ANSWERS
Chapter 1 2.45 − 2.65, 2.65 − 2.85 7/30 16/30 Approx. .68 Approx. .95 Approx. .815 Approx. 0 y¯ = 9.79; s = 4.14 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
1.5 a b c 1.9 a b c d 1.13 a b
1.21 .84 1.23 a 16% b Approx. 95% 1.25 a 177 c y¯ = 210.8; s = 162.17 d k = 1: (48.6, 373); k = 2: 1.27 1.29 1.31 1.33 1.35
(−113.5, 535.1); k = 3: (−275.7, 697.3) 68% or 231 scores; 95% or 323 scores .05 .025 (0.5, 10.5) a (172 − 108)/4 = 16 b y¯ = 136.1; s = 17.1 c 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(E 5 ) = .10; P(E 4 ) = .20 b 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 b 2.17 a b c d 2.19 a b c 2.27 a b c
.09 .19 .08 .16 .14 .84 (V1 , V1 ), (V1 , V2 ), (V1 , V3 ), (V2 , V1 ), (V2 , V2 ), (V2 , V3 ), (V3 , V1 ), (V3 , V2 ), (V3 , V3 ) If equally likely, all have probability of 1/9. P(A) = 1/3; P(B) = 5/9; P(A ∪ B) = 7/9; P(A ∩ B) = 1/9 S = {CC, CR, CL, RC, RR, RL, LC, LR, LL} 5/9 5/9 877
878
Answers
2.29 c 1/15 2.31 a 3/5; 1/15 b 14/15; 2/5 2.33 c 11/16; 3/8; 1/4 2.35 42 2.37 a 6! = 720 b .5 2.39 a 36 b 1/6 2.41 9(10)6 2.43 504 ways 2.45 408,408 2.49 a 8385 b 18,252 c 8515 required d Yes 2.51 a 4/19,600 b 276/19,600 c 4140/19,600 d 15180/19,600 2.53 a 60 sample points b 36/60 = .6
90 10 + 90 70 20 = .111 b 10 6 4 (4 × 12)/1326 = .0362 a .000394 b .00355 364n a 365n b .5005 1/56 5/162 a P(A) = .0605 b .001344 c .00029 a 1/3 b 1/5 c 5/7 d 1 e 1/7 a 3/4 b 3/4 c 2/3 a .40 b .37 c .10 d .67 e .6 f .33 g .90 h .27 i .25 .364 a .1 b .9
2.55 a
2.57 2.59 2.61 2.63 2.65 2.67 2.71
2.73 2.77 2.93 2.95
c .6 d 2/3 2.97 a .999 b .9009 2.101 .05 2.103 a .001 b .000125 2.105 .90 2.109 P(A) ≥ .9833 2.111 .149 2.113 (.98)3 (.02) 2.115 (.75)4 2.117 a 4(.5)4 = .25 b (.5)4 = 1/16 2.119 a 1/4 b 1/3 2.121 a 1/n
1 1 ; n n 3 c 7 1/12 a .857 c No; .8696 d Yes .4 .9412 a .57 b .18 c .3158 d .90 a 2/5 b 3/20 P(Y = 0) = (.02)3 ; P(Y = 1) = 3(.02)2 (.98); P(Y = 2) = 3(.02)(.98)2 ; P(Y = 3) = (.98)3 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 18! .0083 a .4 b .6 c .25 4[ p 4 (1 − p) + p(1 − p)4 ] .313 a .5 b .15 c .10 d .875
b
2.125 2.127 2.129 2.133 2.135
2.137 2.139
2.141 2.145 2.147 2.149 2.151 2.153 2.155
Answers 879
2.157 .021 2.161 P(R ≤ 3) = 12/66 2.163 P(A) = 0.9801 P(B) = .9639
2.165 .916 2.167 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 = .25 2.173 .25 2.177 a .364 b .636 n c (49/50) ≥ .60, so n is at most 25 1 6 = .3125 2 10 1 b 27 2
2.179 a 20
Chapter 3 P(Y = 2) = .3894, P(Y = 3) = .1406 c P(Y = 1) = .3594 d µ = E(Y ) = 1.56, σ 2 = .7488, σ = 0.8653 e (−.1706, 3.2906), P(0 ≤ Y ≤ 3) = 1
3.1 P(Y = 0) = .2, P(Y = 1) = .7, 3.3 3.5 3.7
3.9
3.11
3.13
3.15
P(Y = 2) = .1 2 1 1 p(2) = , p(3) = , p(4) = 6 6 2 3 1 2 p(0) = , p(1) = , p(3) = 6 6 6 6 3 3! = , p(2) = , p(0) = 27 27 27 3 18 6 − = p(1) = 1 − 27 27 27 a P(Y = 3) = .000125, P(Y = 2) = .007125, P(Y = 1) = .135375, P(Y = 0) = .857375 c P(Y > 1) = .00725 12 8 , P(X = 1) = , P(X = 0) = 27 27 1 6 , P(X = 3) = , P(X = 2) = 27 27 2744 , P(Y = 0) = 3375 588 , P(Y = 1) = 3375 14 , P(Y = 2) = 3375 1 , Z = X + Y, P(Y = 3) = 3375 54 27 , P(Z = 1) = , P(Z = 0) = 125 125 8 36 , P(Z = 3) = P(Z = 2) = 125 125 27 1 7 2 , E(Y ) = , E(Y ) = , V (Y ) = 4 4 16 1 cost = 4 a P(Y = 0) = .1106, P(Y = 1) = .3594,
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 3.21 3.23 3.25
C = $85 13,800.388 $.31 Firm I : E (profit) = $60,000 E(total profit) = $120,000
3.27 $510 3.35 .4; .3999 3.39 a .1536; b .9728 3.41 .000 3.43 a .1681 b .5282 3.45 P(alarm functions) = 0.992 3.49 a .151 b .302 3.51 a .51775 b .4914 3.53 a .0156 b .4219 c 25%
880
Answers
3.57 $185,000 3.59 $840 3.61 a .672 b .672 c 8 3.67 .07203 3.69 Y is geometric with p = .59 3.73 a .009 b .01 3.75 a .081 b .81 3.81 2 3.83 3.87 3.91 3.93 3.95 3.97
3.99
9 1 , p(3) = 30 30 3.113 P(Y ≤ 1) = .187 3 1 1 3.115 p(0) = , p(1) = , p(2) = 5 5 5 3.117 a P(Y = 0) = .553 b E(T ) = 9.5, V (T ) = 28.755, σ = 5.362 p(2) =
3.119 .016 3.121 a .090 b .143 c .857 d .241 3.123 .1839 1 n−1 5 3.125 E(S) = 7, V (S) = 700; no n n p ln( p) 1 3.127 .6288 =− E Y 1− p 3.129 23 seconds $150; 4500 3.131 .5578 a .04374 3.133 .1745 b .99144 3.135 .9524 .1 3.137 .1512 a .128 3.139 40 b .049 2 3.141 $1300 c µ = 15, σ = 60 y! 3.149 Binomial, n = 3 and p = .6 pr q y+1−r , p(x) = (r − 1)!(y − r + 1)! 3.151 Binomial, n = 10 and p = .7,
y = r − 1, r , r + 1, . . . 5 3.101 a 11 r b y0 1 3.103 42 3.105 b .7143 c µ = 1.875, σ = .7087
3.107 hypergeometric with N = 6, n = 2, and r = 4.
