Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Unformatted text preview: 36-220 Engineering Stats, Fall 2005 Homework 11 Solutions Due: Dee-her T, EDDE Regular Exercises 4‘]. .. finmmm = 1.52 + .o2[1o] —1.4o[.5] + .o2|[5o] —.ooo5[moj = 1.95 b. ,2me =1.52 + .o2[2-:}] — 1 so; 51+ .o2[5o]— .—0-0-05[3o] _ 1.412 e. 34 =—.ooos; Ion-34 = —.os. 11. There are no intersetion predietors — eg.= x5 = x1114 — in the model. There wouldbe dependents ifinteraetion predietors involving 34 had been included. 2G. 11} e. R2 =l—E= Lil-9'9 FortestingHfl .31: fl: =13} =fi4=fivs.I-I.:atleast one among ,3 ,3 is not zero the test statistic is F — 1% H‘1 will be 1 ----- 4 : — L 2 i? - 1 E {a—k—l} .49 mjecledifferrflms =2.2~5. f =é=sn Because so} 2. 25 ,I-Iuts 5 rejected and the model 15 Judged nsefill [this even though the value ofF‘.2 is not all that impressive}. 41. Ht] 231:}5}: =...=fifi =fl' vs.I-I,:atleastoneamong :31 ..... 13:1 isnot zero. Thetest 1% statistie is F am. Hominberejectediff‘eiju =2.42. a- _ E f— {14%; _24_41_ Because 24_ 41}2_42 ,I-Iuisrejeetedsndthemodelisjndged fl useful. 42. {55. To test Ht] Zfil = ,3: =43 vs. Ha I atleast one )3: ifl', the teststatistic is f = fl = 3-19.31 {from output}. The associated p—value is I}, so at any reasonable M'SE level of significance, He1 should he rejected. There does appear to be a useful linear relationship between temperature difference and at 1eas one of the two predictors. . Thedegreesoffi'eedomforSSE=n—flc+ l]=9—{2—l}=fi[whlch}rouoo1fldsimp1j read in the DF column ofthe printout}, and rm“ 2 2.44? , so the desired confidence m1 e soon i [2.44:4]{4321} = soon i 111578 , or about [1 341441574]. Holding furnace temperature fixed, we estimate that the average change in temperature difl'erenoe on the die surface will be somewhere between 1.943 and 4.351 When 31=13Dfl andrg = T, tbeeetimated average temperature difference is j» = 499.55 + emcee1 + 3.13-3ng = —199_55+.21oo[1341o]+ 3.4mm = 94.44 _ The desired confidence interval e then 9444 i {244?}[353] = 94.44 i .354, or [93.58353fl] _ . Fromthepn'ntcrut: s = 1.058, so theprediction intervalis 94.44i[2.44? [14:6st +353]2 =94_44:2.?29=[91_?1,9?_1?]_ P{satis) = .51 .2 P{mean| sans): — = .3922 .51 P{n1adian | satis) = .2941 P{fllfldfll satis] = .313? 50 Mean (and. not Modal} is 1:11: must likaljr anthur, while Madian is least- 83. I. LetD1= detectiunon 1St fixation: D2 = detectiunon 2nd fixation. P[detectionin atmust 2 fixatiuns}=P[D1) +P(D1' an) =P£D13+Pm1IDr3Pmn =P+P{l-P}=p[l-P}- b. DefineDl,Dg, ,Duas in]. ThnnPfat mustnfizaflnns} =P(D1]+P(D1'HD2]+P(D1'HD2'H D3]+___+ Pml'nflg'n...nDn_1rn Du] =P+P{1-P)+PI[1-P)2+---+P(1-P)“'1 1—{1—p) =1_{1_ 1-{1-P} Alternatively, Hal mnstn fizatinns}= l—P[at1east n+1 arereq’d} =1— P[nu- detection in 1“t n fixations} =l—P(D1'an'n___nDn') =1—<1—p)“ =pl1+(1—p}+u—pf+...+{1—p)“'1]=p- p)” :. P[nu detection in 3 fixatiuns} = [1 — pf 95. When three expatiments are perfmmed, there are 3 different ways in which detection can occur on mu)? 