ANSWERS

Chapter 1

1.5 a 2.45 − 2.65, 2.65 − 2.85b 7/30c 16/30

1.9 a Approx. .68b Approx. .95c Approx. .815d Approx. 0

1.13 a y = 9.79; s = 4.14b k = 1: (5.65, 13.93); k = 2: (1.51,

18.07); k = 3: (−2.63, 22.21)1.15 a y = 4.39; s = 1.87

b k = 1: (2.52, 6.26); k = 2: (0.65,8.13); k = 3: (−1.22, 10)

1.17 For Ex. 1.2, range/4 = 7.35; s = 4.14;for Ex. 1.3, range/4 = 3.04; s = 3.17;for Ex. 1.4, range/4 = 2.32, s = 1.87.

1.19 y − s = −19 < 0

1.21 .841.23 a 16%

b Approx. 95%1.25 a 177

c y = 210.8; s = 162.17d k = 1: (48.6, 373); k = 2:

(−113.5, 535.1); k = 3: (−275.7,697.3)

1.27 68% or 231 scores; 95% or 323 scores1.29 .051.31 .0251.33 (0.5, 10.5)1.35 a (172 − 108)/4 = 16

b y = 136.1; s = 17.1c a = 136.1 − 2(17.1) = 101.9;

b = 136.1 + 2(17.1) = 170.3

Chapter 2

2.7 A = {two males} = {(M1, M2),(M1,M3), (M2,M3)}B = {at least one female} = {(M1,W1),(M2,W1), (M3,W1), (M1,W2), (M2,W2),(M3,W2), (W1,W2)}B = {no females} = A; A ∪ B = S;A ∩ B = null; A ∩ B = A

2.9 S = {A+, B+, AB+, O+, A−, B−,AB−, O−}

2.11 a P(E5) = .10; P(E4) = .20b p = .2

2.13 a E1 = very likely (VL); E2 =somewhat likely (SL); E3 =unlikely (U); E4 = other (O)

b No; P(VL) = .24, P(SL) = .24,P(U) = .40, P(O) = .12

c .48

2.15 a .09b .19

2.17 a .08b .16c .14d .84

2.19 a (V1, V1), (V1, V2), (V1, V3),(V2, V1), (V2, V2), (V2, V3),(V3, V1), (V3, V2), (V3, V3)

b If equally likely, all haveprobability of 1/9.

c P(A) = 1/3; P(B) = 5/9;P(A ∪ B) = 7/9;P(A ∩ B) = 1/9

2.27 a S = {CC, CR, CL, RC, RR, RL,LC, LR, LL}

b 5/9c 5/9

877

878 Answers

2.29 c 1/152.31 a 3/5; 1/15

b 14/15; 2/52.33 c 11/16; 3/8; 1/42.35 422.37 a 6! = 720

b .52.39 a 36

b 1/62.41 9(10)6

2.43 504 ways2.45 408,4082.49 a 8385

b 18,252c 8515 requiredd Yes

2.51 a 4/19,600b 276/19,600c 4140/19,600d 15180/19,600

2.53 a 60 sample pointsb 36/60 = .6

2.55 a(

9010

)b(

204

)(706

)/(9010

)= .111

2.57 (4 × 12)/1326 = .03622.59 a .000394

b .00355

2.61 a364n

365n

b .50052.63 1/562.65 5/1622.67 a P(A) = .0605

b .001344c .00029

2.71 a 1/3b 1/5c 5/7d 1e 1/7

2.73 a 3/4b 3/4c 2/3

2.77 a .40 b .37 c .10d .67 e .6 f .33g .90 h .27 i .25

2.93 .3642.95 a .1

b .9

c .6d 2/3

2.97 a .999b .9009

2.101 .052.103 a .001

b .0001252.105 .902.109 P(A) ≥ .98332.111 .1492.113 (.98)3(.02)2.115 (.75)4

2.117 a 4(.5)4 = .25b (.5)4 = 1/16

2.119 a 1/4b 1/3

2.121 a 1/n

b1

n;

1

n

c3

72.125 1/122.127 a .857

c No; .8696d Yes

2.129 .42.133 .94122.135 a .57

b .18c .3158d .90

2.137 a 2/5b 3/20

2.139 P(Y = 0) = (.02)3;P(Y = 1) = 3(.02)2(.98);P(Y = 2) = 3(.02)(.98)2;P(Y = 3) = (.98)3

2.141 P(Y = 2) = 1/15; P(Y = 3) = 2/15;P(Y = 4) = 3/15; P(Y = 5) = 4/15;P(Y = 6) = 5/15

2.145 18!2.147 .00832.149 a .4

b .6c .25

2.151 4[p4(1 − p) + p(1 − p)4]2.153 .3132.155 a .5

b .15c .10d .875

Answers 879

2.157 .0212.161 P(R ≤ 3) = 12/662.163 P(A) = 0.9801

P(B) = .96392.165 .9162.167 P(Y = 1) = 35/70 = .5;

P(Y = 2) = 20/70 = 2/7;P(Y = 3) = 10/70;P(Y = 4) = 4/70; P(Y = 5) = 1/70

2.169 a (4!)3 = 13,824

b 3456/13,824 = .252.173 .252.177 a .364

b .636c (49/50)n ≥ .60, so n is at most 25

2.179 a 20

(1

2

)6

= .3125

b 27

(1

2

)10

Chapter 3

3.1 P(Y = 0) = .2, P(Y = 1) = .7,P(Y = 2) = .1

3.3 p(2) = 1

6, p(3) = 2

6, p(4) = 1

2

3.5 p(0) = 2

6, p(1) = 3

6, p(3) = 1

6

3.7 p(0) = 3!

27= 6

27, p(2) = 3

27,

p(1) = 1 − 6

27− 3

27= 18

273.9 a P(Y = 3) = .000125,

P(Y = 2) = .007125,P(Y = 1) = .135375,P(Y = 0) = .857375

c P(Y > 1) = .00725

3.11 P(X = 0) = 8

27, P(X = 1) = 12

27,

P(X = 2) = 6

27, P(X = 3) = 1

27,

P(Y = 0) = 2744

3375,

P(Y = 1) = 588

3375,

P(Y = 2) = 14

3375,

P(Y = 3) = 1

3375, Z = X + Y ,

P(Z = 0) = 27

125, P(Z = 1) = 54

125,

P(Z = 2) = 36

125, P(Z = 3) = 8

125

3.13 E(Y ) = 1

4, E(Y 2) = 7

4, V (Y ) = 27

16,

cost = 1

43.15 a P(Y = 0) = .1106,

P(Y = 1) = .3594,

P(Y = 2) = .3894,P(Y = 3) = .1406

c P(Y = 1) = .3594d µ = E(Y ) = 1.56, σ 2 = .7488,

σ = 0.8653e (−.1706, 3.2906),

P(0 ≤ Y ≤ 3) = 1

3.17 µ = E(Y ) = .889,σ 2 = V (Y ) = E(Y 2)−[E(Y )]2 = .321,σ = 0.567, (µ − 2σ ,µ + 2σ) = (−.245, 2.023),P(0 ≤ Y ≤ 2) = 1

