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Unformatted text preview: 82 83 84 85 CHLTEM (a) From(811)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.Ol9l=0. 05mm: 2 =¥=L5 u=.933 7:.866 (a) From(811)with a=l and n=4: x : ozu/Jﬁ'y 203 t lmm (b) From(812l) 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 (84)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, (811l) yields
the interval X t 2975 am = 30 i 112 86 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 ba 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 ba. If f(a) > f(b), we form the interval (a61, b62) (Fig. 86b) and proceed similarly. '(ql WWW; 2H9 1 "will/II” Special case. If f(x)=xe"‘ then (see Problem 4—9) F(x)=le"‘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 4.750.04=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 88 N(0,aw) where aw2=ax2+aiz=oz+03/n. With c=z.9.,saw$ it follows that P(w < c}=.95. Hence
P(ic<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 (811) . 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 (752) it follows that their sum y=351 +    +5 = ng has an Erlang distribution:
N n 115 89 810 9“ my) = (—57; y"‘1 'e’"y Um §y(s) = (05) n and the RV 5:29;: = 21195 has a x’(2n) distribution: nl _‘
W) = 2‘7, f, (gum = e ’2 U(z) Hence,
P{x36/2(2n) < E; < x31_6/z(2n)} = '1 = 16 Since le025( 78) = 54.6, x2_975( 78)=104.4, and 2nx=480, this yields the interval 4.60 < r] < 8.79 minutes From (819)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 .018 116 811 812 813 814 (a) In this problem, X=0.40, n=900 and zu r3 2. From(821) £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(821)with i=0.29 and zu=2z
ZUXInx =004 n>i(l);§ zzu=515
. . p(1p)
The problem 15 to find n such that [see (820)] 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 (836)with k=1
5 .4 < p < .6 .6
= = 5 = ,5 =%
0 otherwise ’4
10p .4 < p < .6 .6
fp(pH)  ﬁ = 10 fpzdp = .5067
.4 117 815 816 817 n 
From Prob. 88: fquﬂ) = x“'1e'"”‘ n+ 1 one(chi )0 From (832): 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.88) . of Bayesian estimation, this means that k
fiat”) = ems k = 11)! = 0,1,... Inserting into (832) , we obtain [see (476)] 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—“ﬁ9o: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
818 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 __
819 From (823)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 > 0.057 820 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
821 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 119 822 823 824 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 828 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
«111 ﬂ=laif_i[.3_f]2 ﬂ+[¢]2_iaﬂ
a9‘fa9 392 £392 {2&9 692 as 'fa92 But =!{%§de=0 hence E{%+ =0 120 825 826 (a) From (8307): 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_anq = 2,9
 _ _ 8.78
"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 (8301) with q = — x = ﬁ in = 27.67 s2 = % {jogx) = 17.6
This yields s=4.2 and q=0.33. Since
q = t“ (nl) = 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 121 827 828 hypothesis no for any no>28. We note, however, that the corresponding OC function ﬁn) is smaller than the function [30(17) obtained from (8301) with no=2s. From (8297) with qu=tu(nl): Critical region litno] > t1_a/2(nl)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 11131 > 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 (8307) 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___04_
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 (8313)]. (b) From (8313) 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=82x2=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 (8301)]. 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 (975) 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 (8325) that 9 (nj100)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 = 632032) is an RV. We wish to find its best bayesian estimate. Using the MS criterion, we obtain 1 .
65 = 13(25} = L 6p2(1p2)r(p)dp If f(p)=l, then 65 = 1/5. Given fv(v) evz/Zaz few) e(000)2/20°2
To show that {0(91x)~ e(o 91)2/2"12
where l __ l n o _ <71z nalz _ 012307”?— 1_a°2 0+7?"
Proof (x4?)2
rxoqo) = f,,(x—0) ~ exp {  202 } l
f(X)0 ~ exp {  $2 2 (Xi9)2 } Since 2 (xit9)2 = Z (xix)2 + n (X0)2, we conclude from (832) omitting factors that do not depend on 0 that Mir0)2
02 + 1 [at0°): 2 002 f(0X) ~ exp { 
} 126 The above bracket equals 1 n 2 00 n l 2
[3+7]o—2[0—2+?]o+.._;F(a2aal)+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: (xin)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. 653). 8—40 k1+k2np1np2 = nn(p1+p2) = 0; Hence, lklnpll = lkz—npzl W+W= l l _ (klnpl)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 ...
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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 UniversityWest Lafayette.
 Spring '08
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