Chapter 1 1.5 a 2.45 − 2.65, 2.65 − 2.85 b 7/30 c 16/30 1.9 a Approx. .68 b Approx. .95 c Approx. .815 d Approx. 0 1.13 a ¯y= 9.79; s = 4.14 b 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 .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: (−113.5, 535.1); k = 3: (−275.7, 697.3) 1.27 68% or 231 scores; 95% or 323 scores 1.29 .05 1.31 .025 1.33 (0.5, 10.5) 1.35 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(E5) = .10; P(E4) = .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 .09 b .19 2.17 a .08 b .16 c .14 d .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 have probability 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 5/9 c 5/9 877
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 2.55 a 90 10 b 20 4 70 6 + 90 10 = .111 2.57 (4 × 12)/1326 = .0362 2.59 a .000394 b .00355 2.61 a 364 n 365n b .5005 2.63 1/56 2.65 5/162 2.67 a P(A) = .0605 b .001344 c .00029 2.71 a 1/3 b 1/5 c 5/7 d 1 e 1/7 2.73 a 3/4 b 3/4 c 2/3 2.77 a .40 b .37 c .10 d .67 e .6 f .33 g .90 h .27 i .25 2.93 .364 2.95 a .1 b .9 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 b 1 n; 1 n c 3 7 2.125 1/12 2.127 a .857 c No; .8696 d Yes 2.129 .4 2.133 .9412 2.135 a .57 b .18 c .3158 d .90 2.137 a 2/5 b 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 .0083 2.149 a .4 b .6 c .25 2.151 4[ p4(1 − p) + p(1 − p)4] 2.153 .313 2.155 a .5 b .15 c .10 d .875
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 c (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 27 3.9 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 4 3.15 a P(Y = 0) = .1106, P(Y = 1) = .3594, P(Y = 2) = .3894, P(Y = 3) = .1406 c P(Y = 1) = .3594 d µ = E(Y ) = 1.56, σ2= .7488, σ = 0.8653 e (−.1706, 3.2906), P(0 ≤ Y ≤ 3) = 1 3.17 µ = E(Y ) = .889, σ2= V (Y ) = E(Y2)−[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)= $60,000 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%
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 1 n n− 1 n 5 3.87 E 1 Y = −p ln(p) 1− p 3.91 $150; 4500 3.93 a .04374 b .99144 3.95 .1 3.97 a .128 b .049 c µ = 15, σ2= 60 3.99 p(x) = y! (r − 1)!(y − r + 1)!prqy+1−r, y= r − 1, r, r + 1, . . . 3.101 a 5 11 b r y0 3.103 1 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 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 30 3.113 P(Y ≤ 1) = .187 3.115 p(0) =1 5, p(1) = 3 5, p(2) = 1 5 3.117 a P(Y = 0) = .553 b E(T ) = 9.5, V (T ) = 28.755, σ = 5.362 3.119 .016 3.121 a .090 b .143 c .857 d .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 2 c 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 6 3.167 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 8 c E(Y ) = 1.5, V (Y ) = .75, σ = .866
3.175 a 38.4 b 5.11 3.177 (61.03, 98.97) 3.179 No, P(Y ≥ 350) ≤ 1 (2.98)2 = .1126. 3.181
p= Fraction defective P(acceptance)
a 0 1 b .10 .5905 c .30 .1681 d .50 .0312 e 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, P(2 < Y < 5) = .289, so not equal b P(2 ≤ Y ≤ 5) = 0.618, P(2 < Y ≤ 5) = 0.316, so not equal
c Y is not a continuous random
variable, so the earlier results do not hold.
