斯托克、沃森著《計量經(jīng)濟學》第二版答案.pdf

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1、PART ONE Solutions to Exercises Chapter 2 Review of Probability Solutions to Exercises 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 d Y 1 1

2、 d Y 2 Y t 2 Probability 0 0.25 0.75 1.0 (c) = ( ) (0 0.25) (1 0.50) (2 0.25) 1.00 Y EYP u u u Using Key Concept 2.3: 22 var( ) ( ) ( ) ,YEY EY and u u u 22 2 2 ()(0 0.25)(10.50)(2 0.25)1.50EY so that 22 2 var( ) ( ) ( ) 1.50 (1.00) 0.50.YEY EY 2. We know from Table 2.2 that Pr( 0)

3、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) 00221078 078, () 0Pr( 0)1Pr( 1) 00301070 070 Y X EY Y Y EX X X P P u u u u u u u u (b) 22 22 22 22 ( ) (0 0.70) Pr( 0) (1 0.70) Pr( 1) (070) 030 030 070 021 ( ) (0 0.78) Pr( 0) (1 0.78) Pr( 1) (078) 02

4、2 022 078 01716 XX YY EX XX EY YY VP VP u u u u u u u u 4 Stock/Watson - Introduction to Econometrics - Second Edition (c) Table 2.2 shows Pr ( 0, 0) 0 15,XY Pr( 0, 1) 015,XY Pr( 1, 0) 007,XY Pr ( 1, 1) 0 63.XY So cov( , ) ( )( ) (0 - 0.70)(0 - 0.78)Pr( 0, 0)

5、(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) (070) (078) 015 (070) 022 015 030 ( 078) 007 0 XY X Y XY E X Y XY XY XY XY V PP u u u u u u 30 0 22 0 63 0084, 0084 (,) 04425 021 01716 XY XY cor X Y V VV u u u 3. For the two new random variabl

6、es 36WX and 20 7 ,VY we have: (a) () (20 7) 20 7() 20 7 078 1454, () (36)36()3607072 EV E Y EY EW E X EX u u (b) 222 2 var (3 6 ) 6 36 0 21 7 56, var (20 7 ) ( 7) 49 0 1716 8 4084 WX VY X Y VV u u (c) (3 6 , 20 7 ) 6( 7) ( , ) 42 0 084 3 528 3528 (,) 04425 756 84084 WV WV WV cov

7、 X Y cov X Y cor W V V V VV u u 4. (a) uu 33 3 ()0(1 )1E Xpp (b) uu ()0(1 )1 kk k E (c) ()0.3EX 22 var( ) ( ) ( ) 0.3 0.09 0.21XEX EX Thus, 0.21 0.46.V To compute the skewness, use the formula from exercise 2.21: 33 2 3 23 ()()3()()2() 0.3 3 0.3 2 0.3 0.084 E XEXEXEXEXP

8、u u Alternatively, uu 33 ()= (10.3)0.3(00.3)0.70.84EX P Thus, skewness 33 3 ()/.084/0.460.87.EX PV Solutions to Exercises in Chapter 2 5 To compute the kurtosis, use the formula from exercise 2.21: 44 3224 234 ()()4()()6()()3() 0.3 4 0.3 6 0.3 3 0.3 0.0777 E XEXEXEXEXEXEXP u u u Alterna

9、tively, uu 44 4 ()= (10.3)0.3(00.3)0.70.77EX P Thus, kurtosis is 44 4 ()/= .077/0.461.76EX PV 5. Let X denote temperature in qF and Y denote temperature in qC. Recall that Y 0 when X 32 and Y 100 when X 212; this implies (100/180) ( 32) or 17.78 (5/9) .YXY X u u Using Key Concept 2.3, q70

10、F X P implies that 17.78 (5/9) 70 21.11 C, Y P u q and q7F X V implies (5/9) 7 3.89 C. Y V u q 6. The table shows that Pr ( 0, 0) 0 045,XY Pr( 0, 1) 0709,XY Pr( 1, 0) 0005,XY Pr ( 1, 1) 0 241,XY Pr( 0) 0754,X Pr( 1) 0 246,X Pr( 0) 0 050,Y Pr( 1) 0 950.Y (a) u u u u

