TestBank for Introduction to Econometrics 3e Stock Watson SM completed Instructors Guide Completed for Exam preparation and Review

TestBank for Introduction to Econometrics 3e Stock Watson SM For Instructors Solutions to End-of-Chapter Exercises©2011 Pearson Education, Inc. Publishing as Addison Wesley Chapter 2 Review of Probability 2.1. (a) 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) Y = ( ) (0 0.25) (1 0.50) (2 0.25) 1.00 E Y        . F Fq   d , . Using Key Concept 2.3: var( ) ( ) [ ( )] , Y E Y E Y   2 2 and ( | ) u X i i so that var( ) ( ) [ ( )] 1.50 (1.00) 0.50. Y E Y E Y      2 2 2 2.2. We know from Table 2.2 that Pr( 0) 0 22, Y    Pr( 1) 0 78, Y    Pr( 0) 0 30, X    Pr( 1) 0 70. X    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 X Y E X 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 X  3 6  and V Y  20 7 ,  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 6 , 20 7 ) 6 ( 7)cov( , ) 42 0 084 3 528 X Y X Y 3 528 corr( , ) 0 4425 WV W V W V               2.4. (a) E X p p p ( ) 0 (1 ) 1 3 3 3       (b) E X p p p ( ) 0 (1 ) 1 k k k       (c) E X ( ) 0.3  , and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus  = 0.21 = 0.46. var( ) ( ) [ ( )] 0.3 0.09 0.21 X E X E X      2 2    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, E X ( ) [(1 0.3) 0.3] [(0 0.3) 0.7] 0.084          3 3 3 Thus, skewness     E X ( ) / 0.084/0.46 0.87.   3 3 3 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, E X ( ) [(1 0.3) 0.3] [(0 0.3) 0.7] 0.0777          4 4 4 Thus, kurtosis is E X ( ) / 0.0777/0.46 1.76      4 4 44 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 X Y X  (100/180) ( 32) or 17.78 (5/9) .       Using Key Concept 2.3, X  70oF implies that Y       17.78 (5/9) 70 21.11 C, and X  7oF implies Y     (5/9) 7 3.89 C. 2.6. The table shows that Pr( 0, 0) 0 037, X Y     Pr( 0, 1) 0 622, X Y     Pr( 1, 0) 0 009, X Y     Pr( 1, 1) 0 332, X Y     Pr( 0) 0 659, X    Pr( 1) 0 341, X    Pr( 0) 0 046, Y    Pr( 1) 0 954. Y    (a) ( ) 0 Pr( 0) 1 Pr( 1) 0 0 046 1 0 954 0 954 E Y Y Y        Y          (b) Unemployment Rate (unemployed) (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( 0| 0) 1 Pr( 1| 0) X Y X Y            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 ( | ) Pr ( ) X  x Y y X x     For example, from part (e) Pr( 0| 0) 0.804, X Y    while from the table Pr(X  0)  0.659. 2.7. Using obvious notation, C M F   ; thus C M F     and    C M F 2 2 2    2cov( , ). M F This implies (a) C    40 45 $85,000 per year. (b) corr( , ) cov( , ) 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)    C M F 2 2 2    2cov( , ), M F so that C2 2 2      .80 813.60, 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. Y   E Y ( ) 1, Y2   var( ) 4. Y With Z Y  12 ( 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 14.77 Y 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( , ) ( ) ( ) ( ) 171.7 5.33 30.15 11.0 X Y E XY E X E Y       corr( , ) cov( , )/( ) 11.0 / (2.60 14.77) 0.286 X Y X Y      X Y  2.10. Using the fact that if Y N        Y Y ,  2 then Y ~ (0,1) Y Y  N   and Appendix Table 1, we have (a) Pr( 3) Pr (1) 2 2 Y Y                 (b) Pr( 0) 1 Pr( 0) 1 Pr 3 0 3 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) Pr(6 8) Pr 6 5 5 8 5 2 2 2 (2 1213) (0 7071) 2229 Y      Y                       2.11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When Y ~ , 10 2 then Y F /10 ~ . 10, (e) Y Z  2, where Z N ~ (0,1), thus Pr( 1) Pr( 1 1) 0.32. Y Z      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) E Y Y E W W ( ) Var( ) 1 0 1; ( ) Var( ) 100 0 100. 2 2 2 2             Y 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 3 ( ) /   E Y Y Y 4 4  ; solving yields E( ) 3; Y 4  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 E S X E S X E S X ( | 0) 0; ( | 0) 100, ( | 0) 0, ( | 0) 3 100          Similarly, E S X E S X E S X E S X ( | 1) 0; ( | 1) 1, ( | 1) 0, ( | 1) 3.         2 3 4 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) S   E S ( ) 0, thus E S E S ( ) ( ) 0  S 3 3   from part (d). Thus skewness  0. Similarly,   S S 2 2 2     E S E S ( ) ( ) 1.99, and E S E S ( ) ( ) 302.97.    