finalmath306solution - Q-l. A random sample X1,Xg,- - 1...

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Unformatted text preview: Q-l. A random sample X1,Xg,- - 1 3X11 of size n is taken from a population with the density function if 9 < :1: < 00 Ha"; 9) = x3 ' 0 elsewhere As an estimator of 8 we decide to use (9 = Y“) = mlII{X1,X2, - - ‘ , Xn} a} Show that Ym is not an unbiased estimator of E). 7pts b) Produce from Y“) an unbiased estimator of (-3. 3pts (2) What is the Maximum Likelihood Estimator of 6? th3 Note: Recall that the probability density function of Y1 = min X1,Xg, ‘ - - , Xn is given by ( ) fyu)($) = nf(z)[1 - F(:r-)]”_1, where = f; f(t;C-))dt is the cumulative distribution function of the distribution). e W LL '2.\‘ 1, FL We 7' “kin—QM “l fileianQ‘rQ(Q-J) 12mg ...... W :Zhe x?@ 943 at}, x 53610-2. .le/ll‘l'l no 55 -1 ‘1‘“! 4 56(0)}: 91109 g mfg-“dye 7’3 2116 \ z “9 (0"@ F) g.) «ind—l “9‘2"” *2 FL“ , ani £3? \(m m] ll be 4m which: 5'95! 95-Hm44r n. L n ‘1“ C) LCKHA Tail—l- EL 2 9 73 Ill 12 l “:53 )9 “'{nj 1—H“) gal/negro“ cLoeaei my} imp-kg Cit 342-444"QO Pefnf . (Elf demr'Yeil-I‘M Gian O )' lil‘btzm‘efirjflf [$th oLg'nL-Jfiéllj M )- 114,363, Ll7511~321hfiQl 6 glwvt'cl have 43 lo-rzu value; as fof-S‘fih'lole’, 12.14% filing-E. his 9g 3*2i_9'>r'h l; PU; 34w.“ @mlani’y ) llme let/1899,36 Valve 5? '6) 5’) Minéiajhjxhgfi W’nce m MLE 515 G '1: again \QUe-mofxiixypymfi we lrtcwe @ Qt“;- (radium)- (2-2. At a certain intersection in the city the car accidents occurred in the past at the rate of A = 2 accidents per week. Recently a tendency towards a smaller number of accidents is observed. It is assumed that the number of accidents/week follows a Poisson (A) distribution. To test if there has really been a decrease in the average number of accidents/week, we decide to observe the intersection in n = 5 different weeks and record the number of accidents. Then we construct a test as follows : Null hypothesis H0 : A = 2 against Alternative hypothesis H1 : /\ < 2 We decide to adopt the following rule of rejection : " Reject H0 if the sum 2:1 X,; of accidents at five different weeks is 5 or less (Xi: : the number of accidents in the i-th week). 3) Calculate the size G: (Le. the probability of type I error } of this test, 4pm b) Calculate ,8 probabilities of the type II error when the true value of A is /\ = 1.53 A = 1.0 and A = 0.5. ths c) Based on the values obtained in a) and b) give a rough sketch of the power function. 4 pts (Notes: - Recall that the sum of independent Poisson random variables is another Poisson variable }. - Use the cumulative Poisson probabilities table given to you for your calculations a) 32% i‘o MAN-tail; u, name/s3 3 l1] s-x “V14”?! Ho ) luv) .Polsien (in) (Ll-[+14 PM Him a ill I; on; a Q. “Magpie \ gait; ME 73,. a 5“ \ 92214 {>( Pobjmflolég) 10.06774 ("Pct-He) L) p-:P(%x;'>am) s . Pl 2 P( gYC> 6 mini,g)=P(Po’issoéf7.§7>Sltl~l7eium (1.5 4’5) q a: 1~ca‘-2/i £4 $47636 [39.“ PlZ.X€>’5l betel-=WWWfs‘l7‘5l:throiaawnrg)es) a 7—0.616’o5013c5940 fig .7. ? 320.;d=?(?9473m(2.s)7éje l-—-%ilkm(33ig)é§ l :Ie ores 5‘30 -: 0.0420 Q-3. To compare the variability of temperature in June at two locations, one a mountaneous terrain the other one is a sea side place n = 8 measurements (centigrade) in June are recorded : Mountain ] Seasicie__ 17 25 21 23 16 19 13 25 22 28 14 2'? 28 26 15 24 Letting of and 6% to denote the variances of temperatures in June at the mountaneous terreain and the seaside place respectiver and assuming normality of temperature distributions a) Test H0 : of 2 J3 against H1 : of b or: at {1’ = 0.05 ievei. ths b) Test H0 : of = 20% against H1:U?> 20% at rt 2 0.01 Ievei. 8pts Warning: This is a test of hypothesis question. DO NOT construct confidence intervais. In both a) and b) give 1°) 3 test statistic 2°) a rejection rule and then carry out the test. 2, - '2. sag: 3993',“ _ (Semi/g : 2.94441? 1 E44222“; 7 ';L L W I [if .12 -M_ (118). _ _ 3;: 1: iii; a “$5946”:- saw; F a v???‘ .1 51 t 1 £1 2.42 MLJ-t z .2.’ a) U'fléfr Pia F117 "gm 2 -2 "L H _. . 