11、 () 0 Pr( 0) 1Pr( 1) 0005010950 0950 Y EY YYP (b) (unemployed) Unemployment Rate (labor force) Pr( 0) 0 050 1 0 950 1 ( ) # # YEY (c) Calculate the conditional probabilities first: Pr ( 0, 0) 0 045 Pr ( 0| 0) 00597, Pr ( 0) 0 754 XY YX X Pr ( 0, 1) 0 709 Pr ( 1| 0) 09403, Pr (

12、 0) 0 754 XY YX X Pr ( 1, 0) 0 005 Pr ( 0| 1) 00203, Pr ( 1) 0 246 XY YX X Pr ( 1, 1) 0 241 Pr ( 1| 1) 09797 Pr ( 1) 0 246 XY YX X The conditional expectations are (| 1) 0 Pr( 0| 1) 1 Pr( 1| 1) 000203109797 09797, (| 0) 0 Pr( 0| 0) 1 Pr( 1| 0) 00059710940309403 EYX

13、 YX YX EYX YX YX u u u u u u u u 6 Stock/Watson - Introduction to Econometrics - Second Edition (d) Use the solution to part (b), Unemployment rate for college grads 1(| 1)10.97970.0203. Unemployment rate for non-college grads 1(| 0)10.94030.0597. EYX EYX (e) The pr

14、obability that a randomly selected worker who is reported being unemployed is a college graduate is Pr ( 1, 0) 0 005 Pr ( 1| 0) 01 Pr ( 0) 0 050 XY XY Y The probability that this worker is a non-college graduate is Pr ( 0| 0) 1 Pr ( 1| 0) 1 0 1 0 9XY XY (f) Educational achievemen

15、t and employment status are not independent because they do not satisfy that, for all values of x and y, Pr ( | ) Pr ( )YyXx Yy For example, Pr ( 0| 0) 0 0597 Pr ( 0) 0 050YX Y z 7. Using obvious notation, ;CMF thus CMF P PP and 22 2 2cov( , ). CMF M FVVV This implies (a) 40 45 $85,0

16、00 C P per year. (b) (,) (,) , MF Cov M F cor M F VV so that ( , ) ( , ). MF Cov M F cor M FV V Thus (,)12180.80172.80,Cov M F uu where the units are squared thousands of dollars per year. (c) 22 2 2cov( , ), CMF M FVVV so that 222 12 18 2 172.80 813.60, C V u and 813.60 28.524 C V thousand

17、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$ is therefore with eE. The mean is therefore eP

18、C (in units of thousands of euros per year), and the standard deviation is eV C (in units of thousands of euros per year). The correlation is unit-free, and is unchanged. 8. () 1, Y EYP 2 var( ) 4. Y YV With 1 2 (1),ZY 22 111 (1) ( 1) (11)0, 222 111 var ( 1) 41 244 ZY EY Y PP VV u

19、 Solutions to Exercises in Chapter 2 7 9. Value of Y 14 22 30 40 65 Probability Distribution 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 Value of X 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 distribut

20、ion is given in the table above. 22 2 2 2 2 22 () 14 0.21 22 0.23 30 0.30 40 0.15 65 0.11 30.15 ()14 0.212 0.2330 0.3040 0.1565 0.1 127.23 Var(Y) ( ) ( ) 218.21 14.77 Y EY EY EY EY V u u u u u u u u u u (b) Conditional Probability of Y|X 8 is given in the table below Value of Y 14 22 30 4

21、0 65 0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39 0.09/0.39 (| 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 E YX u u u u u 22 2 22 (| 8)14 (0.02/0.39)2 (0.03/0.39)30 (0.15/0.39) 40 (0.10/0.39) 65 (0.09/0.39) 1778.7 EY X u u u u u 2 8 Var( ) 1778.7 39.21 241.65

22、 15.54 YX Y V _ (c) ( ) (1 14 0.02) (1 22 : 0.05) (8 65 0.09) 171.7EXY uu u uu L Cov( , ) ( ) ( ) ( ) 171.7 5.33 30.15 11.0XY EXY EXEY u Corr( , ) Cov( , )/( ) 11.0 /(5.46 14.77) 0.136 XY XY XY V V 10. Using the fact that if 2 , YY YNP V : then (0,1) Y Y Y N P V and Appendix Table