4.9 a Y is a discrete random variable
b 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 16 d φ.5= 4 4.11 a c= 1 2 b F(y) = y 2 4, 0≤ y ≤ 2 d .75 e .75 4.13 a F(y) = 0 y< 0 y2 2 0≤ y ≤ 1 y−1 2 1< y ≤ 1.5 1 y> 1.5 b .125 c .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. c b (b + c) d (b + c) (b + d) 4.17 a c= 3 2 b F(y) = y 3 2 + y2 2, for 0≤ y ≤ 1
d F(−1) = 0, F(0) = 0, F(1) = 1 e 3 16 f 104 123 4.19 a f(y) = 0 y≤ 0 .125 0< y < 2 .125y 2 ≤ y < 4 0 y≥ 4 b 7 16 c 13 16 d 7 9 4.21 E(Y ) = .708, V (Y ) = .0487 4.25 E(Y ) = 31/12, V (Y ) = 1.160 4.27 $4.65, .012 4.29 E(Y ) = 60, V (Y ) =1 3 4.31 E(Y ) = 4 4.33 a E(Y ) = 5.5, V (Y ) = .15
b Using Tchebysheff’s theorem, the interval is (5, 6.275). c Yes; P(Y ) = .5781 4.37 E(Y ) = 0 4.39 .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 4.51 1 3 4.53 a 1 8 b 1 8 c 1 4 4.55 a 2 7 b µ = .015, V (Y ) = .00041 4.57 E π 6D 3= .0000065π, V π 6D 3= .0003525π2 4.59 a z0= 0 b z0= 1.10 c z0= 1.645 d z0= 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) = 200,000π2 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, V(L) = 47,664 4.111 d √β α +1 2 / (α) if α > 0 e 1 β(α − 1)ifα > 1, (α −1 2) √ β(α) ifα > 1 2, 1 β2(α − 1)(α − 2) ifα > 2 4.123 a k= 60 b 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 .75 b .2357 4.133 a c= 105
b µ = 3 8 c σ = .1614 d .02972 4.139 mX(t) = exp{t(4−3µ)+(1/2)(9σ2t2)} normal, E(X) = 4 − 3µ, V (X) = 9σ2, uniqueness of moment-generating functions 4.141 m(t) = etθ2− etθ1 t(θ2− θ1) 4.143 αβ, αβ2 4.145 a 2 5 b 1 (t + 1) c 1 4.147 σ = 1 2 4.149 1
4.151 The value 2000 is only .53 standard deviation above the mean. Thus, we would expect C to exceed 2000 fairly often. 4.153 (6.38, 28.28) 4.155 $113.33 4.157 a F(x) = 0, x< 0 (1/100)e−x/100, 0 ≤ x < 200 1, x≥ 200 b 86.47 4.159 a F1(y) = 0 y< 0 .1 .1 + .15 = .4 0 ≤ y < 5 1 y≥ .5 ; F2(y) = 0 y< 0 4y2/3 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 4.161 φ.9= 85.36 4.163 1− (.927)5= .3155 4.165 a c= 4 b E(Y ) = 1, V (Y ) = .5 c m(t) = 1 (1 − .5t)2, t < 2 4.167 E(Yk) = (α + β)(k + α) (α)(k + α + β) 4.169 e−2.5= .082 4.171 a E(W) = 1 2, V(W) = 1 4 b 1− e−6 4.173 f(r) = 2λπre−λπr2, r> 0 4.175 √2= 1.414 4.179 k= (.4)1/3= .7368 4.181 m(t) = exp(t2/2); 0; 1 4.183 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= .082 b .0186
4.189 E(Y ) = 0. Also, it is clear that V(Y ) = E(Y2) = 1 n− 1. 4.191 c 1− e−4 4.193 150 4.195 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 4 y1 3 y2 2 3− y1− y2 9 3 , where 0≤ y1, 0≤ y2, and y1+ y2≤ 3. 5.5 a .1065 b .5 5.7 a .00426 b .8009 5.9 a k= 6 b 31 64 5.11 a 29 32 b 1 4
5.13 a F 1 2, 1 2 = 9 16 b F 1 2, 2 = 13 16 c .65625 5.15 a e−1− 2e−2 b 1 2 c e−1 5.17 .50 5.19 a y1 0 1 2 p1(y1) 4 9 4 9 1 9 b No 5.21 a Hypergeometric with N= 9, n= 3, and r = 4. b 2 3 c 8 15 5.23 a f2(y2) = 3 2− 3 2y 2 2, 0 ≤ y2≤ 1 b Defined over y2≤ y1≤ 1 if y2≥ 0 c 1 3 5.25 a f1(y1) = e−y1, y1> 0; f2(y2) = e−y2, y 2> 0 b P(1 < Y1< 2.5) = P(1 < Y2< 2.5) = e−1− e−2.5= .2858 c y2> 0
d f(y1|y2) = f1(y1) = e−y1, y1> 0
e f(y2|y1) = f2(y2) = e−y2, y2> 0
f same g same 5.27 a f1(y1) = 3(1 − y1)2, 0≤ y 1≤ 1; f2(y2) = 6y2(1 − y2), 0 ≤ y2≤ 1 b 32 63 c 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 e 1 4 5.29 a f2(y2) = 2(1 − y2), 0 ≤ y2≤ 1;
f1(y1) = 1 − |y1|, for −1 ≤ y1≤ 1
b 1 3
5.31 a f1(y1) = 20y1(1 − y1)2, 0 ≤ y1≤ 1 b f2(y2) = 15(1 + y2)2y2 2, −1 ≤ y2< 0 15(1 − y2)2y2 2, 0 ≤ y2≤ 1 c f(y2|y1) = 3 2y 2 2(1 − y1)−3, for y1− 1 ≤ y2≤ 1 − y1 d .5
5.33 a f1(y1) = y1e−y1, y 1≥ 0; f2(y2) = e−y2, y2≥ 0
b f(y1|y2) = e−(y1−y2), y1≥ y 2 c f(y2|y1) = 1/y1, 0 ≤ y2≤ y1 5.35 .5 5.37 e−1 5.41 1 4 5.45 No 5.47 Dependent
5.51 a f(y1, y2) = f1(y1) f2(y2) so that Y1and Y2are independent.
