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Mathematical Statistics witApplicationStudents SolutionsManu

KandethodyM.RamachandDepartmentofMathematicsandSta

UniversityofSouthFl

Tam

ChrisP.TsoDepartmentofMathematicsandSta

UniversityofSouthFl

Tam

AMSTERDAM BOSTON HEIDELBERG LONDON

NEW YORK OXFORD PARIS SAN DIEGO

SAN FRANCISCO SINGAPORE SYDNEYTOKYO

Academic Press is an imprint of Elsevier

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Elsevier Academic Press30 Corporate Drive, Suite 400, Burlington, MA 01803, USA525 B Street, Suite 1900, San Diego, California 92101-4495, USA84 Theobalds Road, London WC1X 8RR, UK

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ISBN 13: 978-0-08-096443-0

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Contents

CHAPTER 1 Descriptive Statistics . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . 1

CHAPTER 2 Basic Concepts from Probability Theory . . . . . . . .. . . . . . . . .. . . . . . . .. . . . . 11

CHAPTER 3 Additional Topics in Probability . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . 21

CHAPTER 4 Sampling Distributions . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . 37

CHAPTER 5 Point Estimation . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . 49

CHAPTER 6 Interval Estimation. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . 59

CHAPTER 7 Hypothesis Testing . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . 69

CHAPTER 8 Linear RegressionModels. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. 81

CHAPTER 9 Design of Experiments. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . 87

CHAPTER 10Analysis of Variance . . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . . . 91

CHAPTER 11Bayesian Estimation and Inference. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

CHAPTER 12Nonparametric Tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

CHAPTER 13Empirical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

CHAPTER 14Some Issues in Statistical Applications: An Overview . . . . . . . . . . . . . . . . . . 127

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Chapter1Descriptive Statisti

EXERCISES 1.2

1.2.1. The suggested solutions:

For qualitative data we can have color, sex, race, Zip code and so on. For quantitative datawe can have age, temperature, time, height, weight and so on. For cross section data we can

have school funding for each department in 2000. For time series data we can have the crude

oil price from 1995 to 2008.

1.2.3. The suggested questions can be

1. What types of data the amount is?

2. Are these Federal Agency get same amount of money? If not, why?

3. Which Federal Agency should get more money? Why?

The suggested inferences we can make is

1. These Federal Agency get different amount of money.

2. The differences of money between the Agencies are kind of big.

EXERCISES 1.3

1.3.1. For stratified sample, we can say suppose we decide to sample 100 college students from the

population of 1000 (that is 10% of the population). We know these 1000 students come

from three different major, Math, Computer Science and Social Science. We have Math 200,

CS 400 and SS 400 students. Then we choose 10% of each of them Math 20, CS 40 and SS

40 by using random sampling within each major.

For cluster sample, we can say suppose we decide to sample some college students from the

population of 2000. We know these 2000 students come from 20 different countries and

we choose 3 out of the 20 countries by random sampling. Then we get all the individual

information from each of the 3 countries.

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2 CHAPTER 1 Descriptive Statistics

EXERCISES 1.4

1.4.1. By minitab

(a) Bar graphBar graph for the percent of road mileage

35.00%

C2

C1

30.00%

25.00%

20.00%

15.00%

10.00%

5.00%

0.00%Poor Mediocre Fair Good Very good

(b) Pie chart

PoorVery goodGoodFairMediocre

Category

Pie chart of the percent of road mileage

1.4.3. (a) Bar graph

40.00%

30.00%

20.00%C2

C1

10.00%

0.00%

Bar graph

Renewable

Energy

PetroliumNyclear

Electric Power

Natural

Gas

Coal

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Students Solutions Manu

(b) Pareto chart

Renewable

Energy

Petrolium Nyclear

Electric

Power

Natural

Gas

Coal

Percentage

Percent

Percentage

Percent

Cum %

0.40

40.0

40.0

0.23

23.0

63.0

0.22

22.0

85.0

0.08

8.0

93.0

0.07

7.0

100.0

C1

1.0

0.8

0.6

0.4

0.2

0.0

100

80

60

40

20

0

Pareto graph

(c) Pie chart

CoalNatural GasNyclear Electric PowerPetroliumRenewable Energy

Category

Pie chart of speciesspecies


6

5

4

3

2

1

0

Count

C1

A B C D F

Bar graph

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(b) Pie chart

A

BC

DF

Category

Pie chartspecies

1.4.7. (a) Pie chart

Mining

Construction

Manufacturing

Transportation

Wholesale

Retail

Finance

Services

Category

Pie chartspecies

(b) Bar graph

Servi

ces

Finan

ce

Retail

Who

lesale

Transp

ortation

Manu

factur

ing

Constru

ction

Minin

g

8000

7000

6000

5000

4000

3000

2000

1000

0

C1

C2

Bar graph

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1.4.9. Bar chart

20001990198019601900

80

70

60

50

40

30

20

10

0

C1

C2

Bar graph


300

250

200

150

100

50

Accid

ents

0

C2

Bar graph

C1

C

hronic C

C

ancer

Diabetes

Heart

Kidney

Pneu

monia

Stroke

S

uicid

e

(b) Pareto graph

268.0

38.7

38.7

119.4

28.8

67.6

58.5

8.5

76.0

42.3

6.1

82.1

35.1

5.1

87.2

34.5

5.0

92.2

23.9

3.5

95.6

30.2

4.4

100.0

Percentage

Percent

Cum %

C1

Othe

r

Diabetes

Accid

ents

Pneu

monia

Heart

Cancer

Stroke C

700

600

500

400

300

200

1000

100

80

60

40

20

0

Percentage

Percent

Pareto graph

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1.4.13.

90807560

9

8

7

6

5

4

3

2

1

0

Histogram

C1

Frequency

1.4.15. (a) Stem and leafStem-and-leaf of C1 N = 20

Leaf Unit = 10

1 4 7

3 4 99

8 5 00011

10 5 22

10 5 4455

6 5 6667

2 5 9

1 6 0

(b) Histogram

600580560540520500480

5

4

3

2

1

0

C1

Frequency

Histogram

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(c) Pie chart

526

542

546

553

558

565

568

572

595

605

475

493

499502

503

506

510

517

525

Category

Pie chartspecies

EXERCISES 1.5

1.5.1. Mean is 165.6667 and standard deviation is 63.15397

1.5.3. Data is 3,3,5,13 and standard deviation is 4.760952

1.5.5. (a) lower quantiles is 80, median is 95, upper quantiles is 115 and inter quantile range

is 35. The lower limit of outliers is 27.5 and upper limit of outliers is 167.5.

(b) The box plot is

(c) Therefore there are no outliers.

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1.5.7.

li=1

fi(mi x) =l

i=1fi(mi)

li=1

x = nx nx = 0

1.5.9. (a) Mean is 33.105, variance is 177.0430 and range is 48.19.

(b) Lower quantile is 24.9225, median is 32 and upper quantiles is 42.985. The inter

quantile range is 18.0625. The lower limit of outliers is 2.17125 and upper limit ofoutliers is 70.07875. Therefore there are no outliers.

(c)

(d)

Histogram of y

Frequency

0

0

2

4

6

8

10 20 30 40 50 60y

1.5.11. (a) Mean is 110, standard deviation is 83.4847.

(b) 68%, 95%, 99.7%.

1.5.13. (a) Mean is 3.7433, variance is 3.501 and standard deviation is 1.871323.

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(b) Mean is 74.0625, median is 74, variance is 7.223892 and standard deviation is

2.68773.

(c)

The lower limit of outliers is 66 and upper limit of outliers is 82. Therefore we have nooutlier.

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Chapter2Basic Concepts from Probability Theo

EXERCISES 2.2

2.2.1. (a) S= {(R, R, R), (R, R, L), (R, L, R), (L, R, R), (R, L, L), (L, R, L), (L, L, R), (L, L, L)}(b) P= 78(c) P= 48(d) P= 38(e) P= 28

2.2.3. (a) P= 536(b) P= 536(c) P= 48(d) P= 38

2.2.5. PAB = {(H, H ), (H, T ), (T, H)} .2.2.7. (a) Probability space is

S= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}(b) P= 636(c) P= 136

2.2.9. (a) Probability space isS= {N, N, N, S, S}Nstands for normal and Sstands for spoiled.

(b) P= 35 24= 620(c) No more than one means no or just one.

P= 35 24+ 2 25 34= .92.2.11. P= p + 2q

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12 CHAPTER 2 Basic Concepts from Probability Theory

2.2.13. (a) SinceA Bthen letA Ac = B, we knowP(A) + P(Ac) = P(B)by the axiom 3. SinceP(Ac) 0 by axiom 1 we knowP(A) + P(Ac) P(A)thenP(A) P(B).

(b) Let C= AB, A= Aa C and B= Ba C, by axiom 3 we know P(AB)=P(Aa) + P(Ba) + P(C).

Since P(A) = P(Aa

)+ P(C) and P(B) = P(Ba

)+ P(C) by axiom 3 again, we knowP(Aa) = P(A) P(C) and P(Ba) = P(B) P(C). Plug them back in previous equa-tion P(A B) = P(Aa) + P(Ba) + P(C) and we get the following equation P(A B)=P(A) + P(B) P(C)=P(A) + P(B) P(A B). IfA B=then by axiom 1 we knowP(A B) = 0 and just plug in we complete the proof.

2.2.15. (a) From 2.2.13 we knowP (A B) = P(A) + P(B) P(A B)and from axiom 2 we knowP(A B) 1, we can see thatP(A) + P(B) P(A B) 1 and that complete the proofP(A) + P(B) 1 P(A B).

(b) From 2.2.13 we knowP(A1 A2)= P(A1) + P(A2) P(A1 A2). From axiom 1we knowP(A1 A2) 0 it meansP(A1 A2) 0 and we can get the following

inequalityP (A1 A2) = P(A1) + P(A2) P(A1 A2) P(A1) + P(A2).2.2.17. (a) P= .24 + .67 .09 = .82

(b) P= 1 .82 = .18(c) P= 1 .09 = .91(d) P= 1 .09 = .91(e) P= 1 .82 = .18

2.2.19. (a) P= .55(b) P= .3(c) P= .7

2.2.21. (a) P= 3

5 2

4+ 2

5 1

4= .4(b) P= 35 24+ 35 24= .6(c) P= 2 35 24+ 25 14= .7(d) P= 35 24= .3

2.2.23. Without loss of generality let us assumeAn is increasing sequence then A1 A2 . . .An . . .. We know that ifA1 A2 . . . An . . . then A1 A2 . . .An . . . =

i=1

Ai=

limn An. From the condition we know limn An=

i=1

Ai and if we take probability on

both sides then limn

P(An) = P

i

=1

Ai = P limnAn

EXERCISES 2.3

2.3.1. (a) 45

(b) 1

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Students Solutions Manual

(c) 10

(d) 5400

(e) 2520

2.3.3. 1024

2.3.5. 53130

2.3.7. 155117520

2.3.9. 440

2.3.11. (a) p = .4313 .44425 .46798 .5263 1 = .04719(b) p = .0001189(c) p = .4313 .21419 .10344 1 1 = .009557

2.3.13. 180

2.3.15.

