IntroductiontoEconometrics (3 U pdatedEdition, Global Edition...Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 3 (This version August 17, 2014) Stock/Watson - Introduction

©2015 Pearson Education, Ltd.

Introduction to Econometrics (3rd Updated Edition, Global Edition)

by

James H. Stock and Mark W. Watson

Solutions to Odd-Numbered End-of-Chapter Exercises:

Chapter 3(This version August 17, 2014)

Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 3 _____________________________________________________________________________________________________


1

3.1. The central limit theorem suggests that when the sample size ( n ) is large, the distribution of the

(b)

(c)(

sample average Y( ) is approximately N ( ,µ σ 2 )Y Y with2

σ 2Y =σYn

. Given a population µY = 75 and

σY2 = 45, we have

(a) n = 50, 2

2 0.9, and50

σ = σY = 45 =Y n75 73 75Pr(Y ≤ 73) = Pr (−2.108)

0.9 0.9(2.108) =1− 0.9821= 0.0179

Y − − < ≈ Φ

=1−Φ

n = 90, 2

2 45 0.5,90

σ = = =σYY nand

76 75− 75 77 75Pr(760.5 0.5 0.5

(2.828) −Φ(1.414) = 0.9977 − 0.9213 = 0.0764

− −≤ ≤

Y< < 49) = Pr

= Φ

Y

nc = 115, 2

2 45 0.375,120

σ = = =σYY nand

75 69 75Pr( 69)> =1 P− r(Y Y 69)< =1 P− r0.375 0.375

1.633) = Φ(1.633) = 0.9491

− Y −≤

=1−Φ(−

Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 3 2 _____________________________________________________________________________________________________


3.3. Denote each voter’s preference by Y. Y 1 if the voter prefers the incumbent and Y 0 if the voter prefers the challenger. Y is a Bernoulli random variable with probability Pr ( 1 )Y p and Pr ( 0 ) 1Y p . From the solution to Exercise 3.2, Y has mean p and variance p(1 p).

(a) 215 0 537 5.400

p̂

(b) The estimated variance of p̂ is 4(1p pˆ ˆ )

0.5375 (1 0.5375)var( p̂)

400n6 2148 10 . The

standard error is SE1

( ) (varp pˆ ˆ( ))2 0 024 9. (c) The computed t-statistic is

0

SE( p̂) 00249act p̂t

p 0 5375 0 5 1 506

Because of the large sample size (n 400), we can use Equation (3.14) in the text to get the p-value for the test 0H p 0 5 vs. 1H p 0 5 :

-value 2 (p t| act |) 2(1506) 20066 0132

(d) Using Equation (3.17) in the text, the p-value for the test 0H p 0 5 vs. 1H p 0 5 is

-value 1 (p tact ) 1 (1506) 10934 0066

(e) Part (c) is a two-sided test and the p-value is the area in the tails of the standard normal distribution outside (calculated t-statistic). Part (d) is a one-sided test and the p-value is the area under the standard normal distribution to the right of the calculated t-statistic.

(f) For the test 0H p 0 5 vs. 1H p 0 5 , we cannot reject the null hypothesis at the 5% significance level. The p-value 0.066 is larger than 0.05. Equivalently the calculated t-statistic 1506 is less than the critical value 1.64 for a one-sided test with a 5% significance level. The test suggests that the survey did not contain statistically significant evidence that the incumbent was ahead of the challenger at the time of the survey.

Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 3 3_____________________________________________________________________________________________________


3.5. (a) (i) The size is given by Pr(| p̂ 0.5| .02), where the probability is computed assuming that p 0.5.

Pr(| 0.5| 0.02) 0.02p pˆ ˆ 0 .5 . 02)0.02 p̂ 0.5 0.02

p̂ 0.51 Pr 1.30 1.300.50.5/1055

0.19

1 P r(

1 Pr 0.5 0.5/1055 0.5 0.5/1055 0.5 0.5/1055

where the final equality using the central limit theorem approximation. (ii) The power is given by Pr(| p̂ 0.5| 0.02), where the probability is computed assuming that p = 0.53.

Pr(| 0.5| 0.02) 0.02 0.5 . 02)p̂p̂0.02 p̂ 0.5 0.02

0.53 0.47/1055 0.53 0.47/1055 0.53 0.47/10550.05 p̂ 0.53 0.01

p̂ 0.531 Pr 3.25

1 P r(

1 Pr

1 Pr 0.53 0.47/1055 0.53 0.47/1055 0.53 0.47/1055

0.65.530.47/1055

0.74

where the final equality using the central limit theorem approximation.