b Yes, the conditional probabilities are the same as the marginal probabilities.
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. 5.63 1 4 5.65 Exponential, mean 1 5.69 a f(y1, y2) = 1 9 e−(y1+y2)/3, y1> 0, y2> 0 b P(Y1+ Y2≤ 1) = 1−4 3e −1/3= .0446 5.71 a 1 4 b 23 144 5.73 4 3 5.75 a 2 b .0249 c .0249 d 2
e They are equal. 5.77 a 1 4; 1 2 b E(Y2 1) = 1/10, V (Y1) = 3 80, E(Y2 2) = 3 10, V(Y2) = 1 20 c −5 4
5.79 0 5.81 1 5.83 1
5.85 a E(Y1) = E(Y2) = 1 (both
marginal distributions are exponential with mean 1) b V(Y1) = V (Y2) = 1 c E(Y1− Y2) = 0 d E(Y1Y2) = 1 −α 4, so Cov(Y1, Y2) = − α 4 e −2 1 2+α 2, 2 1 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 Y2tends to
decrease. 5.91 Cov(Y1,Y2) = 0
5.93 a 0
b Dependent c 0
d Not necessarily independent 5.95 The marginal distributions for Y1
and Y2are y1 −1 0 1 y2 0 1 p1(y1) 1 3 1 3 1 3 p2(y2) 2 3 1 3 Cov(Y1,Y2) = 0 5.97 a 2 b Impossible
c 4 (a perfect positive linear association)
d −4 (a perfect negative linear association) 5.99 0 5.101 a −α 4 5.103 E(3Y1+ 4Y2− 6Y3) = −22, V(3Y1+ 4Y2− 6Y3) = 480 5.105 1 9 5.107 E(Y1+ Y2) = 2/3 and V(Y1+ Y2) = 1 18 5.109 (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 ) = 38.99 c 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 9 c Cov(Y2, Y3) = − n 9 d E(Y2− Y3) = 0, V (Y2− Y3) = 2n 3 5.121 a .0972 b .2; .072 5.123 .08953 5.125 a .046 b .2262 5.127 a .2759 b .8031 5.133 a y2 2 b 1 4 5.135 a 3 2 b 1.25 5.137 3 8 5.139 a nαβ b λαβ 5.141 E(Y2) = λ 2, V(Y2) = 2λ2 3 5.143 mU(t) = (1 − t2)−1/2, E(U) = 0, V(U) = 1 5.145 1 3 5.147 11 36 5.149 a f(y1) = 3y2 1, 0≤ y1≤ 1 f(y2) = 3 2(1 − y 2 2), 0 ≤ y2≤ 1 b 23 44 c f(y1|y2) = 2y1 (1 − y2 2) , y2≤ y1≤ 1 d 5 12 5.157 p(y) = y+ α − 1 y β β + 1 y 1 β + 1 α , y= 0, 1, 2, . . . 5.161 E( ¯Y − ¯X) = µ1− µ2, V ( ¯Y − ¯X) = σ2 1/n + σ 2 2/m
5.163 b F(y1, y2) =
y1y2[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 forα with−1 ≤ α ≤ 1. 5.165 a (p1et1+ p 2et2+ p3et3)n b m(t, 0, 0) c Cov(X1,X2) = −np1p2 Chapter 6 6.1 a 1− u 2 , −1 ≤ u ≤ 1 b u+ 1 2 , −1 ≤ u ≤ 1 c √1 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(Y2) = 1/6 6.3 b $fU(u) = (u + 4)/100, −4 ≤ u ≤ 6 1/10, 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√2u −1/2e−u/2, u≥ 0
b 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) = 2/3
c E(Y1+ Y2) = 2/3
6.11 a fU(u) = 4ue−2u, u≥ 0, a gamma
density withα = 2 andβ = 1/2 b E(U) = 1, V (U) = 1/2 6.13 fU(u) = FU(u) = u β2e −u/β, u > 0 6.15 [−ln(1 − U)]1/2 6.17 a f(y) =αy α−1 θα , 0≤ y ≤ θ b Y = θU1/α