(a) p

= 1 365

364

...(365

20

+1)

36520 = .4114(b) p = 1 .2936 = .7063(c) Ifn = 23 thenp = .4927

2.3.17. p = 1 .27778 .16667 = .55562.3.19. (a) 7776

(b) 3.954 1021(c) 5.36447 1028(d) 3.022285 1012

2.3.21. The question is asking when the cell does the splitting to produce a child. There will be a

cell with half of the chromosomes. According to this understanding we have(a) 223

(b)

239

223

= .097416

EXERCISES 2.4

2.4.1. (a) .999

(b) 13

2.4.3. (a) P(A

|B)

+P(Ac

|B)

= P(AB)

P(B)

+ P (AcB)

P(B)

= P(B|A)P(A)+P(B|Ac)P(Ac)

P(B)

= P(B)P(B)

=1

(b) (i) ifP (A|B) + P(A|Bc)= 1 then we knowP (A|Bc)= 1 P(A|B)= P(Ac|B)thatmeansA andB are symmetric in probability. But it is not always true.

(ii) ifP (A|B) + P(Ac|Bc) = 1 then we knowP (Ac|Bc) = 1 P(A|B) = P(Ac|B)That means B and Bcs conditional probability are same, which is same as A and B are

independent. But that is not always true.

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2.4.5. IfA andB are independent thenP (A B) = P(A) P(B)

(i) P(AcB) = P(B)P(AB) = P(B)P(A)P(B) = (1P(A))(P(B)) = P(Ac)P(B)then we knowAc andB are independent.

(ii) According to(i)

just switchA

andB

and we can prove it.(iii) P(Ac Bc)= P(Bc) P(A Bc)= P(Bc) P(A) P(Bc)= (1 P(A))(P(Bc))=P(Ac) P(Bc).

2.4.7. P(E|F ) = 113= 452= P(E)thenE and Fare independent2.4.9. .1948

2.4.11. (a) P= .031125(b) P= .06

2.4.13. .8

2.4.15. (a) P(a dime is selected) =12

i=2P(a dime is selected|boxi is selected)P(boxi is selected)

=12

i=2i

12 P(the sum of dies= i)

= 2(1)+3(2)+4(3)+5(4)+6(5)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)12(36)= .583333

(b) P(box 4 is selected|a penny is selected)

= P(box 4 is selected & a penny is selected)P(a penny is selected)

= (3/36)(8/12)1 P(a dime is selected)

= .133332.4.17. P= .65752.4.19. P= .609762.4.21. (a) P(Accident rate) = .25 .086 + .257 .044 + .347 .056 + .146 .098 = .066548

(b) P(gourp4|Accident) = .146.098.0665 = .2152.4.23. P= .166672.4.25. P(Working) = P(B, C) + P(A, B,notC) + P(A,notB, C)

= [1 P(notB) P(notC) P(notB,notC)] + P(A)P(B|notC)P(notC)+ P(A) [P(notB) P(notB,notC)]

= [1 0.1 0.05 0.75 0.05] + 0.85 0.25 0.05 + 0.85 [0.1 0.75 0.05]= 0.8875 + 0.010625 + 0.053125 = 0.95

2.4.27. (a) P(same type of blood) = 1840 1739+ 1640 1539+ 440 339+ 240 139= .358974(b) P(type isO|type isB) = 1839= .4615

2.4.29. LetE denoteA ends up with all the money when he starts withi.

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LetF denote A ends up with all the money when he starts with N i. ForA starts withN imeansBstarts withibecauseNis the total moneyAandBhas so if we gotP(E)thenP(F) = 1 P(E).LetHdenote the event that the first flip lands heads andp denote the probability to have

Hon the first flip.P(E) = P(E|H)P(H) + P(E|HC)P(HC )

This probability represents thatA gets a head and combined with the probability ifB win

the first coin.

Now we letP(E) = P(E|H )p + P(E|HC)(1 p) = Piand define this as the first round.New, given that the first flip lands heads, the situation after the first bet is thatA has i + 1units andB has N (i + 1)units.Since the successive flips are assumed to be independent with a common probabilityp of

heads, it follows that, from that point on, As probability of winning all the money is exactly

the same as if the game were just starting with A having an initial fortune ofi + 1 and B

having an initial fortune ofN (i + 1)Therefore,P (E|H) = Pi+1and P (E|HC) = Pi+1. Letq to be 1 p,

Pi= pPi+1 + qPi1 i = 1,2, . . . , N 1

By applying the condition thatP0= 0 andPN= 1

Pi= 1Pi= (p + q)Pi= pPi+1 + qPi1

Pi+1 Pi= q

p(Pi Pi1) i = 1,2, . . . , N 1

After plug ini

We got

P2 P1= q

p(P1 P0) =

q

pP1

P3 P2= q

p(P2 P1) =

q

p

2P1

PN PN1= q

p(PN1 PN2) =

q

p

N1P1

If qp= 1

ThenP2 P1= P1and P2= 2P1P3= 3P1

PN= NP1PN= 1

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Which meansP1= 1NThereforePi= iN

If qp 1, thenPN P1= 1 P1= P1

q

p

1 qp

N11

qp

= P1

qp q

p

N1

qp

1 = P1

qp

qp

N1 qp

+ 1 = P1

1

qp

N1 qp

P1= 1 qp

1 qp N

Add all equations, we got

PN

P1=

P1 q

p + q

p

2

+ + q

pN

1

Add firsti 1 of all equations, we got

Pi P1= P1

q

p+

q

p

2+ +

q

p

i1= P1

q

p

1

qp

i11 qp

Pi=

1 qp1

qp

N

1

qp

i1 qp

=

1

qp

i1

qp

N

Then if we start withN i, just replacep byq and i byN i.

Qi=

1

pq

Ni1

pq

N if p

q 1

Qi= N i

Nif

p

q= 1

Pi + Qi= 1

EXERCISES 2.5

2.5.1. (a) c = e(b) P= e(c) P= 1 e e 2e2

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2.5.3. F(x) =

0, where x 5.2, where 5 x 0.3, where 0 x 3.7, where 3

x

6

1, where 6 6

2.5.5. p(x) =

0, where x 1.2, where 1 x 3.6, where 3 x 9.2, where x 9

2.5.7. (a) c = 19(b) P= .7037

(c) F(x) = 0, where x 0

1

27 x

3

, where 0 x 31, where x 3

2.5.9. f(x) =

0, where x 02x

(1+x)2 , where x 02.5.11. p = .7013 .55809 = .1432

p = .7364

2.5.13. f(t) =

0, where t 0(t)1

e

(t) , where 0 t

EXERCISES 2.6

2.6.1. m(t) = 16

et + e2t + e3t + e4t + e5t + e6tVar(x) = 2.9167

2.6.3. (a) E(Y) = 3.6E(Y2) = 17.2E(Y3) = 95.3

VAR(Y ) = 4.24(b) My(t) = et .1 + .05 + e2t .25 + e5t .4 + e6t .2

2.6.5. E(X)

= x xP(X = x) =

n=12nP(X

=2n)

=

n=12n 1

2n

=

n=11

= 2.6.7. a = 12

b = 12.6.9. (a) E(c) = cf (x) = c

(b) E(cg(x)) = cg(x)f (x) = c g(x)f(x) = cE(g(x))

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(c) E

i

gi(x)

=

i

gi(x)fi(x) =

i

g(x)f(x) =

i

E(gi(x))

(d) V(ax + b)= E((ax + b) E(ax + b))2 = E(ax + b aE(x) b)2 = E(a(x E(x))2= a2V(x)

Plug inb = 0 get another one.2.6.11. E(X) = c 1 + 0 = c

V(X) = Ex2 (E(x))2 = c2 c2 = 0CDF isF (X) = (x c)where is indicator function

2.6.13. Mx(t) = e(et1)E(X) = Mx(0) = ete(e

t1)|t=0= E(X2) = Mx (0) = ete(e

t1) + (et)2e(et1)|t=0= + 2V(X) = E(X2) (E(x))2 = + 2 2 =

2.6.15. (a) p = 1x+1 letq = 1 p = xx+1 wherexstart from 0 to infinity means number of failures

before the 1st success. Therefore the total number of trails is x+

1.

E(x) =

x=0xp(1 p)x = p

x=0

xp(q)xq 1q= qp

x=0

x + 1

x

(q)x = qpp2 = q

p by nega-

tive binomial

Another way to prove is

E(x) ==

x=0xp(1 p)x =p

x=0

x(q)x1q = pq

x=0

d(q)x

dq

= pq

d

x=0(q)x

dq = pqd

11qdq = pq

1

(1 q)2= pq

p2= q

p

E(x2) =

x=0x2p(1 p)x =pq

x=0

x2(q)x1 = pq

x=0

d(xqx)

dq

= pqd

x=0

(xqx)

dq= pq

d

1p

x=0

(xpqx)

dq= pq

d

11q E(x)

dq

=pq

d 1

1qqp

dq =pq

d q

(1q)2dq =

pq1 2q + q2 + 2q 2q2

(1 q)4 =pq(1 q2)

p4 =pq pq3

p4

V(x) = E(x2) (E(x))2 = pq pq3

p4 q

2

p2= pq pq

3 p2q2p4

= pq pq2(q + p)

p4

= pq(1 q)p4

= p2q

p4 = q

p2

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(b) Mx(t) =

x=0extp(1 p)x = p

x=0

(etq)x = p1etq= p

1(1p)et

Whenetq 1 and that isetq 1 take ln both sides give us ln(et) ln 11p

That is whent ln(1 p).

2.6.17. E(x) = 10 x(x2)dx + 21 x6x2x232 dx + 32 x (x3)22 dx =18+ 1 + 38= 1.52.6.19. Mx(t) =

0 e

xt12 e

xdx + 0 ext12 exdx = 1(t+1)(t1)2.6.21. My(t)=

0 e

tyeydy= 0 etyydy= 1(t) 0 e(t)yd(t )y= 1(t) e(t)y|0 = 1

(t)= (t) t and 0 SinceMy(t)= Mx(t)= (t) and MGF uniquely define the PDF therefore we know thatxhas the same distribution as y and

g(x) =

ey, 0 and 0 x

0, otherwise

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Chapter3Additional Topics in Probabili

EXERCISES 3.2

3.2.1. (a) P(X = 7) = 107 (0.5)

7(0.5)107

= 120(0.5)7(0.5)3= 0.117

(b) P(X 7) = 1 P(8) P(9) P(10)= 1 0.044 0.010 0.001= 0.945

(c) P(X >0) = 1 P(0)= 1 0.001= 0.999

(d) E(X) = 10(0.5) = 5Var(X) = 10(0.5)(1 0.5) = 2.5

3.2.3. (a) P(Z >1.645) = 0.05, soz0= 1.645.(b) P(Z 1.645) = 0.95, soz0= 1.645.

3.2.5. (a) P(X 20) = P

Z 20 105

= P(Z 2) = 0.9772

(b) P(X >5) = PZ >5 10

5 = P(Z > 1) = 0.8413(c) P(12 X 15) = P

12 10

5 Z 15 10

5

= P(0.4 Z 1) = P(Z 1) P(Z

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22 CHAPTER 3 AdditionalTopics in Probability

(d) P(|X 12| 15) = P(15 X 12 15) = P(3 X 27)

= P3 10

5 Z 27 10

5

=P(

2.6

Z

3.4)

=P(Z

3.4)

P(Z 0 andx 0.

