# Ch_08 - 8-2 8-3 8-4 8-5 CHLTEM(a From(8-11)with 1.95 u=.975 2375 r 2 0:0.1 and n=9 we obtain-z“0 0066 c_ﬁ=(b From(8 1 1 with c=9l.Ol-9l=0 05mm

This preview shows pages 1–17. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8-2 8-3 8-4 8-5 CHLTEM (a) From(8-11)with 1:.95, u=.975, 2375 r: 2, 0:0.1, and n=9 we obtain -z“0 0066. c_ﬁ=' (b) From (8 - 1 1) with c=9l.Ol-9l=0. 05mm: 2 =¥=L5 u=.933 7:.866 (a) From(8-11)with a=l and n=4: x : ozu/Jﬁ'y 203 t lmm (b) From(8-12l) with 6=.05: c = 0/4675 = 2.236mm We wish to find n such that P{|i—al<0.2) = 0.95 where a=E{i}.From (8-4)it follows with u=.975 and a=0.lmm that zua —n S 0.2, hence, n=l In this problem, x is uniform with E{x}=0 and oz=4/3.We can use, however, the normal approximation for i because n=100. With 'r=.95, (8-11l) yields the interval X t 2975 am = 30 i 112 8-6 We shall show that if f(x) is a density with a single maximum and P{a<x<b)='7, then b—a is minimum if f(a)=f(b). The density xe‘xU(x) is a special case. It suffices to show that b-a is not minimum if f( a)< f( b) or f(a)>f(b). Suppose first that f(a)< f( b) as in figure (a). Clearly. f'(a)> 0 and f'(b)< 0, hence, we can find two constants 61>0 and62>0 such that P{a+61 < x < b+62} = '1 and f(a) < f(a+61) < f(b+62) < f(b) From this it follows that 6, > 62, hence, the length of thenew interval (a+61, mg) is smaller than b-a. If f(a) > f(b), we form the interval (a-61, b-62) (Fig. 8-6b) and proceed similarly. '(ql WWW; 2H9 1 "will/II” Special case. If f(x)=xe"‘ then (see Problem 4—9) F(x)=l-e"‘-xe"‘, hence, 113 Pia < x < b} = e" + ae" - e“’-be‘b = .95 And since f(a)=f( b), the system results. Solving, we obtain a90.04 has. 75. A numerically simpler solution results if we set 0.025 = P{x s a} = F(a) 0.025 = P{x > b} = l - F(b) as in (9—5). This yields the system 0.025 = l - e"—ae" 0.025 = e' +be'b Solving, we obtain a=0.242, b=5.572. However, the length 5.572—0.242=5.33 of the resulting interval is larger than the length 4.75-0.04=4.7l of the optimum interval. We start with the general problem: We observe the n samples xi of an. N(r),10) RV x and we wish to predict the value x of x at a future trial in terms of the average it of the observations. If n is known, we have an ordinary prediction problem. If it is unknown, we must This RV is first estimate it. To do so, we form the RV w=x-:‘Z. 114 8-8 N(0,aw) where aw2=ax2+aiz=oz+03/n. With c=z.9.,saw\$ it follows that P(|w| < c}=.95. Hence P(i-c<x<i+c) =0.95 For n=20 and a=10 the above yields aw=10.25 and c 9 20.5. Thus, we can expect with .95 confidence coefficient that our bulb will last at least 80—20.5=59.5 and at most 80+20=100.5 hours. The time of arrival of the 40th patient is the sum x1 + + x of n=39 RVs with exponential distribution. We shall estimate the mean n=l/0 of x in terms of its sample mean x=240/39=6.15 minutes using two methods. The first is approximate (large n) and is based on (8-11) . Normal approximation. With A=n andzms/ A139.