北师大统计学基础习题答案5

Exercise 5

1. P201: 4.59.Let 54321Y Y Y Y Y <<<

)()()]()][()()][([!

1!1!1!

5),(424242424,2y f y f y F y F y F y F y y g -=

424422))()(1(120y y y y y y e e e e e e --------= Since 12Z Y =，214Z Z Y +=，11

10

1==

J . 2

2111

211

21

2112421)1)(1(120))(1(120),(z z z z z z z z z z z z z e e e e e

e e

e

e e z z g ---------------=--=)1(20)()]'([)](1)][([!

3!1!

5)()(11424322211z z e e y f y F y F y F y g z g ---=-=

= ??∞----∞--==0

1240

121222211)1)(1(120),()(dz e e e e dz z z g z g z z z z

2222112

2

2205042)1(6)1(20

1120)|5|4()1(120z z z z z z z z e e e e e e e

e

----∞-∞----=-?=----= Thus, =),(21z z g )()(2211z g z g .So 1Z and 2Z are independent.

P274: 6.15. Let X be the mean of a random sample of size n from a

distribution that is )9,(μN . Find n such that 90.0)11Pr(=+<<-X X μ, approximately.

Solution: Since ~X )9,(n N μ,)1,0(~3

)

(N X n μ-.

90.01)3

(2)3|3)(Pr(|

)1|Pr(|)11Pr(=-Φ=<-=<-=+<<-n

n X n X X X μμμ Thus )3(

n Φ=0.95, 645.13

=n

and 35.24≈n . Because n must be an integer, so n=24 or 25.

P275: 6.18. Let 121,,,+n n X X X X be a random sample of size 9 from a

distribution that is ),(2σμN .

(a) If σis known, find the length of a 95 percent confidence interval for if this interval is based on the random variable.

(b) If σis unknown, find the expected value of the length of a 95 percent confidence interval for if this interval is based on the random variable S X /)(8μ-. (c) Compare these two answers. Solution:

(a) Since ~9

9

1

∑==

i i

X

X )9

,

(2

σμN ,thus

)1,0(~)

(9N X σ

μ-.

Pr(|

σ

μ)

(9-X |<1.96)=0.9595.0)96.1)

(396.1Pr(=<-<

-?σ

μX

95.0)3

96.1396.1Pr(=+<<-

?σ

μσX X . The 95 percent confidence interval for μis )3

96.1,396.1(σ

σ+-X X and the length of it is

σσ

31.175

98≈ (b) Since σ is unknown, then )8(~)

(81

/t S

X n S X T μμ-=

--=

975.0)P r (95.01)Pr(295.0))

(8Pr(=≤?=-≤?=<-<

-b T b T b S X b μ

From TABLE IV of Appendix B, we know b=2.306, thus

95.0)8

306.28

306.2Pr(=+<<-σμσX X . The 95 percent confidence interval for μ is )8

306.2,8

306.2(σσ+-

X X with the length S L 8

612.4=

?

∞

-Γ===0

2

34

2

121

222

)4(18

3612.4))/9((38612.4)(8612

.4)(dx e x x S E S E L E x

σ

σσ

)29(22)4(183612.422)4(183612

.42

)4(1

83612

.42

9

4027

29

40

2

2

7

4

ΓΓ=Γ=Γ=??

∞-∞

-

σσσdy e y dx e x y x

Then σ49.1)(≈L E

(c) From (a) we know the answer is σ31.1, from (b) we know the answer is

σ49.1. The two methods yield results that are in substantial agreement, which shows the length of the confidence interval for μ is almost the same with the parameter σ known or unknown.

P279: 6.30.Let two independent random samples, each of size 10, from two normal distributions ),(21σμN and ),(22σμN yield 8.4=x ,

64.821=s ,6.5=y ,88.72

2=s . Find a 95 percent confidence interval for.

Solution: Let 1021,,X X X and 1021,,Y Y Y denote, respectively, independent random samples from the two distributions ),(21σμN and

),(2

2σμN , then ~10

10

1

∑==

i i

X

X )10

,

(2

1σμN and ~10

10

1

∑==

i i

Y

Y )10

,

(2

2σμN .

)5

,

(~2

21σμμ--N Y X .)18(~/)1010(222

221χσS S +

)18(~9

)

()

()()

10

1101(18)(10)()(22

21

2122

21

21t S S Y X S S Y X T +---=

++---=

μμμμ.