c y= 4√u. The values are 2.0785,
3.229, 1.5036, 1.5610, 2.403. 6.25 fU(u) = 4ue−2ufor u≥ 0
6.27 a fY(y) =
2
βw e−w 2/β
, w≥ 0, which is Weibull density with m= 2. b E(Yk/2) = k 2+ 1 βk/2 6.29 a fW(w) = 1 3 2 (kT )3/2w 1/2e−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 ≤ 1 6.37 a mY1(t) = 1 − p + pe t b mW(t) = E(et W) = [1 − p + pet]n
6.39 fU(u) = 4ue−2u, u≥ 0
6.43 a Y has a normal distribution¯ with meanµ and variance σ2/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)
6.45 c= $190.27
6.47 P(U > 16.0128) = .025
6.51 The distribution of Y1+ (n2− Y2) is
binomial with n1+ n2trials and success
probability p= .2 6.53 a Binomial (nm, p) where ni = m b Binomial (n1= n2+ · · · nn, p) c Hypergeometric (r= n, N= n1+ n2+ · · · nn) 6.55 P(Y ≥ 20) = .077 6.65 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(Z2 1) = 1
c Not independent since
ρ = 0.
d This is the bivariate normal distribution withµ1= µ2= 0, σ2 1 = 1, σ22= 2, and ρ = 1 √ 2 6.69 a f(y1, y2) = 1 y2 1y22 , y1> 1, y2> 1 e No 6.73 a g(2)(u) = 2u, 0 ≤ u ≤ 1 b E(U2) = 2/3, V (U2) = 1/18 6.75 (10/15)5 6.77 a n! ( j − 1)!(k − 1 − j)!(n − k)! yjj−1[yk− yj]k−1− j[θ − yk]n−k θn , 0≤ yj < 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 .5 6.87 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 2 0≤ u ≤ 1 1 2u2 u> 1 b fU2(u) = ue−u, 0≤ u c 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) = .0002
6.101 fU(u) = 1, 0 ≤ u ≤ 1 Therefore, U has
a uniform distribution on (0, 1) 6.103 1 π(1 + u21),∞ < u1< ∞ 6.105 1 B(α, β)u β−1(1 − u)α−1, 0< u < 1 6.107 fU(u) = 1 4√u 0≤ u < 1 1 8√u 1≤ u ≤ 9 6.109 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 probabilities are (respectively) .8664, .9284, .9642, .9836. c The probabilities increase with n. d Yes 7.11 .8664 7.13 .9876 7.15 a E( ¯X − ¯Y ) = µ1− µ2 b V( ¯X − ¯Y ) = σ2 1/m + σ 2 2/n
c The two sample sizes should be at least 18. 7.17 P6i=1Z2 i ≤ 6 = .57681 7.19 P(S2≥ .065) = .10 7.21 a b= 2.42 b a= .656 c .95 7.27 a .17271 b .23041 d .40312 7.31 a 5.99, 4.89, 4.02, 3.65, 3.48, 3.32 c 13.2767 d 13.2767/3.32≈ 4 7.35 a E(F) = 1.029 b V(F) = .076
c 3 is 7.15 standard deviations above this mean; unlikely value. 7.39 a normal, E(ˆθ) = θ = c1µ1+ c2µ2+ · · · + ckµk V(ˆθ) = c2 1 n1 + c2 2 n2 + · · · + c2 k nk σ2 b χ2with n 1+ n2+ · · · + nk− k df c t with n1+ n2+ · · · + nk− k df 7.43 .9544 7.45 .0548 7.47 153 7.49 .0217 7.51 664 7.53 b ¯Y is approximately normal: .0132.