3.19 C = $85

3.21 13,800.388

3.23 $.31

3.25 Firm I : E (profit) = $60,000E(total profit) = $120,000

3.27 $510

3.35 .4; .3999

3.39 a .1536;b .9728

3.41 .000

3.43 a .1681b .5282

3.45 P(alarm functions) = 0.992

3.49 a .151b .302

3.51 a .51775b .4914

3.53 a .0156b .4219c 25%

880 Answers

3.57 $185,000

3.59 $840

3.61 a .672b .672c 8

3.67 .07203

3.69 Y is geometric with p = .59

3.73 a .009b .01

3.75 a .081b .81

3.81 2

3.831

n

(n − 1

n

)5

3.87 E

(1

Y

)= − p ln(p)

1 − p

3.91 $150; 4500

3.93 a .04374b .99144

3.95 .1

3.97 a .128b .049c µ = 15, σ 2 = 60

3.99 p(x) = y!

(r − 1)!(y − r + 1)!pr q y+1−r ,

y = r − 1, r , r + 1, . . .

3.101 a5

11

br

y0

3.1031

423.105 b .7143

c µ = 1.875,σ = .7087

3.107 hypergeometric with N = 6, n = 2,and r = 4.

3.109 a .0238b .9762c .9762

3.111 a p(0) = 14

30, p(1) = 14

30,

p(2) = 2

30

b p(0) = 5

30, p(1) = 15

30,

p(2) = 9

30, p(3) = 1

303.113 P(Y ≤ 1) = .187

3.115 p(0) = 1

5, p(1) = 3

5, p(2) = 1

53.117 a P(Y = 0) = .553

b E(T ) = 9.5, V (T ) = 28.755,σ = 5.362

3.119 .016

3.121 a .090b .143c .857d .241

3.123 .1839

3.125 E(S) = 7, V (S) = 700; no

3.127 .6288

3.129 23 seconds

3.131 .5578

3.133 .1745

3.135 .9524

3.137 .1512

3.139 40

3.141 $1300

3.149 Binomial, n = 3 and p = .6

3.151 Binomial, n = 10 and p = .7,P(Y ≤ 5) = .1503

3.153 a Binomial, n = 5 and p = .1

b Geometric, p = 1

2c Poisson, λ = 2

3.155 a E(Y ) = 7

3

b V (Y ) = 5

9

c p(1) = 1

6, p(2) = 2

6, p(3) = 3

63.167 a .64

b C = 10

3.169 d p(−1) = 1/(2k2),p(0) = 1 − (1/k2), p(1) = 1(2k2)

3.171 (85, 115)

3.173 a p(0) = 1

8, p(1) = 3

8, p(2) = 3

8,

p(3) = 1

8c E(Y ) = 1.5, V (Y ) = .75,

σ = .866

Answers 881

3.175 a 38.4b 5.11

3.177 (61.03, 98.97)

3.179 No, P(Y ≥ 350) ≤ 1

(2.98)2= .1126.

3.181p = Fraction defective P(acceptance)

a 0 1b .10 .5905c .30 .1681d .50 .0312e 1.0 0

3.185 a .2277b Not unlikely

3.187 a .023b 1.2c $1.25

3.189 1 − (.99999)10,000

3.191 V (Y ) = .4

3.193 .476

3.195 a .982b P(W ≥ 1) = 1 − e−12

3.197 a .9997b n = 2

3.199 a .300b .037

3.201 (18.35, 181.65)

3.203 a E[Y (t)] = k(e2λt − eλt )

b 3.2974, 2.139

3.205 .00722

3.207 a p(2) = .084, P(Y ≤ 2) = .125b P(Y > 10) = .014

3.209 .0837

3.211 3

3.213 a .1192b .117

3.215 a n[1 + k(1 − .95k)]b g(k) is minimized at k = 5 and

g(5) = .4262.c .5738N

Chapter 4

4.7 a P(2 ≤ Y < 5) = 0.591,P(2 < Y < 5) = .289, sonot equal

b P(2 ≤ Y ≤ 5) = 0.618,P(2 < Y ≤ 5) = 0.316, sonot equal

c Y is not a continuous randomvariable, so the earlier resultsdo not hold.

4.9 a Y is a discrete random variableb These values are 2, 2.5, 4, 5.5, 6,

and 7.

c p(2) = 1

8, p(2.5) = 1

16,

p(4) = 5

16, p(5.5) = 1

8,

p(6) = 1

16, p(7) = 5

16d φ.5 = 4

4.11 a c = 1

2

b F(y) = y2

4, 0 ≤ y ≤ 2

d .75e .75

4.13 a F(y) =

0 y < 0

y2

20 ≤ y ≤ 1

y − 1

21 < y ≤ 1.5

1 y > 1.5b .125c .575

4.15 a For b ≥ 0, f (y) ≥ 0; also,∞∫

−∞f (y) = 1

b F(y) = 1 − b

y, for y ≥ b;

0 elsewhere.

cb

(b + c)

d(b + c)

(b + d)

4.17 a c = 3

2

b F(y) = y3

2+ y2

2, for 0 ≤ y ≤ 1

882 Answers

d F(−1) = 0, F(0) = 0, F(1) = 1

e3

16

f104

123

4.19 a f (y) =

0 y ≤ 0.125 0 < y < 2.125y 2 ≤ y < 40 y ≥ 4

b7

16

c13

16

d7

94.21 E(Y ) = .708, V (Y ) = .04874.25 E(Y ) = 31/12, V (Y ) = 1.1604.27 $4.65, .012

4.29 E(Y ) = 60, V (Y ) = 1

34.31 E(Y ) = 44.33 a E(Y ) = 5.5, V (Y ) = .15

b Using Tchebysheff’s theorem,the interval is (5, 6.275).

c Yes; P(Y ) = .57814.37 E(Y ) = 04.39 .5; .25

4.45 a P(Y < 22) = 2

5= .4

b P(Y > 24) = 1

5= .2

4.47 a P(Y > 2) = 3

4

b c0 + c1

[4

3+ 9

]4.49

3

4

4.511

34.53 a

1

8

b1

8

c1

4

4.55 a2

7b µ = .015, V (Y ) = .00041

4.57 E(π

6D3)

= .0000065π ,

V(π

6D3)

= .0003525π2

4.59 a z0 = 0

b z0 = 1.10c z0 = 1.645d z0 = 2.576

4.63 a P(Z > 1) = .1587b The same answer is obtained.

