Test Bank Introduction to Econometrics 3rd Edition By James H. Stock & Mark W. Watson Complete

Test Bank Introduction to Econometrics 3rd Edition By James H. Stock & Mark W. Watson Complete . Probability distribution function for Y Outcome (number of heads) Y  0 Y  1 Y  2 Probability 0.25 0.50 0.25 (b) Cumulative probability distribution function for Y Outcome (number of heads) Y  0 0  Y  1 1  Y  2 Y  2 Probability 0 0.25 0.75 1.0 (c) = ( ) (0 0.25) (1 0.50) (2 0.25) 1.00 Y E Y        . , . d F  Fq  Using Key Concept 2.3: 2 2 var(Y)  E Y( ) [E Y( )] , and ( | ) i i u X so that 2 2 2 var(Y)  E Y( ) [E Y( )] 1.50 (1.00)  0.50. 2.2. We know from Table 2.2 that Pr (Y  0)  0 22, Pr (Y 1) 0 78, Pr (X  0)  0 30, Pr (X 1) 0 70. So (a) ( ) 0 Pr ( 0) 1 Pr ( 1) 0 0 22 1 0 78 0 78, ( ) 0 Pr ( 0) 1 Pr ( 1) 0 0 30 1 0 70 0 70 Y X E Y Y Y E X X X                                  (b) 2 2 2 2 2 2 2 2 2 2 2 2 [( ) ] (0 0.70) Pr ( 0) (1 0.70) Pr ( 1) ( 0 70) 0 30 0 30 0 70 0 21, [( ) ] (0 0.78) Pr ( 0) (1 0.78) Pr ( 1) ( 0 78) 0 22 0 22 0 78 0 1716 X X Y Y E X X X E Y Y Y                                                Solutions to End-of-Chapter Exercises 3 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) cov ( , ) [( )( )] (0 0.70)(0 0.78) Pr( 0, 0) (0 0 70)(1 0 78) Pr ( 0 1) (1 0 70)(0 0 78) Pr ( 1 0) (1 0 70)(1 0 78) Pr ( 1 1) ( 0 70) ( 0 78) 0 15 ( 0 70) 0 22 0 15 0 30 ( 0 78) 0 07 0 XY X Y E X X Y Y X Y X Y X Y X Y                                                             0 084,       0 084 corr ( , ) 0 4425 XY X Y X Y             2.3. For the two new random variables W  3  6X and V  20  7Y, we have: (a) ( ) (20 7 ) 20 7 ( ) 54, ( ) (3 6 ) 3 6 ( ) 3 6 0 70 7 2 E V E Y E Y E W E X E X                      (b) 2 2 2 2 2 2 var (3 6 ) 6 36 0 21 7 56, var (20 7 ) ( 7) W X V Y X Y                         (c) WV  cov (3 6X, 20 7 )Y  6 ( 7) cov (X Y, )  42 0 084   3 528 3 528 corr ( , ) 0 4425 WV W V W V               2.4. (a) 3 3 3 E X( )  0 (1 p) 1  p  p (b) ( ) 0 (1 ) 1 k k k E X    p   p  p (c) E X( )  0.3 , and var(X) = E(X 2 )−[E(X)]2 = 0.3 −0.09 = 0.21. Thus = 0.21= 0.46. 2 2 var (X)  E X( ) [E X( )]  0.30.09 0.21  0.21 0.46. To compute the skewness, use the formula from exercise 2.21: 3 3 2 3 2 3 ( ) ( ) 3[ ( )][ ( )] 2[ ( )] 0.3 3 0.3 2 0.3 0.084 E X    E X  E X E X  E X       Alternatively, 3 3 3 E X(  ) [(10.3) 0.3][(00.3) 0.7] 0.084 Thus, skewness 3 3 3  E X(  ) /   0.084/0.46  0.87. To compute the kurtosis, use the formula from exercise 2.21: 4 4 3 2 2 4 2 3 4 ( ) ( ) 4[ ( )][ ( )] 6[ ( )] [ ( )] 3[ ( )] 0.3 4 0.3 6 0.3 3 0.3 0.0777 E X    E X  E X E X  E X E X  E X         Alternatively, 4 4 4 E X(  ) [(1 0.3) 0.3][(0 0.3) 0.7] 0.0777 Thus, kurtosis is 4 4 4 E X(  ) /   0.0777/0.46 1.76 4 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley 2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y  0 when X  32 and Y 100 when X  212; this implies Y  (100/180) (X  32) or Y  17.78  (5/9)  X. Using Key Concept 2.3, X  70o F implies that 17.78 (5/9) 70 21.11 C, Y       and X  