7.55 a random sample; approximately 1. b .1271
7.57 .0062 7.59 .0062 7.61 n= 51 7.63 56 customers 7.65 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> 9 b n> 14, n > 14, n > 36, n > 36, n> 891, n > 8991 7.73 .8980 7.75 .7698 7.77 61 customers
7.79 a Using the normal approximation: .7486.
b Using the exact binomial probability: .729. 7.81 a .5948
b With p= .2 and .3, the
probabilities are .0559 and .0017 respectively. 7.83 a .36897 b .48679 7.85 .8414 7.87 .0041 7.89 µ = 10.15
7.91 Since X , Y , and W are normally distributed, so are ¯X, ¯Y , and ¯W.
µU = E(U) = .4µ1+.2µ2+.4µ3 σ2 U = V (U) = .16 σ2 1 n1 + .04 σ2 2 n2 + .16 σ2 3 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(ˆθ) = aθ + b − θ = (a − 1)θ + b b Let ˆθ∗= (ˆθ − b)/a 8.5 a MSE(ˆθ∗) = V (ˆθ∗) = V (ˆθ)/a2 8.7 a= σ 2 2 − c σ2 1 + σ22− 2c 8.9 Y¯ − 1 8.11 θ3ˆ − 9ˆθ2+ 54 8.13 b [n2/(n − 1)](Y/n)[1 − (Y/n)] 8.15 a 1 3n− 1 β b MSE( ˆβ) = 2 (3n − 1)(3n − 2)β2 8.17 a (1 − 2p)/(n + 2) b np(1 − p) + (1 − 2p) 2 (n + 2)2 c p will be close to .5. 8.19 MSE( ˆθ) = β2 8.21 11.5 ± .99 8.23 a 11.3 ± 1.54 b 1.3 ± 1.7 c .17 ± .08 8.25 a −.7 b .404 8.27 a .601 ± .031 8.29 a −.06 ± .045 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 2Y 9.48773, 2Y .71072 8.41 a (Y2/5.02389, Y2/.0009821) b Y2/.0039321 c Y2/3.84146 8.43 b [Y(n)](.95)−1/n 8.45 a Y /.05132 b 80% 8.47 c (2.557, 11.864) 8.49 c (3.108, 6.785) 8.57 .51± .04 8.59 a .78 ± .021 8.61 (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 ± .738 8.69 .22 ± .34 or (−.12, .56) 8.71 n= 100
8.73 n= 2847 8.75 n= 136 8.77 n= 497 8.79 a n= 2998 b n= 1618 8.81 60.8 ± 5.701 8.83 a 3.4 ± 3.7 b .7 ± 3.32 8.85 −1 ± 4.72 8.87 (−.624, .122) 8.91 (−84.39, −28.93) 8.93 a 2 ¯X+ ¯Y ± 1.96σ 1 4 n+ 3 m b 2 ¯X+ ¯Y ± tα/2S 1 4 n+ 3 m, where S2= (Yi− ¯Y )2+ 1/3 (Xi− ¯X)2 n+ m − 2 8.95 (.227, 2.196) 8.99 a 0 (n − 1)S2 χ2 1−α b 0 (n − 1)S2 χ2 α 8.101 s2= .0286; (.013 .125) 8.103 (1.407, 31.264); no 8.105 1− 2(.0207) = .9586 8.107 765 seeds 8.109 a .0625 ± .0237 b 563 8.111 n= 38,416 8.113 n= 768 8.115 (29.30, 391.15) 8.117 11.3 ± 1.44 8.119 3± 3.63 8.121 −.75 ± .77 8.123 .832 ± .015 8.125 a S 2 1 S2 2 ×σ22 σ2 1 b S2 2 S2 1Fv2,v1,α/2, S2 2 S2 1 Fv1,v2,α/2 vi = ni − 1, i = 1, 2 8.129 a 2(n − 1)σ 4 n2 8.131 c= 1 n+ 1 8.133 b 2σ 4 n1+ n2− 2 Chapter 9 9.1 1/3; 2/3; 3/5 9.3 b 12n 2 (n + 2)(n + 1)2 9.5 n− 1 9.7 1/n 9.9 a X6= 1
9.23 c need Var(X2i− X2i−1) < ∞
9.25 b .6826 c No 9.31 αβ 9.35 a Y¯nis unbiased forµ. b V( ¯Yn) = 1 n2 n i=1σ 2 i 9.47 n i=1 ln(Yi); no 9.57 Yes 9.59 3 ¯ Y2+ ¯Y 1−1 n 9.61 n+ 1 n Y(n) 9.63 b 3n+ 1 3n Y(n) 9.69 θ =ˆ 2 ¯Y− 1
1− ¯Y , no, not MVUE