23、1, we have (a) 131 Pr( 3) Pr (1) 0 8413 22 Y Y d d ) (b) Pr( 0) 1 Pr( 0) 303 1Pr 1 (1) (1)08413 33 YY Y ! d d ) ) 8 Stock/Watson - Introduction to Econometrics - Second Edition (c) 40 50 50 52 50 Pr (40 52) Pr 555 (0 4) ( 2) (0 4) 1 (2) 06554 1 09772 06326 Y Y dd d d ) ) )

24、 ) (d) 65 5 85 Pr (6 8) Pr 222 (2 1213) (0 7071) 09831 07602 02229 Y Y dd d d ) ) 11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When 2 10 ,Y F then 10, /10 .YF f (e) 2 ,YZ where N(0,1),Z thus Pr( 1) Pr( 1 1) 0.32.YZd dd 12. (a) 0.05 (b) 0.950 (c) 0.953 (d) The t df distribution and N(0,

25、1) are approximately the same when df is large. (e) 0.10 (f) 0.01 13. (a) 22 2 2 ()Var() 101;( )Var() 10 010. YW EY Y EW WPP (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

26、3, so 4 $ () 3; Y Y EY P V solving yields 4 E( ) 3;Y a similar calculation yields the results for W. (d) First, condition on 0,X so that :SW 2342 . (| 0) 0; ( | 0) 100, ( | 0) 0, ( | 0) 3 100ESX ES X ES X ES X u Similarly, 23 4 (| 1) 0; ( | 1) 1, ( | 1) 0, ( | 1) 3.ESX ES X ES X ES X

27、 From the large of iterated expectations () (| 0) Pr(X 0) (| 1) Pr( 1) 0ES ESX ESX X u u 22 2 () (| 0)Pr(X0) (| 1)Pr( 1)1000.0110.991.99ES ES X ES X X u u uu 33 3 () (| 0)Pr(X0) (| 1)Pr( 1)0ES ES X ES X X u u 44 4 2 () (| 0)Pr(X0) (| 1)Pr( 1)3100 0.01310.9 302.97ES ES X ES X X u u

28、 uuuu Solutions to Exercises in Chapter 2 9 (e) ( ) 0, S ESP thus 33 ()()0 S ES ESP from part d. Thus skewness 0. Similarly, 222 ()()1.9, SS ES ESVP and 44 ()()302.97. S ES ESP Thus, 2 kurtosis 302.97 /(1.99 ) 76.5 14. The central limit theorem suggests that when the sample size (n) is

29、large, the distribution of the sample average ()Y is approximately 2 , Y Y N P V with 2 2 . Y nY V V Given 100, Y P 2 43 0, Y V (a) 100,n 2 2 43 100 043, Y nY V V and 100 101 100 Pr( 101) Pr (1 525) 0 9364 043 043 Y Y d d |) (b) 165,n 2 2 43 165 02606, Y nY V V and 100 98

30、100 Pr ( 98) 1 Pr ( 98) 1 Pr 02606 02606 1(39178) (39178)100(roundedtofourdecimalplaces) Y YY ! d d |) ) (c) 64,n 2 2 43 64 64 06719, Y Y V V and 101 100 100 103 100 Pr (101 103) Pr 06719 06719 06719 (3 6599) (1 2200) 0 9999 0 8888 0 1111 Y Y dd d d |) ) 15. (a) 9.6 10

31、10 10.4 10 Pr (9.6 10.4) Pr 4/ 4/ 4/ 9.6 10 10.4 10 Pr 4/ 4/ Y Y nn n Z nn dd d d d d 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/ ZZ nn dd d d (ii) n 100; 9.6 10 10.4 10 Pr Pr( 2.00 2.00) 0.954 4/ 4/nn dd d d (iii) n 1000; 9.6 10 10.4 10 Pr Pr( 6.