3 _ ,. .29,?9 Rej‘e’d- Ho ‘i’ zavficjcgjzj J v‘og)??? Emilee-3422319 Areas/{Ji- Ho L - 5 .2, i . . slwt As- "1,15*43 ré’I if, win,— _: asfiéLz‘e—Jln— ‘. :.i;l ) 1" V HO mg a?! 2_ $22, 2 L , L_ z iéml 3%? 16.459 ms afizéso Arte/pi- Ha I (Ti _Q_6‘L Q-4. We like to test a hypothesis that the life time of a specific type of battery follows a normal distribution. Our past experience indicates that p.- = 3.5 years with a standard deviation or = 0.7 years. The life spans of 40 batteries of this type are recorded to get the following frequency distribution : Class Limits f}- 1.45 — 2.45 i 3 2.45 —2.95 ' 4 2.95 — 3.4-5 15 3.45 — 3.95 10 3.9.5 — 4.95 8 Carry out a goodness of fit test at or = 0.05 level to see if the given frequencies are coming from the normal distribution with ,u- = 3.5, a = 0.7 years. ( If the expected frequency in any class is less than 5 combine it with the next class 171mg P — 1‘ ‘- P‘ $0.4; (#245) ~ P( fixes—:50 (Z < 2-46~3'5°):P(.2.sz (214.5) . 9.? -: £021.93) $05) = 0 4633-0 .4332 c 0-06 5} i’JPCZ *4 §< x< 24m : Pew; < 242%“sz 0.; 2 H151 - {‘(039) = o .4322- 0.2.5.292 = 04490 : Pf-(.§<Z{~av?9) P; =P(2 -%€<X<3~4si=l>(~0~?962(3 .49-2 50 (9.2;. 2" HO‘WJL- {—(omi : 0.2:?52 43.02 :79: 0 £52? l=P{. <9 .%ZE<—O«O?) {94" W3 -4€<)<< ’3 35% 1964 .v'IH'zQ’ Oi szfl— 0'? 9'0? (260.61!) -:. t (0.94) ++ro.o;i= o .1185+v.0276 :- 0.266.? 127. 2 PC3-9;<>< 445$] = P[0.64 <24 4W)=P(3.64<z <2 09) 0.? an ‘ aerawiirom) mews-0.223? two-29:9 as 39:. scam; 1. 45-4.“: 7 406.0551) 2—. 2.904 3 1.46-2-55 4» 409.1430) 29520 gig“ y=4fiofllzg 2-55-— 3-4si u; 400.2533) = ram; 98 - 7 22‘s 3-‘r§w§'5$ ’9 40(o.1659):10-6?2 0'05}; . 2-35'» 9.9; 3 in. (9.9.46): 9.6% 2 L L 7. C .. f — . _ L x 1 '1‘ 8524) +Ls‘ 107.92) +[1u “9.6123 +(g..9,6?6’) :O‘nzgmgé wmfimz‘ao: 3.92.; {0.9.31 run-6:12. 9.6% 9- 7f, - 4515’; _ _ 2. __ .- 7a '24ng myoyfigaams Mowi- l'l‘a 204*th Nomi-2.1%,) Q-5. Given the following joint probability density function of X and Y 2 ‘ _ - . “33:31): 5(214313) 1f O<m<1,0éy<1 0 elsewhere Find My”, i.e. the regression equation of Y on X. [ Note : Recall that 1“.le = E(Y|X = :12) = [01 nyIX (ylmfly 14pm l l '2. 2— 1 a... @XC‘K‘l: ...- gCi‘x-l-ZLaijac—g— (Email-3:62.: = E (2:. +3.): {LI-Pg. ) 047w g o lat-fir: 5 £5316 + ) SLY [5A2 H” 3 M 00-3)“ ’2’ LEI-x43 ‘tx +3 ‘5’ Q-G. ln spray—painting the humidity influences evaporation. A controlled study is conducted to examine the relation between relative humidity X (as Va) and the solvent evaporation Y (‘70 in weight}. [ The knowledge of this relationship will be useful in that it will allow the painter to adjust his spray—gun setting to account for humidity ). In 25 measurements the following summary data are obtained after some rounding : n = 25, {is 11:; = 13153 233%.- : 236, 3353:? = 76308, :35 y? = 2286, 2? my.- 2 11825. Assuming that the linear hypothesis holds a) Find the estimated regression line 3; = d: + 41335 13} Carry out the test Hg 1 i3 = 0 against H1 : ,3 4 0 at the a = 0.05 level. 419155 c) Find a 95% (two-sided ) confidence interval for 0'2, the variance of the error. zipts d) Find a 90% prediction interval for the single predicted value 1/010): i.e. x0 = 40. 4pm 8) Find the sample correlation coefficient r‘. 4pm; f) is there a positive correlation or a negative correlation between X and Y ? Would you say that it is a strong correlation ? Why ? Spts ‘ _. 5:3 <3 .6 6 .., 5y '2. l; 3 J ""1 “31.5 .. (l l )(ZSé g -—--—- .. We . $6308... Q3157L jlzs‘ -"'oo 2/ A - fl" _ fl dab—P 7.‘ 3 EE 40.0%???)213-648 (4-13.649_0.ogx _a 23.531 (42.02) 522-6) : L) 8"“ 95E : %.L£&s,lu“”(?f " l 3 w h-—-’L 52,3 ...l|.0‘3’2- . - L oi-G‘vl‘? S't , . . 13 42% ' 0.05313 SC .30- d ————_.r-—- 5““ Pill?“ "(how'ON'lm “9-24 «wen, we “a 3e. 0‘b93?’ U\ 0) SE <g< 955 ) item ngimm " 33.0?6 new ‘52, 0.07.9)“ ousasn) eassMGléo‘MVZ Al all} "’0: lé’m—z Sal i +-\- + of" "We :3-é4s_o.os(4o)t 1.7-11-i(o-6938lJ H515. + 6433 _l i [0.442 (him ._ ._ . ._ __ “59 5‘13?(‘C54°l(lti s ,. ' ' ' .6 a) P: «j z 51959.5 588 . ¢_0\9134 we, first rem“ 644-26 ...
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finalmath306solution - Q-l. A random sample X1,Xg,- - 1...

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