IE4521_midterm2b

IE4521_midterm2b - IE 4521 Fall Semester 2008(Day 2"d...

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Unformatted text preview: IE 4521 ' Fall Semester 2008 (Day) - 2"d Test Name: _____Hu_ _____ fiction 00' Problem 1 (20 points) The desired percentage of SiOg in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independentiy obtained samples are analyzed. Suppose the percentage of Si02 in a sample is normally distributed with o = .3 and that Y = 5.25 (8 points) a) Does this indicate conclusively that the true average percentage differs-'from 5.5? Carry out the analysis, clearly stating the test statistic used, its sampling distribution and the sequence of steps leading to the inference drawn by you. (2 points) b) Compute the p value of the test. (5 points) 0) It the true average percentage is p = 5.6 and a level of or = 0.01 is used in the above test with n = 16, what is the probability of detecting this departure from HO. (5 points) d) What value of n is required to satisfy 0: = 0.01 and B (5.6) = .01? . ox: I 9 Cf - ') h . 2:3 Saw-n2: ck;- 0‘9 I 7 2%: 2' 5 1'[o_ I O‘/,T'r'\ (F‘s/4r ) :12: 2733 ‘7 z“; midst '75., OI, J I, a . . . +5.9. ‘ [24's L-row‘ '1; -' . -+'I;L.i.g overus- 9-' {£11 - -_ "T :-> W... v. w. - r at W) L)thwa :.- ‘2, xiijgyl = 2! €993 35) ‘— QXVW‘A‘ : 0””? a L;__‘-—S_. to) EASE-1,}; 9; i mm + '5 _ q‘u H..._/ I l fx i Jr tn 5 ‘1 | \— :QL2.53“;'?729 ' ; €90. as in?“ ERNM‘MW' - :»'\:(}-f05é- Prohibit/it)? atlf‘cedi-..r i'l'vlt 041 {’P‘v'fit/‘f‘? 7W9” H” ' (‘— 95'”) a O mang O‘Y‘ie‘ “Meta/r OLA-(Va If: Lays .r/( C. bus . (JV-L J I b - I . — C r— “ I fig _ #Ifi—T—H :> C} I. f) L, —- 20 o! J: tit/fa » a __ e _\ 2. a. :2léT5N25i a 5.5-5.3 ' IE 4521 Fall Semester 2008 (Evening) - 2MI Test ,- Name: Dede-in (7t? 2 Problem 1 ( points) 23: use The yield of a chemical process is being studi . From the previous experience with this process the standard deviation of the yield is known to b 3 The past five days of plant operation have resulted in the foilowing yields: 91.6%, 88.75%. 90.8%, 89.95% and 91.3%. Use 0t = 0.05 in the following analysis. (8 points) a) Is there enough evidence that the yield is not 90%? Formulate and define HQ and HA, and then analyze the situation stating clearly the test statistic being used, its sampling distribution and inference from the data. (4 points) b) What is the P value for the test? {4 points) 0) What sample size is required to detect a true mean yield of 85% with probability 0.95? (5 points) d) What is the Probability of the type ll error if the true mean yield is 92%? ._.- if E40 N' ‘ f9 (I: -- b," a x a»\ do. 42 .1; -— 39/ . % r 5/ w . : )2. E ‘6 ID if": {P‘CBE’E til-alt; =5 /\Jo-i raffles ff.) -.J ‘ . I I-‘A 1‘ :1 7 No“! 2A,!) \? R 2,417- 0694 LQ +f/‘u9tpt. “FL-e, .er l S WOT" i o . “a. J .I, T ' I‘ ‘ I y i” Write. :- ‘“ >< G? (-— i“ ‘ = .2 1 CL :9 9-07: 58» r: 2f 6.4.249 : r9358 - J .x‘ 4 . ,5, ) d, {C}- . I /-~.\ "Ex ,I “flail View}; 0461M U hoing ( 1 .:. r i]; (g: Jr i t N "\I L J / X (M O I j“ 3%.“ w u j“ (I. I _- I» I 9- d v; C .- _...._...