4.65 $425.604.67 µ = 3.000 in.4.69 .26604.71 a .9544

b .82974.73 a .406

b 960.5 mm4.75 µ = 7.3014.77 a 0.758

b 22.24.87 a φ.05 = .70369.

b φ.05 = .351854.89 a β = .8

b P(Y ≤ 1.7) = .88064.91 a .1353

b 460.52 cfs4.93 a .5057

b 19364.97 .36794.99 a .7358

4.101 a E(Y ) = 1.92b P(Y > 3) = .21036d P(2 ≤ Y ≤ 3) = .12943

4.103 E(A) = 200π , V (A) = 200,000π2

4.105 a E(Y ) = 3.2, V (Y ) = 6.4b P(Y > 4) = .28955

4.107 a (0, 9.657), because Y mustbe positive.

b P(Y < 9.657) = .953384.109 E(L) = 276, V(L) = 47,664

4.111 d√

β�

(α + 1

2

)/�(α) if α > 0

e1

β(α − 1)if α > 1,

�(α − 12 )√

β�(α)

if α >1

2,

1

β2(α − 1)(α − 2)if α > 2

4.123 a k = 60b φ.95 = 0.84684

4.125 E(Y ) = 3

5, V (Y ) = 1

25

4.129 E(C) = 52

3, V (C) = 29.96

4.131 a .75b .2357

4.133 a c = 105

Answers 883

b µ = 3

8c σ = .1614d .02972

4.139 m X (t) = exp{t (4−3µ)+(1/2)(9σ 2t2)}normal, E(X) = 4 − 3µ, V (X) = 9σ 2,uniqueness of moment-generatingfunctions

4.141 m(t) = etθ2 − etθ1

t (θ2 − θ1)4.143 αβ, αβ2

4.145 a2

5b

1

(t + 1)c 1

4.147 σ = 1

24.149 14.151 The value 2000 is only .53 standard

deviation above the mean. Thus, wewould expect C to exceed 2000fairly often.

4.153 (6.38, 28.28)4.155 $113.334.157 a F(x) =

0, x < 0(1/100)e−x/100, 0 ≤ x < 2001, x ≥ 200

b 86.474.159 a F1(y) =

0 y < 0.1

.1 + .15= .4 0 ≤ y < 5

1 y ≥ .5

;

F2(y) =

0 y < 04y2/3 0 ≤ y < .5(4y − 1)/3 .5 ≤ y < 11 y ≥ 1

b F(y) = 0.25F1(y) + 0.75F2(y)

c E(Y ) = .533, V (Y ) = .0764.161 φ.9 = 85.364.163 1 − (.927)5 = .31554.165 a c = 4

b E(Y ) = 1, V (Y ) = .5

c m(t) = 1

(1 − .5t)2, t < 2

4.167 E(Y k) = �(α + β)�(k + α)

�(α)�(k + α + β)4.169 e−2.5 = .082

4.171 a E(W ) = 1

2, V (W ) = 1

4b 1 − e−6

4.173 f (r) = 2λπre−λπr2, r > 0

4.175√

2 = 1.4144.179 k = (.4)1/3 = .73684.181 m(t) = exp(t2/2); 0; 14.183 a E(Y ) = 598.74 g

V (Y ) = e22(e16 − 1)10−4

b (0, 3,570,236.1)c .8020

4.187 a e−2.5 = .082b .0186

4.189 E(Y ) = 0. Also, it is clear that

V (Y ) = E(Y 2) = 1

n − 1.

4.191 c 1 − e−4

4.193 1504.195 a 12

b w = 120

Chapter 5

5.1 a y1

0 1 2

0 19

29

19

y2 1 29

29 0

2 19 0 0

b F(1, 0) = 1

3

5.3

4y1

3y2

23 − y1 − y2

9

3

, where

0 ≤ y1, 0 ≤ y2, and y1 + y2 ≤ 3.5.5 a .1065

b .55.7 a .00426

b .80095.9 a k = 6

b31

64

5.11 a29

32

b1

4

884 Answers

5.13 a F

(1

2,

1

2

)= 9

16

b F

(1

2, 2

)= 13

16c .65625

5.15 a e−1 − 2e−2

b1

2c e−1

5.17 .505.19 a

y1 0 1 2

p1(y1)49

49

19

b No5.21 a Hypergeometric with N = 9,

n = 3, and r = 4.

b2

3

c8

155.23 a f2(y2) = 3

2− 3

2y2

2 , 0 ≤ y2 ≤ 1

b Defined over y2 ≤ y1 ≤ 1 if y2 ≥ 0

c1

35.25 a f1(y1) = e−y1 , y1 > 0;

f2(y2) = e−y2 , y2 > 0b P(1 < Y1 < 2.5) = P(1 < Y2 <

2.5) = e−1 − e−2.5 = .2858c y2 > 0d f (y1|y2) = f1(y1) = e−y1 , y1 > 0e f (y2|y1) = f2(y2) = e−y2 , y2 > 0f sameg same

5.27 a f1(y1) = 3(1 − y1)2, 0 ≤ y1 ≤ 1;

f2(y2) = 6y2(1 − y2), 0 ≤ y2 ≤ 1

b32

63c f (y1|y2) = 1

y2, 0 ≤ y1 ≤ y2,

if y2 ≤ 1

d f (y2|y1) = 2(1 − y2)

(1 − y1)2,

y1 ≤ y2 ≤ 1 if y1 ≥ 0

e1

45.29 a f2(y2) = 2(1 − y2), 0 ≤ y2 ≤ 1;

f1(y1) = 1 − |y1|, for−1 ≤ y1 ≤ 1

b1

35.31 a f1(y1) = 20y1(1 − y1)

2, 0 ≤y1 ≤ 1

b f2(y2) ={15(1 + y2)

2 y22 , −1 ≤ y2 < 0

15(1 − y2)2 y2

2 , 0 ≤ y2 ≤ 1

c f (y2|y1) = 32 y2

2 (1 − y1)−3,

for y1 − 1 ≤ y2 ≤ 1 − y1d .5

5.33 a f 1(y1) = y1e−y1 , y1 ≥ 0;f 2(y2) = e−y2 , y2 ≥ 0

b f (y1|y2) = e−(y1−y2), y1 ≥ y2

c f (y2|y1) = 1/y1, 0 ≤ y2 ≤ y1

5.35 .55.37 e−1

5.411

45.45 No5.47 Dependent5.51 a f (y1, y2) = f1(y1) f2(y2) so that

Y1 and Y2 are independent.b Yes, the conditional probabilities

are the same as the marginalprobabilities.

5.53 No, they are dependent.5.55 No, they are dependent.5.57 No, they are dependent.5.59 No, they are dependent.5.61 Yes, they are independent.