S 4 4 Thus, kurtosis 302.97 / (1.99 ) 76.5   2 2.14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average ( ) Y is approximately N       Y ,  Y2 with  Y2  nY2 . Given Y 100, Y2  43 0,  (a) n 100, 2 2 43 Y     nY 100 0 43, and Pr( 101) Pr (1 525) 0 9364 0 43 0 43 Y Y                    (b) n 165, 2 2 43 Y     nY , 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 2 43 0 6719, 64 64 Y Y       and Pr(101 103) Pr 6719 (3 6599) (1 2200) 1111 Y      Y                          2.15. (a) Pr(9.6 10.4) Pr 9.6 10 10 10.4 10 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; Pr Pr( 0.89 0.89) 0.63 9.6 10 10.4 10 4/ 4/ Z Z n n                (ii) n  100; Pr Pr( 2.00 2.00) 0.954 9.6 10 10.4 10 4/ 4/ Z Z n n                (iii) n  1000; Pr Pr( 6.32 6.32) 1.000 9.6 10 10.4 10 4/ 4/ Z Z n n                (b) Pr(10 10 ) Pr 10 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 Y2    0.4 0.6 0.24 (a) (i) P(Y  0.43)  Pr Pr 0.6124 0.27 0.4 0.43 0.4 0.4 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)  Pr Pr 1.22 0.11 0.4 0.37 0.4 0.4 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 1.96 0.41 0.40 24 / n     and 0.39 0.40 1.96. 24 / n    Solving these inequalities yields n  9220. 2.18. Pr( 0) 0 95, Y $    Pr( 20000) 0 05. Y $    (a) The mean of Y is Y        0 Pr( 0) 20,000 Pr( 20000) 1000. 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 12  Y  (1 9 10 ) 4359     7 $ (b) (i) E Y $ ( ) 1000,   Y 2 7 2 5 1.9 10 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 i j 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  x Y y X x Y y i j i j      so10 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 R s b 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 0.07 2(1 ) 0.04 (2 4 ) [0.07 0.04 0.25] 2 2 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 w  0.18, 0.038.  R  2.23. X and Z are two independently distributed standard normal random variables, so      X Z X Z XZ      0, 1, 0. 2 2 (a) Because of the independence between X and Z, Pr( | ) Pr( ), Z  z X x Z z    and E Z X E Z ( | ) ( ) 0.   Thus E Y X E X Z X E X X E Z X X X ( | ) ( | ) ( | ) ( | ) 0  2 2 2 2        (b) E X ( ) 1, 2 2 2      X X and   Y Z  E X Z E X ( ) ( ) 1 0 1 2 2        (c) E XY E X ZX E X E ZX ( ) ( ) ( ) ( ).     3 3 Using the fact that the odd moments of a standard normal random variable are all zero, we have E X ( ) 0. 3  Using the independence between X and Z, we have E ZX ( ) 0.  Z X   Thus E XY E X E ZX ( ) ( ) ( ) 0.  3   (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) E Y ( ) i2 2 2 2       and the result follows directly. (b) (Yi/) is distributed i.i.d. N(0,1), 2 1 W Y   in ( / ) , i  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) Write12 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 i i Y Y n n Y Y V           which follows from dividing the numerator and denominator by . Y1/ ~ N(0,1), 2 2  in ( / ) Yi  ~ n21 , and Y1/ and  in2 ( / ) Yi  2 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)  cov( , )/ cov( , )/ cov( , )/ 2 Y Y Y Y Y Y i j Y Y i j Y Y i j Y  i j         , 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 1 1 2 2 2 Y Y Y   , so that E(Y )  1 1 ( ) ( ) 1 2 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 Y 0  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(V2)  E(W2)  E[h(Z)2]  2  E[W  h(Z)]. Using an argument like that in (b), E[W  h(Z)]  0. Thus, E(V2)  E(W2)  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 N       Y ,  Y2 with 2 2 . Y Y  n   Given a population  Y  100, Y2   43 0, 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 2 43 0 6719, 64 Y Y  n      and Pr(101 103) Pr 6719 (3 6599) (1 2200) 1111 Y      Y                          (c) n 165, 2 2 43 0 2606, 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) Y p i   and Pr ( 0) 1 . Y p i    The random variable Yi has mean E( ) 0 Pr( 0) 1 Pr( 1) , Y Y Y p i        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 (success) ( 1) 1 ˆ n # # Yi i i Y p Y n n n         (b) 1 1 1 1 1 nˆ( ) ( ) 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 ) nˆvar( ) var var( ) (1 ) 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 Y i 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 ( 1) Y p   and Pr ( 0) 1 . Y 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 var( ) 6 2148 10 . ˆ ˆ ˆ (1 ) 0.5375 (1 0.5375) 4 400 p p p n          The standard error is SE 12 ( ) (var( )) 0 0249. p p ˆ ˆ    (c) The computed t-statistic is ˆ 1 506 pˆSE( ) 0 0249 act p t  p            Because of the large sample size ( 400), n  we can use Equation (3.14) in the text to get the p-value for the test H p 0    0 5 vs. H p 1    0 5: p t -value 2 ( | |) 2 ( 1 506)              act (d) Using Equation (3.17) in the text, the p-value for the test H p 0    0 5 vs. H p 