7 o F implies (5/9) 7 3.89 C. Y     2.6. The table shows that Pr (X  0,Y  0)  0 037, Pr (X  0,Y 1) 0 622, Pr (X 1,Y  0)  0 009, Pr (X 1,Y 1) 0 332, Pr (X  0)  0 659, Pr (X 1) 0 341, Pr (Y  0)  0 046, Pr (Y 1) 0 954. (a) ( ) 0 Pr( 0) 1 Pr ( 1) 0 0 046 1 0 954 0 954 E Y  Y   Y    Y           (b) (unemployed) Unemployment Rate (labor force) Pr ( 0) 1 Pr( 1) 1 ( ) 1 0 954 0.046 # # Y Y E Y              (c) Calculate the conditional probabilities first: Pr ( 0, 0) 0 037 Pr ( 0| 0) 0 056, Pr ( 0) 0 659 Pr ( 0, 1) 0 622 Pr ( 1| 0) 0 944, Pr ( 0) 0 659 Pr ( 1, 0) 0 009 Pr ( 0| 1) 0 026, Pr ( 1) 0 341 Pr ( 1, 1) 0 332 Pr ( 1| 1) 0 974 Pr ( 1) 0 341 X Y Y X X X Y Y X X X Y Y X X X Y Y X X                                              The conditional expectations are ( | 1) 0 Pr ( 0| 1) 1 Pr ( 1| 1) 0 0 026 1 0 974 0 974, ( | 0) 0 Pr ( 0| 0) 1 Pr ( 1| 0) 0 0 056 1 0 944 0 944 E Y X Y X Y X E Y X Y X Y X                                    (d) Use the solution to part (b), Unemployment rate for college graduates  1  E(Y|X  1)  1  0.974  0.026 Unemployment rate for non-college graduates  1  E(Y|X  0)  1  0.944  0.056 (e) The probability that a randomly selected worker who is reported being unemployed is a college graduate is Pr ( 1, 0) 0 009 Pr ( 1| 0) 0 196 Pr ( 0) 0 046 X Y X Y Y             The probability that this worker is a non-college graduate is Pr (X  0|Y  0)  1 Pr (X  1|Y  0)    1 0 196  0 804 Solutions to End-of-Chapter Exercises 5 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (f) Educational achievement and employment status are not independent because they do not satisfy that, for all values of x and y, Pr (X  x Y|  y)  Pr (X  x) For example, from part (e) Pr (X  0|Y  0)  0.804, while from the table Pr(X  0)  0.659. 2.7. Using obvious notation, C  M  F; thus C  M  F and 2 2 2 2cov( , ). C  M  F  M F This implies (a) C  40  45  $85,000 per year. (b) cov( , ) corr ( , ) M F M F M F    , so that cov ( , ) corr ( , ). M F   M F M F Thus cov (M F, )  1218 0.80 172.80, where the units are squared thousands of dollars per year. (c) 2 2 2 2cov( , ), C  M  F  M F so that 2 2 2 .80 813.60, C      and 813.60 28.524 C   thousand dollars per year. (d) First you need to look up the current Euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose that this exchange rate is e (say e  0.80 Euros per dollar); each 1 dollar is therefore with e Euros. The mean is therefore e  C (in units of thousands of Euros per year), and the standard deviation is e  C (in units of thousands of Euros per year). The correlation is unit-free, and is unchanged. 2.8. ( ) 1, Y  E Y  2 var ( ) 4. Y  Y  With 1 2 Z  (Y 1), 2 2 1 1 1 ( 1) ( 1) (1 1) 0, 2 2 2 1 1 1 var ( 1) 4 1 2 4 4 Z Y Z Y E Y Y                               2.9. Value of Y Probability Distribution of X Value of X 1 0.02 0.05 0.10 0.03 0.01 0.21 5 0.17 0.15 0.05 0.02 0.01 0.40 8 0.02 0.03 0.15 0.10 0.09 0.39 Probability distribution of Y 0.21 0.23 0.30 0.15 0.11 1.00 (a) The probability distribution is given in the table above. 