Since eachp(x) 0, then

p(x) xp(x) = x exx! = ex x

x!= e

1 + 1

1!+2

2!+

= e(e) = 1 here we apply Taylors expansion one.

This shows thatp(x) 0 and xp(x) = 1.3.2.13. The probability density function is given by

f(x) =

1

10, 0 x 10

0, otherwise

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Hence,

P(5 X 9) =9

5

1

10dx = 0.4.

Hence, there is a 40% chance that a piece chosen at random will be suitable for kitchen use.

3.2.15. The probability density function is given by

f(x) =

1

100, 0 x 100

0, otherwise

(a) P(60 X 80) = 8060 1100 dx = 0.2.(b) P(X >90) =

100

90

1

100dx = 0.1.

(c) There is a 20% chance that the efficiency is between 60 and 80 units; there is 10% chancethat the efficiency is greater than 90 units.

3.2.17. LetX= the failure time of the component. And X follows exponential distribution with rate0.05. Then the p.d.f. ofX is given by

f(x) = 0.05e0.05x, x >0.Hence,

R(10) = 1 F (10) = 1 10

0

0.05e0.05xdx = 1 1 e0.5 = e0.5 = 0.607.3.2.19. The uniform probability density function is given by

f(x) = 1, 0 x 1.

Hence,

P(0.5 X 0.65|X 0.75) = P(0.5 X 0.65 and X 0.75)P(X 0.75)

= P(0.5 X 0.65)P(X

0.75)

=

0.650.5

1dx

0.750 1dx

= 0.150.75

= 0.2

3.2.21. First, findz0such thatP(Z > z0) = 0.15.P(Z >1.036) = 0.15, soz0= 1.036.x0= 72 + 1.036 6 = 78.22

The minimum score that a student has to get to an A grade is 78.22.

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3.2.23. P(1.9 X 2.02) = P

1.9 1.960.04

Z 2.02 1.960.04

= P(1.5 Z 1.5) = 0.866

P(X 2.02) = 1 P(1.9 X 2.02) = 0.13413.4% of the balls manufactured by the company are defective.

3.2.25. (a) P(X >125) = P

Z > 125 11510

= P(Z >1) = 0.16

(b) P(X z2) = 0.5 andP(Z > z3) = 0.8

Using standard normal table, we can find thatz1= 0.842,z2= 0 andz3= 0.842.Then

y1= 0 + (0.842) 0.65= 0.5473 x1= exp(y1)= 0.58, similarly we can obtainx2= 1 andx3= 1.73.For the probability of surviving 0.2, 0.5 and 0.8 the experimenter should choose doses 0.58,

1 and 1.73, respectively.

3.2.29. (a) MX(t) = E(etX) =

0

etx 1

()x1ex/dx

= 1()

0

x1 exp1 t

x

dx

= 1()

0

1 tu1

eu

1 tduby lettingu = 1 t

xwith 1 t >0

= 1

()

1 t

0

u1 exp(

u)du

note that the integrand is the kernel density of(, 1)

= 1()

1 t

() 1

= (1 t)when t < 1

.

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(b) E(X) = MX(0) = (1 t)1()

t=0= , and

E(X2) = M(2)X (0) = ddt[(1 t)1()]

t=0

= (

+1)(1

t)2(

)2

t=0=(

+1)2

Then Var(X) = E(X2) E(X)2 = ( + 1)2 ()2 = 2.3.2.31. (a) First consider the following product

()() =

0

u1eudu

0

v1evdv

=

0

x2(1)ex2

2xdx

0

y2(1)ey2

2ydv by lettingu = x2 and v = y2

= 2

0

|x|21 e

x2

dx

2

0

|y|21 e

y2

dy noting that the integrands are

even functions

=

|x|21 ex2 dx

|y|21 ey2 dy

=

|x|21 |y|21 e(x2+y2)dxdy

Transforming to polar coordinates withx = r cos and y = r sin

()() =2

0

0

|r cos |21 |r sin |21 er2 rdrd

=

0

r2+22er2

rdr

20

(cos )21(sin )21 d

=1

2

0

s+1esds

4

/20

(cos )21(sin )21d

by lettings = r2

= ( + )2/20

(cos )21(sin )21d

= ( + )21

0

t1/2(1 t)1/2 12

t(1 t) dtby lettingt= cos2

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= ( + )1

0

t1(1 t)1dt

= ( + )B(, )

Hence, we have shown that

B(, ) = ()()( + ) .

(b) E(X) = 1B(, )

10

x x1(1 x)1dx = B( + 1, )B(, )

10

x(+1)1(1 x)1B( + 1, ) dx

= B( + 1, )B(, )

1 = ( + 1)()( + + 1)

( + )()()

= ()()

( + )( + )( + )()() =

+ , and

E(X2) = 1B(, )

10

x2 x1(1 x)1dx = B( + 2, )B(, )

10

x(+2)1(1 x)1B( + 2, ) dx

= B( + 2, )B(, )

1 = ( + 2)()( + + 2)

( + )()()

= ( + 1)()()( + )( + + 1)( + )

( + )()()

= ( + 1)( + )( + + 1) .

Then Var(X)

=E(X2)

E(X)2

=

( + 1)

( + )( + + 1)

+

2

=

( + )2

( + + 1).

3.2.33. In this case, the number of breakdowns per month can be assumed to have Poisson

distribution with mean 3.

(a) P(X= 1)= e3311! = 0.1494. There is a 14.94% chance that there will be just onenetwork breakdown during December.

(b) P(X 4) = 1 P(0) P(1) P(2) P(3) = 0.3528. There is a 35.28% chance thatthere will be at least 4 network breakdowns during December.

(c) P(X7)=7

x=0e33x

x! =0.9881. There is a 98.81% chance that there will be at most 7network breakdowns during December.

3.2.35. (a) P(1< X

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The probability that an acid solution made by this procedure will satisfactorily etch a

tray is 0.6442.

(b) P(1< X

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Thus, ifc= 1/4, then 11 11 f (x, y)dxdy= 1. And we also see thatf (x, y) 0 for all xandy. Hence,f (x, y)is a joint probability density function.

3.3.5. By definition, the marginal pdf ofXis given by the row sums, and the marginal pdf ofY is

obtained by the column sums. Hence,

xi 1 3 5 otherwisefX(xi) 0.6 0.3 0.1 0

yi 2 0 1 4 otherwisefY(yi) 0.4 0.3 0.1 0.2 0

3.3.7. FromExercise 3.3.5we can calculate the following.

P(X = 1|Y= 0) = P(X = 1, Y= 0)fY(0)

= 0.10.3

= 0.33.

3.3.9. (a) The marginal ofXis

fX(x) =2

x

f (x, y)dy =2

x

8

9xydy = 4

9(4x x3), 1 x 2.

(b) P(1.5< X 1) =1.75

1.5

2

x

8

9xydy

dx =

1.751.5

4

9

4x x3 dx

= 49 2x2

x4

4 1.75

1.5

= 0.2426.

3.3.11. Using the joint density inExercise 3.3.9we can obtain the joint mgf of(X, Y )as

M(X,Y )(t1, t2) = E(et1X+t2Y) =2

1

2x

et1x+t2y8

9xydydx

=2

1

8

9xet1x

2

x

et2yydy

dx =

21

8

9xet1x

K x

t2et2x + 1

t22et2x

dx

whereK = e2t2

t22(2t2 1)

= 89

K

21

xet1xdx 89t2

21

x2e(t1+t2)xdx + 89t22

21

xe(t1+t2)xdx

= 89

K

x

t1et1x 1

t21et1x

21

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89t2

x2

t1 + t2e(t1+t2)x 2x

(t1 + t2)2e(t1+t2)x + 2

(t1 + t2)3e(t1+t2)x

2

1

+ 89t2

2

x

t1+

t2e(t1+t2)x 1

(t1+

t2)2e(t1+t2)x

2

1

After simplification we then have

M(X,Y) (t1, t2) =t1 + 3t2 t21 3t22 4t1t2 + t21 t2 + 2t1t22+ t32

t22 (t1 + t2)3 et1+t2 + (2t2 1)(1 t1)

t21 t22

et1+2t2

+

t1 3t2 + 2t21+ 6t22+ 8t1t2 4t21 t2 + 8t1t22 4t32t22 (t1 + t2)3

+ (2t2 1)(2t1 1)t21 t

22

e2t1+2t2

3.3.13. (a) fX(x) = n

y=0f (x, y) = n

y=0

6xy

n(n + 1)(2n + 1)

2 = 36x2[n(n + 1)(2n + 1)]2

ny=0

y2

= 6x2

n(n + 1)(2n + 1) , x = 1,2, . . . , n.

fY(y) =n

x=0f (x, y) =

nx=0

6xy

n(n + 1)(2n + 1)2

= 36y2

[n(n + 1)(2n + 1)]2n

x=0x2

= 6y2

n(n + 1)(2n + 1) , y = 1,2, . . . , n.

Giveny = 1,2, . . . , n, we have

f (x|y) = f (x, y)fY(y)

=

6xy

n(n + 1)(2n + 1)

2

6y2

n(n + 1)(2n + 1)

= 6x2

n(n + 1)(2n + 1) , x = 1,2, . . . , n.

(b) Givenx = 1,2, . . . , n, we have

f (y|x) = f (x, y)fX(x)

=

6xy

n(n + 1)(2n + 1)2

6x2

n(n + 1)(2n + 1)

= 6y2

n(n + 1)(2n + 1) , y = 1,2, . . . , n.

3.3.15. (a) E(XY) = x,y

xy f (x, y) =3

x=1

3y=1 xy f (x, y) =

35

12 .

(b) E(X) = x,y

x f (x, y) =3

x=1

3y=1

x f (x, y) = 53

, and

E(Y) = x,y

y f (x, y) =3

x=1

3y=1

y f (x, y) = 116

.

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Then, Cov(X, Y ) = E(XY) E(X)E(Y) = 3512

53 11

6= 5

36.

(c) Var(X) =

x,y[x E(X)]2 f (x, y) =

3

x=13

y=1 x 53

2 f (x, y) = 5

9, and

Var(Y ) = x,y

[y E(Y)]2 f (x, y) = 3x=1

3y=1

y 116

2 f (x, y) = 2336

.

Then,XY= Cov(X, Y )

Var(X)Var(Y )= 5/36

(5/9)(23/36)= 0.233.

3.3.17. Assume thata and c are nonzero.

Cov(U, V ) = Cov(aX + b, cY+ d) = acCov(X, Y ),

Var(U) = Var(aX + b) = a2Var(X), and Var(V ) = Var(cY+ d) = c2Var(Y ).

Then, UV

=

Cov(U, V )

Var(U)Var(V ) =

acCov(X, Y )

a

2Var(X)c

2Var(Y ) =

ac

(ac)

2XY

= ac|ac| XY=

XY, if ac >0

XY, otherwise .