3.109 a .0238 b .9762 c .9762 14 14 , p(1) = , 30 30 2 p(2) = 30 5 15 b p(0) = , p(1) = , 30 30
3.111 a p(0) =
P(Y ≤ 5) = .1503
3.153 a Binomial, n = 5 and p = .1 1
b Geometric, p = 2 c Poisson, λ = 2
7 3 5 V (Y ) = 9 2 3 1 p(1) = , p(2) = , p(3) = 6 6 6 .64 C = 10
3.155 a E(Y ) = b c
3.167 a b 3.169 d p(−1) = 1/(2k 2 ),
p(0) = 1 − (1/k 2 ), p(1) = 1(2k 2 )
3.171 (85, 115) 3 3 1 3.173 a p(0) = , p(1) = , p(2) = , 8 8 1 p(3) = 8 c E(Y ) = 1.5, V (Y ) = .75, σ = .866
8
Answers 881
3.175 a 38.4 b 5.11 3.177 (61.03, 98.97) 3.179 No, P(Y ≥ 350) ≤
1 = .1126. (2.98)2
3.181 a b c d e
p = Fraction defective P(acceptance) 0 1 .10 .5905 .30 .1681 .50 .0312 1.0 0
3.185 a .2277 b Not unlikely 3.187 a .023 b 1.2 c $1.25 3.189 1 − (.99999)10,000 3.191 V (Y ) = .4 3.193 .476
3.195 a .982 b P(W ≥ 1) = 1 − e−12 3.197 a .9997 b n=2 3.199 a .300 b .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) = .125 b P(Y > 10) = .014 3.209 .0837 3.211 3 3.213 a .1192 b .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, b c 4.9 a b
P(2 < Y < 5) = .289, so not equal P(2 ≤ Y ≤ 5) = 0.618, P(2 < Y ≤ 5) = 0.316, so not equal Y is not a continuous random variable, so the earlier results do not hold. Y is a discrete random variable These values are 2, 2.5, 4, 5.5, 6, and 7.
1 1 , 8 16 1 5 , p(5.5) = , p(4) = 16 8 5 1 , p(7) = p(6) = 16 16 d φ.5 = 4 1 4.11 a c = 2 y2 b F(y) = , 0 ≤ y ≤ 2 4
c p(2) = , p(2.5) =
d .75 e .75
4.13 a
b c 4.15 a
0 y<0 2 y 0≤y≤1 2 F(y) = 1 1 < y ≤ 1.5 y− 2 1 y > 1.5 .125 .575 For b ≥ 0, f (y) ≥ 0; also, #∞ f (y) = 1
−∞
b y
b F(y) = 1 − , for y ≥ b; c d 4.17 a b
0 elsewhere. b (b + c) (b + c) (b + d) 3 c= 2 y2 y3 + , for 0 ≤ y ≤ 1 F(y) = 2 2
882
Answers
d F(−1) = 0, F(0) = 0, F(1) = 1 3 e 16 104 f 123
4.19
4.21 4.25 4.27 4.29 4.31 4.33
4.37 4.39 4.45
4.47
4.49 4.51 4.53
4.55 4.57 4.59
0 y≤0 .125 0 24) = = .2 5 3 a P(Y > 2) = 4 4 +9 b c0 + c1 3 3 4 1 3 1 a 8 1 b 8 1 c 4 2 a 7 b µ = .015, V (Y ) = .00041 π 3 D = .0000065π , E 6 π V D 3 = .0003525π 2 6 a z0 = 0
b z 0 = 1.10 c z 0 = 1.645 d z 0 = 2.576 4.63 a P(Z > 1) = .1587 b The same answer is obtained. 4.65 $425.60 4.67 µ = 3.000 in. 4.69 .2660 4.71 a .9544 b .8297 4.73 a .406 b 960.5 mm 4.75 µ = 7.301 4.77 a 0.758 b 22.2 4.87 a φ.05 = .70369. b φ.05 = .35185 4.89 a β = .8 b P(Y ≤ 1.7) = .8806 4.91 a .1353 b 460.52 cfs 4.93 a .5057 b 1936 4.97 .3679 4.99 a .7358 4.101 a E(Y ) = 1.92 b P(Y > 3) = .21036 d P(2 ≤ Y ≤ 3) = .12943 4.103 E(A) = 200π, V (A) = 200,000π 2 4.105 a E(Y ) = 3.2, V (Y ) = 6.4 b P(Y > 4) = .28955 4.107 a (0, 9.657), because Y must be positive.