9.71 σˆ2= m 2= 1 n n i=1Y 2 i. 9.75 With m2= 1 n n i=1Y 2 i , the MOM estimator ofθ is ˆθ = 1− 2m 2 4m2− 1. 9.77 2 3 ¯ Y 9.81 Y¯2 9.83 a θ =ˆ 1 2 Y(n)− 1 b Y(n)2/12 9.85 a θ =ˆ 1 αY¯ b E(ˆθ) = θ, V (ˆθ) = θ2/(nα) d n i=1Yi e 2in=1Yi 31.4104 , 2ni=1Yi 10.8508 9.87 ˆpA= .30, ˆpB = .38 ˆpC= .32; −.08 ± .1641 9.91 Y(n)/2 9.93 a Y(1) c [(α/2)1/2nY (1), (1 − (α/2))1/2nY(1)] 9.97 a 1/ ¯Y b 1/ ¯Y
9.99 ˆp± zα/2 1 ˆp(1 − ˆp) n 9.101 exp(− ¯Y ) ± zα/2 1 ¯ Y exp(−2 ¯Y ) n 9.103 1 n n i=1 Yi2 9.105 σˆ2= (Yi− µ) 2 n 9.107 exp(−t/ ¯Y ) 9.109 a N1ˆ = 2 ¯Y − 1 b N 2− 1 3n 9.111 252± 85.193 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 ii True iii False 10.17 a H0:µ1= µ2, Ha:µ1> µ2 c z= .075 10.21 z= 3.65, reject H0 10.23 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 withα = .10. 10.25 |z| = 1.105, do not reject 10.27 z= −.1202, do not reject 10.29 z= 4.47 10.33 z= 1.50, no 10.35 z= −1.48 (1 = homeless), no 10.37 approx. 0 (.0000317) 10.39 .6700 10.41 .025 10.43 a .49 b .1056 10.45 .22 ± .155 or (.065, .375) 10.47 .5148 10.49 129.146, yes 10.51 z= 1.58 p–value = .1142, do not reject 10.53 a z= −.996, p–value = .0618 b No c z= −1.826, p–value = .0336 d Yes 10.55 z= −1.538; p–value = .0616; fail to reject H0withα = .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 is 358− 2.821√54
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, s12= .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, .03054 10.73 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
10.75 t= −.647, do not reject
10.77 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 10.79 a χ2= 12.6, do not reject b .05 < p–value < .10 c Applet p–value= .08248
10.83 a σ2 1 = σ 2 2 b σ2 1 < σ22 c σ2 1 > σ 2 2 10.85 χ2= 22.45, p–value < .005; applet 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 10.95 a U= β02 4i=1Yi hasχ(24)2
distribution under H0: reject H0
if U> χ2 α b Yes 10.97 d Yes, is UMP 10.99 a n i=1 Yi≥ k
b Use Poisson table to find k such that P(Yi≥ k) = α c Yes 10.101 a n i=1 Yi< c b Yes 10.103 a Reject H0if Y(n)≤ θ0n √α b Yes 10.107 χ2= (n − 1)S 2 1+ (m − 1)S 2 2 σ2 0 has χ2 (n+m−2)distribution under H0; reject ifχ2> χ2 α 10.109 a λ = ( ¯X)m( ¯Y )m m ¯X+ n ¯Y m+ n m+n
b X¯/ ¯Y distributed as F with 2m and 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 = ( ¯X+ ¯Y − ¯W)−(µ1−µ2−µ3) 1+a+b n(3n−3).(Xi− ¯X)2+a1 (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 ˆy= 1.5 − .6x 11.5 ˆy= 21.575 + 4.842x
11.7 a The relationship appears to be proportional to x2.
b No
c No, it is the best linear model. 11.9 b ˆy= −15.45 + 65.17x
d 108.373 11.11 β1ˆ = 2.514
11.13 a The least squares line is ˆy= 452.119 − 29.402x 11.17 a SSE= 18.286;
S2= 18.286/6 = 3.048
b The fitted line is
ˆy= 43.35 + 2.42x∗. The same
answer for SSE (and thus S2) is
found.
