef[SatMath, SatVerbal] =

Cov[ SatMath, SatVerbal ] Var[ SatMath] Var[ SatVerbal ]

1.0 =

Cov[ SatMath, SatVerbal ] 100 u100

Cov[SatMath SatVerbal] = 10,000 Var[SatSum] = Var[SatMath + SatVerbal] = Var[SatMath] + 2Cov[SatMath, SatVerbal] + Var[SatVerbal] = 10,000 + 2u10,000 + 10,000 = 10,000 + 20,000 + 10,000 = 40,000 14

= Var[SatMath  SatVerbal] = Var[SatMath + (1)SatVerbal)] = Var[SatMath]  2Cov[SatMath, SatVerbal] + Var[(1)SatVerbal] = 10,000  2u10,000 + 10,000 = 10,000  20,000 + 10,000 = 0 d. CorrCoef[SatMath, SatVerbal] = .5 Var[SatDiff]

CorrCoef[SatMath, SatVerbal] =

Cov[ SatMath, SatVerbal ] Var[ SatMath] Var[ SatVerbal ]

.5 =

Cov[ SatMath, SatVerbal ] 100 u100

Cov[SatMath SatVerbal] = 5,000 Var[SatSum] = Var[SatMath + SatVerbal] = Var[SatMath] + 2Cov[SatMath, SatVerbal] + Var[SatVerbal] = 10,000 + 2u5,000 + 10,000 = 10,000 + 10,000 + 10,000 = 30,000 Var[SatDiff] = Var[SatMath  SatVerbal] = Var[SatMath + (1)SatVerbal)] = Var[SatMath]  2Cov[SatMath, SatVerbal] + Var[(1)SatVerbal] = 10,000  2u5,000 + 10,000 = 10,000  10,000 + 10,000 = 10,000 e. CorrCoef[SatMath, SatVerbal] = .5 CorrCoef[SatMath, SatVerbal] =

Cov[ SatMath, SatVerbal ] Var[ SatMath] Var[ SatVerbal ]

.5 =

f.

Cov[ SatMath, SatVerbal ] 100 u100

Cov[SatMath SatVerbal] = 5,000 Var[SatSum] = Var[SatMath + SatVerbal] = Var[SatMath] + 2Cov[SatMath, SatVerbal] + Var[SatVerbal] = 10,000 + 2u5,000 + 10,000 = 10,000  10,000 + 10,000 = 10,000 Var[SatDiff] = Var[SatMath  SatVerbal] = Var[SatMath + (1)SatVerbal)] = Var[SatMath]  2Cov[SatMath, SatVerbal] + Var[(1)SatVerbal] = 10,000  2u5,000 + 10,000 = 10,000 + 10,000 + 10,000 = 30,000 0.0_____ 1.0_____ between 0.0 and 1.0 X less than 0.0 _____ Typically, students who do well on one part of the SAT do well on the other. 15

3. a. No b. Yes c. StateDiff = SatMichigan  SatAlaska Mean[StateDiff] = Mean[SatMichigan  SatAlaska] = Mean[SatMichigan + 1uSatAlaska)] = Mean[SatMichigan] + Mean[1uSatAlaska] = Mean[SatMichigan]  Mean[SatAlaska] = 549  489 = 60 SD[SatMichigan] = SD[SatAlaska] = 100 Var[SatMichigan] = Var[SatAlaska] = 10,000 Cov[SatMichigan, SatAlaska] = 0 SatMichigan and SatAlaska are independent Var[StateDiff] = Var[SatMichigan  SatAlaska] = Var[SatMichigan + 1uSatAlaska] = Var[SatMichigan] + 2Cov[SatMichigan, SatAlaska] + Var[SatAlaska] = 10,000 + 0 + 10,000 = 20,000

SD[ StateDiff ]

z

0  60 141.4



Var[ StateDiff ]

45 141.4

.42

Prob[SatMichigan > SatAlaska]

z 0.2 0.3 0.4 0.5

0.00 0.4207 0.3821 0.3446 0.3085

5. a. Mean[MPG] = 32 SD[MPG] = 4

16

20,000 141.4

0.01 0.4168 0.3783 0.3409 0.3050

0.02 0.4129 0.3745 0.3372 0.3015

= Prob[SatMichigan  SatAlaska > 0] = Prob[StateDiff > 0] = .66

0.03 0.4090 0.3707 0.3336 0.2981

0.04 0.4052 0.3669 0.3300 0.2946

0.05 0.4013 0.3632 0.3264 0.2912

0.06 0.3974 0.3594 0.3228 0.2877

0.07 0.3936 0.3557 0.3192 0.2843

0.08 0.3897 0.3520 0.3156 0.2810

0.09 0.3859 0.3483 0.3121 0.2776

z

30  32 4



2 4

.5

Prob[MPH > 110] = .69 = 69 percent z 0.4 0.5 0.6

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451

z 1.1 1.2 1.3

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823

b.

To achieve that objective, z must equal 1.29.

z

30  x 4

1.29

30  x = 4u1.29 30  x = 5.16 30 + 5.16 = x x = 35.16 The mean must be increased by 35.16  32 = 3.16. c.

z 2 x x

30  32 x

1.29

1.29 2 1.55 1.29

The standard deviation must be decreased by 4  1.55 = 2.45.

17

d. Approach #1 Cost = 3.16u100,000 = $316,000 Approach #2 Cost = 2.45u200,000 = $490,000 Approach #1 is less costly.

7. a. i. Recall that when betting on 1st set of twelve numbers, the mean of vi’s probability distribution equals .027 and the variance equals 1.972. = Mean[v1 + v2 + … + v50] = Mean[v1] + Mean[v2] + … + Mean[v50] = .027  .027  …  .027 = 1.35 = Var[v1 + v2 + … + v50] Since the v’s are independent, all the covariances equal 0. = Var[v1] + Var[v2] + … + Var[v50] = 1.972 + 1.972 + … + 1.972 = 98.60

Mean[Your TNW]

Var[Your TNW]

ii. Mean[TNW] = 1.35 SD[TNW] =

Var[TNW ] =

98.60 9.93 10  (1.35) z 9.93

11.35 1.14 9.93

Prob[TNW t 10] = .1271 z 1.0 1.1 1.2 iii.

0.00 0.01 0.02 0.1587 0.1562 0.1539 0.1357 0.1335 0.1314 0.1151 0.1131 0.1112 Mean[TNW] = 1.35 SD[TNW] =

Var[TNW ] =

98.60 9.93 10  (1.35) z 9.93

8.65 9.93

Prob[TNW d 10] = .1922

18

0.03 0.1515 0.1292 0.1093

.87

0.04 0.1492 0.1271 0.1075

0.05 0.1469 0.1251 0.1056

0.06 0.1446 0.1230 0.1038

0.07 0.1423 0.1210 0.1020

0.08 0.1401 0.1190 0.1003

0.09 0.1379 0.1170 0.0985

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.7 0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 b. i. Recall that when betting on red, the mean of vi’s probability distribution equals .027

0.09 0.2148 0.1867 0.1611

and the variance equals .999. Mean[Friend’s TNW] = Mean[v1 + v2 + … + v50] = Mean[v1] + Mean[v2] + … + Mean[v50] = .027  .027  …  .027 = 1.35 Var[Friend’s TNW] = Var[v1 + v2 + … + v50] Since the v’s are independent, all the covariances equal 0. = Var[v1] + Var[v2] + … + Var[v50] = .999 + .999 + … + .999 = 49.50 ii. Mean[TNW] = 1.35 SD[TNW] =

Var[TNW ] =

49.50 7.04 10  (1.35) z 7.04

11.35 1.61 7.04

Prob[TNW t 10] = .0537 z 0.00 0.01 0.02 1.5 0.0668 0.0655 0.0643 1.6 0.0548 0.0537 0.0526 1.7 0.0446 0.0436 0.0427 iii. Mean[TNW] = 1.35 SD[TNW] =

0.03 0.0630 0.0516 0.0418

0.04 0.0618 0.0505 0.0409

0.05 0.0606 0.0495 0.0401

0.06 0.0594 0.0485 0.0392

0.07 0.0582 0.0475 0.0384

0.08 0.0571 0.0465 0.0375

0.09 0.0559 0.0455 0.0367

0.04 0.1271 0.1075 0.0901

0.05 0.1251 0.1056 0.0885

0.06 0.1230 0.1038 0.0869

0.07 0.1210 0.1020 0.0853

0.08 0.1190 0.1003 0.0838

0.09 0.1170 0.0985 0.0823

Var[TNW ] =

97.23 9.86 10  (1.35) z 7.04

8.65 7.04

1.23

Prob[TNW d 10] = .1093

z 1.1 1.2 1.3 c.

0.00 0.1357 0.1151 0.0968

0.01 0.1335 0.1131 0.0951

0.02 0.1314 0.1112 0.0934

0.03 0.1292 0.1093 0.0918

Your friend appears to be more risk averse than you.

19

Chapter 4: Estimation Procedures, Estimates, and Hypothesis Testing Solutions to Chapter 4 Prep Questions 1. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close the actual value is small. 3. a. Jury Finds Defendant Guilty

Jury Finds Defendant Innocent

Defendant Actually Innocent

Jury is correct__ incorrect X

Jury is correct X incorrect__

Defendant Actually Guilty

Jury is correct X incorrect__

Jury is correct__ incorrect X

For each scenario, indicate whether the jury would be correct or incorrect. b. Defendant Actually Innocent and Jury Finds Defendant Guilty Costs: Imprisoning an innocent young man. Defendant Actually Guilty and Jury Finds Defendant Innocent Costs: Allowing a guilty man to go free who then can continue to commit crimes. Solutions to Chapter 4 Review Questions 1. a. x x

The mean of the estimate’s probability distribution. The variance of the estimate’s probability distribution.

x

The mean of the estimate’s probability distribution determines whether or not the estimation procedure is biased or unbiased. If the mean equals the actual value, the estimation procedure does not systematically underestimates or overestimates the actual value.

b.

21

x

When the estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates the reliability of the estimation procedure. As the variance becomes smaller, the estimate becomes more reliable.

3. a. Type I error occurs when we reject the null hypothesis when it is actually true. b. Type II error occurs when we do not reject the null hypothesis when it is actually false. c. As the significance level becomes smaller, it becomes more difficult to reject the null hypothesis thereby decreasing the likelihood of Type I error and increasing the likelihood of Type II error. Solutions to Chapter 4 Exercises 1. a. This statement is false. Even when an estimation procedure is unbiased we can never expect the estimate to equal the actual value. b. This statement is true whenever the estimate’s probability distribution is symmetric. c. This is always true; it is the definition of an unbiased estimation procedure. d. This statement is always true. The relative frequency interpretation of probability tells us that after many, many repetitions the distribution of the actual values mirrors the probability distribution. Accordingly, after many, many repetitions the average of the estimates would equal the mean of the estimate’s probability distribution; the estimation procedure would be unbiased. 3. a. b.

DemVoterFrac 17 17  15

1,885,178 1,885,178  2,290,723

1,885,178 .45144 | .451 4,175,901

17 .53125 | .531 32

c. Yes. d. Cynic’s View: Despite the election results, there is no unfair districting; there is no gerrymandering. “Sure, the election results suggest that there may be gerrymandering, but the results were just a coincidence. The results were just a random occurrence, a consequence of the luck of the draw.” Ÿ Cynic is correct. e. H0: No gerrymandering Ÿ Cynic is incorrect. H1: Gerrymandering present Prob[Results IF H0 True] = Probability that DemCongressFrac would be .531 or more when DemVoterFrac, the fraction of voters statewide voted for the Democratic candidate, equals .451 f. Mean[DemCongressFrac]: .451 Var[DemDemocratFrac]: .007737 SD[DemCongressFrac]: .088 Mean[DemCongressFrac] = p = .451

22

where p = fraction of population statewide voting Democratic

Var[ DemCongressFrac] SD[ DemCongressFrac]

p(1  p) .451u (1  .451) T 32 Var[ DemCongressFrac]

g. Prob[Results IF H0 True] =

z

.531  .451 .088 z 0.8 0.9 1.0

.451u .549 .007737 32 .007737 .08796 | .088

Probability that DemCongressFrac would be .531 or more when DemVoterFrac, the fraction of voters statewide voted for the Democratic candidate, equals .451

.080 .91 .088

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379

Prob[Results IF H0 True] = .1814 There is about one chance in five that the election results were not a consequence of gerrymandering.

5. a.

DemVoterFrac b.

365 365  173

69, 498, 215 69, 498, 215  59,948, 240

69, 498, 215 .5369 129, 446, 455

365 .6784 538

c. Yes. Democratic. d. Cynic’s View: Despite the election results, the Electoral College is not unfair, it does not favor Democrats. “Sure, the election results suggest that the Electoral College favors Democrats, but the results were just a coincidence. The results were just a random occurrence, a consequence of the luck of the draw.” Ÿ Cynic is correct. e. H0: No Electoral College Unfairness Ÿ Cynic is incorrect. H1: Electoral College Does Not Favor Democrats = Probability that DemElectColFrac would be .6784 or Prob[Results IF H0 True] more when DemVoterFrac, the fraction of voters nationwide voted for the Democratic candidate, equals .5369

23

f.

Mean[DemElectColFrac]: .5369 Var[DemElectColFrac]: .0004622 SD[DemElectColFrac]: .0215 Mean[DemElectColFrac] = p = .5369

SD[ DemElectColFrac]

= Probability that DemElectColFrac would be .6784 or more when DemVoterFrac, the fraction of voters nationwide voted for the Democratic candidate, equals .5369

g. Prob[Results IF H0 True]

.6784  .5369 .0215

fraction of Democratic voters nationwide

p(1  p) .5369 u (1  .5369) .5369 u .4631 .0004622 T 538 538 Var[ DemElectColFrac] .0004622 .0215

Var[ DemElectColFrac]

z

where p =

.1415 .0215

6.55

Prob[Results IF H0 True] < .001 7. a. Yes. On average, you would expect her to win a little less than one third of the time. In fact, she won more than two-thirds of the time. b. Cynic’s View: Despite the results, everything was on the “up and up.” “Sure, the results suggest that something was not on the “up and up,” but the results were just a coincidence. The results were just a random occurrence, a consequence of the luck of the draw.” ŸCynic is correct. c. H0: Everything Was on the “Up and Up” Ÿ Cynic is incorrect. H1: Roulette Wheel Was Rigged to Favor the Girlfriend Prob[Results IF H0 True] = Probability that the casino manager’s girlfriend would win $20 or more if everything was on the “up and up.” = Prob[TNW t 20 IF everything were on the “up and up”] d. Mean[TNW] = 1.35 SD[TNW] =

z

Var[TNW ] =

20  (1.35) 9.93

21.35 9.93

98.60

9.93

2.15

Prob[TNW t 20 IF everything on the “up and up”] = .0158 z 2.0 2.1 2.2

24

0.00 0.0228 0.0179 0.0139

0.01 0.0222 0.0174 0.0136

0.02 0.0217 0.0170 0.0132

0.03 0.0212 0.0166 0.0129

0.04 0.0207 0.0162 0.0125

0.05 0.0202 0.0158 0.0122

0.06 0.0197 0.0154 0.0119

0.07 0.0192 0.0150 0.0116

0.08 0.0188 0.0146 0.0113

0.09 0.0183 0.0143 0.0110

Chapter 5: Ordinary Least Squares Estimation Procedure— The Mechanics Solutions to Chapter 5 Prep Questions 1. a.

b. Yes. For the most part, when income increases, savings increase also. c.

25

By choosing two points on this line, we can solve for the equation of the best fitting line. It looks like the points (200, 15) and (1200, 155) are more or less on the line. We can use these two points to estimate the slope:

155  15 1200  200

140 1000

.14

Now, a little algebra:

y  15 x  200

.14

y  15 = .14x  28 y = .14x  13 This equation suggests that if Americans earn an additional $1,000 of income, savings will rise by $140. This is consistent with our theory. d. $140 million since the slope is .14: .14u$1,000,000,000 = $140,000,000 3. a. Before the experiment is conducted, we cannot determine the numerical value of the random variable with certainty. b. Before the experiment is conducted, we can often calculate the random variable’s probability distribution telling us how likely it is for the random variable to equal each of its possible numerical values. 5. Begin with the equation for SSR:

( y1  bConst  bx x1 )2

SSR



( y2  bConst  bx x2 )2



( y3  bConst  bx x3 )2

Differentiate SSR with respect to bConst and set the derivative equal to 0:

dSSR dbConst

2( y1  bConst  bx x1 )

 2( y2  bConst  bx x2 )

 2( y3  bConst  bx x3 )

Dividing by 2.

( y1  bConst  bx x1 )  ( y2  bConst  bx x2 )  ( y3  bConst  bx x3 ) 0 Collecting like terms.

( y1  y2  y3 )  (bConst  bConst  bConst )  (bx x1  bx x2  bx x3 ) 0 Simplifying.

( y1  y2  y3 )  3bConst  bx ( x1  x2  x3 ) 0 Dividing by 3.

y1  y2  y3 x x x  bConst  bx 1 2 3 3 x

0

y1  y2  y3 x  x2  x3 equals the mean of y, y , and 1 equals the mean of x, x : 3 3 y  bConst  bx x 0

Since

26

0

Therefore,

bConst

y  bx x .