b P(Y < 9.657) = .95338
4.109 E(L) = 276, = 47,664 V(L)
√ 1 / (α) if α > 0 β α + 2 (α − 12 ) 1 if α > 1, √ e β(α − 1) β(α) 1 1 if α > , 2 2 β (α − 1)(α − 2) if α > 2 a k = 60 b φ.95 = 0.84684 1 3 E(Y ) = , V (Y ) = 5 25 52 , V (C) = 29.96 E(C) = 3 a .75 b .2357 a c = 105
4.111 d
4.123 4.125 4.129 4.131 4.133
Answers 883
3
b µ= 8 c σ = .1614 d .02972
4.139 m X (t) = exp{t (4−3µ)+(1/2)(9σ 2 t 2 )}
4.141 4.143 4.145
4.147 4.149 4.151
4.153 4.155 4.157
4.159
normal, E(X ) = 4 − 3µ, V (X ) = 9σ 2 , uniqueness of moment-generating functions tθ e 2 − etθ1 m(t) = t (θ2 − θ1 ) αβ, αβ 2 2 a 5 1 b (t + 1) c 1 1 σ = 2 1 The value 2000 is only .53 standard deviation above the mean. Thus, we would expect C to exceed 2000 fairly often. (6.38, 28.28) $113.33 a F(x) = x <0 0, (1/100)e−x/100 , 0 ≤ x < 200 1, x ≥ 200 b 86.47 a F1 (y) = 0 y<0 .1 = .4 0 ≤ y < 5 ; .1 + .15 1 y ≥ .5 F2 (y) =
4.161 4.163 4.165
4.167 4.169 4.171 4.173 4.175 4.179 4.181 4.183
4.187 4.189 4.191 4.193 4.195
0 y<0 2 4y /3 0 ≤ y < .5 (4y − 1)/3 .5 ≤ y < 1 1 y≥1 b F(y) = 0.25F1 (y) + 0.75F2 (y) c E(Y ) = .533, V (Y ) = .076 φ.9 = 85.36 1 − (.927)5 = .3155 a c=4 b E(Y ) = 1, V (Y ) = .5 1 c m(t) = ,t <2 (1 − .5t)2 (α + β)(k + α) E(Y k ) = (α)(k + α + β) e−2.5 = .082 1 1 a E(W ) = , V (W ) = 2 4 b 1 − e−6 2 f (r ) = 2λπr e−λπr , r > 0 √ 2 = 1.414 k = (.4)1/3 = .7368 m(t) = exp(t 2 /2); 0; 1 a E(Y ) = 598.74 g V (Y ) = e22 (e16 − 1)10−4 b (0, 3,570,236.1) c .8020 a e−2.5 = .082 b .0186 E(Y ) = 0. Also, it is clear that 1 V (Y ) = E(Y 2 ) = . n−1 −4 c 1−e 150 a 12 b w = 120
Chapter 5 0
y1 1
2
0
1 9
2 9
1 9
1
2 9
2 9
0
2
1 9
0
0
5.1 a
y2
b F(1, 0) =
5.3
1 3
4 3 2 y1 y2 3 − y1 − y2 , where 9 3
0 ≤ y1 , 0 ≤ y2 , and y1 + y2 ≤ 3. .1065 .5 .00426 .8009 k=6 31 b 64 29 5.11 a 32 1 b 4
5.5 a b 5.7 a b 5.9 a
884
Answers 1 1 , = 2 2 1 , 2 = F 2 .65625 e−1 − 2e−2 1 2 e−1
5.13 a F b c 5.15 a b c 5.17 .50 5.19 a
y1
0
p1 (y1 )
b c 2 4 9
1 9
b No 5.21 a Hypergeometric with N = 9, b c 5.23 a b c 5.25 a b c d e f g 5.27 a b c d e 5.29 a
b 5.31 a
n = 3, and r = 4. 2 3 8 15 3 3 f 2 (y2 ) = − y22 , 0 ≤ y2 ≤ 1 2 2 Defined over y2 ≤ y1 ≤ 1 if y2 ≥ 0 1 3 f 1 (y1 ) = e−y1 , y1 > 0; f 2 (y2 ) = e−y2 , y2 > 0 P(1 < Y1 < 2.5) = P(1 < Y2 < 2.5) = e−1 − e−2.5 = .2858 y2 > 0 f (y1 |y2 ) = f 1 (y1 ) = e−y1 , y1 > 0 f (y2 |y1 ) = f 2 (y2 ) = e−y2 , y2 > 0 same same f 1 (y1 ) = 3(1 − y1 )2 , 0 ≤ y1 ≤ 1; f 2 (y2 ) = 6y2 (1 − y2 ), 0 ≤ y2 ≤ 1 32 63 1 f (y1 |y2 ) = , 0 ≤ y1 ≤ y2 , y2 if y2 ≤ 1 2(1 − y2 ) f (y2 |y1 ) = , (1 − y1 )2 y1 ≤ y2 ≤ 1 if y1 ≥ 0 1 4 f 2 (y2 ) = 2(1 − y2 ), 0 ≤ y2 ≤ 1; f 1 (y1 ) = 1 − |y1 |, for −1 ≤ y1 ≤ 1 1 3 f 1 (y1 ) = 20y1 (1 − y1 )2 , 0 ≤ y1 ≤ 1
15(1 + y2 )2 y22 ,
−1 ≤ y2 < 0
15(1 − y2 )2 y22 , (y2 |y1 ) = 32 y22 (1
0 ≤ y2 ≤ 1
− y1 )−3 , for y1 − 1 ≤ y2 ≤ 1 − y1 .5 f 1 (y1 ) = y1 e−y1 , y1 ≥ 0; f 2 (y2 ) = e−y2 , y2 ≥ 0 f (y1 |y2 ) = e−(y1 −y2 ) , y1 ≥ y2 f (y2 |y1 ) = 1/y1 , 0 ≤ y2 ≤ y1
c f d 5.33 a
1 4 9
b f 2 (y2 ) =
9 16 13 16
5.35 .5 5.37 e−1 5.41 5.45 5.47 5.51
5.53 5.55 5.57 5.59 5.61 5.63 5.65 5.69
1 4 No Dependent a f (y1 , y2 ) = f 1 (y1 ) f 2 (y2 ) so that Y1 and Y2 are independent. b Yes, the conditional probabilities are the same as the marginal probabilities. No, they are dependent. No, they are dependent. No, they are dependent. No, they are dependent. Yes, they are independent. 1 4 Exponential, mean 1 1 −(y1 +y2 )/3 e a f (y1 , y2 ) = , 9 y1 > 0, y2 > 0 b P(Y1 + Y2 ≤ 1) = 4 1 − e−1/3 = .0446 3 1
5.71 a
4 23 b 144 4 5.73 3 5.75 a 2 b .0249 c .0249 d 2 e They are equal. 1 1 5.77 a ; 4 2 3 b E(Y12 ) = 1/10, V (Y1 ) = , 80 3 1 , V (Y2 ) = E(Y22 ) = 10 20 5 c − 4
Answers 885
5.79 5.81 5.83 5.85
5.87 5.89 5.91 5.93
5.95
5.97
5.99 5.101 5.103 5.105 5.107 5.109 5.113
0 1 1
5.115 b V (Y ) = 38.99 √ c The interval is 14.7 ± 2 38.99 or
