Introduction to Econometrics (3 rd Updated Edition) Solutions to Odd- Numbered End- of- Chapter Exercises: Chapter 3

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1 Itroductio to Ecoometrics (3 rd Updated Editio) by James H. Stock ad Mark W. Watso Solutios to Odd- Numbered Ed- of- Chapter Exercises: Chapter 3 (This versio August 17, 014) 015 Pearso Educatio, Ic.

2 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter The cetral limit theorem suggests that whe the sample size ( ) is large, the distributio of the sample average ( ) is approximately Give a populatio µ = 100, σ = 43. 0, we have N µ, σ with σ σ =. (a) = 100, σ 43 σ = = 100 = 043,. ad Pr ( < 101) = Pr < Φ (1.55) = (b) = 64, σ 43 σ = 64 = 64 = 06719,. ad Pr(101 < < 103) = Pr < < Φ ( ) Φ (1. 00) = = (c) = 165, σ 43 σ = = 165 = 0606,. ad Pr ( > 98) = 1 Pr ( 98) = 1 Pr Φ( ) =Φ ( ) = (rouded to four decimalplaces). 015 Pearso Educatio, Ic.

3 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Deote each voter s preferece by. = 1 if the voter prefers the icumbet ad = 0 if the voter prefers the challeger. is a Beroulli radom variable with probability Pr ( = 1) = p ad Pr ( = 0) = 1 p. From the solutio to Exercise 3., has mea p ad variace p(1 p). p ˆ = = (a) 400 (b) The estimated variace of ˆp is p ˆ(1 p var( ˆ) ˆ) ( ) p = = = The 1 stadard error is SE ( pˆ) = (var( pˆ)) = (c) The computed t-statistic is t 400 pˆ µ p, = = =.. SE( pˆ ) Because of the large sample size ( = 400), we ca use Equatio (3.14) i the text to get the p-value for the test H0 : p= 05. vs. H1 : p 05:. p-value = Φ( t ) = Φ( ) = = (d) Usig Equatio (3.17) i the text, the p-value for the test H0 : p= 05. vs. H1 : p> 05. is p-value = 1 Φ ( t ) = 1 Φ (1. 506) = = (e) Part (c) is a two-sided test ad the p-value is the area i the tails of the stadard ormal distributio outside ± (calculated t-statistic). Part (d) is a oe-sided test ad the p-value is the area uder the stadard ormal distributio to the right of the calculated t-statistic. (f) For the test H0 : p= 05. vs. H1 : p> 05,. we caot reject the ull hypothesis at the 5% sigificace level. The p-value is larger tha Equivaletly the calculated t-statistic is less tha the critical value 1.64 for a oe-sided test with a 5% sigificace level. The test suggests that the survey did ot cotai statistically sigificat evidece that the icumbet was ahead of the challeger at the time of the survey. 015 Pearso Educatio, Ic.

4 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) (i) The size is give by Pr( pˆ 0.5 >.0), where the probability is computed assumig that p = 0.5. Pr( pˆ 0.5 >.0) = 1 Pr( 0.0 pˆ 0.5.0) 0.0 pˆ = 1 Pr.5.5/ / /1055 pˆ 0.5 = 1 Pr /1055 = 0.19 where the fial equality usig the cetral limit theorem approximatio. (ii) The power is give by Pr( pˆ 0.5 >.0), where the probability is computed assumig that p = Pr( pˆ 0.5 >.0) = 1 Pr( 0.0 pˆ 0.5.0) 0.0 pˆ = 1 Pr.53.47/ / / pˆ = 1 Pr.53.47/ / /1055 pˆ 0.53 = 1 Pr /1055 = 0.74 where the fial equality usig the cetral limit theorem approximatio. (b) (i) t = =.61, Pr( t >.61) =.01, so that the ull is rejected at the 5% level /1055 (ii) Pr( t >.61) =.004, so that the ull is rejected at the 5% level. (iii) 0.54 ± /1055 = 0.54 ± 0.03, or 0.51 to (iv) 0.54 ± /1055 = 0.54 ± 0.04, or 0.50 to (v) 0.54 ± /1055 = 0.54 ± 0.01, or 0.53 to Pearso Educatio, Ic.

5 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (cotiued) (c) (i) The probability is 0.95 is ay sigle survey, there are 0 idepedet surveys, so the probability if = 0.36 (ii) 95% of the 0 cofidece itervals or 19. (d) The relevat equatio is 1.96 SE( pˆ ) <.01or 1.96 p(1 p) / <.01. Thus 1.96 p(1 p) must be chose so that >, so that the aswer depeds o the value of.01 p. Note that the largest value that p(1 p) ca take o is 0.5 (that is, p = 0.5 makes p(1 p) as large as possible). Thus if of error is less tha 0.01 for all values of p > = 9604, the the margi Pearso Educatio, Ic.

6 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter The ull hypothesis i that the survey is a radom draw from a populatio with p = The t-statistic is t = where SE( pˆ) = pˆ(1 pˆ)/. (A alterative pˆ 0.11, SE( pˆ ) formula for SE( ˆp ) is 0.11 (1 0.11) /, which is valid uder the ull hypothesis that p = 0.11). The value of the t-statistic is -.71, which has a p-value of that is less tha Thus the ull hypothesis p = 0.11(the survey is ubiased) ca be rejected at the 1% level. 015 Pearso Educatio, Ic.