2 2 2 2 2 2 2 2 ( ) 14 0.21 22 0.23 30 0.30 40 0.15 65 0.11 30.15 ( ) 14 0.21 22 0.23 30 0.30 40 0.15 65 0.11 1127.23 var( ) ( ) [ ( )] 218.21 Y 14.77 E Y E Y Y E Y E Y                            6 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley (b) The conditional probability of Y|X  8 is given in the table below Value of Y 0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39 0.09/0.39 2 2 2 2 2 2 ( | 8) 14 (0.02/0.39) 22 (0.03/0.39) 30 (0.15/0.39) 40 (0.10/0.39) 65 (0.09/0.39) 39.21 ( | 8) 14 (0.02/0.39) 22 (0.03/0.39) 30 (0.15/0.39) 40 (0.10/0.39) 65 (0.09/0.39) 1778.7 E Y X E Y X                         2 8 var( ) 1778.7 39.21 241.65 15.54 Y X Y       (c) E XY ( )  (1 14  0.02)  (1 22 : 0.05) (865 0.09) 171.7 cov(X Y, )  E XY ( )  E X E Y ( ) ( ) 171.7  5.3330.15 11.0 corr(X Y, )  cov(X Y, )/(  X Y ) 11.0 / (2.6014.77)  0.286 2.10. Using the fact that if 2 , Y N Y Y        then ~ (0, 1) Y Y Y N    and Appendix Table 1, we have (a) 1 3 1 Pr ( 3) Pr (1) 0 8413 2 2 Y Y                 (b) 3 0 3 Pr( 0) 1 Pr( 0) 1 Pr 3 3 1 ( 1) (1) 0 8413 Y Y Y                         (c) Pr (40 52) Pr 5 5 5 (0 4) ( 2) (0 4) [1 (2)] 0 6326 Y Y                                   (d) 6 5 5 8 5 Pr (6 8) Pr 2 2 2 (2 1213) (0 7071) 2229 Y Y                            2.11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When 2 ~ 10 Y  , then 10, Y/10 ~F .  (e) 2 Y  Z , where Z ~ N(0,1), thus Pr (Y 1)  Pr ( 1 Z 1) 0.32. Solutions to End-of-Chapter Exercises 7 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 2.12. (a) 0.05 (b) 0.950 (c) 0.953 (d) The tdf distribution and N(0, 1) are approximately the same when df is large. (e) 0.10 (f) 0.01 2.13. (a) 2 2 2 2 ( ) Var ( ) 1 0 1; ( ) Var ( ) 100 0 100. E Y  Y  Y    E W  W  W    (b) Y and W are symmetric around 0, thus skewness is equal to 0; because their mean is zero, this means that the third moment is zero. (c) The kurtosis of the normal is 3, so 4 4 3 ( ) /  E Y  Y Y ; solving yields 4 E(Y )  3; a similar calculation yields the results for W. (d) First, condition on X  0, so that S  W: 2 3 4 2 . E S X ( |  0)  0; E S( |X  0) 100,E S X ( |  0)  0, E S( |X  0)  3 100  Similarly, 2 3 4 E S X ( | 1) 0; E S( |X 1)1,E S( |X 1) 0, E S( |X 1) 3. From the law of iterated expectations 2 2 2 3 3 3 4 4 4 2 ( ) ( | 0) Pr (X 0) ( | 1) Pr( 1) 0 ( ) ( | 0) Pr (X 0) ( | 1) Pr( 1) 100 0.01 1 0.99 1.99 ( ) ( | 0) Pr (X 0) ( | 1) Pr( 1) 0 ( ) ( | 0) Pr (X 0) ( | 1) Pr( 1) 3 100 0.01 3 E S E S X E S X X E S E S X E S X X E S E S X E S X X E S E S X E S X X                                             1 0.99  302.97 (e) ( ) 0, S  E S  thus 3 3 ( ) ( ) 0 E S  S  E S  from part (d). Thus skewness  0. Similarly, 2 2 2 ( ) ( ) 1.99, S  E S  S  E S  and 4 4 ( ) ( ) 302.97. E S  S  E S  Thus, 2 kurtosis 302.97 / (1.99 ) 76.5 2.14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average (Y ) is approximately 2 , N Y Y       with 2 2 . Y Y n    Given 100, Y  2 43 0, Y   (a) n 100, 2 2 43 100 0 43, Y Y n       and Pr ( 101) Pr (1 525) 0 9364 0 43 0 43 Y Y                    (b) n 165, 2 2 43 , Y Y n       and Pr ( 98) 1 Pr ( 98) 1 Pr 1 ( 3 9178) (3 9178) 1 000 (rounded to four decimal places) Y Y Y                             8 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) n  64, 2 , 64 64 Y Y       and Pr (101 103) Pr 6719 (3 6599) (1 2200) 1111 Y Y                               2.15. (a) 9.6 10 10 10.4 10 Pr (9.6 10.4) Pr 4/ 4/ 4/ 9.6 10 10.4 10 Pr 4/ 4/ Y Y n n n Z n n                          where Z ~ N(0, 1). Thus, (i) n  20; 9.6 10 10.4 10 Pr Pr ( 0.89 0.89) 0.63 4/ 4/ Z Z n n                (ii) n  100; 9.6 10 10.4 10 Pr Pr( 2.00 2.00) 0.954 4/ 4/ Z Z n n                (iii) n  1000; 9.6 10 10.4 10 Pr Pr( 6.32 6.32) 1.000 4/ 4/ Z Z n n                (b) 10 Pr (10 10 ) Pr 4/ 4/ 4/ Pr . 4/ 4/ c Y c c Y c n n n c c Z n n                          As n get large 4 / c n gets large, and the probability converges to 1. (c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6. 2.16. There are several ways to do this. Here is one way. Generate n draws of Y, Y1, Y2, … Yn. Let Xi  1 if Yi  3.6, otherwise set Xi  0. Notice that Xi is a Bernoulli random variables with X  Pr(X  1)  Pr(Y  3.6). Compute X. Because X converges in probability to X  Pr(X  1)  Pr(Y  3.6), X will be an accurate approximation if n is large. 2.17. Y = 0.4 and 2 0.4 0.6 0.24 Y    (a) (i) P(Y  0.43)  0.4 0.43 0.4 0.4 Pr Pr 0.6124 0.27 0.24/ 0.24/ 0.24/ Y Y n n n                    Solutions to End-of-Chapter Exercises 9 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (ii) P(Y  0.37)  0.4 0.37 0.4 0.4 Pr Pr 1.22 0.11 0.24/ 0.24/ 0.24/ Y Y n n n                     (b) We know Pr(1.96  Z  1.96)  0.95, thus we want n to satisfy 0.41 0.40 0.41 1.96 24 / n     and 0.39 0.40 1.96. 24 / n    Solving these inequalities yields n  9220. 2.18. Pr (Y  $0)  0 95, Pr (Y  $20000)  0 05. (a) The mean of Y is 0 Pr ( 0) 20,000 Pr ( 20000) 1000. Y   Y  $   Y  $  $ The variance of Y is 2 2 2 2 2 2 7 ( ) (0 1000) Pr( 0) () Pr ( 20000) ( 1000) 1 9 10 , Y E Y Y Y Y                             so the standard deviation of Y is 1 2 7 Y  (1 9 10 )  $4359 (b) (i) ( ) 1000, E Y Y    $ 2 7 2 1.9 10 5 1 9 10 . 100 Y Y n         (ii) Using the central limit theorem, 5 5 Pr ( 2000) 1 Pr ( 2000) 000 1 Pr 1 9 10 1 9 10 1 (2 2942) Y Y Y                                2.19. (a) 1 1 Pr ( ) Pr ( , ) Pr ( | )Pr ( ) l j i j i l j i i i Y y X x Y y Y y X x X x             (b) 1 1 1 1 1 1 ( ) Pr ( ) Pr ( | ) Pr ( ) Pr ( | ) Pr ( ) ( | )Pr ( ) i i i k k l j j j j i i j j i l k j j i j i l i E Y y Y y y Y y X x X x y Y y X x X x E Y X x X x                                       (c) When X and Y are independent, Pr ( , ) Pr ( )Pr ( ) X i j i j  x Y  y  X  x Y  y  so 10 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley 1 1 1 1 [( )( )] ( )( ) Pr ( , ) ( )( ) Pr ( ) Pr ( ) XY X Y l k i X j Y i j i j l k i X j Y i j i j E X Y x y X x Y y x y X x Y y                           1 1 ( ) Pr ( ) ( ) Pr ( ( ) ( ) 0 0 0, 0 corr( , ) 0 l k i X i j Y j i j X Y XY X Y X Y x X x y Y y E X E Y X