Solutions to Chapter 5 Review Questions 1. The ordinary least squares (OLS) estimation procedure minimizes the sum of squared residuals to determine the best fitting line. 3. The parameter estimates are random variables. Solutions to Chapter 5 Exercises 1. a.

b. First, calculate the

x y

means:

5  10  30 45 15 3 3 14  44  80 138 46 3 3

Second, for each student calculate the deviation of x from its mean and the deviation of y from its mean: xt Student yt xt  x yt  y x y 1 14 46 32 5 15 10 2 44 46 2 10 15 5 3 80 46 34 30 15 15 Third, calculate the products of the y and x deviations and squared x deviations for each student; then calculate the sums: 27

( xt  x )2

( yt  y )( xt  x )

Student

2

32u10 = 320

1

2u5 =

2

10 = 100 2

5 = 25

10

2

3

34u15 = 510 Sum = 840 Now, apply the equations for bx and bConst:

15 = 225 Sum = 350

T

¦ ( y  y )( x  x ) t

t

t 1

bx

T

¦ (x  x )

2

840 350

2.4

t

t 1

bConst

y  bx x

46  2.4 u15 46  36 10

c. Calculate the sum of squared residuals by filling in the following blanks: Student

yt

Estyt = bConst + bxxt

Rest = yt  Estyt

Rest2

14  22 = 8 64 44  34 = 10 100 80  82 = 2 4 SSR = 168 Economic theory teaches that the supply curve is upward sloping. Hence, an increase in the price should increase oil production. 1 2 3

3. a.

xt 5 10 30

14 44 80

10 + 2.4u5 = 22 10 + 2.4u10 = 34 10 + 2.4u30 = 82

b. Ordinary Least Squares (OLS) Dependent Variable: OilProdBarrels Estimate SE Explanatory Variable(s): Price 92.50159 28.12319 Const 5944.958 478.7771 Number of Observations

t-Statistic 3.289157 12.41696

Prob 0.0028 0.0000

29

EstOilProd = 5,945 + 92.5Price Estimated Equation: c. Yes. The coefficient estimate for the price is positive suggesting that an increase in price increases oil production. d. i. Increase oil production by 92.5 thousand barrels. ii. Increase oil production by 185.0 thousand barrels. iii. Decrease oil production by 462.5 thousand barrels. 5. a. Economic theory teaches that the demand curve is downward sloping. Hence, an increase in the price should decrease gasoline consumption. b. Ordinary Least Squares (OLS) Dependent Variable: GasCons Estimate SE t-Statistic Prob Explanatory Variable(s): PriceDollars 151.6556 47.57295 3.187853 0.0128 Const 516.7801 60.60223 8.527410 0.0000 28

Number of Observations Sum Squared Residuals SE of Regression

10 54.00000 7.348469

Estimated Equation: EstGasCons = 517  151.7PriceDollars c. Yes. The coefficient estimate for the price is negative suggesting that an increase in price decreases gasoline consumption. d. i. Decrease gasoline consumption by 151.7 million gallons. ii. Decrease gasoline consumption by 303.3 million gallons. iii. Decrease gasoline consumption by 75.8 million gallons. iv. Increase gasoline consumption by 151.7 million gallons. v. Increase gasoline consumption by 303.3 million gallons. 7. a. Earmarks are projects initiated by members of Congress that circumvent merit-based criteria. b. Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE t-Statistic Prob Explanatory Variable(s): PartyDem1 2.361883 0.577102 4.092661 0.0001 Const 5.613527 0.424482 13.22441 0.0000 Number of Observations 451 EstNumber = 5.6 + 2.3PartyDem1 Estimated Equation: c. Yes. The coefficient estimate for the Democratic party explanatory variable is positive suggesting that Democratic members of Congress receive more earmarks than their Republican colleagues.

29

Chapter 6: Ordinary Least Squares Estimation Procedure— The Properties Solutions to Chapter 6 Prep Questions 1. NB: Because random influences are actually random, the numerical values that you reported will no doubt differ from those below. Therefore, your means and variances which are based on your numerical values will also differ.

Repetition 1 2 3

Numerical Value of Coefficient Estimate 3.40 4.70 1.44

Your Mean

Calculations Variance

Simulation’s Mean

Calculations Variance

4.05 3.18

.4225 1.7955

4.1 3.2

.4 1.8

Repetition 2 Calculations:

Mean

3.40  4.70 2 Value 3.40 4.70

Variance

8.10 2 Mean 4.05 4.05

4.05 Deviation Squared From Mean Deviation .65 .4225 .65 .4225 Sum = .8450

Sum of Squared Deviations 2

.8450 .4225 2

Repetition 3 Calculations:

Mean

3.40  4.70  1.44 3 Value 3.40 4.70 1.44

Variance

Mean 3.18 3.18 3.18

9.54 3

3.18

Deviation From Mean .22 1.52 1.74 Sum =

Sum of Squared Deviations 3

Squared Deviation .0484 2.3104 3.0276 5.3864

5.3864 1.7955 3

3. a. Variance of a constant times a variable: Var[cx] = c2Var[x] b. Variance of the sum of a variable and a constant: Var[c + x] = Var[x] 31

c. Variance of the sum of two variables: Var[x + y] = Var[x] + 2Cov[x, y] + Var[y] d. Variance of the sum of two independent variables: Var[x + y] = Var[x] + Var[y] Solutions to Chapter 6 Review Questions 1. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close the actual value is small. 3. a. The mean of the coefficient estimate’s probability distribution is unaffected. b. The variance of the coefficient estimate’s probability distribution decreases. c. The coefficient estimate becomes more reliable. When the sample size increases more information is available. 5. When the ordinary least squares (OLS) standard premises are satisfied, the ordinary least squares (OLS) estimation procedure is the best linear unbiased estimation procedure (BLUE). Solutions to Chapter 6 Exercises

1. a.

bSlope

y j  yi x j  xi

b. bSlope

y j  yi x j  xi

E Const  E x x j  e j  ( EConst  E x xi  ei ) x j  xi

E Const  E x x j  e j  E Const  E x xi  ei ) x j  xi

E x x j  E x xi  e j  ei x j  xi

E x ( x j  xi )  (e j  ei ) x j  xi

Ex  Ex 

32

e j  ei x j  xi 1 (e j  ei ) x j  xi

c.

Mean[bSlope ]

[

Mean E x 

1 (e j  ei ) x j  xi

]

[ x 1 x (e  e )]

E x  Mean

j

j

i

i

[

]

Ex 

1 Mean e j  ei ) x j  xi

Ex 

1 Mean[e j ]  Mean[ei ] x j  xi

[

]

Ex d. What does the variance of bSlope‘s probability distribution, Var[bSlope], equal?

Var[bSlope ]

[

Var E x 

1 (e j  ei ) x j  xi

]

[ x 1 x (e  e )]

Var

j

j

i

i

[

]

1 Var e j  ei ) ( x j  xi ) 2

[

]

1 Var e j  (ei )] ( x j  xi ) 2

[

]

[

]

1 Var[e j ]  Var[ei ] ( x j  xi ) 2 1 Var[e j ]  Var[ei ] ( x j  xi ) 2

[

1 Var[e]  Var[e] ( x j  xi ) 2

]

2 Var[e] ( x j  xi ) 2 3. a. There are 10 possible pair wise combinations of the 5 observations: Obs 1 and 2 Obs 1 and 3 Obs 1 and 4 Obs 1 and 5 Obs 2 and 3 Obs 2 and 4 Obs 2 and 5 Obs 3 and 4 Obs 3 and 5 Obs 4 and 5 Since the two observations are chosen randomly the probability of each of these

1 . 10 1 1 = uMean[bSlope for obs 1 and 2] + … + uMean[bSlope for obs 4 and 5] 10 10 1 1 = uEx + … + uE x 10 10 = Ex

combinations is Mean[bx]

33

b. The slopes of the 10 pair wise combinations of observations are independent so we need not worry about the covariances. Var[bx]

1 1 u Var[bSlope for Obs 1 and 2]  u Var[bSlope for Obs 1 and 3] 10 10 1 +...+ u Var[bSlope for Obs 4 and 5] 10 1 2 1 2 u u Var[e]  u u Var[e] 2 10 ( x1  x2 ) 10 ( x1  x3 ) 2

+...+

1 2 u u Var[e] 10 ( x4  x5 ) 2

1 2 1 2 u u 500  u u 500 2 10 ( x1  x2 ) 10 ( x1  x3 ) 2 +...+

1 2 u u 500 10 ( x4  x5 ) 2

1 1000 1 1000 u  u 2 10 ( x1  x2 ) 10 ( x1  x3 ) 2 +...+

1 1000 u 10 ( x4  x5 ) 2

1 1000 1 1000 1 1000 1 1000 u  u 2  u 2  u 2 10 62 10 12 10 18 10 24 1 1000 1 1000 1 1000 + u 2  u 2  u 2 10 6 10 12 10 18 1 1000 1 1000 + u 2  u 2 10 6 10 12 1 1000 + u 2 10 6 100 100 100 100    62 122 182 242 100 100 100 + 2  2 2 6 12 18 100 100 + 2  2 6 12 100 + 2 6 400 300 200 100    62 122 182 242 11.1  2.08  .62  .17 13.98 c. Yes 34

5. a. i. Ordinary Least Squares (OLS) Dependent Variable: OilProdBarrels Estimate SE Explanatory Variable(s): Price 92.50159 28.12319 Const 5944.958 478.7771

t-Statistic 3.289157 12.41696

Prob 0.0028 0.0000

Number of Observations 29 EstOilProdBarrels = 5,945 + 92.5Price Estimated Equation: ii. Ordinary Least Squares (OLS) Dependent Variable: OilProdBarrelsPlus1000 Estimate SE t-Statistic Explanatory Variable(s): Price 92.50159 28.12319 3.289157 Const 6944.958 478.7771 12.41696

Prob 0.0028 0.0000

Number of Observations 29 EstOilProdBarrelsPlus1000 = 6,945 + 92.5Price Estimated Equation: Best fitting line: EstOilProdBarrelsPlus1000 = 6,945 + 92.5Price b. The best fitting line shifts up by 1,000. c. All the points in the scatter diagram have shifted up by 1,000. Consequently, the best fitting line, the line that minimizes the sum of squared residuals has also shifted up by 1,000. 7. a. i. Ordinary Least Squares (OLS) Dependent Variable: GasCons Estimate SE t-Statistic Explanatory Variable(s): PriceDollars 47.57295 151.6556 3.187853 Const 516.7801 60.60223 8.527410

Prob 0.0128 0.0000

Number of Observations 10 Estimated Equation: EstGasCons = 516.8  151.7PriceDollars

35

ii. Ordinary Least Squares (OLS) Dependent Variable: GasCons Estimate SE t-Statistic Explanatory Variable(s): PriceDollarsPlus2 47.57295 151.6556 3.187853 Const 820.0912 155.6644 5.268328 Number of Observations 10 Estimated Equation: EstGasCons = 820.1  151.7PriceDollarsPlus2 b. The estimated constants differ. c. The estimated coefficients are identical. d. All the points in the scatter diagram have shifted to the right by 2.00. Consequently, the best fitting line, the line that minimizes the sum of squared residuals has also shifted to the right by 2.00.

36

Prob 0.0128 0.0008

Chapter 7: Estimating the Variance of an Estimate’s Probability Distribution Solutions to Chapter 7 Prep Questions 1. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other variance, if the variance is large, the probability that the estimate lies close the actual value is small. 3. a.

Rise 90  66 24 Run 25  5 20 y  66 6 x 5 5 5 y  330 6 x  30 5 y 6 x  300 6 y x  60 1.2 x  60 5

b. Slope =

6 5

c. 0 d. i. Slope =

Rise Run

90  70 25  5

20 1 20

y  70 1 x5 y  70 x  5 y x  65 ii.

0

Rise 90  86 Run 25  5 y  86 1 x 5 5 5 y  430 x  5 5 y x  425 1 y x  85 1.2 x  85 5

e. i. Slope =

f.

ii. 0

4 20

1 5

0

37

Solutions to Chapter 7 Review Questions 1.

Var[e ]

Var[bx ]

T

¦ (x  x )

2

t

t 1

3. While this is an unbiased estimation procedure for the variance of the error term’s probability distribution, it does not help us. We can never compute the values of the error terms because the error terms are unobservable. 5. We divide by the degrees of freedom because they equal the number of pieces of information that are available to estimate the variance of the error term’s probability distribution. Solutions to Chapter 7 Exercises 1. First, calculate the means:

x y

5  10  30 45 15 3 3 14  44  80 138 46 3 3

Second, for each student calculate the deviation of x from its mean and the deviation of y from its mean: xt Student yt xt  x yt  y x y 1 14 46 32 5 15 10 2 44 46 2 10 15 5 3 80 46 34 30 15 15 Third, calculate the products of the y and x deviations and squared x deviations for each student; then calculate the sums:

( yt  y )( xt  x )

Student

32u10 = 320 2u5 = 10 34u15 = 510 Sum = 840 Now, apply the formulas: 1 2 3

T

¦ ( y  y )( x  x ) t

bx

t

t 1

T

¦ (x  x )

2

840 350

2.4

t

t 1

bConst

38

y  bx x

46  2.4 u15 46  36 10

( xt  x )2 102 = 100 52 = 25 152 = 225 Sum = 350

3. Finally, use the quiz data to estimate the variance and standard deviation of the coefficient estimate’s probability distribution, the coefficient estimate’s standard error. Estimated Variance of the Coefficient Estimate’s Probability Distribution:

EstVar[e ]

EstVar[bx ]

168 350

T

¦ (x  x )

2

.48

t

t 1

Standard Error of the Coefficient Estimate:

SE[bx ] 5. a. Yes X

EstVar[bx ]

.48

.69

No ____

SE of Regression

EstVar[e] 78.09 EstVar[e ] 78.092

b. Yes X

No ____

SE[bx ]

EstVar[bx ]

4.49

EstVar[bx ] 4.492 c. Yes X

6, 098

20.16

No ____

EstVar[e ]

EstVar[bx ]

T

¦ (x  x )

2

t

t 1

T

¦ (x  x )

EstVar[e ] EstVar[bx ]

2

t

t 1

d. Yes X

302.48

No ____

T

¦(y

t

bx

6, 098 20.16

 y )( xt  x )

t 1

T

¦ (x  x )

2

t

t 1

T

T

¦ ( yt  y )( xt  x ) bx u ¦ ( xt  x )2 15.26 u 302.48 4, 615.8 t 1

e. Yes

t 1

X

No ____

SSR Degrees of Freedom EstVar[e ] u Degrees of Freedom 6, 098 u (10  2) 6, 098 u 8 48, 784

EstVar[e ] SSR f.

Yes ____

No X

39

Chapter 8: Interval Estimates and Hypothesis Testing Solutions to Chapter 8 Prep Questions 1.

x

5  10  30 3

T

¦ (x  x )

45 15 3

(5  15)2  (15  15) 2  (25  15) 2 102  02  102 100  0  100 200

2

t

t 1

Var[bx ]

Var[e ] T

¦ (x  x )

2

50 200

.25

t

t 1

|68.5% |95.5% |99.7% Actual Values Ex Var[e] 2 2 2

50 50 50

From Value

To Value

Repetitions Between From and to Values

1.5 1.0 .5

2.5 3.0 3.5

|68.5% |95.5% |99.7%

3. z equals the number of standard deviations the value lies from the mean:

z

Value of Random Variable  Distribution Mean Distribution Standard Deviation Number of Stadard Deviations from the Mean

Solutions to Chapter 8 Review Questions 1. Interval Estimate Question: What is the probability that the estimate, _____, lies within ____ of the actual value____? 3. x x x x x x

Step 0: Formulate a model reflecting the theory to be tested. Step 1: Collect data, run the regression, and interpret the estimates. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Step 3: Formulate the question to assess the null hypothesis and the cynic’s view. Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True]. Step 5: Decide on the standard of proof, a significance level.

41

5. x x

Each estimation procedure is unbiased; each estimation procedure does not systematically underestimate or overestimate the actual value. The ordinary least squares (OLS) estimation procedure for the coefficient value is the best linear unbiased estimation procedure (BLUE).

Solutions to Chapter 8 Exercises 1. a. PetroConsPC = PetroCons*1000/Pop PriceReal = PriceNom*100/CPI b. Ordinary Least Squares (OLS) Dependent Variable: PetroConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceReal 540.5367 203.4389 2.656999 Const 1418.403 144.7593 9.798354 Number of Observations

Prob 0.0289 0.0000

10

Estimated Equation: EstPetroConsPC = 1,418  540.5PriceReal Interpretation of Estimates: bPriceReal = 540.5: A $1 increase in the real price of gasoline reduces per capita Nebraska petroleum consumption by 540.5 gallons. c. The Student t-distribution must be used since we must estimate the variance of the coefficient estimate’s probability distribution. d. First, note that the degrees of freedom equal 8: Degrees of Freedom = Sample Size – Number of Estimated Parameters = 10 – 2 =8 i. Convert 400 into standard errors: 400 equals

400 = 1.97 standard errors. 203.4

The probability of the coefficient estimate being within 400 of the actual coefficient value equals the probability of the coefficient estimate being within 1.97 standard errors of the actual coefficient value, that is, between t’s of 1.97 and +1.97.

42

Using the Econometrics Lab: Left tail: Right tail: Degrees of Freedom: 8 Degrees of Freedom: 8 t: 1.97 t: 1.97 Left tail probability = .04 Right tail probability = .04 The probability of the coefficient estimate being within 200 of the actual coefficient value equals .92. 1.00  (.04 + .04) = 1.00  .08 = .92 ii. 300 of the actual coefficient value? .82 Convert 300 into standard errors: 300 equals

300 = 1.47 standard errors. 203.4

The probability of the coefficient estimate being within 300 of the actual coefficient value equals the probability of the coefficient estimate being within 1.47 standard errors of the actual coefficient value, that is, between t’s of 1.47 and +1.47. Using the Econometrics Lab: Left tail: Right tail: Degrees of Freedom: 8 Degrees of Freedom: 8 t: 1.47 t: 1.47 Left tail probability = .09 Right tail probability = .09 The probability of the coefficient estimate being within 200 of the actual coefficient value equals .82. 1.00  (.09 + .09) = 1.00  .18 = .82 iii. 200 of the actual coefficient value? .64. Convert 200 into standard errors: 200 equals

200 = .98 standard errors. 203.4

The probability of the coefficient estimate being within 200 of the actual coefficient value equals the probability of the coefficient estimate being within .98 standard errors of the actual coefficient value, that is, between t’s of .98 and +.98. Using the Econometrics Lab: Left tail: Right tail: Degrees of Freedom: 8 Degrees of Freedom: 8 t: .98 t: .98 Left tail probability = .18 Right tail probability = .18 The probability of the coefficient estimate being within 200 of the actual coefficient value equals .64 1.00  (.18 + .18) = 1.00  .36 = .64 3. a. Economic theory teaches us that the demand curve is downward sloping. Accordingly, theory postulates that an increase in the price should decrease consumption. b. Step 0: Formulate a model reflecting the theory to be tested. Model: GasConst = EConst + EPPriceDollarst + et The theory suggests that EP should be negative. 43

Step 1: Ordinary Least Squares (OLS) Dependent Variable: GasCons Estimate SE t-Statistic Explanatory Variable(s): PriceDollars 47.57295 151.6556 3.187853 Const 516.7801 60.60223 8.527410 Number of Observations

Prob 0.0128 0.0000

10

Estimated Equation: EstGasCons = 516.8  151.7PriceDollars Interpretation of Estimates: bPriceDollars = 151.7: A $1 increase in the gasoline price decreases gasoline consumption by 151.7 million gallons per day. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Cynic’s view: Despite the results, the price of gasoline has no impact on gasoline consumption. The results were just “the luck of the draw.” Now, we construct the null and alternative hypotheses: H0: EP = 0 Cynic is correct: Price has no impact on gasoline consumption H1: EP < 0 Cynic is incorrect: Higher prices reduce gasoline consumption Step 3: Formulate the question to assess the null hypothesis and the cynic’s view. Question for the Cynic: x Generic Question: What is the probability that the results would be like those we actually obtained (or even stronger), if the cynic is correct and price actually has no impact? x Specific Question: The regression’s coefficient estimate was 151.7: What is the probability that the coefficient estimate in one regression would be 151.7 or less if H0 were actually true (if the actual coefficient, EP, equals 0)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True] Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True]. OLS estimation procedure unbiased

If H0were true

Mean[bP] = EP = 0

44

SE column

SE[bP] = 47.57

Number of observations

Number of parameters

DF = 10  2 = 8

Recall that the Prob. column reports the “tails probability:” Prob. Column (Tails Probability): The probability that the coefficient estimate, bP, resulting from one regression would be 151.7 or less, if the actual coefficient, EP, equals 0. Prob[Results IF H0 True] =

.0128 | .006 2

The probability that the coefficient estimate in one regression would be 151.7 or less if H0 were actually true (if the actual coefficient, EP, equals 0) is .006. c. Step 5: Decide on the standard of proof, a significance level If we adopt a 1 or 5 percent significance level, we conclude that this probability is small and reject the null hypothesis thereby supporting the theory. 5. a. Step 0: Formulate a model reflecting the theory to be tested. Model: Numbert = EConst + EDemPartyDem1t + et The allegation suggests that EDem should be positive. Step 1: Collect data, run the regression, and interpret the estimates. Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE Explanatory Variable(s): PartyDem1 2.361883 0.577102 Const 5.613527 0.424482 Number of Observations

t-Statistic 4.092661 13.22441

Prob 0.0001 0.0000

451

EstNumber = 5.6 + 2.36PartyDem1 Estimated Equation: Interpretation of Estimates: bDem = 2.36: A Democratic member of Congress receives 2.36 more earmarks than his/her non-Democratic colleagues. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Cynic’s view: Despite the results, party membership has no impact on earmarks. The results were just “the luck of the draw.” Now, we construct the null and alternative hypotheses: H0: EDem = 0 Cynic is correct: Party affiliation has no impact on earmarks. H1: EDem > 0

Cynic is incorrect: Democratic party affiliation increases earmarks.