____._.:__.. .. .. E g; -- d _ - _, \_ . _. _) [J ----------- C ..1 " L. m. u % $ TV H PL 3 “:9 a 1 3 ) \ I "I n - '“. e «If: : 9673:?— 13) C. n c a if j _ ,J i". f . [EV-1 -'— \, ‘— ‘D J a; 9‘ x I) L a . —, f 3A .- __ /‘— __-_r'__fl | ._ “Hg—m = I, b If C : 4 [ Wat/7f ed: 1 i ['53 55?er or! Problem 2 (20 points) Water samples are taken from water used for cooling, as it is discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most 150°F, there will be no negative effects on the rivers ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above 150°F, 50 water samples were taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses Hozp 3 150° versus HA = p. > 1500. {10 points) a) in the context of this situation describe Type I and Type II errors. Which type of error would you consider more serious? Explain. (10 points) b) The above problem could have been formulated as H0: 1.12150 HA=u<150. Compare the two formulations. Which one do you like? In general, what principle can you state in defining Ho and HA. I E {01) Type] .eror': Erov'n analysis 0+ SMVWf’Ll “(Q-+9” W" (Mafia “gm” M “JV-wiel- irse > {90". lam-r "Lissa. Mien :5 '5 tax? ~.*. _'_'.. arrbv; “magma 3am?” gig-LG! [N2 gig-ELLJJI hELE/Hfi: M ql’ww-J b51- E ’39“, bM‘i’ +Le 4mm, maes- “)1 ‘5”; 2’9- Q/rr. O r L9] a A 1152.2 L0 n.5l 9%: (“,6.‘ —-:’ O LL-Q. -» A “hf/VT”) P 'r-p b Lama“. . II l’V/UY CL GE. r 7 7} out {. awn 'ert Mia-:4. L i/V’Q, «01 We a"? '53 {icfi‘ ’3 m3 9"? ‘: V 2 Q i,- oJrl'fi CL «2 on ‘1'; far ’2. law 3y stat“ lb) 1 I .. +Lszri tax: a :Ff*mg...c-t"’ mil/l M [MAE-"M f“ / WW oh mar- M i W N“ 1 i l ~...q'..’§f’rrro£51. (st-URL 0" T‘J-L l gyro" will rim???" N" a 9me . f 4% Fro 5w“ 5'33“- ’4 v52 r‘g 5mm,“ , I' ~ r - w \+ S VIA-01's}. earl-'0 veg w p In ‘H/O'S ‘Pi'obuaw {row .1".- WJZ larva-V ' l 'H ‘ . ‘ j:- 1 w- i‘ — 5 Qua —it.;p.Q. ii Q,[email protected] rp #Qfflfl,i64_r or +A ;. fr, « 3 Op atiu- {f/Ofvt of”: x \H.‘ {poef’ErfJ é-‘r'LrOVSL 341*?” “XS 7/0 M'Mk'q' Ft "L‘ -. -. \. \ - M . ‘9 f "" _ .— QJPV in fig fig; pM{\ 9—- : ’3" Wl+k If“, 1 I30 9 Inf}. i £.Imr i s W gm r a wig g. r/i- a v:— iw-ras-n n Wm!" “Wu P”? “3" V ’ "" V 9 .. 9 _ _ H a ‘ E ‘ r We W9” 5 pr; "ifl 1f— fivr n-1,..,.t_'\9.f-l‘-t’3 r- !fi ‘ /U\ 3 i5 0 W" 'T’v’ui; "éflfm l . gill/true,» 61 L . we. 6 LVO VK { 91 3&9 H 0 0| nai- flfr’o ""j‘ffig 3 " I r G .. 7 +{1J r {f s; r‘I' 9 via-5 Q/r r O r— l\ S 7 —I' 9/" (—9 r- Saddam 0.02 Problem 2 (20 points) ; A manufacturer of car batteries guarantees that his batteries will last on the average, three years with a standard deviation of one year. If five of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years: (10 points) a) Is the manufacturer still convinced that the average of the batteries is 3 or more years? (10 points) b) Is the manufacturer still convinced that the standard deviation of life is not more than 1 year? (State your analysis steps clearly). ' a —_ i-‘ira-waoHs-taa (a) sz. . )4' 5.? =3 Ho : M2 3 H9: /'{-'" 4 3 90‘? tit-O 0" tam =~ twin. = 37am _ K— M. — _ ts _.#___. _. é» E -. , rm ‘ "37;; h 9 “‘7 <. “hoard-.7) [Up-s. rmjar’ "E" :\ ‘ s' I .