Y                                      2.20. (a) 1 1 Pr ( ) Pr ( | , ) Pr ( , ) l m i i j h j h j h Y y Y y X x Z z X x Z z           (b) 1 1 1 1 1 1 1 1 1 ( ) Pr ( ) Pr ( ) Pr ( | , ) Pr ( , ) Pr ( | , ) Pr ( , ) ( | , ) Pr ( , ) k i i i i k l m i i j h j h i j h l m k i i j h j h j h i l m j h j h j h E Y y Y y Y y y Y y X x Z z X x Z z y Y y X x Z z X x Z z E Y X x Z z X x Z z                                          where the first line in the definition of the mean, the second uses (a), the third is a rearrangement, and the final line uses the definition of the conditional expectation. 2.21. (a) 3 2 3 2 2 2 2 3 3 2 2 3 3 2 2 3 3 2 3 ( ) [( ) ( )] [ 2 2 ] ( ) 3 ( ) 3 ( ) ( ) 3 ( ) ( ) 3 ( )[ ( )] [ ( )] ( ) 3 ( ) ( ) 2 ( ) E X E X X E X X X X X E X E X E X E X E X E X E X E X E X E X E X E X E X                                 (b) 4 3 2 2 3 4 3 2 2 3 3 2 2 3 4 4 3 2 2 3 4 4 3 2 2 4 ( ) [( 3 3 )( )] [ 3 3 3 3 ] ( ) 4 ( ) ( ) 6 ( ) ( ) 4 ( ) ( ) ( ) ( ) 4[ ( )][ ( )] 6[ ( )] [ ( )] 3[ ( )] E X E X X X X E X X X X X X X E X E X E X E X E X E X E X E X E X E X E X E X E X E X                                    Solutions to End-of-Chapter Exercises 11 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 2.22. The mean and variance of R are given by 2 2 2 2 0.08 (1 ) 0.05 0.07 (1 ) 0.042 2 (1 ) [0.07 0.04 0.25] w w w w w w                    where 0.07 0.04 0.25 ( , )    Cov R s Rb follows from the definition of the correlation between Rs and Rb. (a)  0.065;  0.044 (b)  0.0725;  0.056 (c) w  1 maximizes  ;  0.07 for this value of w. (d) The derivative of  2 with respect to w is 2 2 2 2 0.07 2(1 ) 0.04 (2 4 ) [0.07 0.04 0.25] 0.0102 0.0018 d w w w dw w              Solving for w yields w 18 / 102 0.18. (Notice that the second derivative is positive, so that this is the global minimum.) With 0.18, 0.038. w  R  2.23. X and Z are two independently distributed standard normal random variables, so 2 2 0, 1, 0. X  Z  X  Z  XZ  (a) Because of the independence between X and Z, Pr (Z  z X|  x)  Pr (Z  z), and E Z X ( | )  E Z( )  0. Thus 2 2 2 2 E Y X ( | )  E X(  Z X| )  E X X ( | )  E Z X ( | )  X  0  X  (b) 2 2 2 ( ) 1, E X  X  X  and 2 2 ( ) ( ) 1 0 1 Y  E X  Z  E X  Z     (c) 3 3 E XY ( )  E X(  ZX)  E X( )  E ZX ( ). Using the fact that the odd moments of a standard normal random variable are all zero, we have 3 E X( )  0. Using the independence between X and Z, we have ( ) 0. E ZX   Z X  Thus 3 E XY ( )  E X( )  E ZX( )  0. (d) cov ( ) [( )( )] [( 0)( 1)] ( ) ( ) ( ) 0 0 0 0 corr ( , ) 0 X Y XY X Y X Y XY E X Y E X Y E XY X E XY E X X Y                          2.24. (a) 2 2 2 2 ( ) E Yi       and the result follows directly. (b) (Yi/ ) is distributed i.i.d. N(0,1), 2 1 ( / ) , n i i W Y     and the result follows from the definition of a 2 n random variable. (c) 2 2 2 2 1 1 ( ) . n n i i i i Y Y E W E E n          (d) Write 12 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley 2 2 2 2 1 1 ( / ) 1 1 / n n i Yi i Y n n Y Y V           which follows from dividing the numerator and denominator by . Y1/ ~ N(0,1), 2 2 ( / ) n i i Y    ~ 2 n1 , and Y1/ and 2 2 ( / ) n i i Y    are independent. The result then follows from the definition of the t distribution. 