5.631

45.65 Exponential, mean 1

5.69 a f (y1, y2) =(

1

9

)e−(y1+y2)/3,

y1 > 0, y2 > 0b P(Y1 + Y2 ≤ 1) =

1 − 4

3e−1/3 = .0446

5.71 a1

4

b23

144

5.734

35.75 a 2

b .0249c .0249d 2e They are equal.

5.77 a1

4;

1

2b E(Y 2

1 ) = 1/10, V (Y1) = 3

80,

E(Y 22 ) = 3

10, V (Y2) = 1

20c −5

4

Answers 885

5.79 05.81 15.83 15.85 a E(Y1) = E(Y2) = 1 (both

marginal distributions areexponential with mean 1)

b V (Y1) = V (Y2) = 1c E(Y1 − Y2) = 0

d E(Y1Y2) = 1 − α

4, so

Cov(Y1, Y2) = −α

4

e(

−2

√2 + α

2, 2

√2 + α

2

)5.87 a E(Y1 + Y2) = ν1 + ν2

b V (Y1 + Y2) = 2ν1 + 2ν2

5.89 Cov(Y1,Y2) = −2

9. As the value of Y1

increases, the value of Y2 tends todecrease.

5.91 Cov(Y1,Y2) = 05.93 a 0

b Dependentc 0d Not necessarily independent

5.95 The marginal distributions for Y1

and Y2 are

y1 −1 0 1 y2 0 1

p1(y1)1

3

1

3

1

3p2(y2)

2

3

1

3

Cov(Y1,Y2) = 05.97 a 2

b Impossiblec 4 (a perfect positive linear

association)d −4 (a perfect negative linear

association)5.99 0

5.101 a −α

45.103 E(3Y1 + 4Y2 − 6Y3) = −22,

V (3Y1 + 4Y2 − 6Y3) = 480

5.1051

95.107 E(Y1 + Y2) = 2/3 and

V (Y1 + Y2) = 1

185.109 (11.48, 52.68)5.113 E(G) = 42, V (G) = 25; the value $70

is70 − 42

5= 7.2 standard deviations

above the mean, an unlikely value.

5.115 b V (Y ) = 38.99c The interval is 14.7 ± 2

√38.99 or

(0, 27.188)5.117 p1 − p2,

N − n

n(N − 1)[p1 + p2 − (p1 − p2)

2]

5.119 a .0823

b E(Y1) = n

3, V (Y1) = 2n

9c Cov(Y2, Y3) = −n

9

d E(Y2 − Y3) = 0, V (Y2 − Y3) = 2n

35.121 a .0972

b .2; .0725.123 .089535.125 a .046

b .22625.127 a .2759

b .8031

5.133 ay2

2

b1

4

5.135 a3

2b 1.25

5.1373

85.139 a nαβ

b λαβ

5.141 E(Y2) = λ

2, V (Y2) = 2λ2

35.143 mU (t) = (1 − t2)−1/2, E(U ) = 0,

V (U ) = 1

5.1451

3

5.14711

365.149 a f (y1) = 3y2

1 , 0 ≤ y1 ≤ 1

f (y2) = 3

2(1 − y2

2 ), 0 ≤ y2 ≤ 1

b23

44c f (y1|y2) = 2y1

(1 − y22 )

, y2 ≤ y1 ≤ 1

d5

125.157 p(y) =(

y + α − 1y

)(β

β + 1

)y ( 1

β + 1

)α

,

y = 0, 1, 2, . . .

5.161 E(Y − X) = µ1 − µ2, V (Y − X) =σ 2

1 /n + σ 22 /m

886 Answers

5.163 b F(y1, y2) =y1 y2[1 − α(1 − y1)(1 − y2)]

c f (y1, y2) =1 − α[(1 − 2y1)(1 − 2y2)],0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1

d Choose two different values for α

with −1 ≤ α ≤ 1.5.165 a (p1et1 + p2et2 + p3et3)n

b m(t , 0, 0)c Cov(X1, X2) = −np1 p2

Chapter 6

6.1 a1 − u

2, −1 ≤ u ≤ 1

bu + 1

2, −1 ≤ u ≤ 1

c1√u

− 1, 0 ≤ u ≤ 1

d E(U1) = −1/3, E(U2) =1/3, E(U3) = 1/6

e E(2Y −1) = −1/3, E(1−2Y ) =1/3, E(Y 2) = 1/6

6.3 b fU (u) ={(u + 4)/100, −4 ≤ u ≤ 61/10, 6 < u ≤ 11

c 5.5833

6.5 fU (u) = 1

16

(u − 3

2

)−1/2

,

5 ≤ u ≥ 53

6.7 a fU (u) = 1√π

√2

u−1/2e−u/2,

u ≥ 0b U has a gamma distribution with

α = 1/2 and β = 2 (recall that�(1/2) = √

π).6.9 a fU (u) = 2u, 0 ≤ u ≤ 1

b E(U ) = 2/3c E(Y1 + Y2) = 2/3

6.11 a fU (u) = 4ue−2u , u ≥ 0, a gammadensity with α = 2and β = 1/2

b E(U ) = 1, V (U ) = 1/2

6.13 fU (u) = F ′U (u) = u

β2e−u/β , u > 0

6.15 [−ln(1 − U )]1/2

6.17 a f (y) = αyα−1

θα, 0 ≤ y ≤ θ

b Y = θU 1/α

c y = 4√

u. The values are 2.0785,3.229, 1.5036, 1.5610, 2.403.

6.25 fU (u) = 4ue−2u for u ≥ 0

6.27 a fY (y) = 2

βwe−w2/β , w ≥ 0, which

is Weibull density with m = 2.

b E(Y k/2) = �

(k

2+ 1

)βk/2

6.29 a fW (w) =1

�(

32

)(kT )3/2

w1/2e−w/kT w > 0

b E(W ) = 3

2kT

6.31 fU (u) = 2

(1 + u)3, u ≥ 0

6.33 fU (u) = 4(80 − 31u + 3u2),4.5 ≤ u ≤ 5

6.35 fU (u) = − ln(u), 0 ≤ u ≤ 16.37 a mY1(t) = 1 − p + pet

b mW (t) = E(etW ) = [1− p + pet ]n

6.39 fU (u) = 4ue−2u , u ≥ 06.43 a Y has a normal distribution

with mean µ and variance σ 2/nb P(|Y − µ| ≤ 1) = .7888c The probabilities are .8664, .9544,

.9756. So, as the sample sizeincreases, so does the probabilitythat P(|Y − µ| ≤ 1)

6.45 c = $190.276.47 P(U > 16.0128) = .0256.51 The distribution of Y1 + (n2 − Y2) is

binomial with n1 + n2 trials and successprobability p = .2

6.53 a Binomial (nm, p) whereni = m

b Binomial (n1 = n2 + · · · nn , p)c Hypergeometric (r = n,

N = n1 + n2 + · · · nn)6.55 P(Y ≥ 20) = .0776.65 a f (u1, u2) =

1

2πe−[u2

1+(u2−u1)2]/2 =1

2πe−(2u2

1−2u1u2+u22)/2

b E(U1) = E(Z1) = 0,E(U2) = E(Z1 + Z2) = 0,V (U1) = V (Z1) = 1,V (U2) = V (Z1 + Z2) =V (Z1) + V (Z2) = 2,Cov(U1, U2) = E(Z 2

1) = 1

Answers 887

c Not independent sinceρ = 0.

d This is the bivariate normaldistribution with µ1 = µ2 = 0,

σ 21 = 1, σ 2

2 = 2, and ρ = 1√2

6.69 a f (y1, y2) = 1

y21 y2

2

, y1 > 1,

y2 > 1e No

6.73 a g(2)(u) = 2u, 0 ≤ u ≤ 1b E(U2) = 2/3, V (U2) = 1/18

6.75 (10/15)5

6.77 an!