3.3.19. We fist state the famous CauchySchwarz inequality:

|E(XY)|

E(X2)E(Y2)and the equality holds if and only if there exists some constant

and, not both zero, such thatP ( |X|2 = |Y|2) = 1.Now, consider

|XY| = 1

Cov(X, Y )Var(X)Var(Y )

= 1 |Cov(X, Y )| =

Var(X)Var(Y )

|E(X X)(Y Y)| =

E(X X)2E(Y Y)2

By the CauchySchwarz inequality we have

P

|X X|2 = |Y Y|2 = 1

P(X X= K(Y Y)) = 1 for some constantK P(X = aY+ b) = 1 for some constantsa andb.

3.3.21. (a) First, we compute the marginal densities.

fX(x) = x

f (x, y)dy = x

eydy = ex, x 0, and

fY(y) =y

0

f (x, y)dx =y

0

eydx = yey, y 0.

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For giveny 0, we have the conditional density as

f (x|Y= y) = f (x, y)fY(y)

= ey

1

yey

= 1y

, 0 x y.

Then,(X|Y= y)followsUniform(0, y). Thus,E(X|Y= y) = y2

.

(b) E(XY) = y xxy f (x, y)dxdy = 0y

0 xyeydx

dy = 0 12 y3eydy = 3,

E(X) = x

x fX(x)dx =

0 xexdx = 1, and

E(Y) = y

y fY(y)dy =

0 y2eydy = 2.

Then, Cov(X, Y ) = E(XY) E(X)E(Y) = 3 1 2 = 1.(c) To check for independence ofX andY

fX(1)fY(1) = e2 = e1 = f (1, 1).

Hence,X and Yare not independent.

3.3.23. Let2 = Var(X) = Var(Y ). SinceX and Yare independent, we have Cov(X, Y )= E(XY) E(X)E(Y) = E(X)E(Y) E(X)(Y) = 0. Then, Cov(X, aX+Y ) = aCov(X, X)+Cov(X, Y ) =aVar(X)= a2, and Var(aX + Y )= a2Var(X) + Var(Y )= (a2 + 1)2. Thus, X, aX+Y=

Cov(X, aX + Y )Var(X)Var(aX + Y ) =

a22(a2 + 1)2

aa2 + 1

.

EXERCISES 3.4

3.4.1. The pdf ofX is fX(x) = 1a if 0 < x < aand zero otherwise.

FY(y) = P(Y < y) = P(cX + d < y) = P

X 0.

3.4.7. LetU= X + YandV= Y.ThenX = U V andY= V, and

J=

x

u

x

v

y

u

y

v

=

1 10 1 = 1.

Then the joint pdf ofUand Vis given by

fU,V(u, v) = fX,Y(u v, v) |J| = fX,Y(u v, v).

Thus, the pdf of U is given byfU(u) = fX,Y(u v, v)dv.

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3.4.9. (a) Here let g(x) = x

, and hence, g1(z) = z+ . Thus, ddz

g1(z) = . Also,

fX(x) = 12

e 12

x

2, < x < . Therefore, the pdf ofZ isfZ(z) = fX(g1(z))

ddz g1(z) = 12 e 12 z2 , < z < , which is the pdf ofN (0, 1).(b) The cdf ofUis given by

FU(u) = P(U u) = P

(X )22

u

= Pu X

u

= Pu Z u

=

u

u

12

e12 z

2dz = 2

u

0

12

e12 z

2dz, since the integrand is an even function.

Hence, the pdf ofUis fU(u)=

d

duFU(u)

= 2

2e

u2

1

2u = 1

2u

12 e

u2 , u >0 and

zero otherwise, which is the pdf of2(1).

3.4.11. Since the support of the pdf ofV isv > 0, then g(v)= 12 mv2 is a one-to-one function onthe support. Hence, g1(y)=

2ym

. Thus, dde

g1(y)= 12

m2y

2m

. Therefore, the pdf ofE is

given by

f(y) = fV(g1(y)) ddy g1(y)

= c 2ym e2ym

12my

= c

2ym3

e2y

m , y >0.

3.4.13. LetU=

X2 + Y2 andV= tan1

YX

. HereUis considered to be the radius and Vis the

angle. Hence, this is a polar transformation and hence is one-to-one.ThenX = Ucos V andY= Usin V, and

J=

x

u

x

v

y

u

y

v

=cos v u sin vsin v u cos v

= u cos2 v + u sin2 v = u.

Then the joint pdf ofUand Vis given by

fU,V(u, v) = fX,Y(uv, v) |J| = 1

22exp

1

22

u2 cos2 v + u2 sin2 v

u.

= u22

e u2

22 , u >0, 0 v 0.

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Apply the result in Exercise 3.4.14with = 2. We have the joint pdf ofU= XY2 and V= YasfU,V(u, v) = 12 e(u+v), v > 2u, v >0. Thus, the pdf ofUis given by

fU(u) = v

fU,V(u, v)dv

=

0

1

2e(u+v)dv = 1

2eu, u 0

2u

1

2e(u+v)dv = 1

2eu, u

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3.5.5. Apply Chebyshevs theorem we have

P

Xn

n p

2) = P

X100 22/

100

> 2 22/

100

P(Z >0) = 0.5, whereZ N(0, 1).3.5.13. First note thatE(Xi) = 1/2 and Var(Xi) = 1/12. Then, by CLT we know thatZn= Snn/2n/12 =

X1/21/12/

n

approximately followsN (0, 1)for largen.

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Chapter

4Sampling Distributio

EXERCISES 4.1

4.1.1. (a) There are 53 = 10 equally likely possible samples of size 3, so the probability for each

is1 10without replacement:

X M S

(2, 1, 0) 1 1 1(2, 1, 1) 2 3 1 21 3(2, 1, 2) 1 3 1 41 3(2,0,1) 1 3 0 21 3(2,0,2) 0 0 4(2,1,2) 1 3 1 41 3(1,0,1) 0 0 1(1,0,2) 1 3 0 21 3(1,1,2) 2 3 1 21 3(0,1,2) 1 1 1

(i)

X 1 2/3 1/3 0 1/3 2/3 1p(X) 1/10 1/10 2/10 2/10 2/10 1/10 1/10

(ii)

M 1 0 1p(M) 3/10 4/10 3/10

(iii)S 1

7 3 2

13 3

p(S) 3/10 4/10 1/10 2/10

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38 CHAPTER 4 Sampling Distributions

(iv)E

X = (1)1 10+ 2 31 10+ 1 31 10+ 0 1 10+ 1 32 10

+

23

1

10

+ (1)

1

10

= 0

E

X2 = (1)2 1 10+ 2 32 1 10+ 1 32 1 10+ 02 2 10

+

13

2 210

+

23

2 110

+ 12

1

10

= 1 3

Var

X = EX2 EX2 = 1

3 02 = 1

3

(b) We can get 53 = 125 samples of size 3 with replacement4.1.3. Population: {1,2,3}.p(x) = 1 3, forx in {1,2,3}

(a) = 1N

Ni=1

ci= 2, 2 = 1NN

i=1(ci )2 = 2 3

(b)

Sample X Sample X Sample X

(1,1,1) 1 (2,1,1) 11 3 (3,1,1) 12 3

(1,1,2) 11 3 (2,1,2) 12 3 (3,1,2) 2

(1,1,3) 12 3 (2,1,3) 2 (3,1,3) 21 3

(1,2,1) 11 3 (2,2,1) 12 3 (3,2,1) 2

(1,2,2) 12 3 (2,2,2) 2 (3,2,2) 21 3

(1,2,3) 2 (2,2,3) 21 3 (3,2,3) 22 3

(1,3,1) 12 3 (2,3,1) 2 (3,3,1) 21 3

(1,3,2) 2 (2,3,2) 21 3 (3,3,2) 22 3

(1,3,3) 21 3 (2,3,3) 22 3 (3,3,3) 3

X 1 11 3 12 3 2 21 3 22 3 3

p(X) 1/27 1/9 2/9 7/27 2/9 1/9 1/27

(c) E

X

=X

x p(x) = 2, E

X2

=X

x2 p(x) = 42 9, then VarX

= 2 9

4.1.5. Since

i

(xi x)2 =

i

x2i nx2, we have E(S)2 = 1n

i

EX2i nEX2nAssuming the sampling from a population with mean and variance2, we have

E

S2 = 1

nn

2 + 2

2

n+ 2

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= 2 2

n

=

n 1n

2 < 2 = E

S2

4.1.7. LetXbe the weight of sugarX N( = 5lb, = 0.2lb)

ThenX = 1n

ni=1

Xiis the mean weight, wheren = 15.

By Corollary 4.2.2,E(X) = , and Var(X) = 2n

. ThenX N(5, 0.22 15), and X50.22 15

= ZN(0, 12). Therefore, the probability requested is P(0.2< X 5< 0.2) = P(

15< Z

170) = P(Z >226) = 04.1.11. Let X be the time. X

N(

=95, 2

=102). Then X9510

=Z

N(0, 12). Therefore,

P( X< 85) = P( Z< 1) = 0.8413, or 84.13% of measurement times will fall below85 seconds.

4.1.13. According the information, = 215 and= 35.(a) Ifn = 55, we can assumeX N(, ), thenPX >230 = PZ > 2302153555

= 0.0007

(b) Ifn = 200, we can assumeX N(, ), thenPX >230 = PZ > 23021535200

= 0

(c) Ifn = 35, we can assumeX N(, ), thenPX >230 = PZ > 2302153535

= 0.0056(d) Increasing the sample size, decrement the probability

4.1.15. LetTbe the temperature.

Sincen=60, we assumeT N (98.6, 0.952). ThenT N (98.6, 0.952 60). Therefore, P (T99.1) = 0

EXERCISES 4.2

4.2.1. We have thatY 2(15)(a) We can see, for example in a table, thatP (Y 6.26) = 0.025. Theny0= 6.26(b) Choosing upper and lower tail area to 0.025, and since P(Y 27.5)= 0.975, and

P(Y 6.26)= 0.025, then P (a < Y < b )= 0.95, then b= 20.975,15 = 27.5, a=20.025,15= 6.26

(c) P(Y

22.307)=

1

P(Y 2) = 0.4232

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4.2.5. SinceX1, X2, . . . , X5are i.i.d.N (55, 223), thenY=5

i=1(Xi55)2

223 2(5)

(a) Since Z = Y n

X552223 , and

n

X552223 2(1), Zis Chi-square distributed with 4 degrees

of freedom andYis Chi-square distributed with 5 degrees of freedom

(b) Yes

(c) (i)P (0.62 Y 0.76) = 0.0075 (ii)P (0.77 Z 0.95) = 0.0251

4.2.7. Since the random sample comes from a normal distribution, (n1)S2

2 2(n1).

Setting the upper and lower tail area equal to 0.05, even this is not the only choices,and using

a Chi-square table withn 1 = 14 degrees of freedom, we have (n1)b2

= 20.95,14= 23.68,and (n1)a

2 = 20.05,14= 6.57. Then, with= 1.41,b = 3.36, anda = 0.93

4.2.9. SinceT t8(a) P(T 2.896) = 0.99(b) P(T

1.860)

=0.05

(c) Sincet-distribution is symmetric, we finda such thatP(T > a) = 0.012 . Thena = 3.35

4.2.11. According with the information,=11.4, n=20, y= 11.5, ands= 2, thent= ys

n=

0.224. The degrees of freedom are n 1 = 19, so the critic value is 1.328 at = 0.05-level.Then, the data tend to agree with the psychologist claims.