1    0 5 is p t -value 1 ( ) 1 (1 506)            act (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 H p 0    0 5 vs. H p 1    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 is16 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley p SE p ˆ ˆ      1.96 ( ) 0.5375 1.96 0.0249 (0.4887,0.5863). (b) 99% confidence interval for p is p SE p ˆ ˆ      2.57 ( ) 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(| 0.5| .02), pˆ   where the probability is computed assuming that p  0.5. Pr(| 0.5| 0.02) 1 Pr( 0.02 0.5 .02) ˆ ˆ 0.02 0.5 0.02 ˆ 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 by Pr(| 0.5| 0.02), pˆ   where the probability is computed assuming that p  0.53. Pr(| 0.5| 0.02) 1 Pr( 0.02 0.5 .02) ˆ ˆ 0.02 0.5 0.02 ˆ 1 Pr 0.53 0.47/1055 0.53 0.47/1055 0.53 0.47/1055 0.05 0.53 0.01 ˆ 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 0.74           where the final equality using the central limit theorem approximation. (b) (i) 0.54 0.50 2.61, and Pr(| | 2.61) 0.01, (0.54 0.46) /1055 t t       so that the null is rejected at the 5% level. (ii) Pr( 2.61) .004, t   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.    Solutions to End-of-Chapter Exercises 17 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) (i) The probability is 0.95 is any single survey, there are 20 independent surveys, so the probability if 0.95 0.36 20  (ii) 95% of the 20 confidence intervals or 19. (d) The relevant equation is 1.96 SE( ) .01or 1.96 (1 ) / .01.      p p p n ˆ Thus n must be chosen so that 2 2 1.96 (1 ) , 0.01 p p 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 if 2 2 1.96 0.25 9604, 0.01 n    then the margin of error is less than 0.01 for all values of p. 3.6. (a) No. Because the p-value is less than 0.05 ( 5%),   5 is rejected at the 5% level and is therefore not contained in the 95% confidence interval. (b) No. This would require calculation of the t-statistic for   6, which requires Y and SE ( ). Y Only the p-value for test that   5 is given in the problem. 3.7. The null hypothesis is that the survey is a random draw from a population with p = 0.11. The tstatistic is ˆ 0.11, SE( ) ˆ p t p   where SE( ) (1 )/ . pˆ ˆ ˆ  p p n  (An alternative formula for SE( pˆ ) is 0.11 (1 0.11) / ,   n which is valid under the null hypothesis that p  0.11). The value of the tstatistic is 2.71, which has a p-value of that is less than 0.01. Thus the null hypothesis p  0.11 (the survey is unbiased) can be rejected at the 1% level. 3.8 1110 1.96 123 1000        or 1110 7.62.  3.9. Denote the life of a light bulb from the new process by Y. The mean of Y is  and the standard deviation of Y is 200 Y  hours. Y is the sample mean with a sample size n 100. The standard deviation of the sampling distribution of Y is 200 20 100 Y Y n      hours. The hypothesis test is H0: 2000   vs. H1   2000 . The manager will accept the alternative hypothesis if Y  2100 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 7 size Pr( 2100| 2000) 1 Pr( 2100| 2000) 1 Pr | 2000 20 20 1 (5) 1 0 2 87 10 , Y Y Y                                18 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley where Pr( 2100| 2000) Y    means the probability that the sample mean is greater than 2100 hours when the new process has a mean of 2000 hours. (b) The power of a test is the probability of correctly rejecting a null hypothesis when it is invalid. We calculate first the probability of the manager erroneously accepting the null hypothesis when it is invalid: Pr( 2100| 2150) Pr | 2150 20 20 ( 2 5) 1 (2 5) Y          Y                       The power of the manager’s testing is 9938.        (c) For a test with 5%, the rejection region for the null hypothesis contains those values of the t-statistic exceeding 1.645. 2000 20 2032 9 20 act act Y act t Y             The manager should believe the inventor’s claim if the sample mean life of the new product is greater than 2032.9 hours if she wants the size of the test to be 5%. 3.10. (a) New Jersey sample size n1 100, sample average Y1  58, and sample standard deviation s1  58. The standard error of Y1 is SE(Y1 )  s n 1 1 / 8/ 100   0.8. The 95% confidence interval for the mean score of all New Jersey third graders is 1 1 1              Y Y 1 96SE( ) 58 1 96 0 8 () (b) Iowa sample size n2  200, sample average Y2  62, sample standard deviation s2 11. The standard error of Y Y 1 2  is SE 2 2 1 2 1 2 1 2 64 121 ( ) 1 1158. 