( j − 1)!(k − 1 − j)!(n − k)!y j−1

j [yk − y j ]k−1− j [θ − yk]n−k

θ n,

0 ≤ y j < yk ≤ θ

b(n − k + 1) j

(n + 1)2(n + 2)θ 2

c(n − k + j + 1)(k − j)

(n + 1)2(n + 2)θ 2

6.81 b 1 − e−9

6.83 1 − (.5)n

6.85 .56.87 a g(1)(y) = e−(y−4), y ≥ 4

b E(Y(1)) = 5

6.89 fR(r) = n(n − 1)rn−2(1 − r),0 ≤ r ≤ 1

6.93 f (w) = 2

3

(1√w

− w

), 0 ≤ w ≤ 1

6.95 a fU1(u) =

1

20 ≤ u ≤ 1

1

2u2u > 1

b fU2(u) = ue−u , 0 ≤ uc Same as Ex. 6.35.

6.97 p(W = 0) = p(0) = .0512,p(1) = .2048, p(2) = .3264,p(3) = .2656, p(4) = .1186,p(5) = .0294, p(6) = .0038,

p(7) = .00026.101 fU (u) = 1, 0 ≤ u ≤ 1 Therefore, U has

a uniform distribution on (0, 1)

6.1031

π(1 + u21)

, ∞ < u1 < ∞

6.1051

B(α, β)uβ−1(1 − u)α−1, 0 < u < 1

6.107 fU (u) =

1

4√

u0 ≤ u < 1

1

8√

u1 ≤ u ≤ 9

6.109 P(U = C1 − C3) = .4156;P(U = C2 − C3) = .5844

Chapter 7

7.9 a .7698b For n = 25, 36, 69, and 64, the

probabilities are (respectively).8664, .9284, .9642, .9836.

c The probabilities increase with n.

d Yes7.11 .86647.13 .98767.15 a E(X − Y ) = µ1 − µ2

b V (X − Y ) = σ 21 /m + σ 2

2 /nc The two sample sizes should be at

least 18.7.17 P

(∑6i=1 Z 2

i ≤ 6)

= .57681

7.19 P(S2 ≥ .065) = .107.21 a b = 2.42

b a = .656c .95

7.27 a .17271b .23041d .40312

7.31 a 5.99, 4.89, 4.02, 3.65, 3.48, 3.32c 13.2767d 13.2767/3.32 ≈ 4

7.35 a E(F) = 1.029b V (F) = .076c 3 is 7.15 standard deviations above

this mean; unlikely value.7.39 a normal, E(θ) = θ =

c1µ1 + c2µ2 + · · · + ckµk

V (θ) =(

c21

n1+ c2

2

n2+ · · · + c2

k

nk

)σ 2

b χ2 with n1 + n2 + · · · + nk − k dfc t with n1 + n2 + · · · + nk − k df

7.43 .95447.45 .05487.47 1537.49 .02177.51 6647.53 b Y is approximately normal: .0132.7.55 a random sample; approximately 1.

b .1271

888 Answers

7.57 .00627.59 .00627.61 n = 517.63 56 customers7.65 a Exact: .91854; normal

approximation: .86396.7.67 a n = 5 (exact: .99968;

approximate: .95319); n = 10(exact: .99363; approximate:.97312); n = 15 (exact: .98194;approximate: .97613); n = 20(exact: .96786; approximate:.96886)

7.71 a n > 9b n > 14, n > 14, n > 36, n > 36,

n > 891, n > 89917.73 .89807.75 .76987.77 61 customers7.79 a Using the normal approximation:

.7486.b Using the exact binomial

probability: .729.7.81 a .5948

b With p = .2 and .3, theprobabilities are .0559 and .0017respectively.

7.83 a .36897b .48679

7.85 .84147.87 .00417.89 µ = 10.157.91 Since X , Y , and W are normally

distributed, so are X , Y , and W .

µU = E(U ) = .4µ1+.2µ2+.4µ3

σ 2U = V (U ) = .16

(σ 2

1

n1

)+ .04

(σ 2

2

n2

)+ .16

(σ 2

3

n3

)7.95 a F with num. df = 1, denom. df = 9

b F with num. df = 9, denom. df = 1c c = 49.04

7.97 b .15877.101 .84137.103 .15877.105 .264

Chapter 8

8.3 a B(θ) = aθ + b − θ = (a − 1)θ + bb Let θ∗ = (θ − b)/a

8.5 a MSE(θ∗) = V (θ∗) = V (θ)/a2

8.7 a = σ 22 − c

σ 21 + σ 2

2 − 2c8.9 Y − 1

8.11 θ3 − 9θ2 + 548.13 b [n2/(n − 1)](Y/n)[1 − (Y/n)]

8.15 a(

1

3n − 1

)β

b MSE(β) = 2

(3n − 1)(3n − 2)β2

8.17 a (1 − 2p)/(n + 2)

bnp(1 − p) + (1 − 2p)2

(n + 2)2

c p will be close to .5.8.19 MSE(θ ) = β2

8.21 11.5 ± .998.23 a 11.3 ± 1.54

b 1.3 ± 1.7c .17 ± .08

8.25 a −.7b .404

8.27 a .601 ± .0318.29 a −.06 ± .045

8.31 a −.03 ± .0418.33 .7 ± .2058.35 a 20 ± 1.265

b −3 ± 1.855, yes8.37 1020 ± 645.1

8.39(

2Y

9.48773,

2Y

.71072

)8.41 a (Y 2/5.02389, Y 2/.0009821)

b Y 2/.0039321c Y 2/3.84146

8.43 b [Y(n)](.95)−1/n

8.45 a Y /.05132b 80%

8.47 c (2.557, 11.864)8.49 c (3.108, 6.785)8.57 .51 ± .048.59 a .78 ± .0218.61 (15.46, 36.94)8.63 a .78 ± .026 or (.754, .806)8.65 a .06 ± .117 or (−.057, .177)8.67 a 7.2 ± .751

b 2.5 ± .7388.69 .22 ± .34 or (−.12, .56)8.71 n = 100

Answers 889

8.73 n = 28478.75 n = 1368.77 n = 4978.79 a n = 2998

b n = 16188.81 60.8 ± 5.7018.83 a 3.4 ± 3.7

b .7 ± 3.328.85 −1 ± 4.728.87 (−.624, .122)8.91 (−84.39, −28.93)