4.2.13. If X 2(v), then X

= v2 , = 2

, then E(X)= v2 (2)= v and Var(X)= v2 (2)2 = 2v

4.2.15. IfX1, X2, . . . , Xnis fromN (, 2)

then, by Theorem 4.2.8,

(n

1)S2

2 is from2

(n1)then, by Exercise 4.2.13, Var

(n1)S2

2

= 2(n 1)

Since Var(aX) = a2Var(X), (n1)24

Var(S2) = 2(n 1)Simplifying after multiplying by

4

(n1)2 , we obtain Var(S2) = 24

n1

4.2.17. IfX and Yare independent random variables from an exponential distribution with com-

mon parameter = 1, then using 4.2.16 with n= 1, 2X 2(2) and 2Y 2(2) thenXY= 2X2Y F (2, 2)

4.2.19. IfX

F (9,12)

(a) P(X 3.87) = 0.9838(b) P(X 0.196) = 0.01006(c) F0.975(9,12) = 0.025 thenF0.975= 3.4358.

0.025 = P(X < F0.975) = P

1X

> 1F0.975

, where 1

X F (12,9)

Then 1F0.975

= 3.8682 andF0.975= 0.258518. Thus,a = 0.2585, b = 3.4358

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4.2.21. IfX F (n1, n2)the PDF is given by

f(x) =

[(n1 + n2)/2](n1/2)(n2/2)

n1

n2

n1

n2x

n121

1 + n1n2

x

(n1+n2)/2, 0< x <

0, otherwise

Then

EX =

0

x[(n1 + n2)/2]

(n1/2)(n2/2)

n1

n2

n1

n2x

n121

1 + n1n2

x

(n1+n2)/2dx

= [(n1 + n2)/2](n1/2)(n2/2)

n1

n2

n1

n2

n121

0

xn12

1 + n1

n2x

(n1+n2)/2dx

Let y

=1

1

+ n1n2

x1

then x

= n1n2

1y(1

y)1 and dx

= n1n2

1(1

y)2dy and

limx

1

1 + n1

n2x1= 1, then

EX = [(n1+n2)/2](n1/2)(n2/2)

n1n2

n22

n2n1

n22+1 1

0

yn22 (1 y)n12 dy, which converges forn1 >2.

For > 0, > 01

0

y1(1 y)1dy = ()()(+) , where () =

0 x

1exdx with the

property() = ( 1)( 1).

ThenEX = [(n1 + n2)/2](n1/2)(n2/2)

n1

n2

n22

n2

n1

n22+1 (n1/2 + 1)(n2/2 1)

[(n1 + n2)/2]

= n2n1

[(n1 + n2)/2](n1/2)(n2/2 1)(n2/2 1)

n12

(n1/2)(n2/2 1)[(n1 + n2)/2]

= n2n2 2

, n2 >2

Similarly,

EX2 = [(n1 + n2)/2](n1/2)(n2/2)

n1

n2

n12

n2

n1

n22+2

10

yn12 (1 y)

n223dy, which converges forn2 >4

= [(n1 + n2)/2](n1/2)(n2/2

1)(n2/2

2)(n2/2

2)

n2

n12 (n1/2 + 1)(n1/2)(n1/2)(n2/2 2)

[(n1

+n2)/2

]=

n2

n1

2 n1(n1 + 2)(n2 2)(n2 4)

, n2 >4.

Now, Var(X) = EX2 (EX)2. Therefore,

EX = n2n2 2

, n2 >2 and Var X =n22(2n1 + 2n2 4)

n1(n2 2)2(n2 4)

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4.2.23. IfX1, X2, . . . , Xn1 is a random sample from a normal population with mean 1 and vari-

ance2 and ifY1, Y2, . . . , Yn2 is a random sample from an independent normal population

with mean2 and variance 2, then X N

1,

2

n1

, Y N

2,

2

n2

,

(n1 1)S212

2(n11), and (n

2 1)S2

22

2(n21).

ThenX Y N

1 2, 2n1 + 2

n2

and

(n11)S212

+ (n21)S222

2(n1+n22)then XY(12)

2n1

+ 2n2

N(0, 12)and (n11)S21

2 + (n21)S22

2 2(n1+n22)

Then, since the samples are independent, we have by definition that

X Y (1 2)2

n1+

2

n2

(n1 1)S212

+ (n2 1)S22

2

[n1 + n2 2]

T(n1+n22)

This after simplification becomes:

X Y (1 2)(n1 1)S21+ (n2 1)S22

n1 + n2 2

1

n1+ 1

n2

T(n1+n22) Q.E.D.

4.2.25. IfX 2(v)withv >0, then the pdf ofX is given by

f(x) =

1

(v/2)2v/2ex/2xv/21, 0 < x <

0, x 0

Then, by definition of MGF,

MX(t) = 1

(v/2)2v/2

0

ex(t1/2)xv/21dx =

1

1 2tv/2

0

ewwv/21(v/2)

dw

= (1 2t)v/2, t 0, then the cumulative distribution of1is

F1 (t) =t

0

1

10ex/10dx = ex/10

t

0= 1 et/10.

Let Y represent the life length of the system, then Y=min(1, 2) and FY(y) = 1[1 Fi (y)]2, then the pdf ofYis of the formfY(y) = 2fi (y)[1 Fi (y)] and is given by

fy(y) =

1

5ey/5, 0 < y <

0, otherwise

4.3.3. X1, X2take values 0, 1;X3take values 1,2, 3, andY1= min {X1, X2, X3}Since the values ofX1, X2are less or equal to the values forX3, Y1take values 0, 1

Since the values ofX3are greater than the values forX1, X2, thenY3= max{X1, X2, X3}take values 1,2, 3

SinceY1 Y2 Y3, Y2take values 0, 14.3.5. LetX1, X2, . . . , Xn be a random sample from exponential distribution with mean , then

the common pdf is given byf(x) = 1

ex , ifx >0

Using Theorem 4.3.2, the pdf of thek-th order statistic is given by

fk(y) = fYk (y) = n!

(k 1)!(n k)! f(y)(F(y))k1(1 F(y))nk, whereF(y) = 1 e y

thenfk(y) = n!(k1)!(nk)! f(y)

1 e yk1

ey

nk

Then, the pdf ofY

1isf

1(y)

=nf (y) e

yn1 = ne

ny

, which is the pdf of an exponentialdistribution with mean n

, and the pdf ofYn is

fn(y) = nf (y)[F(y)]n1

= n

ey

1 e ny

n1

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4.3.7. X1, . . . , Xna random sample are i.i.d with pdff (x) = 12 , 0 x 2thenF(x) = x0 12 dx = x2 , if 0 x 2

thenF(x) = 1, x >2

x

2 , 0 x 20, x 10) = 1 P(Yn 10)

The CDF Fn(y)ofYnis [F(y)]n, whereF(y)is the cdf ofXevaluated inyThen,P (Yn

y)

=Fn(y)

= [F(y)

]n

= [P(X

y)

]n

andP (X y) = P

Z y102

ThenP (Yn y) =

P

Z y102n

ThenP(Yn > y) = 1 P(Yn < y)= 1

P

Z y102n

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ThereforeP (Yn >10) = 1 [P(Z 0)]n= 1 (0.5)n

4.3.11. X1, . . . , Xnis a random sample from Beta(x = 2, = 3)The joint pdf ofY1and Yn, according Theorem 4.3.3, is given by

fY1,Yn (x, y) = n!

(1 1)!(n 1 1)!(n n)! [F(x)]11[F(y) F(x)]n11[1 F(y)]nnf(x)f(y)

= n(n 1)[F(y) F(x)]n2f(x)f(y), if x < y

SinceXi Beta(X = 21 = 3)fori = 1,2, . . . , n, the pdf is

f(x) = ( + )()()

1(1 x)1, x [0, 1]

and, the DF is

F(x) =

0, x 0( + )()()

x0 t

1(1 t)1dt, 0 x 1

1, x 1In our case,

f(x) = (5)(2)(3)

x21(1 x)31 = 4!1!2! x(1 x)

2 = 12x(1 x)2, if 0 x 1

and

F(x) =

x

0

12t(1 t)2

dt= 12x2

2 2x3

3 +x4

4

, if 0 x 1

Then, the joint pdffY1,Yn (x, y), using the 4.3.3, is given by

fY1,Yn (x, y) = n(n 1)

12

y2

2 2y

3

3 + y

4

4

10

x2

2 2x

3

3 + x

4

4

n212x(1 x)212y(1 y)2

= 12nn(n 1)

1

2

y2 x2 2

3

y3 x3+ 1

4

y4 x4n2 xy(1 x)2(1 y)2,

if 0 x y 1, andfY1,Yn (x, y) = 0, otherwise

EXERCISES 4.4

4.4.1. X1, X2, . . . , Xn, wheren = 150, = 8, 2 = 4, thenx= 8 and2x= 4

By Theorem 4.4.1: limn P

Z X x

x

= 1

2k

z e

2 2du

thenP (7.5< X

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4.4.3. LetTbe the time spent by a customer coming to certain gas station to fill up gas

SupposeT1, T2, . . . , Tn are independent random variables, with t= 3 minutes, 2t = 1.5minutes, andn = 75

ThenY

=

n

i=1

Tiis the total time spent by then customers

Since,Y= 3 hours = 180 minutes,P(Y

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4.4.13. SIDS occurs between the ages 28 days and one year

Rate of death due to SIDS is 0.0013 per year.

Randon sample of 5000 infants between the ages 28 day and are your

LetX be the number of SIDS related deaths

p = 0.00103,n = 5000ThenX Bin(n = 5000, p = 0.00103)The probability requested is

P(X >10) P

Z >10 np 0.5

np(1 p)

= 0.0274

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50 CHAPTER 5 Point Estimation

5.2.11. We have

E(X) = = xVariance(2) = 1

n

Xi X

2

2

= 1

n x2i x2

=

1n

Xi

2

Xin

2The method of moments estimator for i s given by T (X1, . . . . . . .Xn) =

1n

Xi

2

Xin

2

5.2.13. The method of moments estimator for = T(X1, . . . . . . .Xn) = E(X) = XT(X1, . . . . . . .Xn) = XSince and2 both are unknown.2 = 1

n

Xi X

2This implies2

= (n1).1

n.(n

1) Xi

X2,2 = n1n s2Lets2 = n1

n s2

method of moments estimator for2 is given by:

s2 = 1n

ni=1

Xi X

2

EXERCISES 5.3

5.3.1. f(x) = nx

px(1 p)nx

L(p, x1, x2, . . . .xn) = log

n

i=1n

xi

+

Xi logp + n

Xi

log(1 p)

plogL(p, X1, X2, . . . .Xn) =

Xi

p+ n

Xi

1 p (1)

plogL(p, X1, X2, . . . .Xn) =

Xi

p n

Xi

1 p

For maximum likelihood estimator ofp

plogL(p, X1, X2, . . . .Xn) = 0

Xi

p

n

Xi

1 p

=0

(1 p)Xi pn Xi = 0. This implies Xi= pXi + n Xip =

Xi

n ,

p = X.Hence, MLE ofp = p = X.By invariance propertyq = 1 p is MLE ofq.