45

Step 3: Formulate the question to assess the null hypothesis and the cynic’s view. Question for the Cynic: x Generic Question: What is the probability that the results would be like those we actually obtained (or even stronger), if the cynic is correct and party affliliation actually has no effect on earmarks? x Specific Question: The regression’s coefficient estimate was 2.36: What is the probability that the coefficient estimate in one regression would be 2.36 or more if H0 were actually true (if the actual coefficient, EDem, equals 0)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True] Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True].

OLS estimation procedure unbiased

If H0were true

Mean[bDem] = EDem = 0

SE column

SE[bDem] = .577

Number of observations

Number of parameters

DF = 451  2 = 449

Recall that the Prob. column reports the “tails probability:” Prob. Column (Tails Probability): The probability that the coefficient estimate, bDem, resulting from one regression would be 2.36 or more, if the actual coefficient, EDem, equals 0. Prob[Results IF H0 True] =

.0001 < .0001 2

The probability that the coefficient estimate in one regression would be 2.36 or more if H0 were actually true (if the actual coefficient, EDem, equals 0) is less than .0001. b. Step 5: Decide on the standard of proof, a significance level If we adopt a 1 or 5 percent significance level, we conclude that this probability is small and reject the null hypothesis thereby supporting the allegation. 7. a. Economic theory teaches us that the supply curve is upward sloping. Accordingly, theory postulates that an increase in the price should increase production. b. Step 0: Formulate a model reflecting the theory to be tested. Model: OilProdBarrelst = EConst + E PPricet + et Theory suggests that EP should be positive.

46

Step 1: Collect data, run the regression, and interpret the estimates. Ordinary Least Squares (OLS) Dependent Variable: OilProdBarrels Estimate SE Explanatory Variable(s): Price 92.50159 28.12319 Const 5944.958 478.7771 Number of Observations

t-Statistic 3.289157 12.41696

Prob 0.0028 0.0000

29

EstOilProdBarrels = 5,945 + 92.5Price Estimated Equation: Interpretation of Estimates: bP = 92.5: A 1 dollar increase in the price increases oil production by 92.5 thousand barrels. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Cynic’s view: Despite the results, price has no impact on oil production. The results were just “the luck of the draw.” Now, we construct the null and alternative hypotheses: Cynic is correct: Price has no impact on oil production. H0: EP = 0 Cynic is incorrect: A higher price increases oil production. H1: EP > 0 Step 3: Formulate the question to assess the null hypothesis and the cynic’s view. Question for the Cynic: x Generic Question: What is the probability that the results would be like those we actually obtained (or even stronger), if the cynic is correct and price actually has no impact on oil production? x Specific Question: The regression’s coefficient estimate was 92.5: What is the probability that the coefficient estimate in one regression would be 92.5 or more if H0 were actually true (if the actual coefficient, EP, equals 0)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True] Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True]. OLS estimation procedure unbiased

If H0 were true

Mean[bP] = EP = 0

SE column

SE[bP] = 28.1

Number of observations

Number of parameters

DF = 29  2 = 27

Recall that the Prob. column reports the “tails probability:” Prob. Column (Tails Probability): The probability that the coefficient estimate, bP, resulting from one regression would be 92.5 or more, if the actual coefficient, EP, equals 0. 47

Prob[Results IF H0 True] =

.0028 = .0014 2

The probability that the coefficient estimate in one regression would be 92.5 or more if H0 were actually true (if the actual coefficient, E P, equals 0) is .0014. c. Step 5: Decide on the standard of proof, a significance level The Prob[Results IF H0 True] equals .0014, less than .01. So, even with a 1 percent significance level, we would reject the null hypothesis and thereby support the theory.

48

Chapter 9: One-Tailed Tests, Two-Tailed Tests, and Logarithms Solutions to Chapter 9 Prep Questions 1. a.

EConst P E

Q

dQ dP

P

EConst E P P E

b.

P Q

c.

dQ P dP Q

P 1

P EConst P EP

1 EConst P EP 1

EConst E P P E

P 1

1 EConst P EP 1

EP

3. a. PuQ = 1,000 b. i. EConst = 1,000 ii. EP = 1 PuQ = 1,000 Q = 1,000P 1 5.

d log( z ) dz

1 z

Solutions to Chapter 9 Review Questions 1. a.

Theory:

b.

Theory:

E>c H0: E = c H1: E > c E=c H0: E = c H1: E z c

or

E 1

1 Cynic’s view is incorrect: Demand is inelastic; the elasticity of

demand is greater than 1 Step 3: Formulate the question to assess the cynic’s view. Question for the Cynic: x Generic Question: What is the probability that the results would be like those we actually obtained (or even stronger), if the cynic is correct and actual price elasticity of demand equals 1? x Specific Question: The regression’s coefficient estimate was .365: What is the probability that the coefficient estimate, bP, in one regression would be .365 or more, if H0 were actually true (if the actual coefficient, EP, equals 1)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True]

50

Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True]. If the null hypothesis were true, the actual coefficient would equal 1. Since ordinary least squares (OLS) estimation procedure for the coefficient value is unbiased, the mean of the probability distribution of coefficient estimates would be 1. The SE column of the printout provides us with the standard error of the coefficient estimates, .132. The degrees of freedom equal the number of observations, 10, less 2, since we are estimating two parameters, the constant and the coefficient; consequently the degrees of freedom equal 8. OLS estimation procedure unbiased

Assume H0 is true

Mean[bP] = E P = 1

SE column

SE[bP] = .132

Number of observations

Number of parameters

DF = 10  2 = 8

Econometrics Lab: Prob[Results IF H0 True] = .0007 Clever Algebraic Manipulations to Calculate Prob[Results IF H0 True]. Alternatively, we can use a clever algebraic manipulation approach by defining EClever so that it equals 0 when E P equals 1: EClever = E P + 1 and EP = EClever  1 Next, consider the model and perform some algebra: EPLogPriceReal LogPetroConsPC = c + LogPetroConsPC

= c +

(EClever  1)LogPriceReal

LogPetroConsPC + LogPriceReal

= c +

ECleverLogPriceReal

ECleverLogPriceReal yClever = c + where yClever = LogPetroConsPC + LogPriceReal Expressing the null and alternative hypotheses in terms of EClever: H0: EClever = 0 Ÿ EP = 1 Cynic’s view is correct: Actual price elasticity H1: EClever > 0

Ÿ EP > 1

of demand equals 1 Cynic’s view is incorrect: Demand is inelastic

51

Ordinary Least Squares (OLS) Dependent Variable: yClever Estimate SE Explanatory Variable(s): LogPriceReal 0.634546 0.131853 Const 6.813011 0.048429 Number of Observations

Prob[Results IF H0 True] =

t-Statistic 4.812516 140.6798

10

.0013 | .0007 2

We have obtained the same value for Prob[Results IF H0 True]. e. Step 5: Decide on the standard of proof, a significance level Since Prob[Results IF H0 True] equals .0007, we reject the null hypothesis that demand is unit elastic even at the 1 percent significance level. This supports the theory that demand is inelastic. 3. a. Step 0: Formulate a model reflecting the theory to be tested. Constant price elasticity model: OilProdBarrels = EConst PriceEP

where EP = Price elasticity of demand Taking natural logarithms of both sides. log(OilProdBarrels) = log(EConst) + EPlog(Price) LogQ = c +EPLogP where LogQ = log(QilProdBarrels) c = log(EConst) LogP = log(Price) Theory asserts that EP equals .10. Step 1: Collect data, run the regression, and interpret the estimates. We must generate the two variables: the logarithm of quantity and the logarithm of price: x LogQ = log(OilProdBarrels) x LogP = log(Price)

52

Prob 0.0013 0.0000

Ordinary Least Squares (OLS) Dependent Variable: LogQ Estimate SE Explanatory Variable(s): LogP 0.213620 0.069544 Const 8.324825 0.187512 Number of Observations

t-Statistic 3.071739 44.39626

Prob 0.0048 0.0000

29

EstLogQ = 8.32 + .2136LogP Estimated Equation: Interpretation of Estimates: bP = .2136: A 1 percent increase in the price increases oil production by .2136 percent. That is, the estimate of the price elasticity of supply equals .2136. The evidence suggests that the elasticity of supply does not equal .10. More specifically, the estimate is .1136 from .10. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Cynic’s view: Sure the coefficient estimate from regression suggests that the price elasticity of supply does not equal .10, but this is just “the luck of the draw.” In fact, the actual price elasticity of supply equals .10. H0: EP = .10 Cynic’s view is correct: Actual price elasticity of supply equals .10 Cynic’s view is incorrect: Actual price elasticity of supply does not equal .10 H1 EP z .10 Step 3: Formulate the question to assess the cynic’s view. Question for the Cynic: x Generic Question: What is the probability of obtaining a result like the one obtained from the regression (or even stronger), if the cynic is correct and actual price elasticity of supply equals .10? x Specific Question: The regression’s coefficient estimate was .2136: What is the probability that the coefficient estimate, bP, in one regression would be at least .1136 from .10, if H0 were actually true (if the actual coefficient, EP, equals .10)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True] Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True].

53

If the null hypothesis were true, the actual coefficient would equal .10. Since ordinary least squares (OLS) estimation procedure for the coefficient value is unbiased, the mean of the probability distribution of coefficient estimates would be .10. The SE column of the printout provides us with the standard error of the coefficient estimates. The degrees of freedom equal the number of observations less 2, since we are estimating two parameters, the constant and the coefficient. OLS estimation procedure unbiased

Assume H0 is true

Mean[bP] = E P = .10 i.

Number of observations

SE column

Number of parameters

DF = 29  2 = 27

SE[bP] = .0695

We can use the Econometrics Lab to compute Prob[Results IF H0 True]: We shall calculate this probability in two steps: x First, calculate the probability of the estimated lying in the right hand tail. Calculate the probability that the estimate lies .1136 or more above .10; that is, the probability that the estimate lies at or above .2136. x Second, calculate the probability of the estimated lying in the left hand tail. Calculate the probability that the estimate lies .1136 or more below .10; that is, the probability that the estimate lies at or below .0136. Left Right Tail Tail p p Prob[Results IF H0 True] = .0569 + .0569 | .114.

ii. Alternatively, we can use a clever algebraic manipulation approach by defining EClever so that it equals 0 when E P equals .10: EClever = E P  .10 and EP = EClever + .10 Next, consider the model and perform some algebra: EPLogP LogQ = c + LogQ

= c +

(EClever + .10)LogP

LogQ  .10LogP

= c +

ECleverLogP

yClever

= c +

ECleverLogPriceReal

where yClever = LogQ  .10LogP Expressing the null and alternative hypotheses in terms of EClever:

54

H0:

EClever = 0

Ÿ

EP = .10 Cynic’s view is correct: Actual price elasticity

of supply equals .10 H1: EClever z 0 Ÿ EP z .10 Cynic’s view is incorrect: Actual price elasticity of supply does not equal .10 Ordinary Least Squares (OLS) Dependent Variable: yClever Estimate SE t-Statistic Explanatory Variable(s): LogP 0.113620 0.069544 1.633795 Const 8.324825 0.187512 44.39626

Prob 0.1139 0.0000

Number of Observations 29 Next, calculate Prob[Results IF H0 True] focusing on EClever: Question: What is the probability of obtaining a result like the one calculated from the regression (a coefficient estimate, bClever, of .1136, .1136 from 0), if the cynic’s view and the null hypothesis were correct (that is, if the actual coefficient, EClever, equals 0)? Answer: Prob[Results IF H0 True] =

.1139 .1139 + | .114. 2 2

This is the same answer as before. By a clever algebraic manipulation, we can get the statistical software to perform the calculations. b. Step 5: Decide upon the standard of proof, what constitutes proof beyond a reasonable doubt. Decide on the significance level, the dividing line between small and large probability: Even at a 10 percent significance level, we do not reject the null hypothesis that the elasticity of supply equals .10. That is, at a 10 percent significance level, the estimate of .2136 is not statistically different from .10. 5. a. Most regard cigarettes as addictive. Accordingly, the price elasticity of demand for cigarettes should be inelastic. b. Step 0: Formulate a model reflecting the theory to be tested: Constant price elasticity model: CigConsPC = EConst PriceConsumerEP where EP = Price elasticity of demand

55

Taking natural logarithms of both sides. log(CigConsPC) = log(EConst) + EPlog(PriceConsumer) .

= c + EPLogPriceConsumer where LogCigConsPC = log(CigConsPC) c = log(EConst) LogCigConsPC = log(PriceConsumer) Theory suggests that the absolute value of EP should be less than 1; since EP is LogCigConsPC

negative, theory suggests that EP should be greater than 1. Step 1: Collect data, run the regression, and interpret the estimates. Generate two new variables: x LogCigConsPC = log(CigConsPC) x LogPriceConsumer = log(PriceConsumer) Ordinary Least Squares (OLS) Dependent Variable: LogCigConsPC Estimate SE t-Statistic Explanatory Variable(s): LogPriceConsumer 0.237416 1.443914 6.081787 Const 6.379758 0.381990 16.70135 Number of Observations

Prob 0.0000 0.0000

51

Estimated Equation: EstLogCigConsPC = 6.38  1.44 LogPriceConsumer Interpretation of Estimates: bP = 1.44: A 1 percent increase in the price decreases gasoline consumption by 1.44 percent. That is, the estimate of the price elasticity of demand equals 1.44. c. No. The estimate for the price elasticity of demand suggests that demand is elastic, not inelastic. 7. a. Step 0: Formulate a model reflecting the theory to be tested: Constant price elasticity model: HoursPerWeek =EConst WageEW where EW = Wage elasticity of supply Taking natural logarithms of both sides. log(HoursPerWeek) = log(EConst) + EWlog(Wage) LogHoursPerWeek = c + EWLogWage where LogHoursPerWeek

= log(HoursPerWeek) c = log(EConst)

LogWage = log(Wage) Theory suggests that the absolute value of EP should be less than 1. Step 1: Collect data, run the regression, and interpret the estimates. Generate two new variables: x LogHoursPerWeek = log(HoursPerWeek) x LogWage = log(Wage)

56

Ordinary Least Squares (OLS) Dependent Variable: LogHoursPerWeek Estimate SE t-Statistic Explanatory Variable(s): LogWage 0.768171 0.294727 2.606380 Const 0.037095 1.093466 0.033925 Number of Observations

Prob 0.0107 0.9730

92

Estimated Equation: LogHoursPerWeek = .037 + .768 LogWage Interpretation of Estimates: bW = .768: A 1 percent increase in the wage increases hours worked by .768 percent. That is, the estimate of the wage elasticity of supply equals .768. Step 2: Play the cynic, challenge the evidence, and construct the null and alternative hypotheses. Cynic’s view: Sure, the coefficient estimate from regression suggests that the supply is inelastic, but this is just “the luck of the draw.” In fact, the wage elasticity of supply is not less than 1, it equals 1. Cynic’s view is correct: Actual wage elasticity of supply equals 1 H0: EP = 1 H1: EP < 1

Cynic’s view is incorrect: Supply is inelastic; the wage elasticity of supply is less than 1 Step 3: Formulate the question to assess the cynic’s view. Question for the Cynic: x Generic Question: What is the probability that the results would be like those we actually obtained (or even stronger), if the cynic is correct and actual wage elasticity of supply equals 1? x Specific Question: The regression’s coefficient estimate was .768: What is the probability that the coefficient estimate, bP, in one regression would be .768 or less,

if H0 were actually true (if the actual coefficient, EP, equals 1)? Answer: Prob[Results IF Cynic Correct] or Prob[Results IF H0 True] Step 4: Use the general properties of the estimation procedure, the probability distribution of the estimates, to calculate Prob[Results IF H0 True]. If the null hypothesis were true, the actual coefficient would equal 1. Since ordinary least squares (OLS) estimation procedure for the coefficient value is unbiased, the mean of the probability distribution of coefficient estimates would be 1. The SE column of the printout provides us with the standard error of the coefficient estimates, .295. The degrees of freedom equal the number of observations, 92, less 2, since we are estimating two parameters, the constant and the coefficient; consequently the degrees of freedom equal 90.

57

OLS estimation procedure unbiased

Assume H0 is true

Mean[bWage] = E Wage = 1

SE column

SE[bP] = .295

Number of observations

Number of parameters

DF = 92  2 = 90

Econometrics Lab: Prob[Results IF H0 True] = .2168 b. Step 5: Decide on the standard of proof, a significance level. Even at the 10 percent significance level we cannot reject the null hypothesis casting suspicion on the theory.