2 uffln: experiments: (i) #1 and #2 and nut #3 {ii} #1 and not #2 and #3; {iii} [Infill anfl#2 andifl. che impurity,r is prawn, the probabiliw of emfljr 2 detections in three {mt} experiment is{_flj{.81[.-2}+[_8}[_2)[_E}+ LEXEKE} = 334. Elbe lmpflfltj' 15 absent, flu: analogous probath is 3[_1}(_l}[_9) = .1321 Thus PW | delaectedinexacfly 2 nutnf3}= P{det ecmdingxacrfitfl fl presmr) P{det ectm’ingxacrfitfl) {.334}{.4} = —= .9135 {.334}{_4) + {in T}[_fi} 12. 64. .. Graph of 1131) = 3133314 — :2) 3 2 e. PfK}fl)= L2.fi93?5{4—x2)dk=.fl93?5{4x—x?)] =.3 I] e P(.l-::X=:1)= [11.39333{4—x2}dx=.5333 d. P(x<:-jt1Rx:».5}=l—P{—55X£j)=l—I;:.{193T5{4—x2}afi: =1-3d32=.d323 a. Pchfl)=F[fl)=j I]. Pf-liXil}=F{l)—F(—1)= }fil=.fiBT5 c. me.3)=1—P(x53)=1—F(.33=1-33343=3164 ii 1 3 x3 3 3x2 2 d Fx=F'x=—— _ __ ={}+— 4—— =.fl-93T54— U U afr[2+32[4x 3]] 32[ 3] [ x) e. FILE] = .5 by definition. Fm) = .3 £13m . abm'e, which is as desired. Let X1, .. ., K5 dmote taming times and Kg, .. ., Km denote evenJhg times. I- E1X1+---+X1:1)=E[X11+ +E13£131=5E1X11+5E1361 = S{4}+5{Sj= 45 b. VarfiXl + __ _+ Km) = veecxl) + __ _ Harm“) = 3 veal) + 3mm) :5 E+E =fl=ds33 12 12 12 Ein—Xa)=Er;x1)-Ecm=4—S=—1 .54 um 154 v — =v +1.? =_+_=_=13_57 at(511%) an[(511) “(3:5) 12 12 12 - EKXI+——-+X5)-(Xd+---+X1u)]=5{4)-5(5}=-5 VHIEXI + ---+X5)-G‘:d + ---+X1u)] =VaI[X1+ ___+X5)+VEI(X5 + ___+Xm]]=15833 Extra Credit 1'34. .15 1~=cR2-=cR3 i .0525 R1dR3dR2 .15 .EISTS LIT-'5 “WHERE .15 .525 .15 _ . .. PIGIR1=1R2<R3= —=-5T,P{BIR1<RQfiRa)=-33,d3551f3rflfigwmm .15 + .flTfi .fifiZS _ b. P[G|R1 £123 :: R2]= — = 2941:: .GIS, 50 classflfifasbasalt .2125 HG | R; q: 111:: F4) = % = .fl'fifi? , so classify as basalt. t. Harm £13311] = PIER classifas G} + P[G classifas- B} =P[classifasG| BFfiB)+P[classifasB |G)P[G) =PI;R1-=:R2<R; IBX.?5}+HR1cR;cR2mR;cR1=:R2 IGI-25} = {JDXJfl + [25 + .15125) = .175 15. Forwhatwlnes ufpwfllP[G|R1-=:R2:‘JR3]::_5,P{G|R1{Rgdllflbi PIGIR3<R1<R2)}-5? P[G|R1=:Rng3)=fi—P= 'fip 3.534931 .fip+_l{l—p) _l+_5pl T Henna; c112): i:- .5 ifl-‘p 3. i .25p+_2{l —p) 9 P[G|R3~=:R1~::Rg)= it: .5 if}? 3'- E(mm!11:5:ttil:l;i'.n::} .15p+_?{l—p} 1? Ep 2:- % always chasifyrasgmnitt FfK)=flfiarx£fl,=1foerl,andforD-=:X<: l, F£XJ=Lflyw=Lgay“{1—y)ay=9fiL{y“—y9w 9*}ny _fiy1u I =lflx9 _9x1u "Ill Fit} 5 F(.5)= 10:5? 4:31“ z .010? P1325 5 x5 .5) = 13.15) — FILES) : mm — [1011239 — 3231“] x mm — mm x mm The Tim peroenfile is fine value Dfx for whicth} = 35 2:: .15 =1qxf — 9931“ 2:: x: 30345 3. E{X}= Ex-ftjxjmc {333350—3133 =33Ex"{1—x)sx = 331“ —%x“]; = fi :3 .3132 srmmmrdx _fl 11 _fl 12E —11x 12}: 35.6313 3:133} 3 .3313 — {3132)2 =11124, on= .11134. I. [.1 4; a=(.3333,.3333}.11m_ H11 -3331: H+c)=F(.9295)—F[.TDEB) =34455—.lfit}2=.6363 11- 3334:3432): Effie-fir: L1132[§][1—{13—r)2 )3 3 11 3 11 2(1le r2[l—{lflfl—2flr+r2}}i =13L —33r2 +2333 —r“sr=13o2m 43. 4’1". 3:; =13, s = 2513131133; =gm3m=333= so 3 =133.43—43.33[3.3]—3.3313(333]+3.3333[333]= 43.313 No, it is not legitimate to interpret ’31 in this way. It is not possible to increase by 1 unit the cobalt content 3.1, while keeping the interaction predictor, a, fixed. Whenal changes= so does 333, since u = 3133;. Yes, there appears to he a useful h'near relationship between y and the predictors. We determine this by observing that the p—value corresponding to the model utility test is 31' .