a E(Y1 ) = E(Y2 ) = 1 (both
5.117 p1 − p2 ,
(0, 27.188)
N −n [ p 1 + p 2 − ( p 1 − p 2 )2 ] n(N − 1) 5.119 a .0823 n 2n b E(Y1 ) = , V (Y1 ) = 3 9 n c Cov(Y2 , Y3 ) = − 9 2n d E(Y2 − Y3 ) = 0, V (Y2 − Y3 ) = 3 5.121 a .0972 b .2; .072 5.123 .08953 5.125 a .046 b .2262 5.127 a .2759 b .8031 y 5.133 a 2 2 1 b 4 3 5.135 a 2 b 1.25 3 5.137 8 y1 −1 0 1 y2 0 1 5.139 a nαβ 1 1 1 2 1 p1 (y1 ) p2 (y2 ) b λαβ 3 3 3 3 3 λ 2λ2 5.141 E(Y2 ) = , V (Y2 ) = Cov(Y1, Y2 ) = 0 2 3 a 2 5.143 m U (t) = (1 − t 2 )−1/2 , E(U ) = 0, b Impossible V (U ) = 1 1 c 4 (a perfect positive linear 5.145 association) 3 11 d −4 (a perfect negative linear 5.147 36 association) 5.149 a f (y1 ) = 3y12 , 0 ≤ y1 ≤ 1 0 3 α f (y2 ) = (1 − y22 ), 0 ≤ y2 ≤ 1 a − 2 4 E(3Y1 + 4Y2 − 6Y3 ) = −22, 23 b V (3Y1 + 4Y2 − 6Y3 ) = 480 44 1 2y1 c f (y1 |y2 ) = , y2 ≤ y1 ≤ 1 9 (1 − y22 ) E(Y1 + Y2 ) = 2/3 and 5 d 1 12 V (Y1 + Y2 ) = 18 5.157 p(y) = y α (11.48, 52.68) β 1 y+α−1 , E(G) = 42, V (G) = 25; the value $70 y β +1 β +1 70 − 42 y = 0, 1, 2, . . . is = 7.2 standard deviations 5 5.161 E(Y¯ − X¯ ) = µ1 − µ2 , V (Y¯ − X¯ ) = above the mean, an unlikely value. σ12 /n + σ22 /m marginal distributions are exponential with mean 1) b V (Y1 ) = V (Y2 ) = 1 c E(Y1 − Y2 ) = 0 α d E(Y1 Y2 ) = 1 − , so 4 α Cov(Y1 , Y2 ) = − 14 1 α α −2 , 2 2 + 2 + e 2 2 a E(Y1 + Y2 ) = ν1 + ν2 b V (Y1 + Y2 ) = 2ν1 + 2ν2 2 Cov(Y1, Y2 ) = − . As the value of Y1 9 increases, the value of Y2 tends to decrease. Cov(Y1, Y2 ) = 0 a 0 b Dependent c 0 d Not necessarily independent The marginal distributions for Y1 and Y2 are
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 ( p1 et1 + p2 et2 + p3 et3 )n b m(t, 0, 0) c Cov(X 1, X 2 ) = −np1 p2
Chapter 6 1−u , −1 ≤ u ≤ 1 2 u+1 , −1 ≤ u ≤ 1 b 2 1 c √ − 1, 0 ≤ u ≤ 1 u 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 b $fU (u) = (u + 4)/100, −4 ≤ u ≤ 6 1/10, 6 < u ≤ 11 c 5.5833 1 u − 3 −1/2 fU (u) = , 16 2 5 ≤ u ≥ 53 1 a fU (u) = √ √ u −1/2 e−u/2 , π 2 u≥0 b U has a gamma distribution with α = 1/2 and β = 2 (recall that √ (1/2) = π ). a fU (u) = 2u, 0 ≤ u ≤ 1 b E(U ) = 2/3 c E(Y1 + Y2 ) = 2/3 a fU (u) = 4ue−2u , u ≥ 0, a gamma density with α = 2 and β = 1/2 b E(U ) = 1, V (U ) = 1/2 u fU (u) = FU (u) = 2 e−u/β , u > 0 β [−ln(1 − U )]1/2 αy α−1 a f (y) = ,0≤ y ≤θ θα 1/α b Y = θU √ c y = 4 u. The values are 2.0785, 3.229, 1.5036, 1.5610, 2.403. fU (u) = 4ue−2u for u ≥ 0 2 2 a f Y (y) = we−w /β , w ≥ 0, which β is Weibull density with m = 2. k b E(Y k/2 ) = + 1 β k/2 2
6.1 a
6.3
6.5 6.7
6.9 6.11
6.13 6.15 6.17
6.25 6.27
6.29 a f W (w) =
1 w 1/2 e−w/kT w > 0 (kT )3/2 2 3 b E(W ) = kT 2 2 fU (u) = ,u ≥0 (1 + u)3 fU (u) = 4(80 − 31u + 3u 2 ), 4.5 ≤ u ≤ 5 fU (u) = − ln(u), 0 ≤ u ≤ 1 a m Y1 (t) = 1 − p + pet b m W (t) = E(et W ) = [1 − p + pet ]n fU (u) = 4ue−2u , u ≥ 0 a Y¯ has a normal distribution with mean µ and variance σ 2 /n b P(|Y¯ − µ| ≤ 1) = .7888 c The probabilities are .8664, .9544, .9756. So, as the sample size increases, so does the probability that P(|Y¯ − µ| ≤ 1) c = $190.27 P(U > 16.0128) = .025 The distribution of Y1 + (n 2 − Y2 ) is binomial with n 1 + n 2 trials and success probability p = .2 a Binomial (nm, p) where ni = m b Binomial (n 1 = n 2 + · · · n n , p) c Hypergeometric (r = n, N = n1 + n2 + · · · nn ) P(Y ≥ 20) = .077 a f (u 1 , u 2 ) = 1 −[u 2 +(u 2 −u 1 )2 ]/2 e 1 = 2π 1 −(2u 2 −2u 1 u 2 +u 2 )/2 1 2 e 2π b E(U1 ) = E(Z 1 ) = 0, E(U2 ) = E(Z 1 + Z 2 ) = 0, V (U1 ) = V (Z 1 ) = 1, V (U2 ) = V (Z 1 + Z 2 ) = V (Z 1 ) + V (Z 2 ) = 2, Cov(U1 , U2 ) = E(Z 12 ) = 1
6.31 6.33 6.35 6.37 6.39 6.43
6.45 6.47 6.51 6.53
6.55 6.65
3
Answers 887
c Not independent since
6.89 f R (r ) = n(n − 1)r n−2 (1 − r ),
d This is the bivariate normal
6.93 f (w) =
ρ = 0.