58

Chapter 10: Multiple Regression Analysis—Introduction Solutions to Chapter 10 Prep Questions 1. a.

Q

EConst P E I E ChickP E P

I

CP

EConst P E I E ChickP  E P

I

P EI

EConst P E I E ChickP  E ChickP  E P

EConst EConst

I

P

EP

I

EI

P I (ChickP )(ChickP ) EP

(

P ChickP

EI

EP

)(

I ChickP

)

EI

b.

P is unchanged ChickP I is unchanged ChickP Q is unchanged c. If EP + EI + ECP = 0, then ECP = EP  EI. In part b, we showed that if ECP = EP  EI, then Q is unchanged when the good’s own price, income, and the price of other goods all double. 3. log(Q) = log(EConst) + E Plog(P) + EIlog(I) + ECPlog(ChickP) Since EClever = EP + EI + ECP, ECP = EClever  EP  EI; now, substitute for ECP = log(EConst) + EPlog(P) + EIlog(I) + (EClever  EP  EI)log(ChickP) Multiplying out the last term = log(EConst) + EPlog(P) + EIlog(I) + ECleverlog(ChickP)  EPlog(ChickP)  EIlog(ChickP) Rearranging terms = log(EConst) + EPlog(P)  EPlog(ChickP) + EIlog(I)  EIlog(ChickP)+ ECleverlog(ChickP) Factoring EP from the second and third terms and EI from the fourth and fifth terms = log(EConst) + EP[log(P)  log(ChickP)] + EI[log(I)  log(ChickP)] + ECleverlog(ChickP)

59

Solutions to Chapter 10 Review Questions 1. Multiple regression analysis includes more than one explanatory variables; simple regression analysis includes only a single explanatory variable. 3. No. 5. Yes, if the exponents sum to 0. Solutions to Chapter 10 Exercises 1. a. x x x

Additional labor increases value added. Additional land under cultivation increases value added. Additional machinery increases value added.

x x

ELabor > 0. ELand > 0. EMachinery > 0.

x

H0: ELabor = 0

b. x c. H1: ELabor > 0 x

H0: ELand = 0 H1: ELand > 0

x

H0: EMachinery = 0 H1: EMachinery > 0

d. Ordinary Least Squares (OLS) Dependent Variable: ValueAdded Estimate SE Explanatory Variable(s): Labor 267.6604 11.53321 Land 2013.179 794.9466 Machinery 21917.88 1012.361 Const 5.28E+08 5.20E+08 Number of Observations

t-Statistic 23.20780 2.532471 21.65027 1.015321

Prob 0.0000 0.0125 0.0000 0.3119

133

EstValueAdded = 528,000,000 + 268Labor + 2,013Land + 21,918Machinery Interpretation of Estimates: bLabor = 268: 1 more farm worker, while the amount of cultivated land and machinery remain constant, increases value added by $268. bLand = 2,013: 1 more square kilometer of cultivated land, while the amount of labor Estimated Equation:

60

and machinery remain constant, increases value added by $2,013. bMachinery = 21,918: We estimate that 1 more tractor, while the amount of labor and cultivated land remain constant, increases value added by $21,918. e. Note that one-tail tests are appropriate. The Prob column indicates that we would not reject any of our null hypotheses. Consequently all our theories are supported. 3. a. Taking natural logarithms and adding an error term: log(ValueAddedt) = log(EConst) + ELaborlog(Labort) + ELandlog(Landt) + EMachinerylog(Machineryt) + et b. Estimate the parameters of the model. Ordinary Least Squares (OLS) Dependent Variable: LogValueAdded Estimate SE Explanatory Variable(s): LogLabor 0.405212 0.051607 LogLand 0.028024 0.062265 LogMachinery 0.421532 0.029000 Const 11.27838 0.680996 Number of Observations c. H0: ELabor + ELand + EMachinery = 1

t-Statistic 7.851944 0.450084 14.53552 16.56159

Prob 0.0000 0.6534 0.0000 0.0000

133

H1: ELabor + ELand + EMachinery z 1 5. a. i.

Downward Sloping Demand Theory: As the price of cigarettes rise, per capita cigarette consumption decreases. ii. Conventional Wisdom: Cigarettes are an inferior good; as per capita income rises, per capita cigarette consumption decreases. b. CigConsPCt = EConst + EPPriceConsumert + EIIncPCt + et c. Theory: EP < 0.

Theory: EI < 0.

d. H0: EP = 0

H0: EI = 0

H1: EP < 0

H1: EI < 0

e. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceConsumer 3.668014 16.04703 4.374856 IncPC 0.318916 0.588471 0.541939 Const 155.0632 20.12211 7.706110 Number of Observations

Prob 0.0001 0.5904 0.0000

51

Estimated Equation: EstCigConsPC = 155  16.0PriceConsumer  .32IncPC Interpretation of Estimates: bP = 16.0: A $1 increase in the price of a pack of cigarettes, while per capita income remains constant, decreases per capita cigarette consumption by 16 packs per year.

61

f.

bI = .32: A $1,000 increase in per capita income, while the price of cigarettes remains constant, decreases per capita cigarette consumption by .32 packs per year. One-tail tests are appropriate. Both theories suggest that the actual coefficient is negative. x The regression results lend support to the downward sloping demand theory. As the Prob column indicates we easily reject the null hypothesis at the 1 percent significance level. x The conventional wisdom asserting that cigarettes are an inferior is shaky. While the coefficient is negative we cannot come close to rejecting the null hypothesis at the 1, 5, or even 10 percent significance level.

7. a. i. As the price of cigarettes rise, the youth smoking rate should decrease. ii. As per capita income rises, the youth smoking rate should decrease. b. SmokeRateYoutht = EConst + EPPriceConsumert + EIIncPCt + et c. Theory: EP < 0. Theory: EI < 0. d. H0: EP = 0

H0: EI = 0

H1: EP < 0

H1: EI < 0

e. Ordinary Least Squares (OLS) Dependent Variable: SmokeRateYouth Estimate SE t-Statistic Explanatory Variable(s): PriceConsumer 1.455381 0.665547 2.186743 IncPC 0.106776 0.202018 1.891985 Const 33.45011 3.651081 9.161700 Number of Observations

Prob 0.0337 0.0645 0.0000

51

EstSmokeRateYouth = 33.5  1.46PriceConsumer  .20IncPC Interpretation of Estimates: bP = 1.46: A $1 increase in the price of a pack of cigarettes, while per capita income remains constant, decreases the youth smoking rate by 1.46 percentage points. bI = .20: A $1,000 increase in per capita income, while the price of cigarettes remains constant, decreases the youth smoking rate by .20 percentage points. f. One-tail tests are appropriate. Both theories suggest that the actual coefficient is negative. In both cases, we reject the null hypothesis at the 5 percent significance level, but not at the 1 percent significance level. Estimated Equation:

62

Chapter 11: Hypothesis Testing and the Wald Test Solutions to Chapter 11 Prep Questions 1. log(Q)

= log(EConst) + EPlog(P) + EIlog(I) + ECPlog(ChickP) + et Since EP + EI + ECP = 0, ECP =  (EP + EI) = log(EConst) + EPlog(P) + EIlog(I)  (EP + EI)log(ChickP) + et Multiplying through = log(EConst) + EPlog(P) + EIlog(I)  EPlog(ChickP)  EIlog(ChickP) + et Rearranging terms = log(EConst) + EPlog(P)  EPlog(ChickP) + EIlog(I)  EIlog(ChickP) + et Factoring EP from the second and third terms and EI from the fourth and fifth terms = log(EConst) + EP[log(P)  log(ChickP)] + EI[log(I)  log(ChickP)] + et

3. a. Before the experiment is conducted, we cannot determine the numerical value of the random variable with certainty. b. Before the experiment is conducted, we can often calculate the random variable’s probability distribution telling us how likely it is for the random variable to equal each of its possible numerical values. Solutions to Chapter 11 Review Questions 1. No, the restricted sum of squared residuals cannot be less than the unrestricted sum. This is true because the ordinary least squares (OLS) estimation procedure “chooses” the parameter estimates that minimize the sum of squared residuals. Placing a restriction on one or more of the parameters can only increase, not decrease, the sum of squared residuals. 3.

A two-tailed t-test is a special case of the Wald test in which the value of a single coefficient is restricted to equal 0.

Solutions to Chapter 11 Exercises 1. a

H0: E Labor + E Land + EMachinery = 1

H1: E Labor + E Land + EMachinery z 1 b. See the solutions to exercises 1 and 3 from Chapter 10. 3. Assess the constant returns to scale theory using the Wald approach the “easy way” using statistical software.

63

Ordinary Least Squares (OLS) Dependent Variable: LogValueAdded Estimate SE Explanatory Variable(s): LogLabor 0.405212 0.051607 LogLand 0.028024 0.062265 LogMachinery 0.421532 0.029000 Const 11.27838 0.680996 Number of Observations Sum Squared Residuals

F-statistic

Value 7.223722

t-Statistic 7.851944 0.450084 14.53552 16.56159

Prob 0.0000 0.6534 0.0000 0.0000

133 90.50033 Degrees of Freedom Num Dem 1 129

Prob 0.0081

Prob[Results IF H0 True] = .0081 5. a. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceSupplier 34.16907 10.66315 3.204408 Tax 4.040626 12.69334 3.141428 Const 207.6718 38.53283 5.389477 Number of Observations

Prob 0.0024 0.0029 0.0000

51

Estimated Equation: EstCigConsPC = 207.7  34.2PriceSupplier  12.7Tax b. Interpret the coefficient estimates. x A $1 increase in the price of cigarettes received by suppliers decreases per capita cigarette consumption by 34.2 packs. x A $1 increase in the cigarette tax decreases per capita cigarette consumption by 12.7 packs. 7. a. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceSupplier 34.16907 10.66315 3.204408 Tax 4.040626 12.69334 3.141428 Const 207.6718 38.53283 5.389477 Number of Observations Sum Squared Residuals SSRU = 19,665.41 DFU = 51  3 = 48 64

51 19665.41

Prob 0.0024 0.0029 0.0000

b.

CigConsPCt

= EConst + EPriceSupplierPriceSuppliert + ETaxTaxt + et H0: EPriceSupplier = ETax = EConst + ETaxPriceSuppliert + ETaxTaxt + et = EConst + ETax(PriceSuppliert + Taxt) + et Since PriceConsumert = PriceSuppliert + Taxt = EConst + ETaxPriceConsumert + et

c. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceConsumer 17.02657 3.168679 5.373398 Const 148.6762 16.19163 9.182288 Number of Observations Sum Squared Residuals SSRR = 20,822.81 d.

51 20822.81

DFR = 51  2 = 49 SSRR = 20,822.81 SSRU = 19.665.41

DFR = 49 DFU = 48

SSRR  SSRU = 1,157.40

DFR  DFU = 1

F

Prob 0.0000 0.0000

( SSRR  SSRU ) / ( DFR  DFU ) 1,157.40 1,157.40 /1 = = = 2.825 SSRU / DFU 19, 665.41/ 48 409.69

e. Prob[Results IF H0 True] = .0993.

65

Chapter 12: Model Specification and Development Solutions to Chapter 12 Prep Questions 1. 0. 3. a. EUnemCurrent < 0 b. H0: EUnemCurrent = 0 H1: EUnemCurrent < 0 5. a. 1912 election stands out because third parties received more than a third of the vote. b. Theodore Roosevelt, a former Republican President, split from his party and formed a third party called the Bull Moose Party. One of his opponents was his former vice president. Solutions to Chapter 12 Review Questions 1. An artificial model is associated with an “original” model. The original model is designed to assess a theory. The artificial model is not designed to assess the theory; it assesses the properties of the original model. 3. a. i. ii. b. i. ii.

No. No. Yes. Yes.

Solutions to Chapter 12 Exercises 1. a. x x x

Higher real GDP growth rates should increase the vote for the incumbent’s party. Higher inflation rates should decrease the vote for the incumbent’s party. More consecutive terms should decrease the vote for the incumbent’s party (throw the ****s out). b. VotePresPartyTwot = EConst + EGdpGrowthRealGdpGrowtht + EInflGdpCurrentInflGdpCurrentt + E PresTermsPresPartyTermst + et The theories imply: x EGdpGrowth > 0. x EInflGdpCurrent < 0. x EPresTerms < 0.

67

c. Ordinary Least Squares (OLS) Dependent Variable: VotePresPartyTwo Estimate SE t-Statistic Explanatory Variable(s): RealGdpGrowth 0.435612 0.221348 1.967995 InflGdpCurrent –0.656971 0.293145 –2.241111 PresPartyTerms –3.144539 1.078933 –2.914488 Const 58.38144 2.595120 22.49663 Number of Observations

Prob 0.0598 0.0338 0.0072 0.0000

30

EstVotePresPartyTwo = 58.38 + .436RealGdpGrowth  .657InflGdpCurrent  3.145PresPartyTerms Interpretation of Estimates: bGdpGrowth = .436: A 1 percentage point increase in the GDP growth rate increases the popular vote of the incumbent President’s party by .436 percentage points. bInflGdpCurrent = .657: A 1 percentage point increase in inflation as measure by the GDP price deflator decreases the popular vote of the incumbent President’s party by .657 percentage points. bPresTerms = 3.145: One additional term the incumbent President’s party occupies the White House decreases the popular vote of the incumbent President’s party by 3.145 percentage points. GDP Growth Inflation Presidential Terms H0: EInflGdpCurrent = 0 H0: EPresTerms = 0 H0: EGdpGrowth = 0 Estimated Equation:

d.

e.

H1: EGdpGrowth > 0 GDP Growth

H1: EInflGdpCurrent < 0 Inflation

H1: EPresTerms < 0 Presidential Terms

.0598 | .030 2

.0338 | .017 2

.0072 | .004 2

We reject the null hypotheses for the GDP growth and inflation theories at the 5 percent significance level, but do not reject the null hypotheses at the 1 percent significance level. We reject the presidential terms null hypothesis at the 1 percent significance level. 3. a. Artificial Model: Tax = EConst + ETobProdPCTobProdPCt + JEstTax2 EstTaxt2 + et Ramsey RESET Test Dependent Variable: Tax Explanatory Variable(s): TobProdPC Const Fitted^2 Number of Observations

Estimate 0.076634 3.387954 2.693214

SE 0.050998 2.322631 1.324261

t-Statistic 1.502701 1.458670 2.033749

Prob 0.1395 0.1512 0.0475

51

Critical Result: The Fitted^2 coefficient is 2.69. The estimate does not equal 0; the estimate is 2.69 from 0. This evidence suggests that the new form of the information adds explanatory power. 68

H0: JEstTax2 = 0

Ÿ

New form of the information adds NO explanatory power

H1: JEstTax2 z 0 Ÿ New form of the information adds explanatory power Prob[Results IF H0 True] =.0475 We do not reject the null hypothesis at the 1 percent significance level, but we do reject the null hypothesis at the 5 percent significance level. So, perhaps we could do better by using the information in a different form. b. Ordinary Least Squares (OLS) Dependent Variable: Tax Explanatory Variable(s): SqrtTobProdPC Const Number of Observations H0: ESqrtTobProdPC = 0

Estimate 0.187790 1.389676

SE 0.068316 0.110588

t-Statistic 2.748863 12.56625

Prob 0.0083 0.0000

51

H1: ESqrtTobProdPC < 0 Prob[Results IF H0 True] =

.0083 = .0042 2

We do reject the null hypothesis at the 1 percent significance level. c. Artificial Model: Taxt = E Const + E TobProdPCSqrtTobProdPCt + JEstTax2 EstTaxt2 + et Ramsey RESET Test Dependent Variable: Tax Explanatory Variable(s): SqrtTobProdPC Const Fitted^2 Number of Observations

Estimate 0.139110 0.549077 1.025663

SE 0.302547 1.751680 0.924855

t-Statistic 0.459796 0.313457 1.108999

Prob 0.6477 0.7553 0.2730

51

Critical Result: The Fitted^2 coefficient is 1.03. The estimate does not equal 0; the estimate is 1.03 from 0. This evidence suggests that the new form of the information adds explanatory power. H0: JEstTax2 = 0 Ÿ New form of the information adds NO explanatory power H1: JEstTax2 z 0 Ÿ New form of the information adds explanatory power Prob[Results IF H0 True] =.2730 We do not reject the null hypothesis at the 10 percent significance level. That is, we do not reject the notion that the new form of the information adds no explanatory power. Hence, there is no compelling need to consider a different specification of the model.

69

5. a. x

The downward sloping theory of demand suggests that higher prices reduce the youth smoking rate. x Conventional wisdom suggests that higher incomes reduce the youth smoking rate. b. SmokeRateYoutht = EConst + EPPriceConsumert + EIIncPCt + et EP < 0. x EI < 0. x c. Ordinary Least Squares (OLS) Dependent Variable: SmokeRateYouth Estimate SE t-Statistic Prob Explanatory Variable(s): PriceConsumer 1.455381 0.665547 2.186743 0.0337 IncPC 0.106776 0.0645 0.202018 1.891985 Const 33.45011 3.651081 9.161700 0.0000 Number of Observations Estimated Equation:

51 EstSmokeRateYouth = 33.45  1.46PriceConsumer  .20IncPC

Interpretation of Estimates: bP = 1.46: A $1 per pack increase in cigarette prices decreases the youth smoking rate by 1.46 percentage points. bI = .20: A 1 thousand dollar increase in per capita income decreases the youth smoking rate by .20 percentage points. d. Price Income H0: EI = 0 H0: EP = 0 H1: E P < 0 e. Prob[Results IF H0 True]: Price

.0337 | .017 2

f.

H1: EI < 0 Income

.0645 | .032 2

We reject both the price and income null hypotheses at the 5 percent significance level, but not at the 1 percent significance level. Artificial Model: SmokeRateYoutht = EConst + E PPriceConsumert + EIIncPCt + JEstYouthRate EstYouthRatet2 + et Ramsey RESET Test Dependent Variable: SmokeRateYouth Estimate SE Explanatory Variable(s): PriceConsumer 4.059803 5.996945 IncPC 0.540266 0.809220 Const 99.14360 58.23491 Fitted^2 0.103279 0.111604 Number of Observations

70

51

t-Statistic 0.676979 0.667638 0.587379 0.925399

Prob 0.5017 0.5076 0.5598 0.3595

Critical Result: The Fitted^2 coefficient is .103. The estimate does not equal 0; the estimate is .103 from 0. This evidence suggests that the new form of the information adds explanatory power. H0: JEstYouthRate = 0 Ÿ New form of the information adds NO explanatory power H1: JEstYouthRate z 0 Ÿ New form of the information adds explanatory power Prob[Results IF H0 True] = .3595 We do not reject the null hypothesis at the 1, 5, or 10 percent significance level. That is we do not reject the notion that the new form of the information adds no explanatory power. Hence, there is no compelling need to consider a different specification of the model.