8.93 a 2X + Y ± 1.96σ

√4

n+ 3

m

b 2X + Y ± tα/2 S

√4

n+ 3

m, where

S2 =∑

(Yi − Y )2 + 1/3∑

(Xi − X)2

n + m − 28.95 (.227, 2.196)

8.99 a

√(n − 1)S2

χ 21−α

b

√(n − 1)S2

χ 2α

8.101 s2 = .0286; (.013 .125)

8.103 (1.407, 31.264); no8.105 1 − 2(.0207) = .95868.107 765 seeds8.109 a .0625 ± .0237

b 5638.111 n = 38,4168.113 n = 7688.115 (29.30, 391.15)8.117 11.3 ± 1.448.119 3 ± 3.638.121 −.75 ± .778.123 .832 ± .015

8.125 aS2

1

S22

× σ 22

σ 21

b(

S22

S21 Fv2,v1,α/2

,S2

2

S21

Fv1,v2,α/2

)vi = ni − 1, i = 1, 2

8.129 a2(n − 1)σ 4

n2

8.131 c = 1

n + 1

8.133 b2σ 4

n1 + n2 − 2

Chapter 9

9.1 1/3; 2/3; 3/5

9.3 b12n2

(n + 2)(n + 1)2

9.5 n − 19.7 1/n9.9 a X6 = 1

9.23 c need Var(X2i − X2i−1) < ∞9.25 b .6826

c No9.31 αβ

9.35 a Yn is unbiased for µ.

b V (Yn) = 1

n2

∑n

i=1σ 2

i

9.47n∑

i=1

ln(Yi ); no

9.57 Yes

9.59 3

[Y 2 + Y

(1 − 1

n

)]9.61

(n + 1

n

)Y(n)

9.63 b3n + 1

3nY(n)

9.69 θ = 2Y − 1

1 − Y, no, not MVUE

9.71 σ 2 = m ′2 = 1

n

∑n

i=1Y 2

i .

9.75 With m ′2 = 1

n

∑n

i=1Y 2

i , the MOM

estimator of θ is θ = 1 − 2m ′2

4m ′2 − 1

.

9.772

3Y

9.81 Y 2

9.83 a θ = 1

2

(Y(n) − 1

)b(Y(n)

)2/12

9.85 a θ = 1

αY

b E(θ) = θ , V (θ) = θ2/(nα)

d∑n

i=1 Yi

e(

2∑n

i=1 Yi

31.4104,

2∑n

i=1 Yi

10.8508

)9.87 pA = .30, pB = .38

pC = .32; −.08 ± .16419.91 Y(n)/29.93 a Y(1)

c [(α/2)1/2nY(1), (1 − (α/2))1/2nY(1)]9.97 a 1/Y

b 1/Y

890 Answers

9.99 p ± zα/2

√p(1 − p)

n

9.101 exp(−Y ) ± zα/2

√Y exp(−2Y )

n

9.1031

n

n∑i=1

Y 2i

9.105 σ 2 = �(Yi − µ)2

n9.107 exp(−t/Y )

9.109 a N1 = 2Y − 1

bN 2 − 1

3n9.111 252 ± 85.193

Chapter 10

10.3 a c = 11b .596c .057

10.5 c = 1.68410.7 a False

b Falsec Trued Truee Falsef i True

ii Trueiii False

10.17 a H0: µ1 = µ2, Ha : µ1 > µ2

c z = .07510.21 z = 3.65, reject H0

10.23 a-b H0: µ1 − µ2 = 0 vs.Ha : µ1 − µ2 = 0, whichis a two–tailed test.

c z = −.954, which doesnot lead to a rejectionwith α = .10.

10.25 |z| = 1.105, do not reject10.27 z = −.1202, do not reject10.29 z = 4.4710.33 z = 1.50, no10.35 z = −1.48 (1 = homeless), no10.37 approx. 0 (.0000317)10.39 .670010.41 .02510.43 a .49

b .105610.45 .22 ± .155 or (.065, .375)10.47 .514810.49 129.146, yes10.51 z = 1.58 p–value = .1142, do not

reject10.53 a z = −.996, p–value = .0618

b Noc z = −1.826, p–value = .0336d Yes

10.55 z = −1.538; p–value = .0616; fail toreject H0 with α = .01

10.57 z = −1.732; p–value = .083610.63 a t = −1.341, fail to reject H0

10.65 a t = −3.24, p–value < .005, yesb Using the Applet, .00241c 39.556 ± 3.55

10.67 a t = 4.568 and t.01 = 2.821 soreject H0.

b The 99% lower confidence bound

is 358 − 2.82154√10

= 309.83.

10.69 a t = −1.57, .10 < p–value <.20,do not reject; using applet,p–value = .13008

i −t.10 = −1.319 and−t.05 = −1.714;.10 < p–value < .20.

ii Using the Applet,2P(T < −1.57) =2(.06504) = .13008.

10.71 a y1 = 97.856, s21 = .3403,

y2 = 98.489, s22 = .3011,

t = −2.3724, −t.01 = −2.583,−t.025 = −2.12, so .02 < p–value< .05

b Using Applet, .0305410.73 a t = 1.92, do not reject

.05 < p–value < .10; appletp–value = .07084

b t = .365, do not reject p–value> .20; applet p–value = .71936

10.75 t = −.647, do not reject10.77 a t = −5.54, reject, p–value < .01;

applet p–value approx. 0b Yesc t = 1.56, .10 < p–value < .20;

applet p–value = .12999d Yes

10.79 a χ2 = 12.6, do not rejectb .05 < p–value < .10c Applet p–value = .08248

Answers 891

10.83 a σ 21 = σ 2

2

b σ 21 < σ 2

2

c σ 21 > σ 2

2

10.85 χ 2 = 22.45, p–value < .005; appletp–value = .0001

10.89 a .15b .45c .75d 1

10.91 a Reject if Y >= 7.82.b .2611, .6406, .9131, .9909

10.93 n = 1610.95 a U = 2

β0

∑4i=1 Yi has χ 2

(24)

distribution under H0: reject H0

if U > χ2α

b Yes10.97 d Yes, is UMP

10.99 an∑

i=1

Yi ≥ k

b Use Poisson table to find k suchthat P(

∑Yi ≥ k) = α

c Yes

10.101 an∑

i=1

Yi < c

b Yes10.103 a Reject H0 if Y(n) ≤ θ0

n√

α

b Yes

10.107 χ 2 = (n − 1)S21 + (m − 1)S2

2

σ 20

has

χ2(n+m−2) distribution under H0;

reject if χ2 > χ2α

10.109 a λ = (X)m(Y )m(m X + nY

m + n

)m+n

b X/Y distributed as F with 2m and2n degrees of freedom

10.115 a Trueb Falsec Falsed Truee Falsef Falseg Falseh Falsei True

10.117 a t = −22.17, p–value < .01b −.0105 ± .001c Yesd No

10.119 a H0: p = .20, Ha : p > .20b α = .0749

10.121 z = 5.24, p–value approx. 010.123 a F = 2.904, no

b (.050, .254)10.125 a t = −2.657, .02 < p–value < .05

b −4.542 ± 3.04610.127 T =

(X+Y−W )−(µ1−µ2−µ3){(1+a+bn(3n−3)