6.69
6.73 6.75 6.77
6.81 6.83 6.85 6.87
distribution with µ1 = µ2 = 0, 1 σ12 = 1, σ22 = 2, and ρ = √ 2 1 a f (y1 , y2 ) = 2 2 , y1 > 1, y1 y2 y2 > 1 e No a g(2) (u) = 2u, 0 ≤ u ≤ 1 b E(U2 ) = 2/3, V (U2 ) = 1/18 (10/15)5 n! a ( j − 1)!(k − 1 − j)!(n − k)! j−1 y j [yk − y j ]k−1− j [θ − yk ]n−k θn 0 ≤ y j < yk ≤ θ (n − k + 1) j 2 θ b (n + 1)2 (n + 2) (n − k + j + 1)(k − j) 2 θ c (n + 1)2 (n + 2) −9 b 1−e 1 − (.5)n .5 a g(1) (y) = e−(y−4) , y ≥ 4
6.95
6.97
,
6.101 6.103 6.105 6.107 6.109
b E(Y(1) ) = 5
0≤r ≤1 2 1 √ −w ,0≤w ≤1 3 w 1 0≤u≤1 2 a fU1 (u) = 1 u>1 2u 2 −u b fU2 (u) = ue , 0 ≤ u c Same as Ex. 6.35. 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) = .0002 fU (u) = 1, 0 ≤ u ≤ 1 Therefore, U has a uniform distribution on (0, 1) 1 , ∞ < u1 < ∞ π(1 + u 21 ) 1 u β−1 (1 − u)α−1 , 0 < u < 1 B(α, β) 1 0≤u<1 √ 4 u fU (u) = 1 √ 1≤u≤9 8 u P(U = C1 − C3 ) = .4156; P(U = C2 − C3 ) = .5844
Chapter 7 7.9 a .7698 b For n = 25, 36, 69, and 64, the
7.11 7.13 7.15
7.17 7.19 7.21 7.27
probabilities are (respectively) .8664, .9284, .9642, .9836. c The probabilities increase with n. d Yes .8664 .9876 a E( X¯ − Y¯ ) = µ1 − µ2 b V ( X¯ − Y¯ ) = σ12 /m + σ22 /n c The two sample sizes should be at least 6 18.2 P Z ≤ 6 = .57681 i=1 i P(S 2 ≥ .065) = .10 a b = 2.42 b a = .656 c .95 a .17271 b .23041 d .40312
7.31 a c d 7.35 a b c 7.39
7.43 7.45 7.47 7.49 7.51 7.53 7.55
5.99, 4.89, 4.02, 3.65, 3.48, 3.32 13.2767 13.2767/3.32 ≈ 4 E(F) = 1.029 V (F) = .076 3 is 7.15 standard deviations above this mean; unlikely value. a normal, E(θˆ ) = θ = c1 µ1 + c 2 µ2 + · · · + ck µk c2 c2 c2 V (θˆ ) = 1 + 2 + · · · + k σ 2 n1 n2 nk b χ 2 with n 1 + n 2 + · · · + n k − k df c t with n 1 + n 2 + · · · + n k − k df .9544 .0548 153 .0217 664 b Y¯ is approximately normal: .0132. a random sample; approximately 1. b .1271
888
Answers
7.57 7.59 7.61 7.63 7.65 7.67
7.71 7.73 7.75 7.77 7.79
7.81
.0062 .0062 n = 51 56 customers a Exact: .91854; normal approximation: .86396. 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) a n>9 b n > 14, n > 14, n > 36, n > 36, n > 891, n > 8991 .8980 .7698 61 customers a Using the normal approximation: .7486. b Using the exact binomial probability: .729. a .5948
b With p = .2 and .3, the 7.83 7.85 7.87 7.89 7.91
probabilities are .0559 and .0017 respectively. a .36897 b .48679 .8414 .0041 µ = 10.15 Since X , Y , and W are normally ¯. distributed, so are X¯ , Y¯ , and W µU = E(U ) = .4µ1 +.2µ2 +.4µ3 2 σ1 σU2 = V (U ) = .16 n1 2 2 σ3 σ2 + .16 + .04 n2 n3
7.95 a F with num. df = 1, denom. df = 9 b F with num. df = 9, denom. df = 1 c c = 49.04 7.97 b .1587 7.101 .8413 7.103 .1587 7.105 .264
Chapter 8 8.3 a B(θˆ ) = aθ + b − θ = (a − 1)θ + b b Let θˆ ∗ = (θˆ − b)/a 8.5 a MSE(θˆ ∗ ) = V (θˆ ∗ ) = V (θˆ )/a 2 σ22 − c σ12 + σ22 − 2c Y¯ − 1 θˆ3 − 9θˆ2 + 54 2 b [n 1)](Y /n)[1 − (Y /n)] /(n − 1 β a 3n − 1 2 ˆ = β2 b MSE(β) (3n − 1)(3n − 2) a (1 − 2 p)/(n + 2) np(1 − p) + (1 − 2 p)2 b (n + 2)2 c p will be close to .5. MSE(θˆ ) = β 2 11.5 ± .99 a 11.3 ± 1.54 b 1.3 ± 1.7 c .17 ± .08 a −.7 b .404 a .601 ± .031 a −.06 ± .045
8.7 a = 8.9 8.11 8.13 8.15
8.17
8.19 8.21 8.23 8.25 8.27 8.29