71

Chapter 13: Dummy and Interaction Variables Solutions to Chapter 13 Prep Questions 1.

dSSR dbConst

2( y1  bConst )  2( y2  bConst )  2( y3  bConst )

0

y1  bConst



y2  bConst



y3  bConst

0

y1



y2



y3

3bConst

y1  y2  y3 3

bConst

NB: bConst equals the mean of y. 3. a. The annual increase in internet use expressed in percentage terms. b. x Year: The internet is an emerging technology; consequently, we would expect users to increase over time. E Year > 0. x

CapitalHuman: Citizens in nations with more human capital will be able to take better advantage of the internet; hence, more human capital would increase internet use. ECapHum > 0.

x

CapitalPhysical: Nations with more human capital are better able to operate the internet; hence, more physical capital would increase internet use. ECapPhy > 0.

x

GdpPC: Citizens in nations with higher per capital GDP will be able to afford the internet; hence, higher per capital GDP would increase internet use. EGDP > 0.

x

Auth: Authoritarian nations will discourage internet use because it is difficult for governments to control its content; hence, more authoritarian nations will decrease internet use. EAuth < 0.

Solutions to Chapter 13 Review Questions 1. A dummy variable takes on only two values: 0 and 1. The dummy variable divides all the observations into two disjoint groups. 3. When we use averages to draw conclusions we implicitly assume that the only a single factor affects the dependent variable. Typically, this is not a realistic assumption. For example, when we use average salaries for men and women to investigate gender discrimination, we implicitly assume that an individual’s gender is the only factor affecting his/her salary.

73

Solutions to Chapter 13 Exercises 1. a. Additional articles should increase a faculty member’s salary: EArt > 0. b. Ordinary Least Squares (OLS) Dependent Variable: Salary Estimate SE t-Statistic Explanatory Variable(s): SexF1 10584.60 5492.306 1.927168 Experience 2435.660 211.0136 11.54267 Exper_SexF1 1086.915 397.0372 2.737564 Articles 1783.056 839.0052 2.125202 Const 37992.42 3862.495 9.836240 Number of Observations

Prob 0.0554 0.0000 0.0068 0.0348 0.0000

200

EstSalary = 37,992 + 10,585SexF1 + 2,436Experience  1,087Exper_SexF1 + 1,783Articles Interpretation of Estimates: bArt = 1,783: 1 additional article increases a faculty member’s salary by $1,783. c. H0: EArt = 0 Estimated Equation:

H1: EArt > 0 d. Prob[Results IF H0 True] =

.0348 | .017 2

We reject both the published articles null hypotheses at the 5 percent significance level, but not at the 1 percent significance level. 3. a. x

x

x

The seniority system provides long time members of Congress more influence than newly elected members; hence additional terms should increase the number of earmarks. E Terms > 0. Conventional wisdom suggests that liberals are more supportive of government projects than conservatives; hence, liberal members of Congress should be more active in acquiring earmarks. ELiberal > 0. Equity suggests that low income states should receive more Federal government assistance than high income states; hence, high income states should receive fewer earmarks. E Income < 0.

b. Use the ordinary least squares (OLS) estimation procedure to estimate the coefficients. Interpret the coefficient estimates.

74

Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE t-Statistic Explanatory Variable(s): Terms 0.260593 0.068817 3.786763 ScoreLiberal 0.052310 0.011216 4.663870 IncPC 0.047239 0.083213 1.761533 Const 6.445098 2.010177 3.206235 Number of Observations

Prob 0.0002 0.0000 0.0789 0.0014

425

EstNumber = 6.45 + .261Terms + .052ScoreLiberal  .083IncPC Interpretation of Estimates: bTerms = .261: We estimate that an additional term in Congress increases a member’s earmarks by .261. bLiberal = .052: We estimate that a 1 point (on a 0-100 scale) increase in a member liberal score increases the member’s earmarks by .052. bIncPC = .083: We estimate that a 1 thousand dollar increase in a state’s per capita income reduces a member earmarks by .083. Seniority Liberal Equity H0: ELiberal = 0 H0: EIncome = 0 H0: ETerms = 0 Estimated Equation:

c.

d.

H1: ETerms > 0 Seniority

H1: ELiberal > 0 Liberal

H1: EIncome < 0 Equity

.0002 | .0001 2

 .0001 | 0. b. Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE t-Statistic Explanatory Variable(s): Terms 0.266513 0.068549 3.887917 ScoreLiberal 0.048719 0.011280 4.319022 IncPC 0.056707 0.153651 2.709577 RegionNorthEast 2.061917 0.927902 2.222127 Const 9.300573 2.377948 3.911176 Number of Observations Estimated Equation:

Prob 0.0001 0.0000 0.0070 0.0268 0.0001

425 EstNumber = 9.30 + .267Terms + .049ScoreLiberal  .154IncPC + 2.06RegionNorthEast 75

Interpretation of Estimates: bNE = 2.06: Members of Congress from the Northeast receive 2.06 more earmarks than their colleagues from other regions of the nation. c. H0: ENE = 0 H1: ENE > 0 d. Prob[Results IF H0 True] =

.0268 | .013 2

We reject the allegation’s null hypothesis at a 5 percent significance level, but not at the 1 percent significance level.

76

Chapter 14: Omitted Explanatory Variables, Multicollinearity, and Irrelevant Explanatory Variables Solutions to Chapter 14 Prep Questions 1. Goal of Multiple Regression Analysis: Multiple regression analysis attempts to sort out the individual effect of each explanatory variable. An explanatory variable’s coefficient estimate allows us to estimate the change in the dependent variable resulting from a change in that particular explanatory variable while all other explanatory variables remain constant. 3. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close to the actual value is small. 5. a. The value of one variable does not allow us to predict the value of the other. b. We use a scatter diagram plotting each variable’s deviation from its mean illustrates independence. Loosely speaking, when the variables are independent their scatter diagram points are spread “evenly” across all four quadrants. c. It equals 0. 7. a. 1.00 b. Yes, something unusual occurs: EViews error message: “Near singular matrix.” Solutions to Chapter 14 Review Questions 1. a. The ordinary least squares (OLS) estimation procedure for an included explanatory variable’s value can be biased. b. An omitted explanatory variable bias problem arises whenever the omitted explanatory variable: x influences the dependent variable; x is correlated with an included explanatory variable. 3. a. The ordinary least squares (OLS) estimated procedure for the included explanatory variables’ values will be unbiased. b. The variance of the probability distributions for the values of the relevant explanatory variables will be larger when the irrelevant explanatory variable is included than when the irrelevant variable is excluded.

77

Solutions to Chapter 14 Exercises 1. a. x x x

The downward sloping demand curve theory suggests that higher prices decrease per capita cigarette consumption. E Price < 0. Conventional wisdom suggests that college educated individuals smoke less. EEduColl < 0. Tobacco is important to the economy of those states that produce large quantities of tobacco; hence, tobacco use will be more acceptable in those states and residents will be more receptive to smoking. ETobProd > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceConsumer 11.18450 3.464492 3.228322 EducCollege 0.580736 1.235767 2.127934 TobProdPC 0.725932 0.310685 2.336554 Const 150.4259 17.75788 8.470934 Number of Observations

Prob 0.0023 0.0386 0.0238 0.0000

51

EstCigConsPC = 150.4  11.2PriceConsumer  1.24EducCollege + .726TobProdPC Interpretation of Estimates: bPrice = 11.2: A $1 rise in the price of a pack of cigarettes decreases per capita cigarette consumption by 11.2 packs per year. bEduColl = 1.24: A 1 percentage point increase in college educated individuals decreases per capita cigarette consumption by 1.24 packs per year. bTobProd = .726: A 1 pound rise in a state’s per capita tobacco production increases per capita cigarette consumption by .736 packs per year. c. Demand Education Production H0: EEduColl = 0 H0: ETobProd = 0 H0: EPrice = 0 Estimated Equation:

d.

H1: EPrice < 0 Demand

H1: EEduColl < 0 Education

H1: ETobProd > 0 Production

.0023 | .001 2

.0386 | .019 2

.0238 | .012 2

x x

78

At the 5 percent significance level we reject the demand, education, and production null hypotheses. At the 1 percent significance level, we reject the demand null hypothesis, but not education and production null hypothesis.

3. a. x x x x

The downward sloping demand curve theory suggests that higher prices decrease per capita cigarette consumption. E Price < 0. Conventional wisdom suggests that college educated individuals smoke less. EEduColl < 0. Conventional wisdom suggests that higher income individuals smoke less. EI < 0. Tobacco is important to the economy of those states that produce large quantities of tobacco; hence, tobacco use will be more acceptable in those states and residents will be more receptive to smoking. ETobProd > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: CigConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceConsumer 12.30477 3.417315 3.600714 EducCollege 0.881137 2.543609 2.886736 IncPC 1.582120 0.818423 1.933132 TobProdPC 0.737257 0.302072 2.440662 Const 134.8625 19.04746 7.080337 Number of Observations

Prob 0.0008 0.0059 0.0594 0.0186 0.0000

51

EstCigConsPC = 134.9  12.3PriceConsumer  2.54EducCollege + 1.58IncPC + .737TobProdPC Interpretation of Estimates: bPrice = 12.3: A $1 rise in the price of a pack of cigarettes decreases per capita cigarette consumption by 12.3 packs per year. bEduColl = 2.54: A 1 percentage point increase in college educated individuals decreases per capita cigarette consumption by 2.54 packs per year. bI = 1.58: A 1 thousand dollar increase in per capita income increases per capita cigarette consumption by 1.58 packs per year. bTobProd = .737: A 1 pound rise in a state’s per capita tobacco production increases per capita cigarette consumption by .737 packs per year. The estimate of the income coefficient is surprising. It is positive rather than negative thereby contradicting the conventional wisdom. c. Since the income results refute the theory we only formulate null and alternative hypotheses for the three other theories. Demand Education Production H0: EEduColl = 0 H0: ETobProd = 0 H0: EPrice = 0 Estimated Equation:

d.

H1: EPrice < 0 Demand

H1: EEduColl < 0 Education

H1: ETobProd > 0 Production

.0008 | .0004 2

.0059 | .003 2

.0186 | .009 2

79

x x

At the 1 percent significance level, we reject the demand, education, and production null hypotheses. Since the income results refute the theory perhaps we should rethink the role of income.

5. a. x x

As a member of Congress has been in the House longer, he/she gains influence. Consequently, he/she is more able to garner earmarks. ETerms > 0. Liberal members of Congress have a more activist view of government’s role than conservative members. Consequently, members of Congress who have high liberal ratings should be more active in their pursuit of earmarks. E Liberal > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE Explanatory Variable(s): Terms 0.256432 0.068947 ScoreLiberal 0.045667 0.010589 Const 3.122135 0.696198 Number of Observations

t-Statistic 3.719247 4.312564 4.484549

Prob 0.0002 0.0000 0.0000

425

Number = 3.12 + .256Terms + .046ScoreLiberal Estimated Equation: Interpretation of Estimates: bTerms = .256: An additional term in Congress increases a member’s earmarks by .256. bLiberal = .046: A 1 point (on a 0-100 scale) increase in a member liberal score increases the member’s earmarks by .046. c. Terms Liberal H0: ELiberal = 0 H0: ETerms = 0 H1: ETerms > 0 d. Terms

.0002 | .0001 2 x

80

H1: ELiberal > 0 Liberal

.0001 | 0. Liberal members of Congress have a more activist view of government’s role than conservative members. Consequently, members of Congress who have high liberal ratings should be more active in their pursuit of earmarks. E Liberal > 0. Democratic members of Congress have a more activist view of government’s role than their non-Democratic colleagues. Consequently, Democratic members of Congress should be more active in their pursuit of earmarks. EDem > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: Number Estimate SE Explanatory Variable(s): Terms 0.259824 0.070022 ScoreLiberal 0.039769 0.023079 PartyDem1 0.367293 1.276776 Const 3.198703 0.746051 Number of Observations

t-Statistic 3.710597 1.723173 0.287672 4.287513

Prob 0.0002 0.0856 0.7737 0.0000

425

EstNumber = 3.20 + .260Terms + .040ScoreLiberal + .367PartyDem1 Interpretation of Estimates: bTerms = .260: An additional term in Congress increases a member’s earmarks by .260. bLiberal = .040: A 1 point (on a 0–100 scale) increase in a member liberal score increases the member’s earmarks by .040. bDem = .367: A Democratic member of Congress increases the number of earmarks by .367. c. Terms Liberal Democrat H0: ELiberal = 0 H0: EDem = 0 H0: ETerms = 0 Estimated Equation:

d.

H1: ETerms > 0 Terms

H1: ELiberal > 0 Liberal

H1: EDem > 0 Democrat

.0002 | .0001 2

.0856 | .043 2

.7737 | .387 2

x x

At the 1 percent significance level we reject the terms null hypothesis, but we do not reject the liberal or Democrat null hypotheses. At the 5 percent significance level we reject the terms and liberal null hypotheses, but we do not reject the Democrat null hypothesis.

81

Chapter 15: Other Regression Statistics and Pitfalls Solutions to Chapter 15 Prep Questions 1. a. Ordinary Least Squares (OLS) Dependent Variable: LogUsersTV Estimate SE t-Statistic Explanatory Variable(s): Year 0.022989 0.015903 1.445595 CapitalHuman 0.036302 0.001915 18.95567 CapitalPhysical 0.001931 0.000510 3.789394 GdpPC 0.058877 0.012338 4.772051 Auth 0.063345 0.012825 4.939278 Const 31.77155 44.95755 1.415025 Number of Observations Estimated Equation:

Prob 0.1487 0.0000 0.0002 0.0000 0.0000 0.1575

742 EstLogUsersTV = 45.0 + .023Year + .036CapitalHuman + .002CapitalPhysical + .059GdpPC + .063Auth

TV b. H0: EYear = .010 TV H1: EYear z .010

A two-tailed test is appropriate. c. Prob[Results IF H0 True] = Probability that the coefficient estimate would be at least .033 from .010, if H0 were true (if the actual coefficient TV , .010) equals, EYear

Prob[Results IF H0 True]

=

Left Tail p .0191

Right Tail p + .0191 | .038.

83

3. a. i. ii. iii. b. i. ii. c. i. ii. d.

Error message: “Near singular matrix.” 1 No Error message: “Near singular matrix.” No Error message: “Near singular matrix.” No

Ordinary Least Squares (OLS) Dependent Variable: TotalSalary Estimate SE Explanatory Variable(s): HomeSalary 1.000000 8.58E-17 VisitSalary 1.000000 8.61E-17 Const 0.000000 4.24E-15

t-Statistic 1.17E+16 1.16E+16 0.000000

Prob 0.0000 0.0000 1.0000

Ordinary Least Squares (OLS) Dependent Variable: Attendance Estimate SE t-Statistic Explanatory Variable(s): PriceTicket 590.7836 184.7231 3.198211 HomeSalary 783.0394 45.23955 17.30874 Const 9246.429 1529.658 6.044767

Prob 0.0015 0.0000 0.0000

Number of Observations 588 The standard errors are very small, approximately 0. e.

Number of Observations

585

EstAttendance = 9,246  591PriceTicket + 783HomeSalary bPriceTicket = 591. We estimate that a $1.00 increase in the price of tickets decreases attendance by 591 per game. bHomeSalary = 783. We estimate that a $1 million increase in the home team salary increases attendance by 783 per game.

Estimated Equation: Interpretation:

i. 591. ii. 20.232. iii. 20232. No. Ordinary Least Squares (OLS) Dependent Variable: Attendance Estimate SE t-Statistic Explanatory Variable(s): PriceTicket 1896.379 142.8479 13.27552 HomeSalary 0.484536 0.088467 0.182580 Const 3697.786 1841.286 2.008263 Number of Observations

84

585

Prob 0.0000 0.8552 0.0451

Estimated Equation: EstAttendance = 3,698 + 1,896PriceTicket  .088HomeSalary iv. x The estimate of the ticket price coefficient changes dramatically, from 591 to 1,896. x The estimate of the home salary coefficient changes dramatically, from 783.0 to .088. 5. No. Solutions to Chapter 15 Review Questions 1. Our approach thus far has followed two steps: x First, we present the theory. x Second, we analyze the data and determine whether the data are consistent with the theory. We have started with the theory and then decide whether the data is compatible with the theory. The confidence interval approach reverses this process. Confidence intervals indicate the range of theories with which the data are compatible. x First, we analyze the data. x Second, we consider the theories and determine which theories are consistent with the data. We start with the data and then decide what theories are compatible with the data. 3. Goal of Multiple Regression Analysis: Multiple regression analysis attempts to sort out the individual effect of each explanatory variable. An explanatory variable’s coefficient estimate allows us to estimate the change in the dependent variable resulting from a change in that particular explanatory variable while all other explanatory variables remain constant. Solutions to Chapter 15 Exercises 1. a. Ordinary Least Squares (OLS) Dependent Variable: LogUsersInternet Estimate SE Explanatory Variable(s): Year 0.449654 0.017078 CapitalHuman 0.023725 0.002470 CapitalPhysical 0.002056 0.000480 GdpPC 0.118177 0.011461 Auth 0.013999 0.095836 Const 899.3201 34.17432 Number of Observations

t-Statistic 26.32965 9.606597 4.286193 10.31146 –6.845761 –26.31567

Prob 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

566

85

Estimated Equation:

b.