11.19 a The least squares line is: ˆy= 3.00 + 4.75x c s2= 5.025 11.23 a t= −5.20, reject H0 b .01 < p–value < .02 c .01382 d (−.967, −.233) 11.25 a t= 3.791, p–value < .01 b Applet p–value= .0053 c Reject d .475 ± .289
11.29 T = β1ˆ − ˆγ1 S 1 1 Sx x + 1 Scc , where S = (SSEY+ SSEW)/(n + m − 4). H0is rejected in favor of Hafor large
values of|T |. 11.31 t= 73.04, p–value approx. 0, H0is rejected 11.33 t= 9.62, yes 11.35 x∗= ¯x. 11.37 (4.67, 9.63) 11.39 25.395 ± 2.875 11.41 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 < .01 11.53 a r2= .682 b .682 c t= 4.146, reject d Applet p–value= .00161 11.57 a sign for r b r and n 11.59 r = −.3783 11.61 .979 ± .104 11.63 a β1ˆ = −.0095, ˆβ0= 3.603 and ˆ α1= −(−.0095) = .0095, ˆ α0= exp(3.603) = 36.70. Therefore, the prediction equation is ˆy= 36.70e−.0095x. b The 90% CI for α0is e3.5883, e3.6171= (36.17, 37.23) 11.67 ˆy= 2.1 − .6x 11.69 a ˆy= 32.725 + 1.812x b ˆy= 35.5625 + 1.8119x − .1351x2 11.73 t= 1.31, do not reject 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 11.83 a F= 21.677, reject b SSER= 1908.08 11.85 a F= 40.603, p–value < .005 b 950.1676 11.87 a F= 4.5, F1= 9.24, fail to reject H0 c F= 2.353, F1= 2.23, reject H0 11.89 a True b False c False 11.91 F= 10.21 11.93 90.38 ± 8.42 11.95 a ˆy= −13.54 − 0.053x b t= −6.86 c .929 ± .33 11.97 a ˆy= 1.4825+.5x1+.1190x2−.5x3 b ˆy= 2.0715 c t= −13.7, reject d (1.88, 2.26) e (1.73, 2.41) 11.99 If−9 ≤ x ≤ 9, choose n/2 at x = −9 and n/2 at x = 9. 11.101 a ˆy= 9.34+2.46x1+.6x2+.41x1x2 b 9.34 , 11.80
d For bacteria A, ˆy= 9.34. For bacteria B, ˆy= 11.80. The observed growths were 9.1 and 12.2, respectively. e 12.81 ± .37 f 12.81 ± .78 11.107 a r= .89 b t= 4.78, p–value <.01, reject Chapter 12 12.1 n1= 34, n2= 56 12.3 n= 246, n1= 93, n2= 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 whenρ > 0.
b This occurs whenρ = 0. c This occurs whenρ < 0. d Paired better whenρ > 0,
independent better whenρ < 0
12.15 a t= 2.65, reject 12.17 a µi 12.31 a µi b µi, 1 n[σ 2 P+ σ 2] c µ1− µ2, 2σ2/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
Chapter 13 13.1 a F= 2.93, do not reject b .109 c |t| = 1.71, do not reject, F = t2 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 13.13 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 forµA: 76± 8.142 or (67.868, 84.142) b 95% CI forµB: 66.33 ± 10.51 or (55.82, 76.84) c 95% CI forµA− µB: 9.667 ± 13.295 13.29 a 6.24 ± .318 b −.29 ± .241 13.31 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± 2.652 13.37 a µ b Overall mean 13.39 b (2σ2)/b 13.41 a F= 3.11, do not reject b p–value> .10 c p–value= .1381 d s2 D= 2MSE 13.45 a F= 10.05; reject b F= 10.88; reject 13.47 Source df SS MS F Treatments 3 8.1875 2.729 1.40 Blocks 3 7.1875 2.396 1.23 Error 9 17.5625 1.95139 Total 15 32.9375 F= 1.40, do not reject 13.49 F= 6.36; reject 13.53 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 nA≥ 3 13.61 b= 16; n = 48
13.63 Sample sizes differ.
13.69 a β0+ β3is the mean response to
treatment A in block III. b β3is the difference in mean
responses to chemicals A and D in block III.
13.71 F= 7; H0is rejected
13.73 As homogeneous as possible within blocks.
13.75 b F= 1.05; do not reject
13.77 a A 95% CI is .084± .06 or (.024, .144).