100 200 s s Y Y n n        The 90% confidence interval for the difference in mean score between the two states is 1 2 1 2 1 2 ( ) 1 64SE( ) (58 62) ( )         Y Y Y Y            Solutions to End-of-Chapter Exercises 19 ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) The hypothesis tests for the difference in mean scores is H H 0 1 2 1 1 2             0 vs 0 From part (b) the standard error of the difference in the two sample means is SE ( ) Y Y 1 2   1 1158.  The t-statistic for testing the null hypothesis is 1 2 1 2 58 62 3 5849 SE( ) 1 1158 act Y Y t Y Y           Use Equation (3.14) in the text to compute the p-value: p t               value 2 ( | |) 2 ( 3 5849) act Because of the extremely low p-value, we can reject the null hypothesis with a very high degree of confidence. That is, the population means for Iowa and New Jersey students are different. 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 2 1 2 1 2 2 2 2 1 1 3 1 3 ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 3 2 2 2 2 1 1 9 1 9 var( ) var( ) var( ) var( ) var( ) 4 4 4 4 1 1 9 1 25 4 2 4 2 n n Y Y Y n n Y Y Y E Y E Y E Y E Y E Y n n n n Y Y Y Y Y n n n n n                                                                 3.12. Sample size for men n1 100, sample average Y1 =3100 sample standard deviation s1 = 200. Sample size for women n2  64, sample average Y 2  2900, sample standard deviation s2  320. The standard error of Y Y 1 2  is 2 2 2 2 1 2 1 2 1 2 200 320 SE( ) 44 721. 100 64 s s Y Y n n        (a) The hypothesis test for the difference in mean monthly salaries is H H 0 1 2 1 1 2             0 vs 0 The t-statistic for testing the null hypothesis is 1 2 1 2 4 4722 SE( ) 44 721 act Y Y t Y Y          Use Equation (3.14) in the text to get the p-value: p t -value 2 ( | |) 2 ( 4 4722) 2 (3 8744 10 ) 7 7488 10                act 6 6  The extremely low level of p-value implies that the difference in the monthly salaries for men and women is statistically significant. We can reject the null hypothesis with a high degree of confidence.20 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley (b) From part (a), there is overwhelming statistical evidence that mean earnings for men differ from mean earnings for women, and a related calculation shows overwhelming evidence that mean earning for men are greater that mean earnings for women. However, by itself, this does not imply gender discrimination by the firm. Gender discrimination means that two workers, identical in every way but gender, are paid different wages. The data description suggests that some care has been taken to make sure that workers with similar jobs are being compared. But, it is also important to control for characteristics of the workers that may affect their productivity (education, years of experience, etc.). If these characteristics are systematically different between men and women, then they may be responsible for the difference in mean wages. (If this is true, it raises an interesting and important question of why women tend to have less education or less experience than men, but that is a question about something other than gender discrimination by this firm.) Since these characteristics are not controlled for in the statistical analysis, it is premature to reach a conclusion about gender discrimination. 3.13 (a) Sample size n  420, sample average Y  646.2 sample standard deviation sY   19 5. The standard error of Y is SE ( ) 0 9515. 19.5 420 sY Y n     The 95% confidence interval for the mean test score in the population is                Y Y 1 96SE( ) 9515 () (b) The data are: sample size for small classes n1  238, sample average Y 1  657 4,  sample standard deviation 1 s   19 4; sample size for large classes n2 182, sample average Y 2   650 0, sample standard deviation s2 17 9.  The standard error of Y Y 1 2  is 2 2 2 2 1 2 1 2 1 2 19.4 17.9 ( ) 1 8281. 238 182 s s SE Y Y n n        The hypothesis tests for higher average scores in smaller classes is H H 0 1 2 1 1 2             0 vs 0 The t-statistic is 1 2 1 2 4 0479 SE( ) 1 8281 act Y Y t Y Y            The p-value for the one-sided test is: p t -value 1 ( ) 1 (4 0479)             act 5 With 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. 