LB EYear

.4161

E

.4832

UB Year

EstLogUsersInternet = 899.3 + .450Year + .024CapitalHuman + .002CapitalPhysical + .118GdpPC  .096Auth

The 95 percent confidence interval is from .4161 to .4832. 3. a. Ordinary Least Squares (OLS) Dependent Variable: PetroConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceReal 124.7236 77.09510 1.617789 Const 609.9538 58.12769 10.49334 Number of Observations Estimated Equation: b. 610.0 c. 124.7

Prob 0.1444 0.0000

Mass1 = 1

10

EstPetroConsPCMass = 610.0  124.7PriceReal

5. a. i. EstPetroConsPCMass = bMass + bPMassPriceReal ii. bMass iii. bPMass b. i. EstPetroConsPCNeb = bNeb + bPNebPriceReal ii. bNeb iii. bPNeb bPMass = 124.7 bNeb = 1,418.4 bPNeb = 540.5 c. bMass = 610.0 d. Ordinary Least Squares (OLS) Dependent Variable: PetroConsPC Estimate SE t-Statistic Explanatory Variable(s): Mass1 609.9538 143.2589 4.257704 Neb 1418.403 106.8526 13.27438 PriceReal_Mass 124.7236 190.0051 0.656422 PriceReal_Neb 150.1664 540.5367 3.599585 Number of Observations Estimated Equation:

20 EstPetroConsPC = 610.0Mass1 + 1,418.4Neb  124.7PriceReal_Mass  540.5PriceReal_Neb

e. Yes. 7. a. i. EstPetroConsPCMass = bConst + bMass + (bP + bPMass)PriceReal ii. bMass iii. bP + bPMass

86

Prob 0.0006 0.0000 0.5209 0.0024

b. i. EstPetroConsPCNeb = bConst + bNeb + (bP + bPNeb)PriceReal ii. bNeb iii. bP + bPNeb c. No, we have 5 unknowns, but only 4 equations: bP + bPMass = 124.7 bMass = 610.0 bP + bPNeb = 540.5 bNeb = 1,418.4 d. Error message: Near singular matrix e. Yes

87

Chapter 16: Heteroskedasticity Solutions to Chapter 16 Prep Questions 1. x

x

x

Error Term Equal Variance Premise: The variance of the error term’s probability distribution for each observation is the same; all the variances equal Var[e]: Var[e1] = Var[e2] = … = Var[eT] = Var[e] Error Term/Error Term Independence Premise: The error terms are independent: Cov[ei, ej] = 0. Knowing the value of the error term from one observation does not help you predict the value of the error term for any other observation. Explanatory Variable/Error Term Independence Premise: The explanatory variables, the xt’s, and the error terms, the et’s, are not correlated. Knowing the value of an observation’s explanatory variable does not help you predict the value of that observation’s error term.

3. The error term equal variance premise played an important role. If this premise were violated the variance would differ from observation to observation. Var[e] would not exist. The expression for Var[bx] would be much more complicated. 5. Model:

LogUsersInternett = EConst + EGDPGdpPCt + et Divide both sides of the equation by

GdpPCt .

Tweaked Model :

LogUsersInternett GdpPCt

EConst GdpPCt

EConst GdpPCt

 EGDP

GdpPCt  GdpPCt

 EGDP GdpPCt 

where H t

et GdpPCt

Ht

et GdpPCt

89

Var[εt ]

Var[

et ] GdpPCt p

Arithmetic of variances: Var[cx ] c 2 Var[x ]

1 Var[et ] GdpPCt p

Var[et ] =V u GpcPCt where V equals a constant

1 V u GpcPCt GdpPCt V Solutions to Chapter 16 Review Questions 1. Heteroskedasticity occurs whenever the variance of the error terms’ probability distributions is not identical; that is, whenever the variance of the error terms’ probability distributions differ from observation to observation. 3. a. The ordinary least squares (OLS) estimation procedure for the coefficient value is unbiased. b. x The ordinary least squares (OLS) estimation procedure for the variance of the error term’s probability distribution is flawed because it is based on a false premise; hence, the standard errors and call calculations based on the standard errors are flawed. x The ordinary least squares (OLS) estimation procedure is not the best linear unbiased estimation procedure (BLUE). Solutions to Chapter 16 Exercises 1. a. x x

A state with a higher crime rate should have more court cases and hence would spend more on its judicial system: ECrimes > 0. A state with a higher per capita GDP has a larger tax base and hence, can afford to spend more on its judicial system: EGDP > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: JudExp Estimate SE t-Statistic Explanatory Variable(s): CrimesAll 0.321427 0.408012 0.787787 GdpPC 0.298469 0.063159 4.725673 Const –2425.185 2705.168 –0.896501 Number of Observations Estimated Equation:

90

50 EstJudExp = 2,425 + .321CrimesAll + .298GdpPC

Prob 0.4348 0.0000 0.3746

Interpretation of Estimates: bCrimesAll = .321: A 1 per 100,000 persons rise in the crime rate increases judicial expenditures by $.321 per 100,000 persons. bGDP = .298: A $1 rise in per capita state income increases judicial expenditures by .298 per 100,000 persons. Critical Result: CrimesAll coefficient estimate equals .321. This evidence, the positive sign of the coefficient estimate, suggests that higher crime rates increase judicial expenditures thereby supporting the theory. GdpPC coefficient estimate equals .298. This evidence, the positive sign of the coefficient estimate, suggests that higher per capita GDP increases judicial expenditures thereby supporting the theory. c. Crime Rate GDP H0: EGDP = 0 H0: ECrimes = 0 H1: ECrimes > 0 H1: EGDP > 0 d. Crime Rate GDP

.4348 | .217 2 x x

 .0001 | < .0001 2

At the 1 percent significance level we reject the GDP null hypothesis. Even at a 10 percent significance level we cannot reject the crime rate null hypothesis.

3. a. Original Model: JudExpt = EConst + ECrimesCrimesAllt + EGDPGdpPCt + et Divide both sides of the equation by

GdpPCt

.

JudExpt Tweaked Model : GdpPCt

EConst GdpPCt

EConst GdpPCt

 ECrimes

CrimesAllt GdpPCt  EGDP  GdpPCt GdpPCt

et GdpPCt

 ECrimes

CrimesAllt  EGDP GdpPCt  GdpPCt

Ht

where H t

et GdpPCt

91

Var[εt ]

Var[

et ] GdpPCt p

Arithmetic of variances: Var[cx] c 2 Var[x]

1 Var[et ] GdpPCt p

Var[et ] =V u GpcPCt where V equals a constant

1 V u GpcPCt GdpPCt V b. Ordinary Least Squares (OLS) Dependent Variable: AdjJudExp Estimate SE t-Statistic Explanatory Variable(s): AdjCrimesAll 0.308650 0.375938 0.821013 AdjGdpPC 0.299664 0.062667 4.781868 AdjConst 2413.696 2517.482 0.958774 Number of Observations

Prob 0.4158 0.0000 0.3426

50

5. a. x

A state with a higher unemployment rate should increase its poverty rate: EUnem > 0.

x

A state with a higher burglary rate should depress economic conditions increasing the poverty rate: EBurg > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: PovRate Estimate SE t-Statistic Explanatory Variable(s): UnemRate 0.446262 0.380817 1.171854 Burglaries 0.006463 0.001780 3.630155 Const 4.949594 2.049562 2.414952 Number of Observations

Prob 0.2472 0.0007 0.0197

50

EstPovRate = 4.95 + .446UnemRate + .00646Burglaries Estimated Equation: Interpretation of Estimates: bUnem = .446: A 1 percentage point rise in the unemployment rate increases the poverty rate by .446 percentage points. bBurglaries = .00646: A 1 per 100,000 persons rise in the burglary rate increases the poverty rate by .00646 percentage points. A 100 per 100,000 persons rise in the burglary rate increases the poverty rate by .646 percentage points.

92

Critical Result: Unem coefficient estimate equals .446. This evidence, the positive sign of the coefficient estimate, suggests that higher unemployment rates increase the poverty rate thereby supporting the theory. Burglaries coefficient estimate equals .00646. This evidence, the positive sign of the coefficient estimate, suggests that higher a higher burglary rate increases the poverty rate thereby supporting the theory. c. UnemRate Burglaries H0: EBurg = 0 H0: EUnem = 0

d.

x x

H1: EUnem > 0

H1: EBurg > 0

UnemRate

Burglaries

.2472 | .124 2

.0007 | .0004 2

At the 1 percent significance level we reject the Burglaries null hypothesis. Even at a 10 percent significance level we cannot reject the unemployment rate null hypothesis.

7. a. Original Model: PovRatet = EConst + EUnemUnemRatet + E BurgBurglariest + et Multiply both sides of the equation by

Popt .

Tweaked Model : PovRatet Popt

EConst Popt  EUnemUnemRatet Popt  E Burg Burglariest Popt  et Popt EConst Popt  EUnemUnemRatet Popt  E Burg Burglariest Popt  where H t

Var[εt ]

Var ª¬et Popt º¼ p Popt Var[et ] p Popt

et Popt

Arithmetic of variances: Var[cx] c 2 Var[x]

Var[et ] =

V where V equals a constant Popt

V Popt

V

93

Ht

b. Ordinary Least Squares (OLS) Dependent Variable: AdjPovRate Estimate SE Explanatory Variable(s): AdjUnemRate 0.814434 0.331685 AdjBurglaries 0.005746 0.001291 AdjConst 3.059615 2.097694 Number of Observations

94

50

t-Statistic 2.455442 4.451748 1.458561

Prob 0.0178 0.0001 0.1513

Chapter 17: Autocorrelation (Serial Correlation) Solutions to Chapter 17 Prep Questions 1. x

x

x

Error Term Equal Variance Premise: The variance of the error term’s probability distribution for each observation is the same; all the variances equal Var[e]: Var[e1] = Var[e2] = … = Var[eT] = Var[e] Error Term/Error Term Independence Premise: The error terms are independent: Cov[ei, ej] = 0. Knowing the value of the error term from one observation does not help you predict the value of the error term for any other observation. Explanatory Variable/Error Term Independence Premise: The explanatory variables, the xt’s, and the error terms, the et’s, are not correlated. Knowing the value of an observation’s explanatory variable does not help you predict the value of that observation’s error term.

3. The error term/error term independence premise played an important role. If this premise were violated, the error term covariance terms would not equal 0. The expression for Var[bx] would be much more complicated. 5. a. The value of one variable does not allow us to predict the value of the other. b. We use a scatter diagram plotting each variable’s deviation from its mean illustrates independence. Loosely speaking, when the variables are independent that scatter diagram points are spread “evenly” across all four quadrants. c. Both are approximately 0. Rest =

7.



yt

Estyt

Substituting for yt yt = EConst + Exxt + et = EConst Substituting for et et = Uet1 + vt

+

= EConst +

Exxt +

et  Estyt

Exxt + Uet1 + vt  Estyt

Substituting for Estyt Estyt = bConst + bxxt = EConst + Exxt + Uet1 + vt  (bConst + bxxt) Rearranging terms = (EConstbConst) + (Exbx)xt + Uet1 + vt

95

Solutions to Chapter 17 Review Questions 1. Autocorrelation exists whenever one observation’s error term is correlated with the previous observation’s error term. 3. a. The ordinary least squares (OLS) estimation procedure for the coefficient value is unbiased. b. x The ordinary least squares (OLS) estimation procedure for the variance of the error term’s probability distribution is flawed because it is based on a false premise; hence, the standard errors and call calculations based on the standard errors are flawed. x The ordinary least squares (OLS) estimation procedure is not the best linear unbiased estimation procedure (BLUE). Solutions to Chapter 17 Exercises 1. a. The downward sloping theory of demand postulates that a rise in the real price of petroleum should decrease petroleum consumption. E P < 0. b. Ordinary Least Squares (OLS) Dependent Variable: PetroConsPC Estimate SE t-Statistic Explanatory Variable(s): PriceReal 18.00446 53.63777 2.979138 Const 563.4323 16.26989 34.63038 Number of Observations

Prob 0.0054 0.0000

35

Estimated Equation: EstPetroConsPC = 563.4  53.6PriceReal Interpretation of Estimates: bP = 53.6: A $1 increase in the real price of petroleum decreases per capital petroleum consumption by 53.6 gallons. Critical Result: Price coefficient estimate equals 53.6. This evidence, the negative sign of the coefficient estimate, suggests that higher petroleum prices reduce petroleum consumption. c. H0: E P = 0 H1: E P < 0 d.

.0054 | .003 2 At the 1 percent significance level we reject the null hypothesis.

96

3. a. Autocorrelation model: vt’s are independent et = Uet1 + vt or vt = et  Uet1 Start with the original model: PetroConsPCt = EConst + E I PriceRealt + et Lag it one year and multiply by U: U PetroConsPCt–1 = UEConst + EIU PriceRealt–1 + Uet–1 Subtract PetroConsPCt  U PetroConsPCt–1 = (1U)EConst + EI(PriceRealt  UPriceRealt–1) + et  Uet–1 Substituting vt for et  Uet–1: PetroConsPCt  U PetroConsPCt–1 = (1U)EConst + EI(PriceRealt  UPriceRealt–1) + vt vt’s are independent b. Use the estimated value of U to generate two new variables: AdjPetroConsPC = PetroConsPC  .77PetroConsPC(1) AdjPriceReal = PriceReal  .77PriceReal(1) Ordinary Least Squares (OLS) Dependent Variable: AdjPetroConsPC Estimate SE t-Statistic Explanatory Variable(s): AdjPriceReal 21.48296 47.58822 2.215161 Const 131.0316 4.954810 26.44533 Number of Observations

Prob 0.0340 0.0000

34

5. a. x x

An increase in per capita income should reduce the crime rate. EI < 0. An increase in the unemployment rate should increase the crime rate. EUn > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: CrimesAll Estimate SE t-Statistic Explanatory Variable(s): IncPC 0.693721 0.116784 5.940206 UnemRate 114.1572 111.4697 1.024109 Const 14897.77 2369.554 6.287163 Number of Observations

Prob 0.0000 0.3201 0.0000

20

Estimated Equation: CrimesAll = 14,898  .694IncPC + 114.2UnemRate Interpretation of Estimates: bI = .694: A $1 increase in per capita income decreases the crime rate by .694 per 100,000. bUn = 114.2: A 1 percentage point increase in the unemployment rate increases the crime rate by 114.2 per 100,000.

97

c.

d.

Critical Result: IncPC coefficient estimate equals .694. This evidence, the negative sign of the coefficient estimate, suggests that higher per capita income reduces crime thereby supporting the theory. UnemRate coefficient estimate equals 114.2. This evidence, the positive sign of the coefficient estimate, suggests that higher unemployment rates increase crime thereby supporting the theory. Income Unemployment H0: EUn = 0 H0: E I = 0 H1: EUn > 0 H1: E I < 0

 .0001 | < .0001 2 x x

.3201 | .160 2

At the 1 percent significance level we reject the income null hypothesis. Even at the 10 percent significance level we do not reject the unemployment null hypothesis.

7. a. Autocorrelation model: vt’s are independent et = Uet1 + vt or vt = et  Uet1 Start with the original model: CrimesAllt = EConst + EIIncPCt + EUnUnemRatet + et Lag it one year and multiply by U: UCrimesAllt–1 = UEConst + EIUIncPCt–1 + EUnUUnemRatet–1 + Uet–1 Subtract CrimesAllt  UCrimesAllt–1 = (1U)EConst + E I(IncPCt  UIncPCt–1) + EUn(UnemRatet  UUnemRatet–1) + et Uet–1 Substituting vt for et  Uet–1: CrimesAllt  UCrimesAllt–1 = (1U)EConst + E I(IncPCt  UIncPCt–1) + EUn(UnemRatet  UUnemRatet–1) + vt vt’s are independent b. Use the estimated value of U to generate three new variables: AdjCrimesAll = CrimesAll  .79CrimesAll(1) AdjIncPC = IncPC  .79IncPC(1) AdjUnemRate = UnemRate  .79UnemRate (1) Ordinary Least Squares (OLS) Dependent Variable: AdjCrimesAll Estimate SE t-Statistic Prob Explanatory Variable(s): AdjIncPC 0.324473 0.122196 2.655344 0.0173 AdjUnemRate 144.7625 55.71745 2.598153 0.0194 AdjConst 1747.274 467.3090 3.739012 0.0018 Number of Observations

98

19

Chapter 18: Explanatory Variable/Error Term Independence Premise, Consistency, and Instrumental Variables Solutions to Chapter 18 Prep Questions 1. a. i. ii. Decrease. iii. Negatively. b. i. et yt ii. Observationxt 1 2 4 11 2 6 2 11 3 10 3 14 4 14 2 11 5 18 1 14 6 22 4 13 iii. Plot the x and y values for each of the 6 observations. c. See above. d. The best fitting line is less steeply sloped than the actual line. Solutions to Chapter 18 Review Questions 1. a. Upward bias. b. Downward bias. c. Unbiased. 3. Only the mean of the estimate’s probability distribution is important. An estimation procedure is unbiased if and only if the mean of the estimate’s probability distribution equals the actual value: Mean[Est] = Actual Value 5. No. An unbiased estimation procedure will not be consistent whenever the variance of the estimate’s probability distribution remains constant or increases as the sample size increases. 7. The instrument and the x “problem” explanatory variable must be correlated. x error term must be independent.

99

Solutions to Chapter 18 Exercises 1. a. .50 b. .0156 c. 32 percent 3. Random Sampling Procedure 5. Comparing the answers: x The unbiased procedure results in 32 percent of the repetitions being “close to” the actual value when the sample size is 16. x The biased procedure results in 36 percent of the repetitions being “close to” the actual value when the sample size is 25. An unbiased procedure is not always better than a biased procedure. When x the magnitude of the bias for the biased procedure is sufficiently small and x the variance of the biased procedure is smaller than the variance of the unbiased procedure The biased procedure can produce more reliable estimates in the sense that a greater proportion of the estimates will be close to the actual value.

100

Chapter 19: Measurement Error and the Instrumental Variables Estimation Procedure Solutions to Chapter 19 Prep Questions 1. a. Sometimes you will be a little premature in clicking the stop watch button. Other times you will be a little late. It is humanly impossible to measure the actual elapsed time perfectly. No matter how careful you are, sometimes the measured value will be a little low and other times a little high. b. i. Half ii. Half iii. 0 Solutions to Chapter 19 Review Questions 1. a. The ordinary least squares (OLS) estimation procedure for the coefficient value will still be unbiased. b. The variance of the coefficient estimate’s probability distribution will be greater as a consequence of dependent variable measurement error. This occurs because measurement error introduces more “more uncertainty” into the mix. 3. a. No. The ordinary least squares (OLS) estimated procedure for the coefficient value is biased when measurement error is present (and the actual value of the coefficient is nonzero). b. While the instrumental variable (IV) estimation procedure for the coefficient value is still biased, it is consistent. Solutions to Chapter 19 Exercises 1. Explanatory variable measurement error would be introduced. The ordinary least squares (OLS) estimation procedure for the value of the SATScores coefficient would be biased toward 0 if SAT scores do have an impact on exam scores. 3. Dependent variable measurement error would be introduced. The ordinary least squares (OLS) estimation procedure x for the value of the ProbScores and SATScores coefficients would not be biased. But x the variance of the estimates’ probability distributions would be greater as a consequence of the randomization.