)[∑(Xi −X)2+ 1

a∑

(Yi −Y )2+ 1b∑

(Wi −W )2]}1/2

with (3n − 3) degrees of freedom

10.129 λ =(∑n

i=1 (yi − y(1))

nθ1,0

)n

×

exp

[−∑n

i=1 (yi − y(1))

θ1,0+ n

].

Chapter 11

11.3 y = 1.5 − .6x11.5 y = 21.575 + 4.842x11.7 a The relationship appears to be

proportional to x2.b Noc No, it is the best linear model.

11.9 b y = −15.45 + 65.17xd 108.373

11.11 β1 = 2.51411.13 a The least squares line is

y = 452.119 − 29.402x11.17 a SSE = 18.286;

S2 = 18.286/6 = 3.048b The fitted line is

y = 43.35 + 2.42x∗. The same

answer for SSE (and thus S2) isfound.

11.19 a The least squares line is:y = 3.00 + 4.75x

c s2 = 5.02511.23 a t = −5.20, reject H0

b .01 < p–value < .02c .01382d (−.967, −.233)

11.25 a t = 3.791, p–value < .01b Applet p–value = .0053c Rejectd .475 ± .289

892 Answers

11.29 T = β1 − γ1

S

√(1

Sxx+ 1

Scc

) , where S =

(SSEY + SSEW )/(n + m − 4).H0 is rejected in favor of Ha for largevalues of |T |.

11.31 t = 73.04, p–value approx. 0, H0 isrejected

11.33 t = 9.62, yes11.35 x∗ = x .11.37 (4.67, 9.63)11.39 25.395 ± 2.87511.41 b (72.39, 75.77)11.43 (59.73, 70.57)11.45 (−.86, 15.16)11.47 (.27, .51)11.51 t = 9.608, p–value < .0111.53 a r 2 = .682

b .682c t = 4.146, rejectd Applet p–value = .00161

11.57 a sign for rb r and n

11.59 r = −.378311.61 .979 ± .10411.63 a β1 = −.0095, β0 = 3.603 and

α1 = −(−.0095) = .0095,α0 = exp(3.603) = 36.70.Therefore, the prediction equationis y = 36.70e−.0095x .

b The 90% CI for α0 is(e3.5883, e3.6171

) = (36.17, 37.23)11.67 y = 2.1 − .6x11.69 a y = 32.725 + 1.812x

b y = 35.5625 + 1.8119x − .1351x2

11.73 t = 1.31, do not reject

11.75 21.9375 ± 3.0111.77 Following Ex. 11.76, the 95%

PI = 39.9812 ± 213.80711.79 21.9375 ± 6.1711.83 a F = 21.677, reject

b SSER = 1908.0811.85 a F = 40.603, p–value < .005

b 950.167611.87 a F = 4.5, F1 = 9.24, fail to

reject H0

c F = 2.353, F1 = 2.23, reject H0

11.89 a Trueb Falsec False

11.91 F = 10.2111.93 90.38 ± 8.4211.95 a y = −13.54 − 0.053x

b t = −6.86c .929 ± .33

11.97 a y = 1.4825+ .5x1 + .1190x2 − .5x3

b y = 2.0715c t = −13.7, rejectd (1.88, 2.26)e (1.73, 2.41)

11.99 If −9 ≤ x ≤ 9, choose n/2 at x = −9and n/2 at x = 9.

11.101 a y = 9.34+2.46x1 + .6x2 + .41x1x2

b 9.34 , 11.80d For bacteria A, y = 9.34. For

bacteria B, y = 11.80. Theobserved growths were 9.1 and12.2, respectively.

e 12.81 ± .37f 12.81 ± .78

11.107 a r = .89b t = 4.78, p–value <.01, reject

Chapter 12

12.1 n1 = 34, n2 = 5612.3 n = 246, n1 = 93, n2 = 15412.5 With n = 6, three rats should receive

x = 2 units and three rats shouldreceive x = 5 units.

12.11 a This occurs when ρ > 0.b This occurs when ρ = 0.c This occurs when ρ < 0.d Paired better when ρ > 0,

independent better when ρ < 0

12.15 a t = 2.65, reject12.17 a µi

12.31 a µi

b µi ,1

n[σ 2

P + σ 2]

c µ1 − µ2, 2σ 2/n, normal12.35 a t = −4.326, .01 < p–value

< .025b −1.58 ± 1.014c 65 pairs

12.37 k1 = k3 = .25; k2 = .50

Answers 893

Chapter 13

13.1 a F = 2.93, do not rejectb .109c |t | = 1.71, do not reject, F = t2

13.7 a F = 5.2002, rejectb p–value = .01068

13.9 SSE = .020; F = 2.0, do notreject

13.11 SST = .7588; SSE = .7462;F = 19.83, p–value < .005, reject

13.13 SST = 36.286; SSE = 76.6996;F = 38.316, p–value < .005, reject

13.15 F = 63.66, yes, p–value < .00513.21 a −12.08 ± 10.96

b Longerc Fewer degrees of freedom

13.23 a 1.568 ± .164 or (1.404, 1.732); yesb (−.579, −.117); yes

13.25 .28 ± .10213.27 a 95% CI for µA: 76 ± 8.142

or (67.868, 84.142)b 95% CI for µB : 66.33 ± 10.51 or

(55.82, 76.84)c 95% CI for µA − µB :

9.667 ± 13.29513.29 a 6.24 ± .318

b −.29 ± .24113.31 a F = 1.32, no

b (−.21, 4.21)13.33 (1.39, 1.93)13.35 a 2.7 ± 3.750

b 27.5 ± 2.65213.37 a µ

b Overall mean13.39 b (2σ 2)/b13.41 a F = 3.11, do not reject

b p–value > .10c p–value = .1381d s2

D = 2MSE13.45 a F = 10.05; reject

b F = 10.88; reject13.47

Source df SS MS FTreatments 3 8.1875 2.729 1.40Blocks 3 7.1875 2.396 1.23Error 9 17.5625 1.95139Total 15 32.9375

F = 1.40, do not reject

13.49 F = 6.36; reject13.53 The 95% CI is 2 ± 2.83.13.55 The 95% CI is .145 ± .179.13.57 The 99% CI is −4.8 ± 5.259.13.59 n A ≥ 313.61 b = 16; n = 4813.63 Sample sizes differ.13.69 a β0 + β3 is the mean response to

treatment A in block III.b β3 is the difference in mean

responses to chemicals A and D inblock III.

