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Unformatted text preview: 36220 Engineering Stats, Fall 2005 Homework 11 Solutions Due: Deeher 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]— .—0005[3o] _ 1.412
e. 34 =—.ooos; Ion34 = —.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= Lil9'9 FortestingHﬂ .31: ﬂ: =13} =ﬁ4=ﬁvs.II.: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 ,IIuts
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.II,: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 ,IIuisrejeetedsndthemodelisjndged
ﬂ useful. 42. {55. To test Ht] Zﬁl = ,3: =43 vs. Ha I atleast one )3: iﬂ', the teststatistic is f = ﬂ = 319.31 {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 + 3.133ng = —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{lP}=p[lP} 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{1P)+PI[1P)2++P(1P)“'1 1—{1—p) =1_{1_ 1{1P} 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:
=13d32=.d323 a. Pchﬂ)=F[ﬂ)=j
I]. PfliXil}=F{l)—F(—1)= }ﬁl=.ﬁBT5 c. me.3)=1—P(x53)=1—F(.33=133343=3164 ii 1 3 x3 3 3x2 2
d Fx=F'x=—— _ __ ={}+— 4—— =.ﬂ93T54—
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[GR1 £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 IGI25}
= {JDXJﬂ + [25 + .15125) = .175 15. Forwhatwlnes ufpwﬂlP[GR1=:R2:‘JR3]::_5,P{GR1{Rgdllﬂbi
PIGIR3<R1<R2)}5? P[GR1=: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[GR3~=: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}= Exftjxjmc {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): Efﬁeﬁr: 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 pvslue ofﬂﬂlﬂ. Since the pvalue 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(3:13]+ 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}. SoaDSEVCJ 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 =955  SST=EF2 [email protected]= 2545 : 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 pvalue = 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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 Fall '05
 Shalizi
 Statistics

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