hw4_solution - [Mus 3’5" HwfiLf §uuwng knew—fit...

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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 Admfi-fii WUSQHN (“Mk-fl? X£N{ 1‘— (9 FTUTWb‘xm of WWHA MU” 8mm on hand {MW/MW {3+ was + 5443+ tum-r 35+ 5+)? N 1) _ ( 6.9.)111cavengc number offnilures beforcthcfim success is 7(a). Thus. on Ives-age 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 = 2.1-1.1(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) SinceX-mmflifandonlyiflnx~hlma“).fi= fullnXJnandér-‘l ;.'=l(1nx.--fi)2/n. (g) mystic-4+ 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)"'[£,‘)PA0-Plat-A, 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: 21-9,]; 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 lfistogm-n and box plot are both approximately symmetric. The maximum-likelihood 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. fink-1y good fit. uccpl between x = —0.5 and x = 1.0: 0.2‘ -3.0 ‘.5 The following 0-0 and P-P plots are both quiu: linear. except for small quantiles in the Q~Q plot: 4.01 Horn-n1 quanti 1- I -1.72 ’1.” Ola-won M11“ LBJ, 0.9981 Horn 1 prob-bi litte- 0 . 0001' 0 n 0007 Dbl.“ problbil “:1“ 0 .9901 A box-plot 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 achi-sqttare [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 A-D 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 goodness-of-fit tests. we have no reason to believe that the normal distribution is not a good model fan-our data. ...
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