hw4_solution - [Mus 3’5"?) HwfiLf §uuwng...

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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 Mud-3i WUSQHN (“wad-n? 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) snicex-massfliiiandonlyinnx—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: 21-9,]; 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 maximum-likelihood 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 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“ Lo; 0.9981 Horn 1 prob-bi litte- 0 . 0001' 0 n 0007 Dbl.“ problbil “:1” 0 .9901 A box-plot 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 achi-sqnal'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 dish-ibnfim 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 A-D 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 goodness-of-fil tests. we have no reason to believe that the normal distribution is not a good model fan-our 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.

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hw4_solution - [Mus 3’5"?) HwfiLf §uuwng...

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