13.71 F = 7; H0 is rejected13.73 As homogeneous as possible within

blocks.13.75 b F = 1.05; do not reject13.77 a A 95% CI is .084 ± .06 or

(.024, .144).13.79 a 16

b 135 degrees of freedom left forerror.

c 14.1413.81 F = 7.33; yes; blocking induces loss in

degrees of freedom for estimating σ 2;could result in sight loss of informationif block to block variation is small

13.83 a

Source df SS MS FTreatments 2 524,177.167 262,088.58 258.237Blocks 3 173,415 57,805.00 56.95Error 6 6,089.5 1,014.9167Total 11 703,681.667

b 6c Yes, F = 258.19, p–value < .005d Yes, F = 56.95, p–value < .005e 22.527f −237.25 ± 55.13

13.85 a SST = 1.212, df = 4SSE = .571, df = 22F = 11.68; p–value < .005

b |t | = 2.73; H0 is rejected; 2(.005)< p–value < 2(.01).

13.87 Each interval should have confidencecoefficient 1 − .05/4 = .9875 ≈ .99;µA − µD : .320 ± .251µB − µD : .145 ± .251µC − µD : .023 ± .251µE − µD : −.124 ± .251

894 Answers

13.89 b σ 2β

c σ 2β = 0

13.91 a µ; σ 2B + 1

k σ 2ε

b σ 2β + ( b

k−1

)∑ki=1 τ 2

i

c σ 2ε + kσ 2

B

d σ 2ε

Chapter 14

14.1 a X 2 = 3.696, do not rejectb Applet p–value = .29622

14.3 X 2 = 24.48, p–value < .00514.5 a z = 1.50, do not reject

b Hypothesis suggested by observeddata

14.7 .102 ± .04314.9 a .39 ± .149

b .37 ± .187, .39 ± .182, .48 ± .15314.11 X 2 = 69.42, reject14.13 a X 2 = 18.711, reject

b p–value < .005c Applet p–value = .00090

14.15 b X 2 also multiplied by k14.17 a X 2 = 19.0434 with a p–value of

.004091.b X 2 = 60.139 with a p–value of

approximately 0.c Some expected counts < 5

14.19 a X 2 = 22.8705, rejectb p–value < .005

14.21 a X 2 = 13.99, rejectb X 2 = 13.99, rejectc X 2 = 1.36, do not reject

14.25 b X 2 = 19.1723, p-value =0.003882, reject

c −.11 ± .13514.27 X 2 = 38.43, yes14.29 a X 2 = 14.19, reject14.31 X 2 = 21.51, reject14.33 X 2 = 6.18, reject; .025 < p–value

< .0514.35 a Yes

b p–value = .00226314.37 X 2 = 8.56, df = 3; reject14.41 X 2 = 3.26, do not reject14.43 X 2 = 74.85, reject

Chapter 15

15.1

Rejection region α

M ≤ 6 or M ≥ 19 P(M ≤ 6) + P(M ≥ 19) = .014M ≤ 7 or M ≥ 18 P(M ≤ 7) + P(M ≥ 18) = .044M ≤ 8 or M ≥ 17 P(M ≤ 8) + P(M ≥ 17) = .108

15.3 a m = 2, yesb Variances not equal

15.5 P(M ≤ 2 or M ≥ 8) = .11, no15.7 a P(M ≤ 2 or M ≥ 7) = .18, do

not rejectb t = −1.65, do not reject

15.9 a p–value = .011, do not reject15.11 T = min(T +, T −), T = T −.15.13 a T = 6, .02 < p–value < .05

b T = 6, 0.1 < p–value < .02515.15 T = 3.5, .025 < p–value < .0515.17 T = 11, reject15.21 a U = 4; p–value = .0364

b U = 35; p–value = .0559c U = 1; p–value = .0476

15.23 U = 9, do not reject15.25 z = −1.80, reject15.27 U = 0, p–value = .009615.29 H = 16.974, p-value < .00115.31 a SST = 2586.1333; SSE =

11,702.9; F = 1.33, do notreject

b H = 1.22, do not reject15.33 H = 2.03, do not reject15.37 a No, p–value = .6685

b Do not reject H0

15.39 Fr = 6.35, reject15.41 a Fr = 65.675, p–value < .005,

rejectb m = 0, P(M = 0) = 1/256,

p–value = 1/12815.45 The null distribution is given by

P(Fr = 0) = P(Fr = 4) = 1/6 andP(Fr = 1) = P(Fr = 3) = 1/3.

15.47 R = 6, no

Answers 895

15.49 a .0256b An usually small number of runs

(judged at α = .05) would imply aclustering of defective items intime; do not reject.

15.51 R = 13, do not reject15.53 rS = .911818; yes.15.55 a rS = −.8449887

b Reject15.57 rS = .6768, use two-tailed test, reject15.59 rS = 0; p–value < .005

15.61 a Randomized block designb Noc p–value = .04076, yes

15.63 T = 73.5, do not reject, consistent withEx. 15.62

15.65 U = 17.5, fail to reject H0

15.67 .015915.69 H = 7.154, reject15.71 Fr = 6.21, do not reject15.73 .10

Chapter 16

16.1 a β(10, 30)

b n = 25c β(10, 30), n = 25d Yese Posterior for the β(1, 3) prior.

16.3 c Means get closer to .4, std devdecreases.

e Looks more and more like normaldistribution.

16.7 aY + 1

n + 4

bnp + 1

n + 4;

np(1 − p)

(n + 4)2

16.9 bα + 1

α + β + Y;

(α + 1)(β + Y − 1)

(α + β + Y + 1)(α + β + Y )

16.11 e Y

(nβ

nβ + 1

)+ αβ

(1

nβ + 1

)

16.13 a (.099, .710)b Both probabilities are .025c P(.099 < p < .710) = .95h Shorter for larger n.

16.15 (.06064, .32665)16.17 (.38475, .66183)16.19 (5.95889, 8.01066)16.21 Posterior probabilities of null and

alternative are .9526 and .0474,respectively, accept H0.

16.23 Posterior probabilities of null andalternative are .1275 and .8725,respectively, accept Ha .

16.25 Posterior probabilities of null andalternative are .9700 and .0300,respectively, accept H0.