~ . / 7 {Mi MD "\ 1”?“ c tit-r21“ cum-4. { cl st~ '_ '. £2»? 5. '3 ‘ i}; a t. E1; +£me ‘7!" kg or Va,- 0 J C9?“ Ha:ch jfimr; f. _ g 1 2. _:“-— Yr d. \l ‘L 3. _‘ 3 _ ., kg S - f_.(;-__._..>: a, a 95? ‘5.“ U-Q-~ 5% has)» it __ in"! 4 '“—“—“W"” = i8‘lb, its a“ a 1., Li; 5*)! g} N O“ :- h-‘BJ‘ N "r is“? out V?“ K ml 1 joeyq. 1' 2'77 "fax-t ;'-‘m-1’-.-:c.“:.x . —- fix ’ ' t ' 5.. ~ ‘———-_, = .3, 2654:3277 r) M+ f9 it "u c two JUM ‘5‘”?- "' viii-1" is 6-H :99" U1“ ck" ‘Hmfi"? L1 ‘.;‘5!‘»- rg/ Ohm} Off 3 " “- {Tl Q ‘i 1 7-27.34. i-'u“.,-=Tiri: his :r\ 1 .. £3.24 r QZW 00! Problem 3 (16 points) 0 A research engineer for a tire manufacturer is investigating tire life for a new rubber compound. He has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are 60. 139.70 and 3,645.94 km. 59% ‘51}? 9. MT M “H,” ,1 619M}! whterfrefiifiu (8 points) a) The engineer ' ' this new tir is in excess of 60,000 km. Formulate and test appropriate hypotheses and draw oonclu ons using 0L: 0.05. i (8 points) b) Find 95% confidence interval on mean tire file. Hemfy'e‘ffi'd 3:5: mm “curls mm a.) - g“ H° "fl-figoow , Hfiifl7fi'ouo Jimmie" fiaé’”; Newtfa mLa. WW Rzéofi'? {firgMflflWfi/fim évwa- -7,/":50m 5:354:94, we WM”me "hi “it “51% M JEN in L gem-71W rum Mae “flak: a1 fe-r 5wl¢hm isn —;o_53 an “‘9‘” " //‘ 3645.?«t/flz I is.» wefe-E‘ffre Ho fiLeuflggww"f Hf flaw“ d20‘05. “Hutf‘IZéNmOUd‘nWi/Ihtgfifi MW "('01 firt " MAMMJfi‘Lfi‘cM At: My, ,5— : [I 753 I T <13 n. =7 12, ~ awn—I” 1 r m Wen? HE => H. ,3 =? 031W (mggmffmte Hurt fie Menu. life if? new tan? is 434 eater; if 50, Ma km it {ELer = mm; "-1131 (“E :9_£i5 P(éC%/§‘<U):9.TS P(f-Ufi<fl<i¢7%_) “‘75 X- 1);? door? - mix 35”” - 53! t7 3 .v x if: = (ai3i'.7+2.l3i xiigg- g2 52 i — O . 31, . 3% m: fife/VJ {5(5'9H7J16’2u324) 56(Wm 002. Problem 3 (10 points) E(€1)=E(é2)=9 V(()1)=c712 and Vflé‘2 ) = 522 Suppose 93 is flamed from 5'1 and 633 by 63 =0 é1+(l—a)é2 A new estimator (2 points) a) Show that Q, is an unbiased estimator of 6. (4 points) b) How should the constant a be chosen to minimize the variance of 9}? Assume that 5'1 and ég are independent. (4 points) 0) How should the constant a be chosen to minimize the variance of when 671 and $2 are not independent, and cov ((631, 92 ) = c. A A milflflkt[fig]Hiring];aE{éj)+sz)1E[§z):49+9_a9:9 :7 6-3; in" Mhbdmflrfyg 1' (-i 12 [am flow“ (iii/“(aim ‘ "‘ " — a i (I— ' ~ 2 A wt 6’ t 508,.th meant-aiwrél): 2 I -. - 12% z 2.5!- Me Ae-mefii/E; A MT am 62 we w :42“? “i‘léa‘a’réz i 2 A 2 3a ’2“ (daszz-ILJ‘Je-a W: 2 {64 3 .r 3&1 11-61. :7 (oTLW/‘T Qwa‘) -—--———--—J_ :0 Z.)2fl(gll+62) 2 9/23 3A 2 ‘362 :03) a: {#65 (UV {é‘ . z A A M 5)‘a¢({9r)f'fl“fl)zyflv{3a 3+2qua)/ (A A I {01/ we ) -4021; A 1 i 4 +(l-aa+az_)}gw(gzl+2m[_flc m 5:3 +(I 24+ l a : fa fl)6’2+{gah2“¢)c if 624 ac)fll+(26 6’1 "2 l )g-lf- z (Ink? Jpfi'gafiiw 62 ‘ 3%«(éh "“ 3 24({3 2_ - t 321/ (9“) 3“ «+64 2012M 2A) W 3 {f 1 L : 1+63'2()?0:’)K0W aim?) 3“ m ‘3'? a :0 a) Ea (9127451 2()+2((-26;1]:o=? 