32、32 6.32) 1.000 4/ 4/ ZZ nn dd d d 10 Stock/Watson - Introduction to Econometrics - Second Edition (b) 10 Pr (10 10 ) Pr 4/ 4/ 4/ Pr . 4/ 4/ cY c cY c nnn cc Z nn dd d d d d As n get large 4/ c n gets large, and the probability converges to 1. (c) This follows from (b) and the definition

33、 of convergence in probability given in Key Concept 2.6. 16. There are several ways to do this. Here is one way. Generate n draws of Y, Y 1 , Y 2 , Y n . Let X i 1 if Y i 3.6, otherwise set X i 0. Notice that X i is a Bernoulli random variables with P X Pr(X 1) Pr(Y 3.6). Compute .X Because

34、 X converges in probability to P X Pr(X 1) Pr(Y 3.6), X will be an accurate approximation if n is large. 17. P Y 0.4 and 2 0.4 0.6 0.24 Y V u (a) (i) P(Y t 0.43) 0.4 0.43 0.4 0.4 Pr Pr 0.6124 0.27 0.24/ 0.24/ 0.24/ YY nn n t t (ii) P(Y d 0.37) 0.4 0.37 0.4 0.4 Pr Pr 1.22 0.11 0.24/

35、0.24/ 0.24/nn n d d (b) We know Pr(1.96 d Z d 1.96) 0.95, thus we want n to satisfy 0.41 0.4 0.24/ 0.41 1.96 n ! and 0.39 0.4 0.24/ 1.96. n Solving these inequalities yields n t 9220. 18. Pr( 0) 0 95,Y$ Pr( 20000) 005.Y$ (a) The mean of Y is 0Pr( 0)20,000Pr( 20000) 1000. Y Y$ Y$ $P u

36、 u The variance of Y is 2 2 2 2 22 7 Pr 0 (20000 1000) Pr ( 20000) (0 1000) (1000) 095 19000 005 19 10, Y Y E Y YY V P u u u u u so the standard deviation of Y is 1 2 7 (1 9 10 ) 4359 Y $V u (b) (i) () 1000, Y EY $P 2 7 25 19 10 100 19 10. Y nY V V u u (ii) Using the centra

37、l limit theorem, 55 Pr ( 2000) 1 Pr ( 2000) 1000 2 000 1 000 1Pr 19 10 19 10 1(22942)10989100109 YY Y ! d d u u |) Solutions to Exercises in Chapter 2 11 19. (a) 1 1 Pr ( ) Pr ( , ) Pr ( | )Pr ( ) l jij i l j ii i Yy XxYy YyXx Xx (b) 111 1 1 1 () Pr( ) Pr( | )Pr( ) Pr ( | )

38、 Pr ( ) (| )Pr( ) i ii kkl j jj ji i jji k l jji j i l i E YyYy y YyXxXx yYyXx Xx EYX x X x (c) When X and Y are independent, Pr ( , ) Pr ( )Pr ( ) ij i j X xY y X x Y y so 11 11 ( )( ) ()()Pr(,) ()()Pr()Pr() ()Pr()()Pr( ()()000, XY XY lk iXjY i j ij lk iXjY i j ij iX

39、i jY j XY EX Y xy XxYy xy XxYy x Xx y Yy EX EY VPP PP PP PP PP u 0 (,) 0 XY XY XY cor X Y V VV VV 20. (a) 11 Pr ( ) Pr ( | , ) Pr ( , ) lm iij h j h jh Yy YyXxZz XxZz (b) 1 111 11 1 11 () Pr( )Pr( ) Pr ( | , ) Pr ( , ) Pr ( | , ) Pr ( , ) (| , )Pr( , ) k

40、 ii i i klm iij hjh ijh lm k iijh jh jh i lm jh jh jh EY y Y y Y y yYyXxZzXxZz yYyXxZz XxZz EYX x Z z X x Z z 12 Stock/Watson - Introduction to Econometrics - Second Edition where the first line in the definition of the mean, the second uses (a), the third is a rearr

41、angement, and the final line uses the definition of the conditional expectation. 21. (a) 32 322223 32 2332 23 32 3 ()()()2 2 ()3() 3() ()3()()3()()() ()3()()2() EX E X X EX X X X X EX EX EX EX EX EX EX EX EX EX EX EX EX PPP PPPPP PPP (b) 432 23 43 22 33 22 34 43 22 34 24 ()(3 3 )