3.14. We have the following relations: 1 0 0254 in m   (or 1 39 37 ), m in   1 0 4536 lb k   g (or 1 2.2046 ). kg  lb The summary statistics in the using metric measurements are X m       79 ; Y kg      ; sX  0457 ;      m sY       4411 ; kg sXY          , m kg and rXY   0 85.Solutions to End-of-Chapter Exercises 21 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 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, pˆa  Ya , pˆb  Yb , Ya 2  pa(1–pa) and Yb 2  pb(1–pb). The answers to (a) follow from this. For part (b), note that var( pˆa – pˆb )  var( pˆa )  var( pˆb ) – 2cov( pˆa , pˆb ). But, they are independent (and thus have cov( , ) pˆ ˆ a b p  0 because pˆa and pˆb are independent (they depend on data chosen from independent samples). Thus var( pˆa – pˆb )  var( pˆa )  var( pˆb ). For part (c), use equation 3.21 from the text (replacing Y with pˆ 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)    or 0.485 ± 0.017. 3.16. (a) The 95% confidence interval if 1.96 SE( ) or 1013 1.96 or 1013 9.95. 108 453 Y Y     (b) The confidence interval in (a) does not include   1000, so the null hypothesis that   1000 (Florida students have the same average performance as students in the U.S.) can be rejected at the 5% level.) (c) (i) The 95% confidence interval is Y Y SE Y Y prep Non prep prep Non prep      1.96 ( ) where 2 2 2 2 95 108 SE( ) 6.61; 503 453 prep non prep prep Non prep prep non prep s s Y Y n n          the 95% confidence interval is () 12.96 6 12.96.    or (ii) No. The 95% confidence interval includes  prep non prep     0. (d) (i) Let X denote the change in the test score. The 95% confidence interval for X is 60 1.96 ( ), where SE( ) 2.82; 453 X SE X X    thus, the confidence interval is 9 5.52.  (ii) Yes. The 95% confidence interval does not include X  0. (iii) Randomly select n students who have taken the test only one time. Randomly select one half of these students and have them take the prep course. Administer the test again to all of the n students. Compare the gain in performance of the prep-course second-time test takers to the non-prep-course second-time test takers. 3.17. (a) The 95% confidence interval is Y Y Y Y m m m m , 2008 , 1992 , 2008 ,1992    1.96 SE( ) where 2 2 2 2 ,2008 ,1992 , 2008 , 1992 ,2008 ,1992 11.78 10.17 SE( ) 0.37; m m m m m m s s Y Y n n       the 95% confidence interval is (24.98  23.27) ± 0.73 or 1.71 ± 0.73. (b) The 95% confidence interval is Y Y Y Y w w w w , 2008 , 1992 , 2008 , 1992    1.96 SE( ) where 2 2 2 2 ,2008 ,1992 , 2008 , 1992 ,2008 ,1992 9.66 7.78 SE( ) 0.31; w w w w w w s s Y Y n n       the 95% confidence interval is (20.87 20.05) 0.60   or 0.82  0.60.22 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) The 95% confidence interval is ( ) ( ) 1.96 SE[( ) Y Y Y Y Y Y m m w w m m , 2004 ,1992 , 2004 ,1992 , 2008 ,1992       ( )], Y Y w w , 2008 , 1992  where SE[( ) ( )] Y Y Y Y m m w w , 2008 , 1992 , 2008 ,1992     2 2 2 2 ,2008 ,1992 ,2008 ,1992 ,2008 ,1993 ,2008 ,1992 m m w w m m w w s s s s n n n n     11.78 10.17 9.66 7.78 2 2 2 2     0.48. The 95% confidence interval is (24.98-23.27) − (20.87−20.05) ± 1.96  0.48 or 0.89 ± 0.95. 3.18. Y … Y 1  n are i.i.d. with mean Y and variance Y2. The covariance cov ( , ) 0, Y Y j i  j  i. The sampling distribution of the sample average Y has mean Y and variance var 2 2Y( ) .YYn   (a) 2 2 2 2 2 2 [( ) ] {[( ) ( )] } [( ) 2( )( ) ( ) ] [( ) ] 2 [( )( )] [( ) ] var( ) 2cov( , ) var( ). i i Y Y i Y i Y Y Y i Y i Y Y Y i i E Y Y E Y Y E Y Y Y Y E Y E Y Y E Y Y Y Y Y                                 (b) 1 1 2 2 2 cov( ) [( )( )] ( ) ( ) 1 1 ( ) [( ) [( )( )] 1 1 cov( , ) n j j Y i Y Y i Y n j j Y i Y j Y i Y I Y j i Y Y j i j i Y Y Y E Y Y E Y n Y E Y E Y E Y Y n n n Y Y n n n                                                                                     (c) 2 2 2 1 1 1 2 2 2 2 2 1 1 1 1 1 ( ) [( ) ] [var( ) 2cov( ) var( )] 1 1 1 1 1 1 2 . 1 1 n n n Y i i i i i i i n n Y Y Y Y Y i i E s E Y Y E Y Y Y Y Y Y n n n n n n n n n                                                                3.19. (a) No. E( ) and ( ) for . Y E YY i j i Y Y i j Y 2 2 2 2        Thus 2 2 2 2 2 2 2 1 1 1 1 1 1 1 ( ) ( ) ( ) n n n i i i j Y Y i i i j i E Y E Y E Y E YY n n n n                     (b) Yes. If Y gets arbitrarily close to Y with probability approaching 1 as n gets large, then Y 2 gets arbitrarily close to Y2 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.) 3.20. Using analysis like that in equation (3.29) 1 1 1 ( )( ) 1 1 ( )( ) ( )( ) 1 1 n XY i i i n i X i Y X Y i s X X Y Y n n n X Y X Y n n n                                Solutions to End-of-Chapter Exercises 23 ©2011 Pearson Education, Inc. Publishing as Addison Wesley because p X X and Y p Y the final term converges in probability to zero. Let W X Y i i x i Y    ( )( ).   