2.25. (a) 1 2 3 1 2 3 1 1 ( ) ( ) n n i n n i i i ax ax ax ax ax a x x x x a x              (b) 1 1 2 2 1 1 2 1 2 1 1 ( ) ( ) ( ) ( ) n i i n n i n n n n i i i i x y x y x y x y x x x y y y x y                         (c) 1 ( ) n i a a a a a na        (d) 2 2 2 2 2 2 1 1 2 2 2 2 2 1 1 1 1 1 ( ) ( 2 2 2 ) 2 2 2 n n i i i i i i i i i i n n n n n i i i i i i i i i i i a bx cy a b x c y abx acy bcx y na b x c y ab x ac y bc x y                             2.26. (a) corr(Yi ,Yj)  2 cov( , )/ cov( , )/ cov( , )/ i j Y Yi j  Y Y  Y Yi j  Y Y  Y Yi j Y  , where the first equality uses the definition of correlation, the second uses the fact that Yi and Yj have the same variance (and standard deviation), the third equality uses the definition of standard deviation, and the fourth uses the correlation given in the problem. Solving for cov(Yi , Yj) from the last equality gives the desired result. (b) 1 2 1 1 2 2 Y  Y  Y , so that E(Y )  1 2 1 1 ( ) ( ) 2 2 E Y  E Y  Y 2 2 1 2 1 2 1 1 2 var( ) var( ) var( ) cov( , ) 4 4 4 2 2 Y Y Y Y Y Y Y        (c) 1 1 n i i Y Y n    , so that 1 1 1 1 ( ) ( ) n n i Y Y i i E Y E Y n n          Solutions to End-of-Chapter Exercises 13 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 1 1 2 2 1 1 1 1 2 2 2 2 1 1 1 2 2 2 2 2 1 var( ) var 1 2 var( ) cov( , ) 1 2 ( 1) 1 1 n i i n n n i i j i i j i n n n Y Y i i j i Y Y Y Y Y Y n Y Y Y n n n n n n n n n n                                                where the fourth line uses 1 1 1 ( 1) (1 2 3 1) 2 n n i j i an n a a n              for any variable a. (d) When n is large 2 0 Y n   and 1 0 n  , and the result follows from (c). 2.27 (a) E(W)  E[E(W|Z)]  E[E(X X )|Z]  E[E(X|Z)  E(X|Z)]  0. (b) E(WZ)  E[E(WZ|Z)]  E[ZE(W)|Z]  E[ Z  0]  0 (c) Using the hint: V  W  h(Z), so that E(V 2 )  E(W 2 )  E[h(Z) 2 ]  2  E[W  h(Z)]. Using an argument like that in (b), E[W  h(Z)]  0. Thus, E(V 2 )  E(W 2 )  E[h(Z) 2 ], and the result follows by recognizing that E[h(Z) 2 ]  0 because h(z) 2  0 for any value of z. ©2011 Pearson Education, Inc. Publishing as Addison Wesley Chapter 3 Review of Statistics 3.1. The central limit theorem suggests that when the sample size (n ) is large, the distribution of the sample average (Y ) is approximately 2 , N Y Y       with 2 2 . Y Y n    Given a population 100, Y  2 43 0, Y   we have (a) n 100, 2 2 43 0 43, 100 Y Y n       and Pr ( 101) Pr (1.525) 0 9364 0 43 0 43 Y Y                   (b) n  64, 2 , 64 Y Y n       and Pr(101 103) Pr 6719 (3 6599) (1 2200) 1111 Y Y                               (c) n 165, 2 , 165 Y Y n       and Pr ( 98) 1 Pr ( 98) 1 Pr 1 ( 3 9178) (3 9178) 1 0000 (rounded to four decimal places) Y Y Y                             3.2. Each random draw Yi from the Bernoulli distribution takes a value of either zero or one with probability Pr( 1) Yi   p and Pr( 0) 1 . Yi    p The random variable Yi has mean ( ) 0 Pr( 0) 1 Pr( 1) , E Yi   Y    Y   p and variance 2 2 2 2 2 var( ) [( ) ] (0 ) Pr( 0) (1 ) Pr( 1) (1 ) (1 ) (1 ) i i Y i i Y E Y p Y p Y p p p p p p                   Solutions to End-of-Chapter Exercises 15 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (a) The fraction of successes is 1 (success) ( 1) ˆ n i i i # # Y Y p Y n n n         (b) 1 1 1 1 1 ( ˆ) ( ) n n n i i i i i Y E p E E Y p p n n n                    (c) 1 2 2 1 1 1 1 (1 ) var( ˆ) var var( ) (1 ) n n n i i i i i Y p p p Y p p n n n n                      The second equality uses the fact that Y1 , , Yn are i.i.d. draws and cov( , ) 0, Y Yi j  for i  j. 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(Y 1) p and Pr(Y  0)  1 p. From the solution to Exercise 3.2, Y has mean p and variance p(1 p). (a) 215 ˆ 0 5375. 400 p    (b) The estimated variance of pˆ is  4 ˆ(1 ˆ) 0.5375 (1 0.5375) var( ˆ) 6 2148 10 . 400 p p p n          The standard error is SE 1 2 ( pˆ)  (var(pˆ))  0 0249. (c) The computed t-statistic is 0 ˆ 506 SE( ˆ) 0 0249 act p p t p            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 0 H  p  0 5 vs. 1 H  p  0 5 : -value 2 ( | |) 2 ( 1 506) act p    t           (d) Using Equation (3.17) in the text, the p-value for the test 0 H  p  0 5 vs. 1 H  p  0 5 is -value 1 ( ) 1 (1 506) act p    t           (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 0 H  p  0 5 vs. 1 H  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 1 506  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. 3.4. Using Key Concept 3.7 in the text (a) 95% confidence interval for p is 16 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley pˆ 1.96SE p( ˆ)  0.53751.960.0249  (0.4887, 0.5863). (b) 99% confidence interval for p is pˆ  2.57SE p( ˆ)  0.5375 2.57 0.0249  (0.4735,0.6015). (c) Mechanically, the interval in (b) is wider because of a larger critical value (2.57 versus 1.96). Substantively, a 99% confidence interval is wider than a 95% confidence because a 99% confidence interval must contain the true value of p in 99% of all possible samples, while a 95% confidence interval must contain the true value of p in only 95% of all possible samples. (d) Since 0.50 lies inside the 95% confidence interval for p, we cannot reject the null hypothesis at a 5% significance level. 3.5. (a) (i) The size is given by Pr(| pˆ  0.5|  .02),where the probability is computed assuming that p  0.5. Pr(|ˆ 0.5| 0.02) 1 Pr( 0.02 ˆ 0.5 .02) 0.02 ˆ 0.5 0.02 1 Pr 0.5 0.5/1055 0.5 0.5/1055 0.5 0.5/1055 ˆ 0.5 1 Pr 1.30 1.30 0.5 0.5/1055 0.19 p p p p                                      where the final equality using the central limit theorem approximation. (ii) The power is given byPr(| pˆ  0.5|  0.02), where the probability is computed assuming that p  0.53. Pr(|ˆ 0.5| 0.02) 1 Pr( 0.02 ˆ 0.5 .02) 0.02 ˆ 0.5 0.02 1 Pr 0.53 0.47/1055 0.53 0.47/1055 0.53 0.47/1055 0.05 ˆ 0.53 0.01 1 Pr 0.53 0.47/1055 0.53 0.47/1055 0.53 0.47/1055 ˆ 0.53 1 Pr 3.25 p p p p p                                             0.65 .53 0.47/1055