=0.315: r{r§l—5 < n < = .95 4.68 < n < 8.98 minutes Exact solution. The RVs 5 are i.i.d. with exponential distribution. From- this and (7-52) it follows that their sum y=351 + - - - +5 = ng has an Erlang distribution: N n 115 8-9 8-10 9“ my) = (—57; y"‘1 'e’"y Um §y(s) = (0-5) n and the RV 5:29;: = 21195 has a x’(2n) distribution: n-l _‘ W) = 2‘7, f, (gum = e ’2 U(z) Hence, P{x36/2(2n) < E; < x31_6/z(2n)} = '1 = 1-6 Since le025( 78) = 54.6, x2_975( 78)=104.4, and 2nx=480, this yields the interval 4.60 < r] < 8.79 minutes From (8-19)with i=2,550/200=12.75 n=200 and zuez ,\2 — 25.52 A + 12.752 = 0 A1 = 12.255 < A < 13.265 = A2 .52 i- .018 116 8-11 8-12 8-13 8-14 (a) In this problem, X=0.40, n=900 and zu r3 2. From(8-21) £1002 Xlgx 11 = t 3. 27% (b) We wish to find 2“. From (9—21) and Table la: 1002u ZL'ﬁ'll=2 2 =1.225 u 0 otherwise : Margin of error u = .89 This yields the confidence coefficient ’1 = 2u — l = .78 From(8-21)with i=0.29 and zu=2z ZUXIn-x =004 n>i(l);§ zzu=515 . . p(1-p) The problem 15 to find n such that [see (8-20)] zu n _ .02 for every p. Since zu'yZ and p(l—p)sl/4, this is the case if zu l/4n s .02 n2 2,500 , From (8-36)with k=1 5 .4 < p < .6 .6 = = 5 = ,5 =% 0 otherwise ’4 10p .4 < p < .6 .6 fp(pH) - ﬁ = 10 fpzdp = .5067 .4 117 8-15 8-16 8-17 n - From Prob. 8-8: fquﬂ) = x“'1e'"”‘ n+ 1 one-(chi )0 From (8-32): you) = “n: A n+1 oo _ From (8_31) : 9 = (ELIEP 1‘6p+1e-(c+nx)0d0= n+1 ' 0 c+nX The sum n)? is a Poisson RV with mean n0 (see Prob.8-8) . of Bayesian estimation, this means that k fiat”) = ems k = 11)! = 0,1,... Inserting into (8-32) , we obtain [see (4-76)] ni+b+l _ me i b -(n+c)0 {Mm - W a“ + e and (8 -31)yields ni+b+l 3 a me I! nx+b+21 nx+b+l I‘(nx+b+l) (n+c)ni+b+2 = me —-}x n—uoo From (8 - 17 ) with :94 9) =2.26 t xt—“ﬁ-9o:3.s7 M 86.43 < n e 93.57 _ ' 2 - 2 - From (8 24) With x .975( 9)-—l9.02, x '025( 9) —2. 70. 118 In the context W - . W = 83.33 3.44 < 0 < 9.13 8-18 The RVs xi/a are N(0,l), hence, the sum z=(x12 + + x210)/az has a x2( 10) distribution. This yields 2 2 _ P{x 7025(10) < z < x_975(10)} _ .95 x2024 10) = 3.25 < g < x2_975( 10) = 20.43 0.442 < 0 < 1.109 - _ 2 _ 2 __ 8-19 From (8-23)w1th n—4,x _025(4)—0.48, x ‘975(4)-ll.l4 no = .12 + .152 + .052 + .042 = .0366 -°356 > a2 > {ﬁg 0.276 > a > 0.057 8-20 In this problem n=3, x1+x2+x3=9.8 f(X,c) ~ c“x"’e‘cx f(X,c) = c4“(x1...xn)3“e"“" “(gm = 4c—“ - nx]f(x,9) = o e = g = 1.224 8-21 The joint density f(X,c) = cne‘m‘i—x") xi > xo has an interior maximum if 3f(X.c) _ A _ 1 ac — 0 c - x_x0 119 8-22 8-23 8-24 The probability p = 1 - F4200) = e‘2°°° of the event (x > 200} is a monoton decreasing function of c. To find the ML estimate 6 of c it suffices to find the ML estimate [3 of p. From Example 8-28 it follows with k=62 and n=80 that f) = g3- = .775 hence The samples of x are the integers xi and the joint density of the RVs xi equals i - r(x,o) = em" n 3%! = e'"" Hence, f(X,9) is maximum if - n + nit/9 = 0. This yields 3 = X If L = In f(x,0) then «1-11 ﬂ=laif_i[.3_f]2 ﬂ+[¢]2_iaﬂ a9‘fa9 392 £392 {2&9 692 as 'fa92 But =!{%§de=0 hence E{%+ =0 120 8-25 8-26 (a) From (8-307): Critical region it > c = no + 2” = 8+2.326 x g = 8.58 2.. «[3 If n=s.7, then nq = = 2.8 ﬂ (n) = G(2.36 - 2.8) = .32 (b) We assume that a=.01, ﬂ (8.7)-—-.05 and wish to find n and c. G(Zl-,.