13.79 a 16
b 135 degrees of freedom left for error.
c 14.14
13.81 F= 7.33; yes; blocking induces loss in
degrees of freedom for estimatingσ2;
could result in sight loss of information if block to block variation is small 13.83 a Source df SS MS F Treatments 2 524,177.167 262,088.58 258.237 Blocks 3 173,415 57,805.00 56.95 Error 6 6,089.5 1,014.9167 Total 11 703,681.667 b 6 c Yes, F= 258.19, p–value < .005 d Yes, F= 56.95, p–value < .005 e 22.527 f −237.25 ± 55.13 13.85 a SST= 1.212, df = 4 SSE= .571, df = 22 F= 11.68; p–value < .005 b |t| = 2.73; H0is rejected; 2(.005) < p–value < 2(.01).
13.87 Each interval should have confidence coefficient 1− .05/4 = .9875 ≈ .99;
µA− µD: .320± .251 µB− µD: .145± .251 µC− µD: .023± .251 µE− µD:−.124 ± .251
13.89 b σ2 β c σ2 β= 0 13.91 a µ; σ2 B+ 1 kσ 2 ε b σ2 β+ b k−1 k i=1τi2 c σ2 ε + kσB2 d σ2 ε Chapter 14 14.1 a X2= 3.696, do not reject b Applet p–value= .29622 14.3 X2= 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 X2= 69.42, reject 14.13 a X2= 18.711, reject b p–value< .005 c Applet p–value= .00090 14.15 b X2also multiplied by k 14.17 a X2= 19.0434 with a p–value of .004091. b X2= 60.139 with a p–value of approximately 0.
c Some expected counts< 5
14.19 a X2= 22.8705, reject b p–value < .005 14.21 a X2= 13.99, reject b X2= 13.99, reject c X2= 1.36, do not reject 14.25 b X2= 19.1723, p-value = 0.003882, reject c −.11 ± .135 14.27 X2= 38.43, yes 14.29 a X2= 14.19, reject 14.31 X2= 21.51, reject 14.33 X2= 6.18, reject; .025 < p–value < .05 14.35 a Yes b p–value= .002263 14.37 X2= 8.56, df = 3; reject 14.41 X2= 3.26, do not reject 14.43 X2= 74.85, reject Chapter 15 15.1 Rejection region α M≤ 6 or M ≥ 19 P(M ≤ 6) + P(M ≥ 19) = .014 M≤ 7 or M ≥ 18 P(M ≤ 7) + P(M ≥ 18) = .044 M≤ 8 or M ≥ 17 P(M ≤ 8) + 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 U= 9, do not reject 15.25 z= −1.80, reject 15.27 U= 0, p–value = .0096 15.29 H= 16.974, p-value < .001 15.31 a SST= 2586.1333; SSE = 11,702.9; F= 1.33, do not reject b H= 1.22, do not reject 15.33 H= 2.03, do not reject 15.37 a No, p–value= .6685 b Do not reject H0 15.39 Fr= 6.35, reject 15.41 a Fr= 65.675, p–value < .005, reject b m= 0, P(M = 0) = 1/256, p–value= 1/128
15.45 The null distribution is given by
P(Fr = 0) = P(Fr = 4) = 1/6 and P(Fr = 1) = P(Fr = 3) = 1/3.
15.49 a .0256
b An usually small number of runs (judged atα = .05) would imply a clustering of defective items in time; do not reject.
15.51 R= 13, do not reject
15.53 rS= .911818; yes.
15.55 a rS= −.8449887
b Reject
15.57 rS= .6768, use two-tailed test, reject
15.59 rS= 0; p–value < .005
15.61 a Randomized block design b No
c p–value= .04076, yes
15.63 T = 73.5, do not reject, consistent with
Ex. 15.62 15.65 U= 17.5, fail to reject H0 15.67 .0159 15.69 H= 7.154, reject 15.71 Fr = 6.21, do not reject 15.73 .10 Chapter 16 16.1 a β(10, 30) b n= 25 c β(10, 30), n = 25 d Yes
e Posterior for theβ(1, 3) prior. 16.3 c Means get closer to .4, std dev
decreases.
e Looks more and more like normal distribution. 16.7 a Y+ 1 n+ 4 b np+ 1 n+ 4; np(1 − p) (n + 4)2 16.9 b α + 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 .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.