101

5. a. Ordinary Least Squares (OLS) Dependent Variable: IncAnnPC Estimate SE Explanatory Variable(s): Coll 0.783234 0.077323 Const 12.76024 2.140549 Number of Observations Estimated Equation:

t-Statistic 10.12934 5.961198

Prob 0.0000 0.0000

t-Statistic 2.545781 17.55610

Prob 0.0141 0.0000

51 EstIncAnnPC = 12.76 + .783Coll

b. Ordinary Least Squares (OLS) Dependent Variable: Covered Estimate SE Explanatory Variable(s): EstIncAnnPC 0.319072 0.125334 Const 75.42874 4.296441 Number of Observations

51

EstCovered = 75.43 + .32EstIncAnnPC Estimated Equation: Interpretation of Estimates: bEstIncAnnPC = .32: A $1,000 increase in permanent per capita disposable income increases the state’s health coverage by .32 percentage points. Critical Result: The EstIncAnnPC coefficient estimate equals .32. This evidence, the positive sign of the coefficient estimate, suggests that increases in permanent disposable income increase health insurance coverage thereby supporting the theory. 7. a. x

A state with a higher crime rate should have more court case and hence would spend more on its judicial system: ECrimes > 0.

x

A state with a higher per capita GDP has a larger tax base and hence, can afford to spend more on its judicial system: EGDP > 0.

b. Ordinary Least Squares (OLS) Dependent Variable: JudExp Estimate SE t-Statistic Explanatory Variable(s): CrimesAll 0.321427 0.408012 0.787787 GdpPC 0.298469 0.063159 4.725673 Const 2425.185 2705.168 0.896501 Number of Observations

50

Estimated Equation: EstJudExp = 2,425 + .321CrimesAll + .298GdpPC Interpretation of Estimates: bCrimesAll = .321: A 1 per 100,000 persons rise in the crime rate increases judicial

102

Prob 0.4348 0.0000 0.3746

expenditures by $.321 per 100,000 persons. bGDP = .298: A $1 rise in per capita state income increases judicial expenditures by $.298 per 100,000 persons. Critical Result: CrimeAll coefficient estimate equals .321. This evidence, the positive sign of the coefficient estimate, suggests that higher crime rates increase judicial expenditures thereby supporting the theory. GdpPC coefficient estimate equals .298. This evidence, the positive sign of the coefficient estimate, suggests that higher per capita GDP increases judicial expenditures thereby supporting the theory. 9. x

x

First, we used annual disposable income as an explanatory variable and applied the ordinary least squares (OLS) estimation procedure. We estimated that a 1 per 100,000 persons rise in crime rate increases judicial expenses by $.321 per 100,000 persons. But we believe that an explanatory variable measurement error problem is present here. Second, we used an instrumental variable (IV) approach which resulted in a higher estimate for the impact of permanent income. We estimated that a 1 per 100,000 persons rise in crime rate increases judicial expenses by $5.12 per 100,000 persons.

103

Chapter 20: Omitted Variables and the Instrumental Variable Estimation Procedure Solutions to Chapter 20 Prep Questions 1. a. yt

= = =

EConst EConst EConst

+ + +

Ex1x1t Ex1x1t Ex1x1t

+ + +

et Ex2 x2t + (Ex2x2t + et) Ht Ht = Ex2x2t + et

b. Yes, the explanatory variable, x1t, and the second model’s error term, Ht, are positively correlated. x1t and x2t positively correlated Included Typically, omitted o variable x1t up variable x2t up Ht = Ex2x2t + et p p x1t up Ht up Ex2 > 0 x1t and Ht positively correlated c. Since the explanatory variable, x1t, and the error term, Ht, are positively correlated, the ordinary least squares (OLS) estimation procedure for the value of the coefficient estimate, Ex1, will be biased upward. 3. .60 Solutions to Chapter 20 Review Questions 1. The instrument and the x “problem” explanatory variable must be correlated. x error term must be independent. Solutions to Chapter 20 Exercises 1. a. Since George W. Bush, a Republican, was the incumbent President, a rising unemployment should hurt the Republicans and increase the votes received by the Democrats. b. EUnTrend > 0.

105

3. x x

Instrument/”Problem” Explanatory Variable Correlation: The instrument, AdvDegt, must be correlated with the “problem” explanatory variable, PopDent. Instrument/Error Term Independence: The instrument, AdvDegt, and the error term, Ht, must be independent.

5. x

x

Instrument/”Problem” Explanatory Variable Correlation: The correlation coefficient of AdvDegt and PopDent equals .77. This along with the results from IV Regression 1 suggests that this condition is satisfied. Instrument/Error Term Independence: We cannot determine if this condition is met.

7. VoteDemPartyTwot

=

EConst +

EGdpGth RealGdpGrowtht + EPopDenPopDent

+

+ +

ELibLiberalt + et (ELibLiberalt + et) Ht

where Ht = ELibLiberalt + et

Included variable PopDent up

PopDent and Liberalt positively correlated Typically, omitted variable o Liberalt up Ht = E PopDen Liberalt + et p Ht up ELib > 0

PopDent and Ht positively correlated p Ordinary least squares (OLS) estimation procedure for the value of the coefficient will be biased upward.

106

9. a. Ordinary Least Squares (OLS) Dependent Variable: PopDen Estimate SE t-Statistic Explanatory Variable(s): AdvDeg 314.3409 37.54406 8.372587 Const –2748.974 393.1890 –6.991482 Number of Observations Estimated Equation:

Prob 0.0000 0.0000

51 EstPopDen = 2,749.0 + 314.3AdvDeg

b. Ordinary Least Squares (OLS) Dependent Variable: VoteDemPartyTwo Estimate SE t-Statistic Explanatory Variable(s): RealGdpGrowth 0.867041 0.416006 2.084205 EstPopDen 0.008464 0.000957 8.844707 Const 49.19420 1.049180 46.88823 Number of Observations

Prob 0.0425 0.0000 0.0000

51

VoteDemPartyTwo = 49.2  .87 RealGdpGrowth + .008EstPopDen Interpretation of Estimates: bGdpGth = .87: A 1 percentage point rise in the real GDP growth rate decreases the vote of the Democrats by .87 percentage points bEstPopDen = .008: A 1 person increase in a state’s population density increases the state’s Democratic vote by .008 percentage points; a 100 person increase in a state’s population density increases the state’s Democratic vote by .8 percentage points. Estimated Equation:

107

Chapter 21: Panel Data and Omitted Variables Solutions to Chapter 21 Prep Questions 1. Fill in each of the following blanks with a Yes or a No: OLS Bias Question: Is the explanatory Satisfied: Explanatory variable/error term independence premise Variable and Error satisfied or violated? Term Independent ~ p Unbiased

Is the OLS estimation procedure for the value of the coefficient unbiased or biased? OLS Reliability Question: Are the error term equal variance and error term/error term independence premises satisfied or violated?

Satisfied

Violated

~ ~ p

~ ~ p

Can the OLS calculations for the standard error be “trusted?”

Yes

No

Is the OLS estimation procedure for the value of the coefficient BLUE?

Yes

No

Violated: Explanatory Variable and Error Term Correlated ~ p Biased

Solutions to Chapter 21 Review Questions 1. For each cross section entity, the value of the omitted variable does not vary from time period to time period. 3. For each period, the value of the omitted variable does not vary from cross section entity to cross section entity. 5. With heteroskedasticity and autocorrelation we exploited the relationship among the error terms to improve the estimation procedure. The random effects approach uses the same strategy.

109

Solutions to Chapter 21 Exercises 1. a. x x

The downward sloping demand curve theory suggests that a rise in the price of beer decreases per capita beer consumption. E P < 0. Conventional wisdom suggests that beer is an inferior good; as per capita income rises, per capita beer consumption decreases. EI < 0.

b. Ordinary Least Squares (OLS) Dependent Variable: BeerPC Estimate SE t-Statistic Explanatory Variable(s): Price 304.6151 80.51845 0.264329 IncPC 2.414644 1.109782 2.175782 Const 394.2818 122.3245 3.223244 Number of Observations Cross Sections Periods

c.

d.

459 51 9

Estimated Equation: EstBeerPC = 394.3  80.5Price  2.4IncPC Interpretation of Estimates: bP = 80.5: A $1 increase in the real price decreases per capita consumption by 80.5 cans per year. bI = 2.4: A $1,000 increase in real per capita disposable income decreases per capita consumption by 2.4 cans per year. Critical Result: The Price coefficient estimate equals 80.5. This evidence, the negative sign of the coefficient estimate, suggests that higher beer prices decrease per capita beer consumption thereby supporting the theory. The IncPC coefficient estimate equals 2.4. This evidence, the negative sign of the coefficient estimate, suggests that higher per capita income decreases per capita beer consumption thereby supporting the theory. Price Disposable Income H0: E I = 0 H0: E P = 0 H1: E I < 0 H1: E P < 0 Price Disposable Income

.7916 | .3958 2 x x

110

Prob 0.7916 0.0301 0.0014

.0301 | .015 2

Even at the 10 percent significance level we cannot reject the price null hypotheses that price has no impact on beer consumption. At the 5 percent significance level, we reject the null hypothesis that disposable income has no impact on beer consumption. On the other hand, at the 1 percent significance level, we do not reject the null hypothesis.

3. Ordinary Least Squares (OLS) Dependent Variable: LogUsersInternet Estimate SE t-Statistic Explanatory Variable(s): Year 0.449654 0.017078 26.32965 CapitalHuman 0.023725 0.002470 9.606597 CapitalPhysical 0.002056 0.000480 4.286193 GdpPC 0.118177 0.011461 10.31146 Auth 0.013999 –6.845761 0.095836 Const 899.3201 34.17432 –26.31567 Number of Observations Cross Sections Periods

Prob 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

566 110 8

The coefficient estimate of Year is positive suggesting that per capita Internet use is growing over time after accounting for human capital, physical capital, per capita GDP, and the political environment. More specifically, we estimate that per capita Internet use is growing by 45 percent annually after accounting for the other explanatory factors. 5. a.

The downward sloping demand curve theory suggests that a rise in the price of gasoline decreases per capita motor fuel consumption.

b. Ordinary Least Squares (OLS) Dependent Variable: MotorFuelPC Estimate SE t-Statistic Explanatory Variable(s): Price 21.87900 50.72342 2.318361 Const 509.3529 21.98667 23.16644 Number of Observations Cross Sections Periods

Prob 0.0224 0.0000

105 3 35

Estimated Equation: EstMotorFuelPC = 509.4  50.7Price Interpretation of Estimates: bP = 50.7: A $1 increase in the real price decreases per capita consumption by 50.7 gallons per year. Critical Result: The Price coefficient estimate equals 50.7. This evidence, the negative sign of the coefficient estimate, suggests that higher gasoline prices decrease per capita motor fuel consumption thereby supporting the theory. c. H0: E P = 0 H1: E P < 0 d.

.0224 | .011 2 111

At the 5 percent significance level, we reject the null hypothesis that the real price of gasoline has no impact on beer consumption. On the other hand, at the 1 percent significance level, we do not reject the null hypothesis. e.

The first 35 observations are for Arkansas, the second 35 observations for Massachusetts, and the last 35 observations for Washington. The residuals for x Arkansas are positive. x Massachusetts are negative. x Washington approximately zero. This suggests that the inclusion of cross section random effects may be appropriate.

112

Chapter 22: Simultaneous Equations Models—Introduction Solutions to Chapter 22 Prep Questions 1. See Appendix 22.1. 3. a. Multiple regression analysis attempts to sort out the individual effect that each explanatory variable has on the dependent variable. b. Each explanatory variable’s coefficient reveals the individual impact that the explanatory variable has on the dependent variable; that is, each explanatory variable’s coefficient tells us how changes in that explanatory variable affect the dependent variable while all other explanatory variables remain constant. c. 'y = bx1'Vbl1 d. 'y = bx2'Vbl2 e. 'y = bx1'Vbl1 + bx2'Vbl2 5. a.

P = = = =

6 + .01Inc + .4FeedP 6 + .01u4,000 + .4u40 6 + 40 + 16 50

Q = 160,000 + 500Inc  4,000FeedP = 160,000 + 50u4,000  4,000u40 = 160,000 + 200,000  160,000 = 200,000 b. On a sheet of graph paper, plot the demand and supply curves to illustrate the equilibrium. Solutions to Chapter 22 Review Questions 1. Endogenous variables are variables whose values are determined “within” the model. 3. a. No. b. No. 5. a. No. b. Yes.

113

Solutions to Chapter 22 Exercises 1. a.

bID estimates how the quantity of beef demanded changes when income changes while the price of beef remains constant.

QD while P remains constant 'I 'QD = bID 'I while the price of beef remains constant b. i. 'P = 1.0562'FeedP + .018825'Inc ii. 0 = 1.0562'FeedP + .018825'Inc Solving for 'FeedP 1.0562'FeedP =  .018825'Inc .018825 'FeedP =  'Inc 1.0562 = .017233'Inc c. i. 'Q =  332.00'FeedP + 17.347'Inc ii. From b2: 'FeedP = .017233'Inc =  332.00u(.017233'Inc) + 17.347'Inc = 5.917'Inc + 17.347'Inc = 23.264'Inc 'Q 23.264 while P remains constant 'Inc bID = 23.264 bID

3.

log(Qt )

Q Q DConst  D FP log( FeedPt )  D IQ log( Inct )



H tQ

log( Pt )

P P D Const  D FP log( FeedPt )  D IP log( Inct )



H tP

5. a. Own Price Elasticity of Demand = bPD

bPD estimates the own price elasticity of demand, the percent change in quantity demanded resulting from a 1 percent change in the price while income remains constant.

bPD

'LogQ D 'LogP

while LogInc remains constant

While LogInc remains constant: 'LogInc = 0 'LogQ = .08452'LogFeedP + .3595'LogInc = .08452'LogFeedP 'LogP = .3317'LogFeedP + .7503'LogInc = .3317'LogFeedP

bPD

'LogQ D 'LogP

.0852'LogFeedP .3317'LogFeedP

.0852 .3317

.255

Interpretation: We estimate that a 1 percent increase in the price of beef decreases the quantity of beef demanded by .225 percent.

114

b. Own Price Elasticity of Supply = bPS

bPS estimates the own price elasticity of supply, the percent change in quantity supplied resulting from a 1 percent change in the price while the feed price remains constant.

'LogQ S 'LogP

bPS

while LogFeedP remains constant

While LogFeedP remains constant: 'LogFeedP = 0 'LogQ = .08452'LogFeedP + .3595'LogInc = .3595'LogInc 'LogP = .3317'LogFeedP + .7503'LogInc = .7503'LogInc S P

b

'LogQ S 'LogP

.3595'LogInc .7503'LogInc

.3595 .479 .7503

Interpretation: We estimate that a 1 percent increase in the price of beef increases the quantity of beef supplieded by .479 percent. 7. Ordinary Least Squares (OLS) Dependent Variable: Q Explanatory Variable(s): FeedP Inc Const

Estimate 533.6727 10.77606 54222.01

Number of Observations

SE 139.9106 1.628569 12057.98

t-Statistic 3.814384 6.616890 4.496775

Prob 0.0003 0.0000 0.0000

72

Estimated Equation: EstQ = 54,222  533.67FeedP + 10.776Inc Interpretation of Estimates: Q = 533.67: A 1 cent increase in the price of cattle feed decreases the quantity of aFP beef by 533.67 million pounds.

aIQ = 10.776: A 1 billion dollar increase in real disposable income increases the quantity of beef by 10.776 million pounds. Ordinary Least Squares (OLS) Dependent Variable: P Explanatory Variable(s): FeedP Inc Const Number of Observations Estimated Equation:

Estimate 1.342682 0.010833 39.19066

SE 0.126652 0.001474 10.91532

t-Statistic 10.60134 7.347909 3.590428

Prob 0.0000 0.0000 0.0006

120 EstP = 39.19 + 1.3427FeedP + .010833Inc

115

Interpretation of Estimates: P = 1.3427: A 1 cent increase in the price of cattle feed increases the price of beef aFP

by 1.3427 cents.

aIP = .010833: A 1 billion dollar increase in real disposable income increases the price of beef by .010833 cents. a. 'Q = 533.67'FeedP + 10.776'Inc b. 'P = 1.3427'FeedP + .010833'Inc 9. a.

An increase in police expenditures provides the populace with more police protection which decreases the crime rate: E PC < 0.

b. An increase in the unemployment rate increases the chances that crimes will be committed: EUC > 0. 11.

Crimest

c DConst  DUCUnemRatet  D GC GdpPCt  H tC

PoliceExpt

P DConst  DUPUnemRatet  D GP GdpPCt  H tP

13. a.

bPC estimates how the crime rate changes when police expenditures change while the unemployment rate remains constant.

bPC

'Crimes 'PoliceExp

whileUnemRate remains constant

b. While UnemRate remains constant, 'UnemRate = 0: 'Crimes = 23.814'UnemRate  .025271'GdpPC = .025271'GdpPC 'PoliceExp = 12.174'UnemRate + .019163'GdpPC = .019163'GdpPC

bPC

116

'Crimes 'PoliceExp

.025271'GdpPC .019163'GdpPC

.025271 1.319 .019163

Chapter 23: Simultaneous Equations Models—Identification Solutions to Chapter 23 Prep Questions 1. Ordinary Least Squares (OLS) Dependent Variable: P Explanatory Variable(s): FeedP Inc Const

Estimate 1.056242 0.018825 33.02715

Number of Observations Estimated Equation:

SE 0.286474 0.005019 31.04243

t-Statistic 3.687044 3.750636 1.063936

Prob 0.0003 0.0003 0.2895

120 EstP = 33.027 + 1.0562FeedP + .018825Inc

3. a. Ordinary Least Squares (OLS) Dependent Variable: Q Explanatory Variable(s): EstP FeedP Const

Estimate 921.4783 –1305.262 108291.8

Number of Observations Estimated Equation: b. They are identical.

SE 113.2551 121.2969 16739.33

t-Statistic 8.136309 –10.76089 6.469303

Prob 0.0000 0.0000 0.0000

120 EstQ = 108,292  921.5EstP  1,305Inc

Solutions to Chapter 23 Review Questions 1. A model is underidentified whenever we cannot estimate all the parameters of the original model using the reduced form (RF) estimation procedure. 3. a. The two procedures provide identical results whenever the model is identified or underidentified. b. The two procedures provide different results whenever the model is overidentified. The reduced form (RF) estimation provides multiple estimates for one or more of original model’s parameters; two stage least squares (TSLS) provides only a single estimate.

117

Solutions to Chapter 23 Exercises 1. a. Ordinary Least Squares (OLS) Dependent Variable: LogQ Estimate SE t-Statistic Explanatory Variable(s): LogFeedP 0.027126 0.178585 6.583483 Const 12.89211 0.101850 126.5791 Number of Observations

Prob 0.0000 0.0000

120

Q Estimate: aFP = .1786

'LogQ = .1786'LogFeedP Ordinary Least Squares (OLS) Dependent Variable: LogP Explanatory Variable(s): LogFeedP Const Number of Observations

Estimate 0.135433 4.586163

SE 0.074912 0.281270

t-Statistic 1.807907 16.30522

Prob 0.0732 0.0000

120

P Estimate: aFP = .1354

'LogP = .1354'LogFeedP i.