66- gal—C gem,“ 00 Problem 4 20 points) ' A rivet is to be inserted into a hole. It the standard deviation of the hole diameter exceeds 0.01 mm, there is an unacceptabiy high probability that the rivet will not fit. A random sample of n = 15 parts is selected, and the hole diameters are me 0| the data ives s = .008 mm. (8 points) a) (i) Is there . viation of the hole diameter d .Otm'“? = . 1. excee s 0 Use on O aéfd 1‘“ firm {a m (2 points) (ii) State any necessary assu on about the underlying distribution of the data. (4 points) b) Find the P-Value for stest. /0 lower confidence interval for 02. 1 {$3 35 n [ ,HA ; {curl I! 07/ " Her 6 504',HA:6?0.01" (6 points) 0) Constructa it} If! I a) tirade ‘ Ho! 6 2,941 i 24% \ 6+ , Ho! ééflfliflhboml "1 {lie rem“ 1’3 Hat: five (D in " - c n - e «ma Ho-éwmiiimgnai ,th 47% fest-er ewfiéie‘ue to AM‘Wf {>0 0! (If tag 1‘ “I 5 “Do Mt” 'fQ‘evf " {if m»: duff H. b-rfzwééim ' 1 efi‘ IA ” ‘ :. fit/1‘me {a mad/mfg 6?“)! Y9; 1‘ S rejerf {—i m; M NH3 5””? vamae It”? (9wa wth “H I 9: / «0|l . t! J 6) d I HA. (“of ’fim‘ 1%ft‘d‘95f'3’9591’l‘ fillpmtwjecf" {1‘ mm tfiafi-l 13 It - 5 — -n.. _ 5". ‘— 4- —0a SF” 1" —§0I Ann} )1 I! >y- {'l {'I ‘4" ' I ' ‘L '- ' g { 5‘.” Xizv 5 X:o[,j+ = 1 m—i} {a Xahf-v _U5'ii :1:qua X2 z - ~ (I .. __"_’ :a‘q‘ 0“ 114/ "> DOMYtd'Ed H, {a awkwe (ya-a} ‘fl . U z Mfrafrhh . MN? 74, are m'dvefifla “Mi/#4;th 5 “W h N I in) Tent 93» NW4}? {Jami-“fed w-fidx wean/k i M49 52 " j | a. myth,” [as 1) DOM "7“ Ho (wioyfiufjror ' denim/e V=i+,fL¢aWrN4ljo{8~a.85,§o ,M/Jq =[»a Naif ft) 0/1,” “HIS: it iii A :1 ° 2 ‘i (FM fine cn‘fimwf [motifs M >110qu : ztflzfl cg 2C.“ H; {.66 I .11 . tjee $1M {Lore e not (KW 9’” '4” R4 “Fume” kt “J” (lie. 61% (—oo, a)” (drool I accept H» mm r ,n I 1.1-“ (1 365mm 0&3 Problem 4(12 points) (2 points) (i) The following three distributions correspond to t sampling distribution with v = 2, v = 5, and v = m (infinity). Label each curve appropriately. Which of the curves represents Z distribution. 3 .3 .1 D i 2 The r—distribution curves for v = 2, 5, and 00. (2 points) (ii) The following two sampling distributions correspond to the distribution of Y for ,u = go under H0 and ,u = #0 + 6 , and one of position HA . Let a = 0.05 Shade the area denoting ,6. where 5 represents the probability of making type II error in a testing of hypotheses problem. f 5mm 00; (2 points} (iii) The following diagram shows several 95% confidence intervals for ,u . Explain in terms of probability what does 95% confidence interval mean? , ff Wild Lt 5“ 5:0}:4hmgaf {n Jplflwe old-EWVJS 1 1:6- #‘ffi Ma! “we 1"? day-{Wm Flu/«(Ell «41‘ ‘5"? CE fllflwly l——-——-——_l 1" .\ Interval estimates of a for different samples. (2 points) (iv) The following diagram gives sampling distributions of 92, 93, which are three alternative estimates of the unknown parameter (9 of the population. A A a. Which estimators are unbiased? {9,66% by Which estimator would you select? (9/: 3131' as at ‘ L 9. ae' m5 +5 , :5 wt 1' 5w! any! 8 AIM mfl ‘1 f 6f Val/1W Sampling distributions of different estimators of 9. . 2 *— (2 points) (v) How are Z and 25., related? 7ft! $42231 h-rg: aim 2,, 2, . .33, (we 3M4”. uWMJcI 94’wa: E-J‘tini» . . 