42、() 3 3 3 3 ()4()()6()()4()() () ()4()()6()()3() EX E X X X X EX X X X X X X E XEXEXEXEXEXEXEX EX EX EX EX EX EX PPPPP PPPPPPP 22. The mean and variance of R are given by 22 2 2 0.08 (1 ) 0.05 0.07(1)0.422 (1)0.70.40.25 ww ww P V u u u u uuu u u where 0.07 0.04 0.25 ( , ) s b

43、Cov R Ruu follows from the definition of the correlation between R s and R b . (a) 0.065; 0.044P V (b) 0.0725; 0.056P V (c) w 1 maximizes ; 0.07P V for this value of w. (d) The derivative of V 2 with respect to w is 2 22 2.072(1)0.04(24)0.070.040.25 0.0102 0.0018 d ww w dw w V u u u u u s

44、olving for w yields 18 /102 0.18.w (Notice that the second derivative is positive, so that this is the global minimum.) With 0.18, .038. R w V 23. X and Z are two independently distributed standard normal random variables, so 22 0, 1, 0. XZ XZ XZ PP VV V (a) Because of the independence betw

45、een X and ,Z Pr( | ) Pr( ),Z zX x Z z and (|) () 0.EZX EZ Thus 22 22 (| ) ( | ) ( | ) (|) 0E YX E X ZX E X X E ZX X X (b) 222 () 1, XX EX VP and ()()101 YZ EX Z EXPP (c) 33 () ( ) () ().E XY E X ZX E X E ZX Using the fact that the odd moments of a standard normal random variable are

46、all zero, we have 3 ()0.EX Using the independence between X and ,Z we have ( ) 0. ZX EZX P P Thus 3 () () ()0.EXY EX EZX Solutions to Exercises in Chapter 2 13 (d) () ( )( ) ( 0)( 1) ()()() 000 0 (,) 0 XY XY XY XY Cov XY E X Y E X Y EXY X EXY EX cor X Y PP V VV VV 24. (a) 2222 ()

47、i EY V PV and the result follows directly. (b) (Y i /V) is distributed i.i.d. N(0,1), 2 1 (/), n i i WYV and the result follows from the definition of a 2 n F random variable. (c) E(W) 22 22 11 () . nn ii ii YY E WE E n VV (d) Write 22 11 (/ ) 11 / nn ii i YY YY V V V which follo

48、ws from dividing the numerator and denominator by V. Y 1 /V N(0,1), 2 2 (/) n i i Y V 2 1n F , and Y 1 /V and 2 2 (/) n i i Y V are independent. The result then follows from the definition of the t distribution. Chapter 3 Review of Statistics Solutions to Exercises 1. The central limit theor

49、em suggests that when the sample size ( n ) is large, the distribution of the sample average (Y ) is approximately 2 , Y Y N P V with 2 2 . Y nY V V Given a population 100, Y P 2 43 0, Y V we have (a) 100,n 2 2 43 100 043, Y nY V V and 100 101 100 Pr ( 101) Pr (1.525) 0 9364 043 043 Y

50、Y |) (b) 64,n 2 2 43 64 64 06719, Y Y V V and 101 100 100 103 100 Pr(101 103) Pr 06719 06719 06719 (3 6599) (1 2200) 0 9999 0 8888 0 1111 Y Y |) ) (c) 165,n 2 2 43 165 02606, Y nY V V and 100 98 100 Pr ( 98) 1 Pr ( 98) 1 Pr 02606 02606 1(39178) (39178)100(roun

51、dedtofourdecimalplaces) Y YY ! d d |) ) 2. Each random draw i Y from the Bernoulli distribution takes a value of either zero or one with probability Pr (1) i Yp and Pr (0)1. i Yp The random variable i Y has mean () 0Pr( 0)1Pr( 1) , i E YY Yp u u and variance 2 22 var( ) ( ) (0 ) P

52、r( 0) (1 ) Pr( 1) (1 ) (1 ) (1 ) iiY YEY pY pY ii pp pppp P u u Solutions to Exercises in Chapter 3 15 (a) The fraction of successes is 1 (1)(success) n iii Y#Y# p Y nnn (b) 1 11 () () n nn ii i ii Y E pE EY p p nn n (c) 1 22 11 (1) var( ) var var( ) (1 ) n nn ii i ii