7 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Deote the life of a light bulb from the ew process by. The mea of is µ ad the stadard deviatio of is σ = 00 hours. is the sample mea with a sample size = 100. The stadard deviatio of the samplig distributio of is σ σ = = = 0 hours. The hypothesis test is H : µ = 000 vs. 0 H 1: µ > The maager will accept the alterative hypothesis if > 100 hours. (a) The size of a test is the probability of erroeously rejectig a ull hypothesis whe it is valid. The size of the maager s test is size = Pr( > 100 µ = 000) = 1 Pr( 100 µ = 000) = 1 Pr µ = = 1 Φ (5) = = Pr( > 100 µ = 000) meas the probability that the sample mea is greater tha 100 hours whe the ew process has a mea of 000 hours. (b) The power of a test is the probability of correctly rejectig a ull hypothesis whe it is ivalid. We calculate first the probability of the maager erroeously acceptig the ull hypothesis whe it is ivalid: β = Pr( 100 µ = 150) = Pr µ = =Φ( 5). = 1 Φ (5). = = The power of the maager s testig is 1 β = = (c) For a test with 5%, the rejectio regio for the ull hypothesis cotais those values of the t-statistic exceedig t = >. > = The maager should believe the ivetor s claim if the sample mea life of the ew product is greater tha 03.9 hours if she wats the size of the test to be 5%. 015 Pearso Educatio, Ic.

8 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Assume that is a eve umber. The % is costructed by applyig a weight of 1 to the odd observatios ad a weight of 3 to the remaiig observatios. E(! ) = 1 1 E( ) E( ) +"1 E( ) E( ) = 1 1 µ + 3 µ = µ var(! ) = var( ) var( ) +"1 4 var( ) var( ) = σ σ = 1.5σ. 015 Pearso Educatio, Ic.

9 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) Sample size = 40, sample average = 646. sample stadard deviatio 19 5 s = The stadard error of is SE ( ) s. = = = The 95% cofidece iterval for the mea test score i the populatio is µ = ± 1. 96SE( ) = 646. ± = ( , ). 40 (b) The data are: sample size for small classes 1 = 38, sample average 1 = , sample stadard deviatio s = 19. 4; sample size for large classes = 18, 1 sample average = , sample stadard deviatio s = The stadard error of 1 is s1 s SE( ) = + = + = The hypothesis tests for higher average scores i smaller classes is H : µ µ = 0 vs. H : µ µ > The t-statistic is t SE( 1 ) = = = The p-value for the oe-sided test is: p-value 1 ( ) 1 (4 0479) = Φ t = Φ. =. =.. With the small p-value, the ull hypothesis ca be rejected with a high degree of cofidece. There is statistically sigificat evidece that the districts with smaller classes have higher average test scores. 015 Pearso Educatio, Ic.

10 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter From the textbook equatio (.46), we kow that E( ) = µ ad from (.47) we σ kow that var( ) =. I this problem, because a ad b are Beroulli radom variables, p ˆ a = a, p ˆ b = b, σ a = p a (1 p a ) ad σ = p b (1 p b ). The aswers to (a) follow from this. For part (b), ote that var( p ˆ a p ˆ b ) = var( p ˆ a ) + var( p ˆ b ) cov( p ˆ a, p ˆ b ). But, are idepedet (ad thus have a cov( p ˆ a, p ˆ b ) = 0 because p ˆ a ad pˆ b are idedepet (they deped o data chose from idepedet samples). Thus var( p ˆ a p ˆ b ) = var( p ˆ a ) + var( p ˆ b ). For part (c), this equatio (3.1) from the text (replacig with ˆp ad usig the result i (b) to compute the SE). For (d), apply the formula i (c) to obtai b 95% CI is ( ) ± ( ) 0.374( ) + or ± Pearso Educatio, Ic.

11 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) The 95% cofidece iterval is m, 01 m,199 ±1.96 SE( m, 01 m,199 ) where SE( m, 01 m,199 ) = S m, 01 m, 01 + S m,199 m,199 = = 0.38; the 95% cofidece iterval is ( ) ± 0.75 or 0.47 ± (b) The 95% cofidece iterval is w, 01 w, 199 ±1.96 SE( w, 01 w,199 ) where SE( w, 01 w,199 ) = S w, 01 w, S w,199 w,199 = = 0.3; the 95% cofidece iterval is ( ) ± 0.63 or 0.11 ± (c) The 95% cofidece iterval is ( m, 01 m,199 ) ( w, 01 w,199 ) ±1.96 SE[( m, 01 m,199 ) ( w, 01 w,199 )], where SE[( m, 01 m,199 ) ( w, 01 w,199 )] = S m, 01 m, S m,199 m,199 + S w, 01 w, S w, 004 w, 004 = = The 95% cofidece iterval is ( ) ( ) ± or 0.36 ± Pearso Educatio, Ic.

12 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) No. E = + E = i jthus ( i ) σ µ ad ( i j) µ for = i i i j = + = µ + σ i= 1 i= 1 i= 1 j i E ( ) E E ( ) E ( ) (b) es. If gets arbitrarily close to µ with probability approachig 1 as gets large, the gets arbitrarily close to µ with probability approachig 1 as gets large. (As it turs out, this is a example of the cotiuous mappig theorem discussed i Chapter 17.) 015 Pearso Educatio, Ic.

13 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Set m = w =, ad use equatio (3.19) write the squared SE of m w as 1 1 ( ) ( ) i= 1 mi m i= 1 wi w ( 1) ( 1) m w = + [ SE( )] i= 1( mi m ) + i= 1( wi w ) =. ( 1) Similary, usig equatio (3.3) [ SE ( )] pooled m w = 1 1 i= 1( mi m ) + i= 1( wi w ) ( 1) ( 1) i= 1( mi m ) + i= 1( wi w ) =. ( 1) 015 Pearso Educatio, Ic.