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5.3.3. f(x) = 1

ex impliesL() = 1

ne

Xi

ln L() = n ln

Xi

Now taking the derivatives with respect to and and setting both equal to zero, we have

ln L = n + Xi2 = 0n+Xi= 0.= XFrom the given data:

MLE ofis given by= X = 1+2++7+214 . = 6.07.5.3.5. Here pdf ofX is given by

f(x) =

2x2

ex2

3 if, x >0

0, otherwise

(5.1)

L(, X1, . . . . . . . .Xn) = 2(2)2n

i=1 Xie x2

2

ln L = ln 2 2 ln +ni=1 ln Xi ni=1 Xi22

ln L = 2n

+ 2

3

ni=1 X

2i

= 0 implies, 2n

+ 2

3

ni=1 X

2i= 0

n2 +

X2 = 0

2

= X2i

n

=

X2in

.

5.3.7.

f(x) =

x1e x

if, x 00, otherwise.

L(, , X) = nn i=1 nX

1i e

Xi

L(, , x) = n ln ln + ( 1) ln Xi Xi 2

L(, , x) = n

+ni=1 ln Xi n ln Xi ln Xi

L(, , x) = n Xi (a)1

L(, , x) = n+ (1) Xi

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For maximum likelihood estimator of:

ln L = 0. This implies;n+ni=1 ln Xi n ln Xi ln Xi = 0

similarly, n+ (1) Xi

= 0 n+ (1)

Xi

n

Xi

= 0 solving for we get =

Xin

+1

Hence,

n

2+

Xi

Xi

1 Xin

+1

ln

Xi

ln

Xin

+1

There is no closed form solution for and . In this case, one can use numerical methodssuch as Newton-Raphson method to solve for, and then with this value to solve for.

5.3.9. f(x) = (2)()2

[x(1 x)](1), 0ln L(, x) = n ln (2) ln ()2+ ( 1)ni=1 ln(Xi 1)

ln L(, x) = n

2(2)(2) 2

()2(0)2

+ni=1 ln Xi(Xi 1).

5.3.13.

f(x) =

13+2 if, 0 x 3+ 20, otherwise.

This impliesL(, x) = 1(3+2)2 for 0 x 3+ 2

When 3+2 max(Xi), the likelihood is 1(3+2)2 . Which is positive and decreasing functionof(forfixedn). However, for < max(Xi)23 , thelikelihood drops to 0, creating discontinuityat point max(Xi)23 .Hence we will not be able to find the derivative. The MLE is the largest order statistic=max(Xi)2

3 = Xn.5.3.15. HereX N(, 2)

ln L(, , x) = n2

ln 2 n2

ln 2 n

i=1

(xi )2

22

L

=

ni=1

(Xi )

Similarly L2

= n22

+ 122

ni=1(Xi )2

For maximum likelihood estimates of and ;

(Xi ) = 0 implies = X

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Similarly, for2,

n22

+ 122

ni=1

(Xi )2 = 0

=

(Xi X)n

.

5.3.17. f(x) = 1

e x

It is given that the reliabilityR(x)= 1 F(x). This implies F(x)= f(x). Hence F(x)= 1

2e

x .

Thus,R(x) = 1 F(x)and

L(x, ) =n

i=1

1 + 1

2e

Xi

.

EXERCISES 5.4

5.4.1. E(X) = E

Xin

i.e E(X) = 1

n

E(Xi). Where, E(X) =

xe(x)dx. By integration by

partsE(X) = (1 + ). ThusEX = 1n

(1 + ). This impliesE(X) = 1 + .

5.4.3. Sample standard deviations =

1n1

(xi X)2

E(s) = E

1

n 1

(Xi + X)2

E(s) = E

1

n

(xi )2 (X )2

E(s) =

1

n

E(xi )2 E(X )2

E(s) =

1

n

2

2

n

5.4.5. LetY

=C1X1

+C2X2

+ +CnXn.

For an unbiased estimate, we need to have E(C1E(X1) + + CnE(Xn)) = That is, C1 + Cn = Which is possible if and only ifCis= 1n for all i= 1, 2 . . . . . . . . . n . This implies 1n +

1n

+ + 1n

=

nn= i.e = . Verified.

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5.4.7. Xi U(0.).Yn= maxX1, . . . . . Xn= Ynis the MLE of.(a) By method of moment estimatorE(X)

=X

= +0

2 . This implies

=2X.

Hence the method of moment estimator= 2X(b)E() = E(Yn)E() = E{max(X1, . . . Xn)}E() = n

n+1E() = E(2X).E() = 2.E(/2). That is, E() = . Hence, 2 ia an unbiased estimate of .(c)E(3) = n+1n E()This impliesE(3) = .is an unbiased estimate of.

5.4.9. Here,Xi N(, 2)

f(x) = 122

exp

(x )

2

22

We have E() = E(X) = is an unbiased estimate for. By definition, the unbiased estimatethat minimizes themean square error is called the minimum variance unbiased estimate (MVUE) of.

MSE() = E( )2

That is MSE()= var(). Minimizing the MSE implies that the minimization of Var().

is the MVUE for.

5.4.11. E(M) = E(X) = . Thus, sample median is an unbiased estimate of population mean .Now, Var(X) = 1

n

(X )2

Var(M) = 1n

(M )2

Where Var(X) VarM5.4.13.

f(X) = 1

2exp

|X|

for < x <

The likelihood function is given by:

f (X1, X2, . . . . . . . . Xn, ) = 12n

1n

exp |Xi|

.

Takeg |Xi|, = 1nexp |Xi| andh(X1, X2 . . . . . . . Xn) = 12n

|Xi| is sufficient for.

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5.4.21. The likelihood function is given by

f(x1, . . . . . . xn) =

nn

i=1(xi)1 for 0< x 00 otherwise

LetU= (X1, X2, . . . . . . . . Xn)theng(x1, . . . . . . . Xn, ) = n

ni=1 x

1i andh(x1, . . . . . . xn) = 1. Therefore U is sufficient for.

5.4.23. The likelihood function is given by

f (x1, . . . . . . . . xn) = 2n

n

n

i=1Xi

e

x2

i

Letg

x2i, = 2n

ne

x2

i andh(x1, . . . . . . . . xn) =

ni=1 xi

Hence, x2i is sufficient for the parameter.

EXERCISES 5.5

5.5.1. ln L(p, x) = lnn

i=1

nxi

+ xi ln pni=1(n xi) ln(1 p) For MLE (1 p) xi p

(n xi) = 0. This implies p =

xin2

. Suppose Yn =

xin

. Thus p = Ynn

. Where

E

Ynn

= 1

nE(Yn)= 1n2 nnp=p.pis an unbiased estimate ofp. Similarly Varp=Var

Ynn=

Var

xin2

This implies Varp = nn4

np(1 p). Thus Varp 0 as n . Ynn

is consistent.

5.5.3. E(Xi) = i, E(X2i) = 2and E(Xi) = 4.E(s2) =

1n

E

Xi + X2This impliesE(s2) = n22

n . That isE(s2) = (n1)2

n .

S2 is an biased estimator of2

HereS2 = n1n

S2

Var(S2) = n1n2

2 0 as n .Where (n1)s

22

2(n1).Thus, Var

(n1)s2

2

= 2(n 1). This implies (n1)2

4 var(S2) = 2(n 1).

Finally, Var(S2) = 2n1

2.

Moreover, Bias(s2) = E(s2) 2. This implies2n

0 as n .Thuss2 is an unbiased estimate of2.

5.5.5. Here E(x) = and var(x)= 2 (For exponential distribution). Now E(X = E) andVar() = 2

n 0 as n . X is an unbiased estimate of.

5.5.7. Here, ln(, X) = n ln + 1

ni=1 ln(Xi).

Differentiating above equation and equating to zero we get =n

i=1ln Xin

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5.5.9. Here, var(1) = 112n and Var(2) = 112 .Thus the efficiency of2relative to 1is (2,1) = 1n 1

Similarly,e

2,3 = n+23 >1 ifn >1.

2is efficient than 3ifn >1.

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5.5.17. It can be easily verified thatE

1 = , E2 = and E3 =

Similarly, Var

1 = 3181 2, Var2 = 6n1725(n3) 2.

Var

3

= 2

n.

Now the corresponding efficiencies are given bye 2,1=31775(n3)81

(6

n17

), e 3,1 =

31n

81

.

e

3,2 = (6n17)n25(n3) .

5.5.21. The ratio of the joint density function of two sample points is given by;

L(x1, . . . . xn)

L(y1, . . . yn)= exp

122

ni=1

X2in

i=1Y2i

2

ni=1

Xi n

i=1Yi

.

For this ratio to be free of and 2, we must haven

i=1 Xi=n

i=1 Yi andn

i=1 X2i =n

i=1 Y2i .

Thusn

i=1 Xiandn

i=1 X2i are jointly minimal sufficient statistics for and

2. SinceXis

unbiased estimate forands2 is an unbiased estimate for2. The estimators are functions

of the minimal sufficient statistics. This implies theXand s2 are MVUES for and2.

5.5.23. The ratio of joint density function at two sample points, we have

L(x1, . . . . xn)

L(y1, . . . yn)=n

i=1(Yi)ni=1(Xi)!

Xi

Yi .

For the ratio to be free of we must have

Xi

Yi= 0. Thus

Xi, form the minimal

sufficient statistics for.

5.5.25.

L(x1, . . . . xn)

L(y1, . . . yn) =eXiYi

For the ratio to be independent of, we need to have

Xi=

Yi. Thus

Xiis minimal

sufficient for. NowE

Xi = nXis UMVUE, by RaoBlackwells theorem.

5.5.27.

L(x1, . . . . xn)

L(y1, . . . yn)=n

i=1(Yi)ni=1(Yi)!

e

X2i

Y2i .

The ratio to be free of, we must have

X2i

Y2i =0. Therefore

X2i is MVUE for.

Moreovers2 is an unbiased estimator for2.

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60 CHAPTER 6 Interval Estimation

p

1

21/2<

2

(n1)s2 < 12/2

= 1

p

(n1)s2

2/2< 2 5So the given data can be approximated as a normal distribution.

Here 1 = 0.98 = 0.02/2 = 0.01z/2= z0.01= 2.325

Thus the 98% confidence interval is given byp z/2

p(1p)

n

=

925 2.325

925 1625

50

= (0.202, 0.518)

6.2.7. n = 50

x = 11.4= 4.51 = 0.95 = 0.05z/2= 1.9695% confidence interval is

x z/2 n

=

11.4 1.96

4.550

= (10.153, 12.647)

6.2.9. n = 400p = 0.3np = 120> 5n(1 p) = 280> 595% confidence interval is given by

p z/2

p(1p)n

=

0.3 1.96

0.30.7400

= (0.255,0.345)

6.2.11. Proportion of defectionp = 40500= 2251 = 0.9/2 = 0.05z/2= 1.64590% confidence interval is given by

225 1.645

225 2325500

= (0.06, 0.1)

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6.2.13. x N(, 16)p(x 2< < x + 2) = 0.95z/2

n= 2

n = (z/22 )2 = 1.9642 2 = 15.37 16

6.2.15. n = 425p = 0.45np >425 0.45> 5n(1 p) = 425 0.55> 595% confidence interval is given by

p z/2

p(1p)n

=

0.45 1.96

0.450.55425

= (0.403, 0.497)

For 98% confidence interval

1 = 0.98 = 0.02z/2= 2.335

0.45 2.335

0.450.55425

= (0.394, 0.506)

6.2.19. p = 52601 = 0.95/2

=0.025

z/2= 1.96The 95% confidence interval is given by

p z/2

p(1p)n

=

5260 1.96

5260 860

60

= (0.781, 0.953)

6.2.21. = 35E = 15E = z/2 nn

= z/2

E 2 = 1.9635

15 2 = 20.92 216.2.23. x = 12.07

= 12.911 = 0.98/2 = 0.01z/2= 2.335

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98% confidence interval for mean is given byx z/2 n

=

12.07 2.335 1.9135

= (11.32, 12.82)

EXERCISES 6.3

6.3.1. (a) When standard deviation is not given and there is not enough sample size, we use

t-distribution.