-ﬂq) = ﬂ zl_a-nq = 2,9 - _ _ 8.7-8 "q — 1.99 2.05 ‘ 4 97 2/43 n=129 c=8+—2——z 841 m .99 Our objective is to test the composite null hypothesis r1>no=28 against the hypothesis «no. Consider first the simple null hypothesis 17=n0=28. In this case, we can use (8-301) with q = — x = ﬁ in = 27.67 s2 = % {jog-x) = 17.6 This yields s=4.2 and q=-0.33. Since q = t“ (n-l) = tomué) = - 1.95 < -o.33 u we conclude that the evidence does not support the rejection of the hypothesis n=28. The resulting OC function 190(0) is determined from (9—60c). If no>28, then the corresponding value of q is larger than ~0.33. From this it follows that the evidence does not support the 121 8-27 8-28 hypothesis no for any no>28. We note, however, that the corresponding OC function ﬁn) is smaller than the function [30(17) obtained from (8-301) with no=2s. From (8-297) with qu=tu(n-l): Critical region lit-no] > t1_a/2(n-l)s/~lﬁ t_95(63) = 1.67 132—31 > 1.67 x 1.5/8 = 0.313 Since X=7.7 is in the interval 8 t 0.317, we accept H0 t_995(63) = 2.62 111-31 > 2.62 x 1.5/8 = 0.49 Since X=7.7 is outside the interval 8 t 0.49, we reject H0. We assume that the RVs is and y are normal and independent. We form N the difference w=§—y of their sample means 1 16 l 26 X = — x = "’ '6 iz=:1 “’1 Z 2—6 iz=21 Xi and use as test statistic the ratio 2 2 q = X- 0 2 = ax + 0y N 0., W W 76' The RV q is normal with aq=l and under hypothesis H0, E{q}=0. We can, 122 therefore, use (8-307) because qu=zu. To find q, we must determine aw. Since (7x and 0,, are not specified, we shall use the approximations Ox '3 sx=l.l and U, '2' sy=0.9. This yields 2"}le 0.92__ _ﬂ-___0-4_ 0w _ + 76- — q -— aw — - Since z0.95=l.645 > 1.223, we accept H0. (a) In this problem, n=64, k=22, p0=q0=0.5 k—np0 Cl = npoqo = 2.5 zm/2 = — 2141/2 '2' -2 Since 2.5 is outside the interval (2, -2), we reject the fair coin hypothesis [see (8-313)]. (b) From (8-313) with n=l6, p0=q0=0.5: kl—np0 k2—np0 mm = za/Z np q = ' Za/z o o o o This yields k1=8-2x2=4, k2=8+2x2=12 We.shall use as test statistic the sum q = x + + x n = 22 ~ “’1 “'m 123 The critical region of the test is q<qa where q=xl+---+xn=90 [see (8-301)]. The RV q is Poisson distributed with parameter nA. Under hypothesis H0, A=Ao=5; hence, nq=nAo=llO=aq2. To find qa we shall use the normal approximation. With a=0.05 this yields qa = nAo + za .[nAO = 9047.25 = 72.75 Since 90 > 72.75, we accapt the hypothesis that A=5. From (9-75) with n=102 and p0i=l/6 6(.-17)2 “Xian—=2 i=1 x29; 5) y 11 Since 2<ll, we accept the fair die hypothesis. Uniformly distributed integers from 0 to 9 means that they have the same probability of appearing. With m=10, p01=.1, and n=l,000, it follows from (8-325) that 9 (nj-100)2 _ _ 2 _ q - T - X 95(9) — Since 17.76 > 16.92, we reject the uniformity hypothesis. 124 f(X,€) is maximum for 6L—-9m=x. And since 0m0=00 we conclude that —n9 - o nx e 00 e—ni in)? A(X) = w = - 2 In A = 2n(60—)'<) + )‘t [Mi/90) With n=50, 60:20, X=l,058/50=2|.l6, this yields w=3. Since m0=l, m=1, and X2'95(1)=3.84>3, we accept H0. We form the RVs m n z = [xi—"’12 w= Z [—in‘" )2 ~ . ax ~ . 