101.295.0)Pr(=?=<<-b b T b and

95.0))(3

)()(3)Pr((2

221212221=++-<-<+-

-S S b Y X S S b Y X μμ So the random interval

is )3101.2)(,3101.2)((2

2212221S S Y X S S Y X ++-+--. Let 8.4=x ,64.821=s ,6.5=y ,88.72

2=s , we get a 95 percent confidence

interval (-3.6,2.0).

2. Use Splus or R software to compute the mean, standard deviation, skewness and kurtosis of the following dataset.

-0.4292, 0.0064, 0.1181, -0.6282, 2.2010, -1.7623, 0.0921,

1.8742, 1.4538, 0.3575, -1.5848, 0.4993, 0.7762, -0.2638, -1.1003, -

2.2480, 0.5419, -0.4018, -0.3562, -0.5872. Solution: > x<-c(-0.4292,0.0064,0.1181,-0.6282,2.2010,-1.7623,0.0921,1.8742,1.4538,0.3575,-1.5848,0.4993,0.7762,-0.2638,-1.1003,-2.2480,0.5419,-0.4018,-0.3562,-0.5872) > a=mean(x) > a

[1] -0.072065 > b=sd(x) > b

[1] 1.143067

> c=mean(((x-a)/b)^3) > c

[1] 0.1234913

> d=mean(((x-a)/b)^4)-3 > d

[1] -0.5280221

So the mean of the dataset is -0.072065, the standard deviation is 1.143067, the skewness is 0.1234913 and the kurtosis is -0.5280221.

3*. P203: 4.73. Let n Y Y Y <<< 21be the order statistics of a random sample of size n from the exponential distribution with p.d.f.

∞<<=-x e x f x 0,)(,zero elsewhere..

(a) Show that 11nY Z =,))(1(122Y Y n Z --=,))(2(233Y Y n Z --=,…

1--=n n n Y Y Z , are independent and that each i Z has the exponential distribution.

Solution: Since n y y y n e e e n y y y g ---= 21!),,(21, and

n n n Z Z n Z

n Z Y n Z n Z Y n Z Y ++-+=-+==

-2

1,,1,12121211 and the Jacobian is

!11

2

1

1110

21

111011

101

n n n n n n n n J =---=

n

n z z z z n z n z n z n z n

z n n e e e e

e

e y y y g J z z z g ---++-+--+--

=== 2121211

)1

(

)1

(

2121),,(||),,(k z n k k n k k e dz dz dz dz z z z g z g -+-∞

∞∞

∞

==??

?

? 110

1210

0),,()(

So )()()(),,(221121n n n z g z g z g z z z g = and ∞<<=-k z k k z e z g k 0,)( Which shows that n z z z ,,21are independent and that each i z has the exponential distribution.

(b) Demonstrate that all linear functions of n Y Y Y ,,,21 , such as ∑n

i i Y a 1, can

be expressed as linear functions of independent random variables. Solution:

)())(1())(1(11122112211---++---++--+=+++n n n k k k n n Y Y b Y Y k n b Y Y n b nY b Y a Y a Y a n n Z b Z b Z b +++= 2211. Then ,)2()1(,)1(232121a n b n n b a n b n b =---=-- n n n n n k k k a b a b b a k n b k n b ==-=--+---+)1(,)1()2(,,)()1(,111 . Then n n n

k k n n a b k n a a b n a a a b n a a a b =+-++=-++=++=

,1

,,1,322211

So

1

,1

1

+-+=

=∑∑==i n a a b Z b Y a n

i i n

i i i n

i i i .Since n Z Z Z ,,,21 are independent, then

n n Z b Z b Z b ,,,2211 are independent. So

∑=n

i i i Y a 1

can be expressed as linear

functions of independent random variables.

P275: 6.19.Let 121,,,+n n X X X X be a random sample of size n+1, n>1, from a distribution that is ),(2

σμN . Let n X X n

i /1∑= and

n X X S n

i /)(21

2

-=∑. Find the constant c so that the statistic

σ

1

+-n X X c

has a t-distribution. If n=8, determine k such

that 80.0)Pr(=+<<-kS X kS X μ. The observed interval is ),(ks x ks x +- often called an 80 percent prediction interval for 9X . Solution: )/,(~2n N X σμ，),(~21σμN X n +，),1,

0(~2

1σn

n N X X n +-+ ).1(~11)

1(1

),

1(~).

1,0(~1

122

1

22

2

1--?+?-=

-+--+-+++n t S

X X n n

n n n nS n n X X n nS N n

n X X n n n σσχσσ

Therefore 1

1

+-=

n n c . 80.0)9

7

97Pr()Pr(

)Pr(999=<-=<-=+<<-k S X X k S X X kS X x kS X 60.1,415.19

7

==k k . So k=1.60.