Note Wi is iid with mean XY and second moment E X Y [( ) ( ) ]. i X i Y     2 2 But E X Y E X E Y [( ) ( ) ] ( ) ( ) i X i Y i X i Y          2 2 4 4 from the Cauchy-Schwartz inequality. Because X and Y have finite fourth moments, the second moment of Wi is finite, so that it has finite variance. Thus 1ni i i XY n1W E W   p ( ) .  Thus, sXY XY p  (because the term nn1 1). 3.21. Set nm  nw  n, and use equation (3.19) write the squared SE of Y Y m w  as 2 2 1 1 2 2 2 1 1 1 1 ( ) ( ) ( 1) ( 1) [ ( )] ( ) ( ) . ( 1) n n i mi m i wi w m w n n i mi m i wi w Y Y Y Y n n SE Y Y n n Y Y Y Y n n                     Similarly, using equation (3.23) 2 2 1 1 2 2 2 1 1 1 1 ( ) ( ) 2( 1) ( 1) [ ( )] 2 ( ) ( ) . ( 1) n n i mi m i wi w pooled m w n n i mi m i wi w Y Y Y Y n n SE Y Y n Y Y Y Y n n                         ©2011 Pearson Education, Inc. Publishing as Addison Wesley Chapter 4 Linear Regression with One Regressor 4.1. (a) The predicted average test score is  TestScore  392 36       (b) The predicted change in the classroom average test score is             TestScore ( 5 82 19) ( 5 82 23) 23 28 (c) Using the formula for ˆ 0 in Equation (4.8), we know the sample average of the test scores across the 100 classrooms is 0 1 TestScore CS               ˆ ˆ 4 395 85 (d) Use the formula for the standard error of the regression (SER) in Equation (4.19) to get the sum of squared residuals: SSR n SER         ( 2) (100 2) Use the formula for R2 in Equation (4.16) to get the total sum of squares: 2 2 12961 13044 1 1 0 08 SSR TSS R        The sample variance is sY2  TSS 13044 n1 99    131 8. Thus, standard deviation is s s Y Y    2 11 5. 4.2. The sample size n  200. The estimated regression equation is  2 Weight Height R               (2 15) 99 41 (0 31) 3 94 0 81 SER 10 2 (a) Substituting Height  70, 65, and 74 inches into the equation, the predicted weights are 176.39, 156.69, and 192.15 pounds. (b)            Weight Height 3 94 3 94 1 5 5 91. (c) We have the following relations: 1 2 54 and1 0 4536 . in cm lb k     g Suppose the regression equation in centimeter-kilogram units is  Weight Hei     ˆ ˆ 0 1 ght . The coefficients are ˆ0         092 ; kg ˆ1        kg per cm. The R2 is unit free, so it remains at R2  0 81  . The standard error of the regression is SER kg       6267 .Solutions to End-of-Chapter Exercises 25 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 4.3. (a) The coefficient 9.6 shows the marginal effect of Age on AWE; that is, AWE is expected to increase by $9.6 for each additional year of age. 696.7 is the intercept of the regression line. It determines the overall level of the line. (b) SER is in the same units as the dependent variable (Y, or AWE in this example). Thus SER is measured in dollars per week. (c) R2 is unit free. (d) (i) 696.7 9.6 25 $936.7;    (ii) 696.7 9.6 45 $1,128.7    (e) No. The oldest worker in the sample is 65 years old. 99 years is far outside the range of the sample data. (f) No. The distribution of earning is positively skewed and has kurtosis larger than the normal. (g)   ˆ ˆ 0 1   Y X , so that Y X     ˆ ˆ 0 1 . Thus the sample mean of AWE is 696.7  9.6  41.6  $1,096.06. 4.4. (a) ( ) ( ) , R     R R R u f m f  so that var ( ) var( ) var( ) 2 R R R R u  f m f        2  cov( , ). u R R m f  But cov( , ) 0, u R R m f   thus var( ) var( ) var( ). R      R R R u f m f  2 With  > 1, var(R  Rf) > var(Rm  Rf), follows because var(u)  0. (b) Yes. Using the expression in (a) var( ) var( ) ( 1) var( ) var( ), R         R R R R R u f m f m f  2 which will be positive if var( ) (1 ) var( ). u R R      2 m f (c) R R m f     7.3% 3.5% 3.8%. Thus, the predicted returns are R R R R ˆ       f m f   ˆ ˆ ( ) 3.5% 3.8% Wal-Mart: 3.5%  0.3×3.8%  4.6% Kellogg: 3.5% 0.5 3.8% 5.4%    Waste Management: 3.5% 0.6 3.8% 5.8%    Verizon: 3.5% 0.6 3.8% 5.8%    Microsoft: 3.5% 1.0 3.8% 7.3%    Best Buy: 3.5% 1.3 3.8% 8.4%    Bank of America: 3.5%  2.4  3.8%  11.9% 4.5. (a) ui represents factors other than time that influence the student’s performance on the exam including amount of time studying, aptitude for the material, and so forth. Some students will have studied more than average, other less; some students will have higher than average aptitude for the subject, others lower, and so forth. (b) Because of random assignment ui is independent of Xi. Since ui represents deviations from average E(ui)  0. Because u and X are independent E(ui|Xi)  E(ui)  0. (c) (2) is satisfied if this year’s class is typical of other classes, that is, students in this year’s class can be viewed as random draws from the population of students that enroll in the class. (3) is satisfied because 0  Yi  100 and Xi can take on only two values (90 and 120). (d) (i) 49 0.24 90 70.6; 49 0.24 120 77.8; 49 0.24 150 85.0          (ii) 0.24 10 2.4.  