(b) (i) 0.54 0.50(0.540.46) /1055

t 2.61, and Pr(| | 2.61) 0. 01,t so that the null is rejected at the

5% level. (ii) Pr(t 2.61) .004, so that the null is rejected at the 5% level.

(iii) 0.54 1.96 (0.54 0.46) /1055 0.54 0.03, or 0.51 to 0.57.

(iv) 0.54 2.58 (0.54 0.46) /1055 0.54 0.04, or 0.50 to 0.58.

(v) 0.54 0.67 (0.54 0.46) /1055 0.54 0.01, or 0.53 to 0.55.

(c) (i) The probability is 0.95 is any single survey, there are 20 independent surveys, so the probability if 0.9520 0.36

(ii) 95% of the 20 confidence intervals or 19. (d) The relevant equation is 1.96 SE( ˆ ) .01or 1.96p p (1 ) / .p 01.n Thus n must be

chosen so that2

2

1.96 p(1 p) ,0.01

n so that the answer depends on the value of p. Note that the

largest value that p(1 − p) can take on is 0.25 (that is, p 0.5 makes p(1 p) as large as

possible). Thus if2

2

1.96 0.25 9604,0.01

n then the margin of error is less than 0.01 for all

values of p.



3.7. The null hypothesis is that the survey is a random draw from a population with p = 0.50. The

t-statistic is t = p̂ − 0.50 , where SE( p)ˆ = p(1ˆ − p) /ˆ n.The value of the t-statistic is 3.18, which has aSE p̂( )

p-value of that is less than 0.01. Thus the null hypothesis p = 0.50 (the survey is unbiased) can be rejected at the 1% level.



3.9. Denote the life of a light bulb from the new process by Y. The mean of Y is and the standard

test is H0: 𝜇𝜇 = 1000 vs. H1: 𝜇𝜇 > 1000. The manager will accept the alternative hypothesisif Y > 1100 hours.

(a) The size of a test is the probability of erroneously rejecting a null hypothesis when it is valid.

The size of the manager’s test is

deviation of Y is σ = 100 hours. Y is the sample mean with a sample size n = 50. The standard

deviation of the sampling distribution of Y is 100 14.1450

σ = = =σYY nhours. The hypothesis

13

1000)=µ

1000 1100 100

99999999

1 Pr−14.1

9999

4 14.14

(7.072) =1− 0. 24 = 7.6×10− .

Y − −≤=

0 | µ 1000=

=1−Φ

Pr( 1100 | 1µ> =Y 1000) = Pr(− → >Y + × =Y= −t 1032.9 14.14

The manager should believe the inventor’s claim if the sample mean life of the new product is greater than 1032.9 hours if she wants the size of the test to be 1%.



3.11. Assume that n is an even number. Then Y is constructed by applying a weight of 1/2 to the n/2 “odd” observations and a weight of 3/2 to the remaining n/2 observations.

1 2 1

1 2 12

222

2

E Y( E Y( E Y( E Y() 1 1 ) 3 ) 1 ) 3 )2 2 2 211 3

22 22

var( ) 1 var(Y Y ) 9 var( ) Y 1 va r( ) 9 var( )4 4 4 4

1 1 924 4 2

nn

YY Y

n n

YYY

E( Yn

nnn

Y Yn

nn n

1

n 1 25



3.13 (a)

(b)

2 0 vs H1 1 20 1H 0

The p-value for the one-sided test is:

Sample size n = 400, sample averageY = 712.1, and sample standard deviation s = 23.2. The 23.2400

standard error of Y is SE(Y ) = s = =n

1.16. The 90% confidence interval for the

mean score of all New Jersey third graders is µ Y= ±1.65SE(Y ) = 712.1±1.65×1.16 (= 710.186,714.014)

Small class sample size n1 = 150, sample averageY1 = 721.8, and sample standard deviation s1 = 24.4. Large class sample size n2 = 250, sample averageY2 = 710.9, and

21sample standard deviation s2 = 20.6. The standard error of −Y Y is2 2

211 2

21

s s+ =1.99 SE(Y Y− ) =

n nThe hypothesis test for higher average test scores in smaller classes is

The t-statistic for testing the null hypothesis is

1 2

1 2

721.8 710.9) 1.99

−= = =

−Y YtSE(Y −Y

5.48

-value = 1−Φ(p t) =1−Φ(5.48) ≈ 0.00000With the small p-value, the null hypothesis can be rejected with a high degree of confidence. There is statistically significant evidence that the districts with smaller classes have higher average test scores.



