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Unformatted text preview: [Mus 3’5"?) HwﬁLf §uuwng knew—ﬁt. Emmi“? Md
[um +31+6o+a1+m 9H 53A 35 + 48+ ”+59. )ﬂo '5 67“}
MM“; Qu’gthm, IMW\% JEAN? = 6513
Admﬁﬁi WUSQHN (“Mkﬂ? X£N{ 1‘— (9
FTUTWb‘xm of WWHA MU” 8mm on hand {MW/MW {3+ was + 5443+ tumr 35+ 5+)? N 1) _ ( 6.9.)111cavengc number offnilures beforcthcﬁm success is 7(a). Thus. on Ivesage one out ofevery 7(5) + l
' trials simuldbeasucccss. 4, (WK 01°)
' (d) “11.9):(28)_"fza'”elp[Z?=}(Xj,u}2/(20’2)] and «ii, a) = (—an) In (21}  n In a — {11(202)12?=1(X. — mi. Taking partial derivatives on. we set ai/au = 2.11.1(Xi * mfaz and iii/Jar: '1 (X.  ml [03 . Setting allay = 0. we get ,1 a 350.). immediately. Substituting this ii for 11 into alaaand setting this partial to 0 yields ("I=1! Lips  EMF/n  (e) SinceXmmﬂifandonlyiflnx~hlma“).ﬁ= fullnXJnandér‘l ;.'=l(1nx.ﬁ)2/n. (g) mystic4+ 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.e., i = X“) and i = X01). (1.) MP)=()EIX};2)"'[£,‘)PA0PlatA, when: “Eggs. so lip) = 3.114;!) + A In p + (m a} In (1 p). 1hus.dlldp = Alp  (m  AMI  p). which equals a ir and only if p=§=§(n)/i.
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 width om: interval is l regardless of 9. Thus.
L is maximized for any choice of satisfying the condition 9—0.5 5X; 5 9 + 0.5 for all i; this comiition
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 “I", S i). then any 9 between Km  0.5 and Km + 0.5 maximises L. Thus. the MLE of 9 in
this case is not unique. F (6.11)Thclikelihoodftmctionis 3" Elixi
“A)=e—'f'._
nial X‘!
so :0.) =4... + Z?___1len1 _ 29:11:. th. Thus, dim= q. + 2:1,, Xi/ﬁ. ma must! = —z;'=lx. [12. Since the Xi’s are 11]) Poisson with parameter 1., EGG) = F. and so ﬂd’HdF) = all” = dill 'l‘hetefose. £9)
land an appmximate 1000 a) patent confidence interval for A is An. :I: 219,]; i/n . where A: = 550:}. II 6 (5.15.)Thereisnot,ingeneral.avalueafxsuchtimtﬂx)qfor0<q<l.
7 ( 6.16.)1heU(0.l)distzihution. 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_m_a_g statistic Value Minimum l.720
Maximum 4.010
Mean 1.230
Median LZZO
Variance 0.852
Coefﬁcient of variation 0.750
Shawna“ 41.225 The following all suggest that the normal dism'bulion is th: most likely candidate to hypoihesizc:
' Tllerange ofthedmis—IJZIOAHJL
° 'I‘hcsamplemnandmedianmalmosthual.
 '11.: sample skunks: is close to 0.
 The lﬁstogmn and box plot are both approximately symmetric.
The maximumlikelihood estimators forthe normal distribution are ,6 = 1.230 and 6' = 0.923. The frequency comparison for the ﬁtted normal distribution and the amt data given below indium I. ﬁnk1y
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“ LBJ, 0.9981 Horn 1 probbi litte 0 . 0001'
0 n 0007 Dbl.“ problbil “:1“ 0 .9901 A boxplot comparison for the cl'l’tl’ data and the normal distﬁhution is given below. with the (imam!)
quantilee of the two distillation matching closely: ....,.. w—e
...... e—eClja—e We next perfumed achisqttare [met for the normal distribution with k = 30 equipmbnble intewuls (so up; =
5.133 for alij). Since 32 = 31.065 < 39.08? = 150.90,“ do not reject the normal distribution! at level a =
0.10forthisclniceofiulervals. Snwose.ontheotherhand.that we use I: = 20 intends (so up; = 7.700 for
all n. 1.; this me x2 = 21.553 > 27.204 = 1129.0.90 so we reject the normal distribution. This exalnple
iﬂusuatesmoflherealdtlwbacksofthechi—squmm The K—S statistic is D154, = 0.047, so that. [be adjusted test statistic is [4154 — 0.01 + {0.85N154 ”0.047 =
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)]o.364 = 0.373. Since 0.373 is lees than the modiﬁed critical value 0.632 (from the second row of Tdale 6J6). we do not. reject
the normal distribution at level a: 0J0. Bsed on the above heuristic procedures and goodnessofﬁt tests. we have no reason to believe that the normal
distribution is not a good model fanour data. ...
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 Spring '08
 Badr,H
 Normal Distribution, Juan, Shawna, Thcljkelihoodftmclionis

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