8.31 a −.03 ± .041 8.33 .7 ± .205 8.35 a 20 ± 1.265 b −3 ± 1.855, yes 8.37 1020 ± 645.1 8.39 8.41 8.43 8.45 8.47 8.49 8.57 8.59 8.61 8.63 8.65 8.67 8.69 8.71
2Y 2Y , 9.48773 .71072 a (Y 2 /5.02389, Y 2 /.0009821) b Y 2 /.0039321 c Y 2 /3.84146 b [Y(n) ](.95)−1/n a Y /.05132 b 80% c (2.557, 11.864) c (3.108, 6.785) .51 ± .04 a .78 ± .021 (15.46, 36.94) a .78 ± .026 or (.754, .806) a .06 ± .117 or (−.057, .177) a 7.2 ± .751 b 2.5 ± .738 .22 ± .34 or (−.12, .56) n = 100
Answers 889
8.73 8.75 8.77 8.79 8.81 8.83 8.85 8.87 8.91 8.93
8.95 8.99
8.101
n = 2847 n = 136 n = 497 a n = 2998 b n = 1618 60.8 ± 5.701 a 3.4 ± 3.7 b .7 ± 3.32 −1 ± 4.72 (−.624, .122) (−84.39, −28.93) 1 3 4 + a 2 X¯ + Y¯ ± 1.96σ n m 1 3 4 + , where b 2 X¯ + Y¯ ± tα/2 S n m (Yi − Y¯ )2 + 1/3 (X i − X¯ )2 S2 = n+m−2 (.227, 0 2.196) (n − 1)S 2 a 2 χ1−α 0 (n − 1)S 2 b χα2 2 s = .0286; (.013 .125)
8.103 8.105 8.107 8.109 8.111 8.113 8.115 8.117 8.119 8.121 8.123 8.125
8.129 8.131 8.133
(1.407, 31.264); no 1 − 2(.0207) = .9586 765 seeds a .0625 ± .0237 b 563 n = 38,416 n = 768 (29.30, 391.15) 11.3 ± 1.44 3 ± 3.63 −.75 ± .77 .832 ± .015 σ2 S2 a 12 × 22 S σ1 2 S22 S22 b , F v ,v ,α/2 S12 Fv2 ,v1 ,α/2 S12 1 2 v i = n i − 1, i = 1, 2 2(n − 1)σ 4 a n2 1 c= n+1 2σ 4 b n1 + n2 − 2
Chapter 9 9.1 1/3; 2/3; 3/5 9.3 9.5 9.7 9.9 9.23 9.25 9.31 9.35
9.47
12n 2 b (n + 2)(n + 1)2 n−1 1/n a X6 = 1 c need Var(X 2i − X 2i−1 ) < ∞ b .6826 c No αβ a Y¯ n is unbiased for µ. 1 n b V (Y¯ n ) = 2 σ2 i=1 i n n ln(Yi ); no
9.75
9.77 9.81 9.83 9.85
i=1
9.57 Yes
1 9.59 3 Y¯ 2 + Y¯ 1 −
n n+1 Y(n) n 3n + 1 Y(n) 9.63 b 3n ¯ 2Y − 1 , no, not MVUE 9.69 θˆ = 1 − Y¯
9.61
1 n Y 2. i=1 i n 1 n With m 2 = Y 2 , the MOM i=1 i n 1 − 2m 2 . estimator of θ is θˆ = 4m 2 − 1 2¯ Y 3 2 ¯ Y 1 a θˆ = Y(n) − 1 2 2 b Y(n) /12 1 a θˆ = Y¯ α b E(θˆ ) = θ, V (θˆ ) = θ 2 /(nα) n d Y n n i i=1 Yi 2 i=1 Yi 2 i=1 e , 31.4104 10.8508 pˆ A = .30, pˆ B = .38 pˆ C = .32; −.08 ± .1641 Y(n) /2 a Y(1) c [(α/2)1/2n Y(1) , (1 − (α/2))1/2n Y(1) ] a 1/Y¯ b 1/Y¯
9.71 σˆ 2 = m 2 =
9.87 9.91 9.93 9.97
890
Answers 1
pˆ (1 − pˆ ) n1 Y¯ exp(−2Y¯ ) 9.101 exp(−Y¯ ) ± z α/2 n n 1 9.103 Yi2 n i=1
9.99 pˆ ± z α/2
(Y − µ)2
i 9.105 σˆ 2 = n 9.107 exp(−t/Y¯ ) 9.109 a Nˆ 1 = 2Y¯ − 1
N2 − 1 3n 9.111 252 ± 85.193
b
Chapter 10 10.3 a c = 11 b .596 c .057 10.5 c = 1.684 10.7 a False b False c True d True e False f i True 10.17 10.21 10.23
10.25 10.27 10.29 10.33 10.35 10.37 10.39 10.41 10.43 10.45 10.47 10.49 10.51 10.53
10.55
ii True iii False a H0 : µ1 = µ2 , Ha : µ1 > µ2 c z = .075 z = 3.65, reject H0 a-b H0 : µ1 − µ2 = 0 vs. Ha : µ1 − µ2 = 0, which is a two–tailed test. c z = −.954, which does not lead to a rejection with α = .10. |z| = 1.105, do not reject z = −.1202, do not reject z = 4.47 z = 1.50, no z = −1.48 (1 = homeless), no approx. 0 (.0000317) .6700 .025 a .49 b .1056 .22 ± .155 or (.065, .375) .5148 129.146, yes z = 1.58 p–value = .1142, do not reject a z = −.996, p–value = .0618 b No c z = −1.826, p–value = .0336 d Yes z = −1.538; p–value = .0616; fail to reject H0 with α = .01
10.57 z = −1.732; p–value = .0836 10.63 a t = −1.341, fail to reject H0 10.65 a t = −3.24, p–value < .005, yes b Using the Applet, .00241 c 39.556 ± 3.55 10.67 a t = 4.568 and t.01 = 2.821 so reject H0 .