26 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley 4.6. Using E u X ( | ) 0, i i  we have E( | ) ( | ) ( | ) ( | ) Y X E X u X E X X E u X X i i i i i i i i i i          0 1 0 1 0 1      4.7. The expectation of ˆ0 is obtained by taking expectations of both sides of Equation (4.8): 0 1 0 1 1 1 0 1 1 1 0 1 ( ) ( ) ˆ ˆ ˆ 1 ( ) ( ) ˆ n i i n i i E E Y X E X u X n E X E u n                                        where the third equality in the above equation has used the facts that E(ui)  0 and E[( ˆ1 −1) X ]  E[(E( ˆ1 −1)| X ) X ]  because E X [( ) | ] 0   1 1  ˆ  (see text equation (4.31).) 4.8. The only change is that the mean of ˆ0 is now 0  2. An easy way to see this is this is to write the regression model as Y X u i i i      ( 2) ( 2). 0 1  The new regression error is (ui  2) and the new intercept is (0  2). All of the assumptions of Key Concept 4.3 hold for this regression model. 4.9. (a) With   ˆ ˆ 1 0   0, , Y and Y Y ˆi   ˆ0 . Thus ESS  0 and R2  0. (b) If R2  0, then ESS  0, so that Y Y ˆi  for all i. But Y X ˆi i     ˆ ˆ 0 1 , so that Y Y ˆi  for all i, which implies that ˆ1  0, or that Xi is constant for all i. If Xi is constant for all i, then 2 1  in ( ) 0 X X i   and ˆ1 is undefined (see equation (4.7)). 4.10. (a) E(ui|X  0)  0 and E(ui|X  1)  0. (Xi, ui) are i.i.d. so that (Xi, Yi) are i.i.d. (because Yi is a function of Xi and ui). Xi is bounded and so has finite fourth moment; the fourth moment is non-zero because Pr(Xi  0) and Pr(Xi  1) are both non-zero so that Xi has finite, non-zero kurtosis. Following calculations like those exercise 2.13, ui also has nonzero finite fourth moment. (b) var( ) 0.2 (1 0.2) 0.16 X i     and X  0.2. Also 2 2 2 var[( ) ] [( ) ] [( ) | 0] Pr( 0) [( ) | 1] Pr( 1) i X i i X i i X i i i i X i i i X u E X u E X u X X E X u X X                  where the first equality follows because E[(Xi  X)ui]  0, and the second equality follows from the law of iterated expectations. E X u X E X u X [( ) | 0] 0.2 1, and [( ) | 1] (1 0.2) 4. i X i i i X i i            2 2 2 2 Putting these results together 1 2 2 2ˆ 2 1 (0.2 1 0.8) ((1 0.2) 4 0.2) 1 21.25   n n 0.16        Solutions to End-of-Chapter Exercises 27 ©2011 Pearson Education, Inc. Publishing as Addison Wesley 4.11. (a) The least squares objective function is  in1( ) . Y b X i i  1 2 Differentiating with respect to b1 yields 2 1 1 1 1 1 ( ) 2 ( ). n i i i n i i i i Y b X X Y b X b           Setting this zero, and solving for the least squares estimator yields 1 1 2 1 ˆ . n i i i n i i X Y X       (b) Following the same steps in (a) yields 1 1 2 1 ( 4) ˆ . n i i i n i i X Y X        4.12. (a) Write 2 2 2 0 1 1 1 1 1 2 2 2 1 1 2 1 1 ( ) ( ) [ ( )] ˆ ˆ ˆ ˆ ( )( ) ˆ ( ) . ( ) n n n i i i i i i n n i i i i n i i i ESS Y Y X Y X X X X Y Y X X X X                                  This implies 2 2 1 2 2 2 1 1 1 2 1 1 1 1 1 2 2 1 1 1 1 2 2 ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) n i i i n n n i i i i i i n n i i i n n n n i i i i XY XY X Y ESS X X Y Y R Y Y X X Y Y X X Y Y X X Y Y s r s s                                                 (b) This follows from part (a) because rXY  rYX. (c) Because rXY  XY X Y s s s , rXY Y X ss  1 1 2 1 2 2 1 1 1 ( )( ) ( )( ) ( 1) ˆ 1 ( ) ( ) ( 1) n n i i i i XY i i n n X i i i i X X Y Y X X Y Y s n s X X X X n                     4.13. The answer follows the derivations in Appendix 4.3 in “Large-Sample Normal Distribution of the OLS Estimator.” In particular, the expression for i is now i  (Xi  X)ui, so that var(i)  3 var[(Xi  X)ui], and the term 2 carry through the rest of the calculations. 4.14. Because 0 1 ˆ ˆ   Y X  , Y X   ˆ0 1  . The sample regression line is y x   ˆ0 1  , so that the sample regression line passes through ( X Y , ).©2011 Pearson Education, Inc. Publishing as Addison Wesley Chapter 5 Regression with a Single Regressor: Hypothesis Tests and Confidence Intervals 5.1 (a) The 95% confidence interval for 1 is { 5 82 1 96 2 21},       that is .      