(b) As differences decreases the sample sizen increases which means we are closing in on

the true parameter value of.

(c) The data are normally distributed, and the values ofx and thesamplestandard deviation

are known.

6.3.3. x = 20s = 4

1 = 0.95(a)

x t/2,4 sn

=

20 t0.025,4 45

(b)

20 t0.025,9 410

(c)

20 t0.025,19 420

6.3.5. x = 2.22s = 0.005n

=26

98% confidence interval for isx t/2,25 sn

=2.22 2.485 1.67

26

= (3.03,1.41)

6.3.7. x = 0.905s = 1.671 = 0.98n = 1098% confidence interval for is

0.905 t0.025,90.00510

6.3.9. Similar to 6.3.8

6.3.11. x = 410.93s = 312.87

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95% confidence interval for isx t/2,14 sn

=

410.93 2.145 312.8715

= (237.65, 584.21)

6.3.13. x

=3.12

s = 1.04n = 1799% confidence interval foris

x t/2,4 sn

=

3.12 2.921

1.0417

= (2.40, 3.84)

6.3.15. x = 3.85s = 4.55n = 2098% confidence interval foris

x t/2,19 sn

=

3.85 2.5394.5520

= (2.64, 5.06)

6.3.17. x = 148.18s = 1.91n = 1095% confidence interval foris

x t/2,9 sn

=

148.18 2.262

1.9110

= (147.19, 149.17)

EXERCISES 6.4

6.4.1. x = 2.2s = 1.421 = 0.90 = 0.10n = 2090% confidence interval for2 is given by

(n1)s22/2,19

, (n1)s2

21

/2,19 = 19(1.42)2

32.85 ,19(1.42)2

8.90655 = (1.1663, 4.3015)6.4.3. x = 60.908

s2 = 12.661 = 0.99 = 0.01n = 10

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99% confidence interval for2 is given by(n1)s2

2/2,9, (n1)s

2

21/2,9

= 912.6623.58 , 912.661.73 = (4.8321, 65.8613)

6.4.5. x = 2.27

s2 = 1.021 = 0.99 = 0.01n = 1899% confidence interval for2 is given by

(n1)s22/2,9

, (n1)s2

21/2,9

= 171.0227.58 , 171.025.69 = (0.6287, 3.0475)

6.4.9. From excel or by calculation, sample variance s2 = 148.44, sample mean x = 97.24, n = 2599% confidence interval for population variance is given by

(n1)s22/2,24

, (n1)s2

21/2,24

= 24148.4436.41 , 24148.449.886 = (97.8456, 360.3642)

6.4.11. x = 13.95s2 = 495.0851 = 0.98 = 0.02n = 25

98% confidence interval for2

is given by(n1)s22/2,24

, (n1)s2

21/2,24

= 24495.08542.97 , 24495.08510.85 = (276.5194, 1095.1189)

EXERCISES 6.5

6.5.1. For procedure I,x1= 98.4,s21= 235.6,n1= 10For procedure II, x2= 95.4,s22= 87.15,n2= 10 = 0.02/2 = 0.01z/2= 2.98598% confidence interval for difference of mean is

x1 x2 z/2

s21n1

+ s22n2

=

98.4 95.4 2.985

235.610 + 87.1510

= (13.9580, 19.9580)

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6.5.3. x1= 16.0, s1= 5.6, n1= 42x2= 10.6, s2= 7.9, n2= 45 = 0.01/

2 = 0.005z/2= 2.575

99% confidence interval for difference of mean is

x1 x2 z/2

s21n1

+ s22n2

=

16.0 10.6 2.575

(5.6)2

42 + (7.9)2

45

= (1.6388, 9.1612)

6.5.5. x1= 58, 550, s1= 4, 000, n1= 25x2= 53, 700, s2= 3, 200, n2= 23

Since21= 22 but unknown, we can use pooled estimatorS2p= (n11)s

21+(n21)s22

(n1+n22)

S2p= 24(4,000)2+22(3,200)246

Sp= 3639.398The 90% confidence interval is

x1 x2 t/2,(n1+n22) Sp

1n1

+ 1n2

= 58, 550 53, 700 2.326 3639.398 125+ 123 = (2404, 7296)6.5.7. x1= 28.4, s1= 4.1, n1= 40

x2= 25.6, s2= 4.5, n2= 32(a) MLE of1 2is given by(x1 x2)(b) 99% confidence interval for1 2is

x1 x2 z/2

s21n1

+ s22n2

=

28.4 25.6 2.565

(4.1)2

40 + (4.5)2

32

=(0.1678, 5.4322)

6.5.9. x1= 148, 822, s1= 21, 000, n1= 100x2= 155, 908, s2= 23, 000, n2= 1501 = 0.98 = 0.02

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/2 = 0.01z/2= 2.57598% confidence interval for difference of mean is given by

x1 x2 z/2

s21n1

+ s22n2

=

148, 822 155, 908 2.2

(21,000)2

100 + (23,000)2

150

= (508, 13, 664)6.5.11. x1= 35.18, s21= 19.76, n1= 11

x2= 38.76, s22= 12.69, n2= 131 = 0.9 = 0.1

90% confidence interval for

21

22 is given bys21s22

1Fn11,n21,1/2

, s21s22

1Fn11,n21,/2

=

19.7612.69

1F10,12,0.95

, 19.7612.691

F10,12,0.05

= 19.7612.69 12.75 , 19.7612.69 2.91 = (0.5662, 4.5313)6.5.13. x1= 68.91, s21= 287.17, n1= 12

x2= 80.66, s22= 117.87, n2= 121 = 0.95 = 0.05/2 = 0.025(a) 95% confidence interval of difference of mean is

x1 x2 z/2

s21n1

+ s22

n2

=

68.91 80.66 1.96

281.1712 + 117.8712

= (23.0525, 0.4475)

(b) 95% confidence interval for 2122

is given by

s21s22

1

Fn11,n21,1/2 ,

s21

s22

1

Fn11,n21,/2 = 281.17117.87 1F11,11,0.975 , 281.17117.87 1F11,12,0.025 = 281.17118.87 13.48 , 281.17118.87 3.48 = (0.6855, 8.3055)

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Chapter7Hypothesis Testin

EXERCISES 7.1

7.1.1. (a) H0: = 0H1: > 0

(b) H0: = 0H1: >1.20

7.1.3. H0 : p = 0.5H1 : p >0.5

n = 15(a) = probability of type I error= p(rejectH0|H0is true)

= p(y 10|p = 0.5) = 1 p(y 10|p = 0.5)

=1

15

y=10

c(15, y)(0.5)y(0.5)15y

= 1 0.941= 0.059

(b) = p(acceptH0|H0is false)= p(y 9|p = 0.7)

=9

y=0c(15, y)(0.7)y(0.3)15y

= 0.278(c)

=p(y

9

|p

=0.6)

=9

y=0c(15, y)(0.6)y(0.4)15y

= 0.597

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70 CHAPTER 7 Hypothesis Testing

(d) For = 0.010.01 = p(y k|p = 0.5)From binomial table = 0.01 falls between k = 2 and k = 3. However, fork= 3, =0.018 which exceeds 0.01. If we want to limit to be no more than 0.01, we will take

k= 2. That is we rejectH0ifk 2.For = 0.010.03 = p(y k|p = 0.5)From binomial table = 0.03 falls between k = 3 and k = 4. However, fork= 4, =0.059 which exceeds 0.05. If we want to limit to be no more than 0.05, we will take

k= 3. That is we rejectH0ifk 3.(e) When = 0.01. From part d, rejection region is of the form(y 2).

Forp = 0.7 = p(y 2|p = 0.7)

= 1 p(y 10

(a) = probability of type I error= p(rejectH0|H0is true)= p(x >11.2| = 10)

= P x/n > 11.2104/25| = 10= P(z >1)= 0.1587

(b) = p(acceptH0|H0is false)= p(x 11.2| = 11)= P

x

/

n 11.211

4/

25| = 11

= P(z 0.25)= 0.5787

(c) 0= 10a

=11

z= z0.01= 2.33z= z0.8= 0.525

n = (z+z)22(a0)2

= (2.330.525)242(1110)2

= 52.13 rounded up to 53

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7.1.9. 2 = 16H0 : = 25H1 : = 25n

= (z+z)22

(a

0)2

= (1.645+1.645)216(1)2

= 173.19 rounded up to 174

EXERCISES 7.2

7.2.1. H0 : = 0H1 : = 1L() = 1

2n

nexp

(xi)222

L(0)

= 12nn exp

(xi0)222

L(1) = 12n

nexp

(xi1)222

L(0)L(1)

= exp

(xi0)222

+

(xi1)222

= exp

(xi1)2

(xi0)2

22

ln

L(0)L(1)

=

(xi1)2

(xi0)2

22

=

x2i 2nx1+21

x2i 2nx02022

= 2nx(01)(01)(0+1)

22

= (01)

xi2

(2021)22

Therefore, the most powerful test is to rejectH0if

(0 1)

xi

2

20 21

22

ln k

(0 1)

xi

2 ln k +

20 21

22

xi

2 ln k +

2021

2

(0 1)

= C

Assume0 < 1, rejection region for = 1is given byxi C

WhereC =

2 ln k+

2021

2(01)

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Rejection region for = 0is given by xi C

7.2.5. H0: = 0

H1: < 0f(y) = 2y

2ey2/2 x >0

L() =n

i=1

2yi2

ne

y2i/2

L(0) =n

i=1

2yi20

ne

y2i/20

L(1) =n

i=1

2yi21

ne

y2i/21

L(0)

L(1) =

n

i=12yi20

ni=1

2yi21

n e y2i/21 y2i/20=

10

2ne

y2i/21

y2i/20

ln

L(0)L(1)

= 2n ln

10

+ y2i/21 y2i/20 ln k

y2i CWhereC=

ln k 2n ln

10

20

21

20217.2.7. H0: p = p0

H1: p = p1wherep1 > p0f(p) = px(1 p)1xL(p) = p

xi (1 p)n

xi

L(p0) = p

xi0 (1 p0)n

xi

L(p1) = p

xi1 (1 p1)n

xi

L(p0)L(p1)

=

p0p1

xi 1p01p1

n xi kTaking natural logarithm, we have

xi ln

p0

p1

+

n

xi

ln

1 p01 p1

ln k

lnp

0p1

ln1 p01 p1 xi + n ln1 p

01 p1

ln k

xi

ln k n ln

1 p01 p1

ln

p0

p1

ln

1 p01 p1

To find the rejection region for a fixed value of, write the region asxi C, whereC=

ln k n ln

1p01p1

ln

p0p1

ln

1p01p1

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EXERCISES 7.3

7.3.1. f(x) = 12

exp (x)2

22

L(1)

