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Unformatted text preview: [Mus 3’5"?) HwﬁLf §uuwng knew—ﬁt. Emmi“? Md
[um +31+6o+a1+m 9H §A+ 35 + 48+ (VI+59. )ﬂo '= 62%
W‘W‘i Qu’gthm, Ihmwj JEAN? = 6513
Mud3i WUSQHN (“wadn? X£N{ 1‘— (9
onwa‘m of WWHA MU” 8mm on hand (vaW '7  73.39}, {3+ was +§+4b+ 11+5’M 35+ 5+)? u
x
N 1) _ ( 6.9.)111cavengc number offsilurcs beforclhcﬁm success is 7(a). Thus. on usage one out ofevery 7(5) + l
' trials sinuldbcasucccss. 4, (WK 01°)
' (d) [(11.8)=(2x)'"’za'”eup[Zf=i(xi,u}2/(20'2)] and up, a) = (—an) In (2;)  n In a — {11(202)12?=1(X. — mi. Taking partial derivatives on, we gel ai/au = Egalix; * mfaz and Juan: nlo+ ELIM  m2 [03 . Setting allay = 0. we get ,1 a 350.). immediately. Substituting this ii for 11 ink) alaaand setting this partial to 0 yields Err 1‘ Lips;  EMF/n  (e) snicexmassﬂiiiandonlyinnx—Nusa“).ﬁ= fullnXJnandér‘l ;.'=l(1nx.ﬁ)2/n. (g) umsmjln mimic i 5x, 5} fork = I. 2. n. [.{i,ﬁ is maximized whenj is as small as
possible and i is as large as possible, i.c., = X“) and = X01). (1.) MP)=(£IX£2)...[£‘)FA0_pyﬂ", when: A=):;;,x.~. so up) = 3.114;!) + A In p + (m a} In (1 p). 1hus.dlldp = Alp  (m  AMI  p). which equals a ir and only if p=§=ifin)/:.
This is amaximum. since L(0)=L(1)=0<b(ﬁ). m L(9)={(') " 9 ‘ 0&1”? 5 9+0" “3‘ ‘3' i} since the widlhofthc interval is l regardless of 9. Thus.
L is maximiud for any choice of satisfying the condition 9—0.5 5X; 5 9 + 0.5 for all i; this coalition
is equivalent with, *05 S 95X“, 1 0.5. if this condition can be met (i.e.. if Km 4 0.5 2 X“,  0.5, i.e., if Km “In, S i). then any 9 between Km  0.5 and Km + 0.5 maximizes L. Thus. the MLE of 9 in
this case is not unique. F (6.11)Thcljkelihoodftmclionis 3" E’sixi Hialxi!
so :0.) =4... + Z?___1len1 _ 29:11:. Xgl. Thus, dim= n + 2;, Xi/ﬁ. as «ﬁnd? = —z;'=lx. [12. Since the X35 are 11]) Poisson with parameter 1., EGG) = F. and so ﬂd’HdF) = all” = dill “mfose. £9)
land an appmximaxe 1000 a) patent confidence interval for A is An. :I: 219,]; i/n . where A: = 550:}. II 6 (5.15.)Thereisnot,ingenaral.avalueafxsuchtimtﬂx)qfor0<q<l.
7 ( 6.16.)1‘h¢U(0.l)disuihution. 3 (6.17.) M, ~ bintn. p9. and may) = up}. ﬂ ( 6.21.) The summary statistics, Histogram with bo=—2.0mddb=0.5. and abox plolarc given below: Sum statistic Value Minimum l.720
Maximum 4.010
Mean 1.230
Median l.220
Variance 0.852
Coefﬁcient of variaﬁon 0.750
Shawna: 41.225 The following all suggest that the normal is 01: most likely candidate to hypoihesizc:
' Tllerange ofthedmis—IJZIOAHJL
° 'I‘hcsamplemnandmedianmalmosthual.
 '11.: sample skewness is close to 0.
 The histogram: and box plot are both approximame symmetric.
The maximumlikelihood estimators forthe normal distribution m ,6 = 1.230 and 6' = 0.923. The frequency comparison for the ﬁtted normal distribution and the amt data given below indicates I. fairly
good ﬁt. uccpl between x = —0.5 and x = 1.0: 0.2‘ 3.0 ‘.5 The following 00 and PP plots are both quiu: linear. except for small quantiles in the Q~Q plot: 4.01 Hornn1 quanti 1 I 1.72 ’1.” Olawon M11“ Lo; 0.9981 Horn 1 probbi litte 0 . 0001'
0 n 0007 Dbl.“ problbil “:1” 0 .9901 A boxplot comparison for the ea'mr data and the normal distribution is given below. with the (internal)
quantiles of the two distillation matching closely: m» w—e
w ~—+~Clja—e We next perfumed achisqnal'e lost for the normal distribution with k = 30 equipmthe intewuls (so up; =
5.133 for all j). Since 32 = 31.065 e 39.08? = 150.90,“ do not reject the normal dishibnﬁm at level a =
0.10fnrthlselniceofiulorvals. Snpptsemntheotherhanthm we use I: = 20 intends (so up; = 7.700 for
all n. 1.; this case x2 = 21.553 > 27.204 = 112%”. so we reject the normal distribution. This example
iﬂusuatesmoflherealdllwbacksofthechi—squmm The K—S statistic is D154, = 0047, so that. [be adjusted test statistic is [4154 — 0.01 + {0.85N154)10.047 =
0.579. Since 0.579 < 0.819 2 C630 (from the second row of Table 6.14). We do not Inject the normal distribution at level a: 0.10.
The AD statistic 4,254 = 0.364. so that unadjusted test statistic is n + (41154) — (2511542)]0364 = 0.373. Since 0.373 is less than the modiﬁed critical value 0.632 (from the second row of Tdale 6J6). we do not. reject
the normal distribution at level a: 0.10. Bsed on the above heuristic procedures and goodnessofﬁl tests. we have no reason to believe that the normal
distribution is not a good model fanour data. ...
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This note was uploaded on 01/31/2011 for the course AMS 553 taught by Professor Badr,h during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Badr,H

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