1 (b) Calculate the t-statistic: 1 1 ˆ 0 5 82 2 6335 SE( ) 2 21 ˆ act t             The p-value for the test H0 1    0 vs. H1 1    0 is p t -value 2 ( | |) 2 ( 2 6335)              act The p-value is less than 0.01, so we can reject the null hypothesis at the 5% significance level, and also at the 1% significance level. (c) The t-statistic is 1 1 ˆ ( 5.6) 0 22 0.10 SE( ) 2 21 ˆ act t          The p-value for the test H0 1 : 5.6    vs. H1 1 : 5.6    is p t -value 2 ( | |) 2 ( 0.10) 0.92        act The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significance level. Because 1  5.6 is not rejected at the 5% level, this value is contained in the 95% confidence interval. (d) The 99% confidence interval for 0 is {520.4 2.58 20.4},   that is, 467.7 573.0.   0 5.2. (a) The estimated gender gap equals $2.12/hour. (b) The null and alternative hypotheses are H0 1    0 vs. H1 1    0. The t-statistic is 1 1 ˆ 0 2.12 5.89 ( ) 0 36 ˆ act t SE        Solutions to End-of-Chapter Exercises 29 ©2011 Pearson Education, Inc. Publishing as Addison Wesley and the p-value for the test is p t -value 2 ( | |) 2 ( 5.89)            act (to four decimal places) The p-value is less than 0.01, so we can reject the null hypothesis that there is no gender gap at a 1% significance level. (c) The 95% confidence interval for the gender gap 1 is {2 12 1 96 0 36},      that is, 1 41 2.83.    1 (d) The sample average wage of women is ˆ0   $12 52/hour. The sample average wage of men is 0 1   ˆ ˆ     $12.52 $2.12 $14.64/hour. (e) The binary variable regression model relating wages to gender can be written as either Wage Male u  0 1     i or Wage Female v   0 1     i In the first regression equation, Male equals 1 for men and 0 for women; 0 is the population mean of wages for women and 0 1   is the population mean of wages for men. In the second regression equation, Female equals 1 for women and 0 for men;  0 is the population mean of wages for men and  0 1  is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations: 0 0 1 0 1 0             Given the coefficient estimates 0 ˆ and ˆ1 , we have 0 0 1 1 0 0 1 ˆ ˆ ˆ 14.64 ˆ ˆ ˆ ˆ 2 12                    Due to the relationship among coefficient estimates, for each individual observation, the OLS residual is the same under the two regression equations: u v ˆ ˆ i i  .Thus the sum of squared residuals, 2 1 ˆ , n i i SSR u    is the same under the two regressions. This implies that both 1/2 1 SSR SER n         and R2   1 TSS SSR are unchanged. In summary, in regressing Wages on Female, we will get  Wages Female     14.64 2 12 R SER 2  0 06 4.2     5.3. The 99% confidence interval is 1.5  {3.94  2.58  0.31) or 4.71 lbs  WeightGain  7.11 lbs. 5.4. (a) 5.38 1.76  16  $22.78 per hour (b) The wage is expected to increase by 1.76  2 = $3.52 per hour.30 Stock/Watson • Introduction to Econometrics, Third Edition ©2011 Pearson Education, Inc. Publishing as Addison Wesley (c) The increase in wages for college education is 1  4. Thus, the counselor’s assertion is that 1   10/4 2.50. The t-statistic for this null hypothesis is 1.76 2.50 9.25, 0.08 act t    which has a p-value of 0.00. Thus, the counselor’s assertion can be rejected at the 1% significance level. A 95% confidence for 1  4 is 4  (1.76  1.96  0.08) or $6.41  Gain  $7.67. 5.5 (a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5 of the standard deviation in test scores, a moderate increase. (b) The t-statistic is 13.9 5.56, 2.5 act t   which has a p-value of 0.00. Thus the null hypothesis is rejected at the 5% (and 1%) level. (c) 13.9  2.58  2.5  13.9  6.45. 5.6. (a) The question asks whether the variability in test scores in large classes is the same as the variability in small classes. It is hard to say. On the one hand, teachers in small classes might able so spend more time bringing all of the students along, reducing the poor performance of particularly unprepared students. On the other hand, most of the variability in test scores might be beyond the control of the teacher. (b) The formula in 5.3 is valid for heteroskesdasticity or homoskedasticity; thus inferences are valid in either case. 5.7. (a) The t-statistic is 3.2 2.13 1.5  with a p-value of 0.03; since the p-value is less than 0.05, the null hypothesis is rejected at the 5% level. (b) 3.2  1.96  1.5  3.2  2.94 (c) Yes. If Y and X are independent, then 1  0; but this null hypothesis was rejected at the