b The 99% lower confidence bound
54 is 358 − 2.821 √ = 309.83. 10 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 y¯ 1 = 97.856, s12 = .3403,
10.73
10.75 10.77
10.79
y¯ 2 = 98.489, s22 = .3011, t = −2.3724, −t.01 = −2.583, −t.025 = −2.12, so .02 < p–value < .05 b Using Applet, .03054 a t = 1.92, do not reject .05 < p–value < .10; applet p–value = .07084 b t = .365, do not reject p–value > .20; applet p–value = .71936 t = −.647, do not reject a t = −5.54, reject, p–value < .01; applet p–value approx. 0 b Yes c t = 1.56, .10 < p–value < .20; applet p–value = .12999 d Yes a χ 2 = 12.6, do not reject b .05 < p–value < .10 c Applet p–value = .08248
Answers 891
10.83 a σ12 =
σ22 2 b σ1 < σ22 c σ12 > σ22 10.85 χ 2 = 22.45, p–value < .005; applet
10.109 a λ =
m+n
b X¯ /Y¯ distributed as F with 2m and
p–value = .0001
10.89 a .15 b .45 c .75 d 1 10.91 a Reject if Y¯ >= 7.82. b .2611, .6406, .9131, .9909 10.93 n = 16 4 2 10.95 a U = β20 i=1 Yi has χ(24)
distribution under H0 : reject H0 if U > χα2 b Yes 10.97 d Yes, is UMP n 10.99 a Yi ≥ k i=1
b Use Poisson table to find k such Yi ≥ k) = α
that P(
c Yes 10.101 a
n
Yi < c
i=1
b Yes √ 10.103 a Reject H0 if Y(n) ≤ θ0 n α b Yes (n −
+ (m − has σ02 2 χ(n+m−2) distribution under H0 ; reject if χ 2 > χα2
10.107 χ 2 =
1)S12
1)S22
( X¯ )m (Y¯ )m m+n m X¯ + n Y¯
2n degrees of freedom
10.115 a True b False c False d True e False f False g False h False i True 10.117 a t = −22.17, p–value < .01 b −.0105 ± .001 c Yes d No 10.119 a H0 : p = .20, Ha : p > .20 b α = .0749 10.121 z = 5.24, p–value approx. 0 10.123 a F = 2.904, no b (.050, .254) 10.125 a t = −2.657, .02 < p–value < .05 b −4.542 ± 3.046 10.127 T =
1+a+b n(3n−3)
¯ )−(µ1 −µ2 −µ3 ) ( X¯ +Y¯ −W . 1/2 ¯ )2 (X i − X¯ )2 + a1 (Yi −Y¯ )2 + b1 (Wi −W
with (3n − 3) degrees of freedom n n i=1 (yi − y(1) ) 10.129 λ = × n nθ1,0 (yi − y(1) ) exp − i=1 +n . θ1,0
Chapter 11 11.3 yˆ = 1.5 − .6x 11.5 yˆ = 21.575 + 4.842x 11.7 a The relationship appears to be b c 11.9 b d 11.11 βˆ 1 11.13 a 11.17 a b
proportional to x 2 . No No, it is the best linear model. yˆ = −15.45 + 65.17x 108.373 = 2.514 The least squares line is yˆ = 452.119 − 29.402x SSE = 18.286; S 2 = 18.286/6 = 3.048 The fitted line is yˆ = 43.35 + 2.42x ∗ . The same
11.19 a c 11.23 a b c d 11.25 a b c d
answer for SSE (and thus S 2 ) is found. The least squares line is: yˆ = 3.00 + 4.75x s 2 = 5.025 t = −5.20, reject H0 .01 < p–value < .02 .01382 (−.967, −.233) t = 3.791, p–value < .01 Applet p–value = .0053 Reject .475 ± .289
892
Answers
βˆ − γˆ1
11.29 T = 1 1
11.33 11.35 11.37 11.39 11.41 11.43 11.45 11.47 11.51 11.53
11.57 11.59 11.61 11.63
11.67 11.69 11.73
+
1 Scc
11.75 21.9375 ± 3.01 11.77 Following Ex. 11.76, the 95%
PI = 39.9812 ± 213.807 11.79 21.9375 ± 6.17 (SSEY + SSEW )/(n + m − 4). 11.83 a F = 21.677, reject H0 is rejected in favor of Ha for large b SSE R = 1908.08 values of |T |. 11.85 a F = 40.603, p–value < .005 t = 73.04, p–value approx. 0, H0 is b 950.1676 rejected 11.87 a F = 4.5, F1 = 9.24, fail to t = 9.62, yes reject H0 x ∗ = x¯ . c F = 2.353, F1 = 2.23, reject H0 (4.67, 9.63) 11.89 a True 25.395 ± 2.875 b False b (72.39, 75.77) c False (59.73, 70.57) 11.91 F = 10.21 (−.86, 15.16) 11.93 90.38 ± 8.42 (.27, .51) 11.95 a yˆ = −13.54 − 0.053x t = 9.608, p–value < .01 b t = −6.86 a r 2 = .682 c .929 ± .33 b .682 11.97 a yˆ = 1.4825+.5x1 +.1190x2 −.5x3 c t = 4.146, reject b yˆ = 2.0715 d Applet p–value = .00161 c t = −13.7, reject a sign for r d (1.88, 2.26) b r and n e (1.73, 2.41) r = −.3783 11.99 If −9 ≤ x ≤ 9, choose n/2 at x = −9 .979 ± .104 and n/2 at x = 9. a βˆ 1 = −.0095, βˆ 0 = 3.603 and 11.101 a yˆ = 9.34 + 2.46x1 + .6x2 + .41x1 x2 αˆ 1 = −(−.0095) = .0095, b 9.34 , 11.80 αˆ 0 = exp(3.603) = 36.70. d For bacteria A, yˆ = 9.34. For Therefore, the prediction equation bacteria B, yˆ = 11.80. The is yˆ = 36.70e−.0095x . observed growths were 9.1 and b The 90% CI for α is 0 12.2, respectively. e3.5883 , e3.6171 = (36.17, 37.23) e 12.81 ± .37 yˆ = 2.1 − .6x f 12.81 ± .78 a yˆ = 32.725 + 1.812x b yˆ = 35.5625 + 1.8119x − .1351x 2 11.107 a r = .89 b t = 4.78, p–value <.01, reject t = 1.31, do not reject S
11.31
1 Sx x
, where S =
Chapter 12 12.1 n 1 = 34, n 2 = 56 12.3 n = 246, n 1 = 93, n 2 = 154 12.5 With n = 6, three rats should receive
x = 2 units and three rats should receive 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, reject 12.17 a µi 12.31 a µi 1
b µi , [σ P2 + σ 2 ] n c µ1 − µ2 , 2σ 2 /n, normal 12.35 a t = −4.326, .01 < p–value < .025
b −1.58 ± 1.014 c 65 pairs
12.37 k1 = k3 = .25; k2 = .50
Answers 893
Chapter 13 13.1 a F = 2.93, do not reject b .109 c |t| = 1.71, do not reject, F = t 2 13.7 a F = 5.2002, reject b p–value = .01068 13.9 SSE = .020; F = 2.0, do not reject
13.11 SST = .7588; SSE = .7462;
F = 19.83, p–value < .005, reject
13.49 13.53 13.55 13.57 13.59 13.61 13.63 13.69
13.13 SST = 36.286; SSE = 76.6996;
F = 38.316, p–value < .005, reject
13.15 F = 63.66, yes, p–value < .005 13.21 a −12.08 ± 10.96 b Longer c Fewer degrees of freedom 13.23 a 1.568 ± .164 or (1.404, 1.732); yes b (−.579, −.117); yes 13.25 .28 ± .102 13.27 a 95% CI for µ A : 76 ± 8.142