53、 Y p p pYp p The second equality uses the fact that 1 Y , , Y n are i.i.d. draws and cov( , ) 0, ij YY for .ijz 3. Denote each voters preference by .Y 1Y if the voter prefers the incumbent and 0Y if the voter prefers the challenger. Y is a Bernoulli random variable with probability

54、 Pr ( 1)Yp and Pr ( 0) 1 .Yp From the solution to Exercise 3.2, Y has mean p and variance (1 ).p p (a) 215 400 05375.p (b) (1 ) 0.5375 (1 0.5375) 4 400 var( ) 62148 10 . pp n p u u The standard error is SE 1 2 () (var()) 00249.pp (c) The computed t-statistic is 0 05375 05 1506 SE( )

55、0 0249 pact p t p 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 0 05Hp vs. 1 05:Hpz -value 2 ( | |) 2 ( 1 506) 2 0 066 0 132 act pt ) ) u (d) Using Equation (3.17) in the text, the p-value for the test 0 05Hp vs. 1

56、 05Hp! is -value 1 ( ) 1 (1 506) 1 0 934 0 066 act pt ) ) (e) Part (c) is a two-sided test and the p-value is the area in the tails of the standard normal distribution outside r (calculated t-statistic). Part (d) is a one-sided test and the p-value is the area under the standard normal distrib

57、ution to the right of the calculated t-statistic. (f) For the test 0 05Hp vs. 1 05,Hp! 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 1.645 for a one-sided test wi

58、th 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. 16 Stock/Watson - Introduction to Econometrics - Second Edition 4. Using Key Concept 3.7 in the text (a) 95% con

59、fidence interval for p is 1.96 ( ) 0.5375 1.96 0.0249 (0.4887,0.5863).pSEpr ru (b) 99% confidence interval for p is 2.57 ( ) 0.5375 2.57 0.0249 (0.4735,0.6015).pSEpr ru (c) The interval in (b) is wider because of a larger critical value due to a lower significance level. (d) Since 0.50 lies inside

60、 the 95% confidence interval for p, we cannot reject the null hypothesis at a 5% significance level. 5. (a) (i) The size is given by Pr(| 0.5| .02),p!where the probability is computed assuming that 0.5.p Pr(| 0.5| .02) 1 Pr( 0.02 0.5 .02) 0.02 0.5 0.02 1Pr .5 .5/1055 .5 .5/1055 .5 .5/1055 0.5 1Pr

61、 1.30 1.30 .5 .5/1055 0.19 pp p p ! dd dd uuu d d u where the final equality using the central limit theorem approximation (ii) The power is given by Pr(| 0.5| .02),p! where the probability is computed assuming that p 0.53. Pr(| 0.5| .02) 1 Pr( 0.02 0.5 .02) 0.02 0.5 0.02 1Pr .53 .47/10

62、55 .53 .47/1055 .53 .47/1055 0.05 0.53 0.01 1Pr .53 .47/1055 .53 .47/1055 .53 .47/1055 0.53 1Pr 3.25 .53 .47/1055 pp p p p ! dd d d uuu d d uuu d d u 0.65 0.74 where the final equality using the central limit theorem approximation. (b) (i) 0.54 0.5 0.54 0.46/1055 2.61, Pr(| | 2.6

63、1) .01,tt u ! 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.ru r (iv) 0.54 2.58 0.54 0.46 /1055 0.54 0.04, or 0.50 to 0.58.ru r (v) 0.54 0.67 0.54 0.46 /1055 0.54 0.01,

64、 or 0.53 to 0.55.ru r (c) (i) The probability is 0.95 is any single survey, there are 20 independent surveys, so the probability if 20 0.95 0.36 (ii) 95% of the 20 confidence intervals or 19. Solutions to Exercises in Chapter 3 17 (d) The relevant equation is 1.96 SE( ) .01 or 1.96 (1 ) / .01.pppn

65、u u Thus n must be chosen so that 2 2 1.96 (1 ) .01 , pp 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 .01 9604,n u ! then the margin of error is less than 0.

66、01 for all values of p. 6. (a) No. Because the p-value is less than 5%, P 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 P 5, which requires Y and SE ().Y Only one the p-value for P 5 is given in the problem. 7. The null hypothesis in that the survey is a random draw from a population with p 0.11. The t-statistic is 0.11 () , p SE p t where () (1 )/.SE p p p n (An alternative form