= 12nn

1

exp(xi)2

221 Here0=

20

anda= R

20

Hence

L

20

= max 120

nexp

(xi )2220

Since the only unknown parameter in the parameter space is 2, < 2 < ; maximumlikelihood function is achieved when2 equals to its maximum likelihood estimator

2mle= 1

n

n

i=1

(xi x)2

=

21

20

n/2exp

(xi )2

221

(xi )2220

=

(xi x)2n20

n/2exp

n

(xi )22

(xi x)2

(xi )2

220

The likelihood ratio test has the rejection region:

Reject if k, which is equivalent ton

2

ln(xi x)2

n

2

lnn20+

n(xi)2

2

(xix)2

(xi)2

220

ln k

7.3.3. f(x) = 12

exp (x)2

22

L(1) = 1

2n

n1

exp

(xi)2221

L(2) = 12n

n2

exp

(yi)2222

Let = L(22 )

L

21

= 21

nexp

(yi)2

222

(xi)2221

Thus the likelihood ratio test has the rejection region

RejectH0if kn ln

2

1

+

(yi )2222

(xi )2221

ln k

(yi )2

22

(xi )221

2 ln k 2n ln

2

1

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74 CHAPTER 7 HypothesisTesting

21= 1

n

ni=1

(xi x)2

22= 1

n

n

i=1

(xi x)2

n

(yi 2)2(yi y)2

n

(xi 1)2(xi x)2

2 ln k 2n ln

2

1

The rejection region is n

(yi 2)2(yi y)2

n

(xi 1)2(xi x)2

C

7.3.5. f(x) = 1

ex/ forx >0

L() = 1n

exp

xi

L(0)

= 1n

0

expxi

0

L(1) = 1n1 exp

xi1

L(0)L(1)

= n1n0

exp

xi1

xi0

=

10

nexp

xi

1

xi

0

We reject the null hypothesis if

n ln

1

0

+

1

1 1

0

xi ln k

xi ln k n ln1

0

100 1

Where we reject null hypothesis if

xi m1or

xi m2

Provided m1=

ln k n ln

10

1001

andm2= 1

ln kn ln

10

0110

EXERCISES 7.4

7.4.1. n = 50, = 0.02, x = 62, s = 8H0 : 64H1 : 1.769) = 0.0384

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(c) Smallest = 0.0384p-value > 0.02

We fail to reject the null hypothesis.

7.4.3. H0:

=0.45

H1: 0.85

} =0.1977

(b) Herez = 0.85z/2= z0.005= 2.58Rejection region is {z < z0.005, z > z0.005}, i.e. {z < 2.58, z >2.58}p-value = min{z < 0.85, z >0.85} = 0.3954

(c) Assumptions: even though population standard deviation is unknown, because of the

large sample size, normal distribution is assumed.

7.4.5. x = $58, 800, n = 15, s = $8, 300(a) P(rejectH0|H0is true) = 0

Since the probability of rejecting null hypothesis equals to zero. Therefore, the null

hypothesis is accepted.(b) = 0.01H0: = 55, 648H1: >55.648

t= xs/

n= 58,80055,648

8,300/

15 = 31522143.05= 1.47

Rejection region is {t > t0.01,14}t0.01,14= 2.624Sincet= 2.642 is greater than 1.47, we fail to reject the null hypothesis

7.4.7. H0: p0= 0.3H

1:p

0>

0.3p = 5501500= 0.366z = pp0

p0q0n

= 0.3660.30.30.71500

= 0.0660.0118= 5.593

Rejection region is {z > z0.01}z0.01= 2.33

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i.e. {z >2.33}Yes, customer has preference over ivory color

7.4.9. (a) x = 42.9, s = 6.3674, = 0.1H0 :

=44

H1 : = 44The data is normally distributed.

z =x s/

n

= 42.9 446.3674/

10

=1.12.013

= 0.5644

Rejection region forz is|z| < z0.05}wherez0.05= 1.645

z0.05= 1.645< 0.56444We fail to reject the null hypothesis

(b) 90% confidence interval for is

x z/2 sn , x z/2 s

n

= (42.9 1.645

2.013, 42.9

+1.645

2.013)

=(39.588, 46.21138)

(c) From a, we can see that it is reasonable to take = 44. The argument is supported bythe confidence interval in b.

7.4.11. x = 13.7, s = 1.655, n = 20H0 : = 14.6H1 : z0.01}z0.01= 2.33

We reject the null hypothesis. Thus there is a statistical evidence to support this claim.

7.4.13. x = 32,277, s = 1,200, n = 100, = 0.05H0 : = 30,692H1 : >30,692

z = 32,27730,6921,200/

100

= 13.20Rejection region is

z > z0.05

z0.05= 1.645Hence we reject the null hypothesis that the expenditure per consumer is increased from

1994 to 1995

7.4.15. H0 :

=1.129

H1 : = 1.129t= 1.241.129

0.01/

24 = 0.1110.00204= 54.41

t0.05,23= 2.807Wheret= 54.41> t0.05,23Thus we reject the null hypothesis. That is price of gas is changed recently.

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EXERCISES 7.5

7.5.1. H0: 1 2= 0H1: 1 2= 0

z = (

y1

y2)

(

1

2)

s21n1

+ s22

n2

= 74

71

8150 + 10050 =

31.9026= 1.5767

Rejection region forzis|z| > z0.025

z0.025= 1.96Sincez0.025 >1.57, we fail to reject the null hypothesis. To see the significant difference we

need to have = 0.0582 level of significance.7.5.3. x1= 58, 550, x2= 53, 700, s1= 4000, s2= 3200, n1= 25, n2= 23

H0: 1 2= 0H1: 1 2 >0S2p

=

(n11)s21+(n21)s22(n1

+n2

2)

=

24(4000)2+22(3200)225

+23

2

=13245217

Sp= 3639.3979t= (x1x2)0

Sp

1n1

+ 1n2

= (58,55053,700)03639

125 + 123

= 4.61288

Rejection region is {t > t0.05,46} i.e. {t >1.679}Since1.679 < 4.61288, werejectthe null hypothesis. Thus implies that there existssignificant

evidence to show that the males salary is higher than that of female.

7.5.5. x1= 105.9, x2= 100.5, s21= 0.21, s22= 0.19, n1= 80, n2= 100H0: 1 2= 0H1: 1 2= 0

Uset=

(

x1

x2)

0

s21n1

+ s22n2

and use two sidedt

-test.

7.5.7. x1= 7.65, x2= 9.75, s1= 0.9312, s2= 0.852, n1= 10, n2= 10(a) H0: 1 2= 0

H1: 1 2 >0S2p= (n1

1)s21+(n21)s22(n1+n22) =

9(0.9312)2+9(0.852)218

Sp= 0.892479t= (x1x2)0

Sp

1n1

+ 1n2

= (7.659.75)00.89

110+ 110

= 5.276

Rejection region is {t > t0.05,18} i.e. {t >1.734}Thus fail to reject the null hypothesis.

(b) H0: 21= 22

H1: 21= 22

Test statisticF= s21

s22= (0.9312)2

(0.852)2= 1.1945

From theF-tableF0.025(9,9)= 4.03

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78 CHAPTER 7 HypothesisTesting

F0.95(9,9)= 14.03= 0.248Rejection region isF > 4.03 andF < 0.248

Since the observed value of the statistic 1.1945 < 4.03, we fail to reject the null

hypothesis

(c) H0 : d= 0H1 : d >0

d= 2.1Sd= 1.1670t= dd0

Sd/

n= 2.1

1.670/

10= 3.16

Fromt-tablet0.05,9= 1.833Wereject the null hypothesis, this implies hat the down stream is less than the upstream.

7.5.9. (a) x1= 2.04, x2= 3.55, s1= 0.551, s2= 0.6958, n1= 14, n2= 14H0 : 1 2= 0H1 : 1 2 1.77, we fail to reject the null hypothesis.

(b) Test statisticF= s21

s22= 0.6281

From theF-tableF0.025(13,13)

=3.11

F0.95(13,13)= 13.11= 0.321Rejection region isF >3.11 andF

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Rejection region isF >2.585 andF 9.48

By using contingency table

2 = 43.862 falls in the rejection region at= 0.05, we reject the null hypothesis. That is collectivebargaining is dependent on employee classification.

7.6.3. O1= 12, O2= 14, O3= 78, O4= 40, O5= 6We now computei(i = 1,2,3,4,5)using continuity correction

1= p(x 55)2= p

z 65.5704

3= p

z 75.5704

4= p

z 85.5704

5= p

z 95.5704

Taking above probability we need to findEiand follow 6.6.5.

7.6.5. E1= 950 0.35, E2= 950 0.15, E3= 950 0.20, E4= 950 0.30O1= 950 0.45, O2= 950 0.25, O3= 950 0.02, O4= 950 0.282 = 834.7183From Chi-Square table

20.05,3= 7.81Thus2 = 834.71> 7.81, we reject the null hypothesis, at least one probabilities is differentfrom hypothesized value.

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Chapter8Linear Regression Mode

EXERCISES 8.2

8.2.1. (a) Proof see example 8.2.1

(b) SSE

2 follows central Kai square with degree of freedomn 2E

SSE2

= n 2 and2 is a constant

ThereforeE(SSE) = (n 2)28.2.3. (a) Least-squares regression line isy = 84.1674 + 5.0384x

(b)

30 40 50 60 70

50

100

150

200

250

x

y

8.2.5. (a) Check the proof in Derivation of

0and

1.

(b) We know the line of best fit is y = 0+1x and plug in the point (x, y) we get0= y 1x. Complete the proof.

8.2.7. y = 1x + SSE = n

i=1e2i=

ni=1

[yi (1xi)]2

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82 CHAPTER 8 Linear Regression Models

(SSE)1

=

ni=1

[yi(1xi)]2

1= 2 n

i=1[yi (1xi)]xi

= 2 ni=1

[xiyi (1x2i)] = 0

1=n

i=1 (xiyi)n

i=1(x2i)

y = 40.175 + .9984x8.2.9. Least-squares regression line isy = .62875 + .83994x

20 30 40 50 60

20

25

30

35

40

45

50

x

y

8.2.11. Least-squares regression line isy = 2.2752 + .00578x

50 100 150

1

2

3

4

x

y

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EXERCISES 8.3

8.3.1. (a) Least-squares regression line isy = 57.2383 .4367x(b)

40 50 60 70 80 90

15

20

2

5

30

35

40

45

50

x

y

(c) The 95% confidence intervals for0is (40.5929, 73.8837)

The 95% confidence intervals for1is (.6806, .1928)

8.3.3. 1andyare both normally distributed. In order to show these two are independent we justneed to show the covariance of them is 0. (Property of normal distribution)

Cov

1, y

= E

1 y

E

1

E (y)

= E

Sxy

Sxx y

1 E

1

n

ni=1

yi

=E

ni=1(yi y) (xi x)n

i=1(xi x)2

y 1

1

n

n

i=1

(0

+1

x)

= 1 (0 + 1x) 1 0 21x

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