{lfl'fll [F test statistic = 18.924}. We wishto test Ha :33 =-:} 33. Ha :fi3 st}. 11131351 statistic is 1=3.433,witha corresponding p-vslue offlfllfl. Since the p-value is 3: alpha = .Dl, we reject H‘1 and conclude that the interaction predictor does provide useful information about 3r. A 95% {1.1. for the mean value of surface area under the stated circmnstances requires the following quantities: 3 = 133.43 — 4333(2) — o.3-:113(3-:1-3]+ 3.3333(2I333] = 31.333. Neat, $3125.13 2 2.1241}, so the 95% confidence interval is 31.333 3 [212314.33] = 31.333 33.3423 = [31.3332,41.3433] For a 1% increase in theperceutage plastics= we would expect a 28.9 kcalu'lrg increase in energy content. Also, for a 1% increase in the moisture we would espect a 314 lccah'kg decrease in energy content. 75. 55. Theappropriatehypothesesare Ht] 231 =fi2 =fi3 = 4 =fl' vs. Ha I atleast one ,3: at fl . The value ofthe F test statich is 16TH, with a corresponding p—value that is extremely small. So. we reject He. and conclude that at least one of the four predictors is useful in predicting energj.r content using a linearmodel. Ha :33 =fi vs. Hg: 53 act}. Thevalueofthetteststatisticist=124,witha corresponding p—value of .034, which is less than the significance level offlfi. So we can reject H9 and conclude that percentage garbage provides useful information about energy:T consumption, given that the other threepredictors remain in the model. 5‘» = 2244.s+2s.925[2o]+ ?.544[25]+4.2s?[4o]— 315541145] = 1555.5, and 15.02125 =2.fififi.moteanerrorinthetes:t: Syn =12.4T,nct?.45}. SoaDSEV-CJ for the true average energy,r content under these circumstances is 15455.5 :r [2.455111124le = 15115.5 i 25.59 =[14?s.s,1531.1]. Because the internal is reasonath narrow, we would conclude that the mean energy content has been precisely estimated. A 95% prediction interval for the energy.r content ofa waste sample having the specified characteristicsis 15115.5:r[2.o5o [31.4s]2 +3247?)2 =15115.5J_rss.?5 = [145511532]. .t‘u'flz,6*1=,5*2 ={:- willherejectediufavorof Ha : either file: ,82 it} if R2 f=l:2—t :3 thu+l =Fc123 =9-55 - SST=EF2 [email protected]= 254-5 : 5“ Hi :I—t—l} ” 35' 1:2 = l—fl=.393,3fld f ’35 =3fl_3. seeeuee511s 29.55 Hqisrejectedat 254.5 = W}; significance level .{}1 and the quadratic model isjudged useful. Thehjrpothesesare Ho 2,32 243' vs. Ha: I32 it'll. Theteststatistic value is value 51' Dill. The quadratic predictor should not be eliminated x=l hereand 115.1 = fin +fi1{1]+fi2{1}2 =45.945. 1m. =l.I5,g:ivingtheCl 45.515:{1.ss5]{1.1151]= {4401,4151}. 3639 Yes p-value = Jill}? which is less than .01. No, p—value = 1143- which is less than .115. .1. 14145? i {1441133301} 4.0501234) E. fiflfl = £3915? , using (I = .135, theinlmval is 5305? i(2.44?jI.||'{_4351]2 4.14522]2 45.051545] ...
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