3.15. From the textbook equation (2.46), we know that E(Y ) Y and from (2.47) we know that

var(Y ) 2Y

n

. In this problem, because Ya and Yb are Bernoulli random variables, p̂a a Y , p̂b

Y , 2 p (1–p ) and b Ya a a Yb2 pb(1–pb). The answers to (a) follow from this. For part (b), note that

var( p̂a – p̂b ) var( p̂a ) var( p̂b ) – 2cov( p̂a , p̂b ). But, they are independent (and thus have

a bcov(p ,ˆ p̂ ) 0 because p̂a and p̂b are independent (they depend on data chosen from independent samples). Thus var( p̂a – p̂b ) var( p̂a ) var( p̂b ). For part (c), use equation 3.21 from the text (replacing Y with p̂ and using the result in (b) to compute the SE). For (d), apply the formula in (c) to obtain

95% CI is (.859 – .374) ± 1.96 0.859(1 0.859) 0.374(1 0.374)5801 4249

or 0.485 ± 0.017.



3.17. (a) The 95% confidence interval is Ym, 2012 −Ym,1992 ±1.96 SE(Ym, 2012 −Ym,1992 ) where

SE(Ym, 2012 −Ym,1992 ) =

Sm , 20122

nm , 2012+ Sm ,1992

2

nm ,1992= 12.0922004 +

10.8521594 = 0.38; the 95% confidence

interval is (25.30 – 24.83) ± 0.75 or 0.47 ± 0.75.

(b) The 95% confidence interval is Yw, 2012 −Yw, 1992 ±1.96 SE(Yw, 2012 −Yw,1992 ) where

SE(Yw, 2012 −Yw,1992 ) =

Sw , 20122

nw , 2008+ Sw ,1992

2

nw ,1992= 9.9921951 +

8.3921368 = 0.32; the 95% confidence

interval is (21.50− 21.39) ± 0.63 or 0.11 ± 0.63.

(c) The 95% confidence interval is

(Ym, 2012 −Ym,1992 )− (Yw, 2012 −Yw,1992 ) ±1.96 SE[(Ym, 2012 −Ym,1992 )− (Yw, 2012 −Yw,1992 )],

where

SE[(Ym, 2012 −Ym,1992 )− (Yw, 2012 −Yw,1992 )]

= Sm , 20122

nm , 2008+ Sm ,1992

2

nm ,1992+ Sw , 2012

2

nw , 2008+ Sw , 2004

2

nw , 2004

= 12.0922004 +10.8521594 +

9.9921951 +

8.3921368 = 0.50.

The 95% confidence interval is

(25.30 – 24.83) − (21.50−21.39) ± 1.96 × 0.50 or 0.36 ± 0.98.



3.19. (a) No. 2 2 2 2( ) and ( ) for .i Y Y i j YE Y E YY i jσ µ µ= + = ≠ Thus

22 2 2 2

2 21 1 1

1 1 1 1( ) ( ) ( )n n n

i i i j Y Yi i i j i

E Y E Y E Y E YYn n n n

µ σ= = = ≠

⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∑ ∑ ∑∑

(b) Yes. If Y gets arbitrarily close to µY with probability approaching 1 as n gets

large, then 2Y gets arbitrarily close to 2Yµ with probability approaching 1 as n

gets large. (As it turns out, this is an example of the “continuous mapping

theorem” discussed in Chapter 17.)



3.21. Set nm = nw = n, and use equation (3.19) write the squared SE of m wY Y− as

2 21 1

2

2 21 1

1 1( ) ( )( 1) ( 1)[ ( )]

( ) ( ) .( 1)

n ni mi m i wi w

m w

n ni mi m i wi w

Y Y Y Yn nSE Y Y

n n

Y Y Y Yn n

= =

= =

∑ − ∑ −− −− = +

∑ − +∑ −=−

Similary, using equation (3.23)

2 21 1

2

2 21 1

1 1( ) ( )2( 1) ( 1)

[ ( )]2

( ) ( ) .( 1)

n ni mi m i wi w

pooled m w

n ni mi m i wi w

Y Y Y Yn n

SE Y Yn

Y Y Y Yn n

= =

= =

⎡ ⎤∑ − + ∑ −⎢ ⎥− −⎣ ⎦− =

∑ − +∑ −=−

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IntroductiontoEconometrics (3 U pdatedEdition, Global Edition...Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 3 (This version August 17, 2014) Stock/Watson - Introduction