0y 1:1 1:1 These RVs are x2(m) and x2(n respectively. If ox=ay, then z/m 3:37.1- ~ Hence (see Prob. 6—23), q has a Snedecor distribution. To test the hypothesis ax=o , we use (8—297) where qu=Fu(m,n) is the tabulated u percentile of the Snedecor distribution. This yields the following test: Accept Ho iff Fa/2(m,n) < q < F1_a/2(m,n). (a) Suppose that the probability P(A) that player A wins a set equals p=l—q. He wins the match in five sets if he wins two of the first four sets and the fifth set. Hence, the probability p5(A) that he wins in five equals 6p3q2. Similarly, the probability p5(B) that player B wins in five equals 6p2q3. Hence, p5 = 95(A) + 95(3) = 60%” + 69%:3 = 6122c:2 125 is the probability that the match lasts five sets. If p=q=l/2, then p5=3/8. (b) Suppose now that P(A) = p is an RV with density f(p). In this case, 25 = 6320-32) is an RV. We wish to find its best bayesian estimate. Using the MS criterion, we obtain 1 . 65 = 13(25} = L 6p2(1-p2)r(p)dp If f(p)=l, then 65 = 1/5. Given fv(v) e-vz/Zaz few) e-(0-00)2/20°2 To show that {0(91x)~ e-(o 91)2/2"12 where l __ l n o _ <71z nalz _ 012-307”?— 1_a°2 0+7?" Proof (x4?)2 rxoqo) = f,,(x—0) ~ exp { - 202 } l f(X)|0 ~ exp { - \$2 2 (Xi-9)2 } Since 2 (xi-t9)2 = Z (xi-x)2 + n (X-0)2, we conclude from (8-32) omitting factors that do not depend on 0 that Mir-0)2 02 + 1 [at-0°): 2 002 f(0|X) ~ exp { - } 126 The above bracket equals 1 n 2 00 n l 2 [3+7]o—2[0—2+?]o+-.._;F(a-2aal)+m O and (i) follows. 8—38 The likelihood function of X equals 1 1 mm" 8"” 52 (“*‘"’2} where 0 = 02 is the unknown parameter. Hence f(X,0) = L(X,9) = — l in (21:9) - \$2: (xi-n)z 2 6L(X,0)_ n l 2.. _' 2 69 "WWZO‘HD‘O LTZO‘”) 8—39 The estimators 01 and 02 have the same variance because otherwise one or the other would not be best. Thus Eléll=E{52}=9 Var§1=var52=02 . 1 . .. =T(01+02 ), then A * l 2 2 1 E{9}=9 092=7(o +02+2ra)=-—§—(l+r)02 where a is the correlation coefficient of 0 1 and 92. If r<l then 05<o which is impossible. Hence, r=l' and 51 = ‘ 2 (see Prob. 6-53). 8—40 k1+k2-np1-np2 = n-n(p1+p2) = 0; Hence, lkl-npll = lkz—npzl W+W= l l _ (kl-npl)2 “91 “92 1 “[3192 8.41 It is given that E{T(X)} = [m we fix; 6) dx 2 12(0), so that after differentiating and making use of (8—81) we get 0° (9 X' 6 , LC T(X) (“86’ ) d1: 2 11 (6) (8.41 - 1) Also using (8—80) [0 211(9) gag—2i?) da: 2 07 (8.41 w 2) and the above two expressions give 00 (9 X‘ 9 , [comm—21(6)] “(36’ lda: = «1 <6) (841— 3) But 8f(X; 6) _ 1 810gf(X;0) 0.9 _ f(X; 9) (‘36 so that (841—3) simpliﬁes to f: lmx) — 10(0)} dz) : W9) and application of Cauchy—Sohwarz inequality as in (8—89)—(8—92)7 Text gives 2 E [{T(X) - 19(6)}? 2 WW (6)] U2} 610 f X;9 41—11 128 ...
View Full Document

## This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue University-West Lafayette.

### Page1 / 17

Ch_08 - 8-2 8-3 8-4 8-5 CHLTEM(a From(8-11)with 1.95 u=.975 2375 r 2 0:0.1 and n=9 we obtain-z“0 0066 c_ﬁ=(b From(8 1 1 with c=9l.Ol-9l=0 05mm

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online