13.29 13.31 13.33 13.35 13.37 13.39 13.41
13.45
or (67.868, 84.142) b 95% CI for µ B : 66.33 ± 10.51 or (55.82, 76.84) c 95% CI for µ A − µ B : 9.667 ± 13.295 a 6.24 ± .318 b −.29 ± .241 a F = 1.32, no b (−.21, 4.21) (1.39, 1.93) a 2.7 ± 3.750 b 27.5 ± 2.652 a µ b Overall mean b (2σ 2 )/b a F = 3.11, do not reject b p–value > .10 c p–value = .1381 d s D2 = 2MSE a F = 10.05; reject b F = 10.88; reject
13.47 Source Treatments Blocks Error Total
df SS MS F 3 8.1875 2.729 1.40 3 7.1875 2.396 1.23 9 17.5625 1.95139 15 32.9375
F = 1.40, do not reject
13.71 13.73 13.75 13.77 13.79
13.81
F = 6.36; reject The 95% CI is 2 ± 2.83. The 95% CI is .145 ± .179. The 99% CI is −4.8 ± 5.259. nA ≥ 3 b = 16; n = 48 Sample sizes differ. 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 in block III. F = 7; H0 is rejected As homogeneous as possible within blocks. b F = 1.05; do not reject a A 95% CI is .084 ± .06 or (.024, .144). a 16 b 135 degrees of freedom left for error. c 14.14 F = 7.33; yes; blocking induces loss in degrees of freedom for estimating σ 2 ; could result in sight loss of information if block to block variation is small
13.83 a Source Treatments Blocks Error Total
df SS MS F 2 524,177.167 262,088.58 258.237 3 173,415 57,805.00 56.95 6 6,089.5 1,014.9167 11 703,681.667
b c d e f 13.85 a
6 Yes, F = 258.19, p–value < .005 Yes, F = 56.95, p–value < .005 22.527 −237.25 ± 55.13 SST = 1.212, df = 4 SSE = .571, df = 22 F = 11.68; p–value < .005 b |t| = 2.73; H0 is rejected; 2(.005) < p–value < 2(.01). 13.87 Each interval should have confidence coefficient 1 − .05/4 = .9875 ≈ .99; µ A − µ D : .320 ± .251 µ B − µ D : .145 ± .251 µC − µ D : .023 ± .251 µ E − µ D : −.124 ± .251
894
Answers
13.89 b c 13.91 a b
σβ2 σβ2 = 0 2 µ; σ B2 + k1 σ εk b 2 σβ2 + k−1 i=1 τi
c σε2 + kσ B2 d σε2
Chapter 14 14.1 a X 2 = 3.696, do not reject b Applet p–value = .29622 14.3 X 2 = 24.48, p–value < .005 14.5 a z = 1.50, do not reject b Hypothesis suggested by observed data
14.7 .102 ± .043 14.9 a .39 ± .149 b .37 ± .187, .39 ± .182, .48 ± .153 14.11 X 2 = 69.42, reject 14.13 a X 2 = 18.711, reject b p–value < .005 c Applet p–value = .00090 14.15 b X 2 also multiplied by k 14.17 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
X 2 = 22.8705, reject p–value < .005 X 2 = 13.99, reject X 2 = 13.99, reject X 2 = 1.36, do not reject X 2 = 19.1723, p-value = 0.003882, reject c −.11 ± .135 X 2 = 38.43, yes a X 2 = 14.19, reject X 2 = 21.51, reject X 2 = 6.18, reject; .025 < p–value < .05 a Yes b p–value = .002263 X 2 = 8.56, df = 3; reject X 2 = 3.26, do not reject X 2 = 74.85, reject
14.19 a b 14.21 a b c 14.25 b 14.27 14.29 14.31 14.33 14.35 14.37 14.41 14.43
Chapter 15 15.1 Rejection region M ≤ 6 or M ≥ 19 P(M ≤ 6) + M ≤ 7 or M ≥ 18 P(M ≤ 7) + M ≤ 8 or M ≥ 17 P(M ≤ 8) +
α P(M ≥ 19) = .014 P(M ≥ 18) = .044 P(M ≥ 17) = .108
15.3 a m = 2, yes b Variances not equal 15.5 P(M ≤ 2 or M ≥ 8) = .11, no 15.7 a P(M ≤ 2 or M ≥ 7) = .18, do not reject
b t = −1.65, do not reject 15.9 a p–value = .011, do not reject 15.11 T = min(T + , T − ), T = T − . 15.13 a T = 6, .02 < p–value < .05 b T = 6, 0.1 < p–value < .025 15.15 T = 3.5, .025 < p–value < .05 15.17 T = 11, reject 15.21 a U = 4; p–value = .0364 b U = 35; p–value = .0559 c U = 1; p–value = .0476
15.23 15.25 15.27 15.29 15.31
15.33 15.37 15.39 15.41
15.45 15.47
U = 9, do not reject z = −1.80, reject U = 0, p–value = .0096 H = 16.974, p-value < .001 a SST = 2586.1333; SSE = 11,702.9; F = 1.33, do not reject b H = 1.22, do not reject H = 2.03, do not reject a No, p–value = .6685 b Do not reject H0 Fr = 6.35, reject a Fr = 65.675, p–value < .005, reject b m = 0, P(M = 0) = 1/256, p–value = 1/128 The null distribution is given by P(Fr = 0) = P(Fr = 4) = 1/6 and P(Fr = 1) = P(Fr = 3) = 1/3. R = 6, no
Answers
15.49 a .0256 b An usually small number of runs
15.61 a b c 15.63 T
15.51 15.53 15.55
15.65 15.67 15.69 15.71 15.73
15.57 15.59
(judged at α = .05) would imply a clustering of defective items in time; do not reject. R = 13, do not reject r S = .911818; yes. a r S = −.8449887 b Reject r S = .6768, use two-tailed test, reject r S = 0; p–value < .005
Randomized block design No p–value = .04076, yes = 73.5, do not reject, consistent with Ex. 15.62 U = 17.5, fail to reject H0 .0159 H = 7.154, reject Fr = 6.21, do not reject .10
Chapter 16 16.1 a b c d e 16.3 c e 16.7 a b 16.9 b
16.11 e
β(10, 30) n = 25 β(10, 30), n = 25 Yes Posterior for the β(1, 3) prior. Means get closer to .4, std dev decreases. Looks more and more like normal distribution. Y +1 n+4 np + 1 np(1 − p) ; n + 4 (n + 4)2 α+1 ; α+β +Y (α + 1)(β + Y − 1) (α+ β + Y+ 1)(α+ β + Y ) 1 nβ ¯ + αβ Y nβ + 1 nβ + 1
895
16.13 a (.099, .710) b Both probabilities are .025 c P(.099 < p < .710) = .95 h 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 and alternative are .1275 and .8725, respectively, accept Ha . 16.25 Posterior probabilities of null and alternative are .9700 and .0300, respectively, accept H0 .