2 (2 pomts) (VI) How are 11, and Fvlow2 O x {dark/i .- ___._ .___.- Wig/2. related? F 7f" it, Seem dolfiaal Problem 5 (25 points) a) In the context of 'Testing of hypothesis’ where HD is the null hypothesis and HA is the alternative hypothesis. Name the following events in their terms. (1 point) (i) Ho/HA Twat IL error (1 point) (ii) HA/HD (We I WW Give the traditional symbols for the following Probabilities. (1 point) (iii) WHO/HA) ’9 (1 point) (iv) Prob (HA/Ho) C4 (1 point) (v) Prob (HA/HA) ]. f3 DOW fist b) (4 points) II J 6"], and 62 are two point estimates of the Parameter 6 A We know then [gr 1-5 a“ “£95560, Ei‘h‘rm‘bv! Jim-t (£339, yaw-Ma, ,. Suppose A A {213 a gamer. [nit [4.9 My, mew = 6 V3r(61) = A A g 9 81% (52] = Mania: 5 19;” 2_ A g A .- 5192):; Mag): 4 MW W L-nw (ma). ‘ Mew wants: 2 I ] : 0 ’_" A i. 1'“ A l _ Which estimator is better? MSE { (an ) : wrfigflffifl: 2 / 92/0 fl) t9,isLefter=9MfEf9J<fl$Er6J=3lu<§~lfyg 1:. z A) : 4 + E ‘ A n n {K In what sense is it better? + 3.3912.th 79mfl9flww. 195:4999 (4; c) (2 points) (i) If Y is the mean of a random sample of size n taken from a population with mean fie and "I? - #5 ark/E finite variance 02, then the limiting quantity of follows what sampling distributes? r7 2.: 5W dol Z002— Problem 5 can’t (2 points) (ii) What is 56) = (fl ? .r’ _ 2 and Var (X)= 5/ ? (1 point) (iii) What does (i) say additional to (ii)? HM SMEWs-amdm mmdfléébl/Mrwlfiwbhfizfivf d) (2 points) i) Let 2 be a standard normal ran om variable e a chi-square random var able with V degrees of freedom. lf 2 and V are independent, then what is the name of the distribution of the . 2 quantity — t (2 points) (ii) If If is a mean of a random sample from a normal population with mean ,u, and population variance unknown and s is the sample standard deviation of the data then what distribution does the following quantity follow? X—fl Six/E Tit—l e) (2 points) Both sample mean (denoted by E) and sample median (denoted by X” ) are unbiased estimators of the population mean y , which of the two estimators is more efficient, and why? SS-Elo Mead 1‘5 at Let-fer 65$~wpfifl if 45 WWW :4 five . . . f) ( pomts) For each of the fol owmg assertions, state whet er or not it is a legitimate statistical hypotheSIs and why. .2 Ifg‘wtfl'fp 4L! 9;! [‘5 (i) H: or>100 V/ [Mn MM”! (ii) H: E :45 X (iii)H: ? -? =5;( 9) (2 points) Write a note not exceeding 2 sentences on ‘why is it important for decision makers to learn about inferential Statistics.‘ ' i‘ Wig wflfit "(Lo RULE 51594131336 9h FonM-gm M Lakdf 9“ {Why i. u Den‘a‘ms Lib-“V6! 0% $9M?!“ (5m ygfifl an “10%)” .i £13455th {g Veg a“???pr 4‘”. ’[LflLclt-tlj hark] ts “Ml/LE Wad- igvfig'jfiyms" ".\ ...
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This note was uploaded on 02/19/2011 for the course IE 4521 taught by Professor Staff during the Spring '08 term at Minnesota.

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IE4521_midterm2b - IE 4521 Fall Semester 2008(Day 2"d...

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