Yes. We can still estimate the own price elasticity of demand because the feed price data “tells us” when the supply curve shifts allowing us to use the equilibria to estimate the slope of the demand curve.

'LogQ D 'LogP

bPD

.1786'LogFeedP .1354'LogFeedP

.1786 .1354

1.319

ii. No. Without chicken price and income data we do not know when the demand curve shifts. We cannot use the equilibria to estimate the slope of the supply curve. b. i. Yes. There is one estimate: 1.319. Two Stage Least Squares (TSLS) Dependent Variable: LogQ Instrument(s): LogFeedP Estimate SE t-Statistic Prob Explanatory Variable(s): LogP 0.669960 0.0514 1.318616 1.968203 Const 18.93950 3.412885 5.549410 0.0000 Number of Observations Estimate bPS = 1.319 ii. No.

118

120

3. a. Ordinary Least Squares (OLS) Dependent Variable: LogQ Estimate SE t-Statistic Explanatory Variable(s): LogFeedP 0.028875 0.092794 3.213667 LogInc 0.333895 0.062472 5.344703 LogChickP 0.040721 0.059085 0.689185 Const 9.583209 0.490918 19.52098 Number of Observations Estimates:

Prob 0.0017 0.0000 0.4921 0.0000

120

Q = .0928 aFP

Q = .3339 aInc

Q = .04072 aCP

'LogQ = .0928'LogFeedP + .3339'LogInc + .04072'LogChickP Ordinary Least Squares (OLS) Dependent Variable: LogP Explanatory Variable(s): LogFeedP LogInc LogChickP Const Number of Observations Estimates:

P = .3207 aFP

Estimate 0.320692 0.716095 0.054309 2.373348

SE 0.087094 0.188434 0.178218 1.480750

t-Statistic 3.682121 3.800250 0.304733 1.602802

Prob 0.0004 0.0002 0.7611 0.1117

120 P = .7161 aInc

P = .05431 aCP

'LogP = .3207'LogFeedP + .7161'LogInc + .05431' LogChickP i.

Yes. We can estimate the own price elasticity of demand because the feed price data “tells us” when the supply curve shifts allowing us to use the equilibria to estimate the slope of the demand curve.

bPD

'LogQ D 'LogP

.0928'LogFeedP .3207'LogFeedP

.0928 .3207

.289

ii. Yes. We can estimate the own price elasticity of supply because the chicken price and income data “tell us” when the demand curve shifts allowing us to use the equilibria to estimate the slope of the supply curve. The reduced form (RF) estimation procedure provides two estimates:

bPS

'LogQ S 'LogP

.3339'LogInc .7161'LogInc

bPS

'LogQ S 'LogP

.04072'LogChickP .05431'LogChickP

.3339 .466 .7161 .04072 .450 .05431

119

b. i.

Yes. There is one estimate: .289. Two Stage Least Squares (TSLS) Dependent Variable: LogQ Instrument(s): LogFeedP LogInc LogChickP Estimate SE t-Statistic Explanatory Variable(s): LogP 0.289355 0.044935 6.439347 LogInc 0.541101 0.027894 19.39826 LogChickP 0.056435 0.030582 1.845364 Const 8.896469 0.191751 46.39605 Number of Observations

Prob 0.0000 0.0000 0.0675 0.0000

120

Estimate bPD = .289 ii. Yes. There is one estimate: .480. Two Stage Least Squares (TSLS) Dependent Variable: LogQ Instrument(s): LogFeedP LogInc LogChickP Estimate SE t-Statistic Explanatory Variable(s): LogP 0.480214 0.157434 3.050255 LogFeedP –0.243622 0.057725 –4.220402 Const 10.68977 0.749584 14.26094 Number of Observations

Prob 0.0028 0.0000 0.0000

120

Estimate: bPS = .480 5. a. Ordinary Least Squares (OLS) Dependent Variable: LogQ Estimate SE Explanatory Variable(s): LogInc 0.717732 0.077586 LogPorkP 0.347416 0.052969 Const 3.458030 0.645048 Number of Observations Estimates:

Q = .7177 aInc

72 Q = .3474 aPP

'LogQ = .7177'LogInc + .3474'LogPorkP

120

t-Statistic 9.250758 6.558847 5.360887

Prob 0.0000 0.0000 0.0000

Ordinary Least Squares (OLS) Dependent Variable: LogP Explanatory Variable(s): LogInc LogPorkP Const

Estimate 0.268894 0.044724 2.253325

Number of Observations Estimates:

SE 0.138588 0.094616 1.152212

t-Statistic 1.940241 0.472686 1.955651

Prob 0.0564 0.6379 0.0546

72 P = .04472 aPP

P = .2689 aInc

'LogP = .2689'LogInc  .04472' LogPorkP i.

No. Without feed price data we do not know when the supply curve shifts. We cannot use the equilibria to estimate the slope of the demand curve. ii. Yes. We can still estimate the own price elasticity of supply because the pork price and income data “tell us” when the demand curve shifts allowing us to use the equilibria to estimate the slope of the supply curve. The reduced form (RF) estimation procedure provides two estimates:

bPS

'LogQ S 'LogP

.7177'LogInc .2689'LogInc

bPS

'LogQ S 'LogP

.3474'LogPorkP .044721'LogPorkP

.7177 .2689

2.67 .3474 .04472

7.77

b. i. No. ii. Yes. There is one estimate: 2.67. Two Stage Least Squares (TSLS) Dependent Variable: LogQ Instrument(s): LogInc LogPorkP Estimate SE Explanatory Variable(s): LogP 2.667617 1.593295 Const 6.847074 0.180328 Number of Observations

t-Statistic 1.674277 0.026337

Prob 0.0985 0.9791

72

Estimate: bPS = 2.67 7. None. 9. One.

121

Chapter 24: Binary and Truncated Dependent Variables Solutions to Chapter 24 Prep Questions 1. The more densely populated states tend to have more voters who are registered Democrats than Republicans; consequently, Democrats would be more likely to win states whose population density is high. EDen > 0. 3. a. Ordinary Least Squares (OLS) Dependent Variable: WinDem1 Estimate SE Explanatory Variable(s): PopDen 0.001001 0.000236 Const 0.192315 0.074299 Number of Observations

t-Statistic 4.247280 2.588406

Prob 0.0001 0.0127

50

EstWinDem1 = .192 + .001PopDen Estimated Equation: b. i. PopDen = 1.2 EstWinDem1 = .1922 + .001u1.2 = .1923 + .0012 = .1935 ii. PopDen = 321.0 EstWinDem1 = .1923 + .001u321.0 = .1923 + .3210 = .5133 iii. PopDen = 566.6 EstWinDem1 = .1923 + .001u566.6 = .1923 + .5666 = .7589 iv. PopDen = 821.3 EstWinDem1 = .1923 + .001u821.3 = .1923 + .8213 = 1.0146 v. PopDen = 1,025.0 EstWinDem1 = .1923 + .001u1,025.0 = .1923 + 1.0250 = 1.2173 vi. PopDen = 1,162.0 EstWinDem1 = .1923 + .001u1,162.0 = .1923 + 1.1620 = 1.3543

123

5.

a. The scatter diagram appears to support the theory. b. The values of the dependent variable, Salary, have a lower bound of 414,000. Solutions to Chapter 24 Review Questions 1. The dependent variable represents a probability. 3. Yes. In a probit model, the best fitting is S-shaped which can never be less than 0 or greater than 1. Hence, estimates of the dependent variable always lie between 0 and 1. Solutions to Chapter 24 Exercises 1. a. The party winning a state’s Electoral College vote depends on the state’s population density; as a state becomes more densely populated, the Democrats rather than the Republicans become more likely to win. EDen < 0. b. Ordinary Least Squares (OLS) Dependent Variable: WinRep1 Estimate SE t-Statistic Explanatory Variable(s): PopDen 0.001001 0.000236 4.247280 Const 0.807685 0.074299 10.87080 Number of Observations

124

50

Prob 0.0001 0.0000

Estimated Equation: EstWinRep1 = .808  .001PopDen Interpretation of Estimates: bDen = .001: A 1 person per square mile increase in a state’s population density decreases the probability of a Republican victory by .001. Critical Result: The PopDen coefficient estimate equals .001. This evidence, the negative sign of the coefficient estimate, suggests that higher population density decreases the probability of a Republican victory thereby supporting the theory. c. H0: EDen = 0 H1: EDen < 0 d. Prob[Results IF H0 True] =

.0001 < .0001 2

At the 1 percent significance level, we reject the null hypothesis that population density has no impact of the probability of a Republican victory. e. Estimated Equation: EstWinDem1 = .192 + .001PopDen Yes, the results are consistent. A 1 person per square mile increase in population density increases the probability of a Democratic victory by .001 which is equivalent to decreasing the probability of a Republican victory by .001. 3. a. Probit Dependent Variable: WinDem1 Estimate SE Explanatory Variable(s): PopDen 0.009060 0.003073 Const –0.899796 0.342966

z-Statistic 2.947983 –2.623574

Prob 0.0032 0.0087

Number of Observations 50 Critical Result: The PopDen coefficient estimate equals .009. This evidence, the positive sign of the coefficient estimate, suggests that higher population density increases the probability of a Democratic victory thereby supporting the theory. b. H0: EDen = 0 H1: EDen > 0 c. Prob[Results IF H0 True] =

.0032 | .002 2

At the 1 percent significance level, we reject the null hypothesis that population density has no impact on the probability of a Democratic victory.

125

5. a. Probit Dependent Variable: WinDem1 Estimate SE Explanatory Variable(s): PopDen 0.008500 0.003279 UnemTrend 0.175342 0.389554 Const 0.416943 1.003798

z-Statistic 2.592533 0.450110 2.407520

Prob 0.0095 0.6526 0.0161

Number of Observations 50 Critical Result: The PopDen coefficient estimate equals .0085. This evidence, the positive sign of the coefficient estimate, suggests that higher population density increases the probability of a Democratic victory thereby supporting the theory. The UnemTrend coefficient estimate equals .175. This evidence, the positive sign of the coefficient estimate, suggests that an upward trend in the unemployment rate increases the probability of a Democratic victory thereby supporting the theory. b. Yes. The signs of the coefficient estimates lend support to the theories. c. The precise numerical values differ, but this is hardly surprising. What is a little troubling, however, is the Prob. column value for the UnemTrend coefficient estimate has changed dramatically. Even at the 10 percent significant level we can no longer reject the null hypothesis that the unemployment trend has no effect on the probability of a Democratic victory. 7.

The scatter diagram appears to support the theory. 126

9. Tobit Dependent Variable: HeatDegDays Estimate SE Explanatory Variable(s): HighTemp 0.027690 0.951145 Const 72.71311 1.873004

z-Statistic 34.34919 38.82165

Prob 0.0000 0.0000

Number of Observations 365 a. The absolute value of the Tobit estimate is greater indicating that the Tobit best fitting line is more steeply sloped than the ordinary least squares (OLS) best fitting line. b. The Tobit estimation procedure accounts for the fact that heating degree days are truncated at 0. Those days that require cooling rather than heating are ignored when we just consider heating degree days.

127

Chapter 25: Descriptive Statistics, Probability, and Random Variables—A Closer Look Solutions to Chapter 25 Prep Questions

3 = .6. 5 2 = .4. ii. 5 1 b. i. = .5. 2 1 = .5. ii. 2 3 = .75. c. i. 4 1 = .25. ii. 4

1. a. i.

3. a.

Mean[v ]

¦ v Prob[v]

All v

b.

Var[v ]

¦ (v  Mean[v])

2

Prob[v]

All v

Solutions to Chapter 25 Review Questions 1. a. The mean is the average of the values. b. The mode is the most frequently occurring value. c. The median is the value that is in the middle; at least half the values lie at the median or above and at least half lie at the median or below. 3. An event tree visually displays the mutually exclusive outcomes (events) of a random process. 5. a. The value of one random variable does not affect the probability distribution of the other. b. 0.

129

Solutions to Chapter 25 Exercises 1. a.

b. Mean = 719 Mode = 750 Median = 730 c. The left tail is longer than the right tail. Accordingly, the left tail “drags” the mean down. 3. a.

130

b. Prob[Yellow Guilty AND Yellow Reported] = .9 u .8 = .72 Prob[Yellow Guilty AND Orange Reported] = .9 u .2 = .18 Prob[Orange Guilty AND Orange Reported] = .1 u .8 = .08 Prob[Orange Guilty AND Yellow Reported] = .1 u .2 = .02 c. Prob[Yellow Reported] = .74 Prob[Orange Reported] = .26 d. Prob[Yellow Guilty IF Yellow Reported]

Prob[Yellow Guilty AND Yellow Reported] Prob[Yellow Reported]

.72 .97 .74

Prob[Orange Guilty IF Yellow Reported]

Prob[Orange Guilty AND Yellow Reported] Prob[Yellow Reported]

.02 .03 .74

Prob[Yellow Guilty IF Orange Reported]

Prob[Yellow Guilty AND Orange Reported] Prob[Orange Reported]

.18 .69 .26

Prob[Orange Guilty IF Orange Reported]

Prob[Orange Guilty AND Orange Reported] Prob[Orange Reported]

.08 .31 .26

e. No. Regardless of which cab the eyewitness reports a Yellow cab is still more likely to be guilty. f. There are two pieces of information: x 90 percent of the cabs are Yellow. x Eyewitness testimony is 80 percent accurate. Since 90 percent of the cabs are Yellow and the eyewitness testimony is only 80 percent accurate, relative number of cabs trumps the eyewitness testimony. 5. a.

131

b. i.

Prob[Prize behind Door 2 IF Monty opens Door 1]

Prob[Prize behind Door 2 AND Monty opens Door 1] Prob[Monty opens Door 1] 1 1 3 3 2 1 1 1 3  3 6 2 ii. Prob[Prize behind Door 3 IF Monty opens Door 1]

Prob[Prize behind Door 3 AND Monty opens Door 1] Prob[Monty opens Door 1] 1 1 6 6 1 1 1 1 3  3 6 2 iii. Prob[Prize behind Door 1 IF Monty opens Door 2]

Prob[Prize behind Door 1 AND Monty opens Door 2] Prob[Monty opens Door 2] 1 1 3 3 2 1 1 1 3  3 6 2 iv. Prob[Prize behind Door 3 IF Monty opens Door 2]

Prob[Prize behind Door 3 AND Monty opens Door 2] Prob[Monty opens Door 2] 1 1 3 6 1 1 1 1 3  3 6 2 c. Stay.

132

Chapter 26: Estimating the Mean of a Population Solutions to Chapter 26 Prep Questions 1.

[ T1

Mean

( v 1 + v 2 + … + v T) ] Mean[cx] = cMean[x] =

1 Mean v1 + v2 + … + vT] T

[

Mean[x + y] = Mean[x] + Mean[y] =

1 ( Mean[v1] + Mean[v2] + … + Mean[vT]) T Mean[v1] = Mean[v2] = … = Mean[vT] = ActMean

=

1 ( ActMean + ActMean + … + ActMean) T There are T ActMean terms

=

=

1 (TuActMean) T Simplifying ActMean

3. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close to the actual value is small. Solutions to Chapter 26 Review Questions 1. a. When the mean of the estimate’s probability distribution equals the actual value, the estimation procedure is unbiased, the estimation procedure does not systematically overestimate or underestimate the actual value. b. When an estimation procedure is unbiased, the variance of the estimate’s probability distribution indicates how reliable an estimate is. On the one hand, if the variance is small, there is a high probability that the estimate is close to the actual value. On the other hand, if the variance is large, the probability that the estimate lies close to the actual value is small.

133

3. We should divide by the sample size less 1. Variance equals the average of the squared deviations. To calculate an average we divide by the number of pieces of available information. When estimating the population variance of a population, the number of pieces of available information equals the sample size less 1. 5. We use the Student t-distribution because we must estimate the standard deviation. Solutions to Chapter 26 Exercises 1. a. 8

¦v

t

EstMean

b.

t 1

8 8.9  6.4  6.2  6.7  7.5  5.3  6.4  6.2 8 Laptop 1 2 3 4 5 6 7 8

vi 8.9 6.4 6.2 6.7 7.5 5.3 6.4 6.2

EstMean 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6.7

53.6 8

6.7

Dev Sqr Dev 2.2 4.84 0.3 0.09 0.5 0.25 0 0 0.8 0.64 1.4 1.96 0.3 0.09 0.5 0.25 Sum = 8.12

Sum of Squared Deviations = 8.12 c.

EstVarAll

Sum of Squared Deviations Using Estimated Population Mean Sample Size  1 8.12 8.12 1.16 8 1 7

d. Knowing whether one battery’s life was greater than or less than average would not help us predict the life of any other battery. e.

Var[ EstMean] SE[ EstMean]

134

ActVarAll 1.16 .145 T 8 Var[ EstMean] .145 .3808

3. a. 64

¦v

t

EstMean

t 1

64 8 u 8.9  8 u 6.4  8 u 6.2  8 u 6.7  8 u 7.5  8 u 5.3  8 u 6.4  8 u 6.2 64 8 u 53.6 428.8 6.7 64 64

b. Sum of Squared Deviations = 8u8.12 = 64.96 c.

EstVarAll

Sum of Squared Deviations Using Estimated Population Mean Sample Size  1 64.96 64.96 1.03111... 64  1 63

d.

Var[ EstMean] SE[ EstMean]

ActVarAll 1.03 .016111... T 64 Var[ EstMean] .016111... .1269

e. Mean: 7.0 Standard Error: .1269 Value: 6.7 Degrees of Freedom: 63 Prob[Results IF H0 True] = .0106 f. At the 5 percent significance level, we would reject the null hypothesis, reject the manufacturer’s claim that mean battery life was 7.0. On the other hand, at the 1 percent significance level we would not reject the null hypothesis. 5.

Sum of Squared Deviations Sample from Size EstMean EstMean 8 6.7 8.12 64 6.7 64.96 120 6.7 121.80

Variance Estimate of Population’s Probability Distribution 1.16 1.03111… 1.0235

Variance Estimate of of EstMean’s Probability Distribution .145 .01611 .008529

Prob[Results IF H0 True] .3808 .0106 .0008

As the sample size increases and the estimated mean continues to equal 6.7 our confidence that the mean of the population falls short of the 7.0 hours claimed by the manufacturer is reinforced. 135

7. As the sample size increases, use of the Student t-distribution becomes less critical. As the sample size increases, the difference in the probabilities declines.

136