Solutions manual 2001 design and analysis of experiments

SolutionsMfromMMontgomery,M D.MC.M (2001)MDesignMandMAnalysisMofMExperiments,MWiley
,M NY

Chapter 2 M


Simple Comparative Experiments
M M



Solutions

2-1 TheMbreakingMstrengthMofMaMfiberMisMrequiredMtoMbeMatMleastM150Mpsi.MPastMexperienceMhasMindicat
edMthatMtheMstandardMdeviationMofMbreakingMstrengthMisMM=M3Mpsi.MAMrandomMsampleMofMfourMspecimen
sMisMtested.MTheMresultsMareMy1=145,My2=153,M y3=150MandMy4=147.

(a) StateMtheMhypothesesMthatM youMthinkMshouldMbeMtestedMinM thisMexperiment.

H0:M M=M150 H1:M M>M150

(b) TestMtheseMhypothesesMusingMM=M0.05.M WhatMareMyourM conclusions?

nM=M 4,MMM M=M3,M yM=M 1/4M (145M+M 153M+M150M+M147)M=M 148.75

yM MMoMMM 148.75M150 1.25M
zMo M M M M0.8333
M
M
 3 3
n 4 2

SinceMz0.05M =M1.645,MdoMnotMreject.

(c) FindMtheMP-valueMforM theMtestMinM partM (b).

FromMtheMz- PM M1M0.7967MM2M 30.7995MM0.7967M0.2014
table:

(d) ConstructM aM 95MpercentM confidenceMintervalM onM theMmeanM breakingMstrength.

TheM95%MconfidenceMintervalM is

M M 
 M M M yMM z 

yMMz n
2


n
2



148.75MM1.963M 2MM M M148.75MM1.963M 2

145.81MMMM151.69


2-2 TheMviscosityMofMaMliquidMdetergentMisMsupposedMtoMaverageM800McentistokesMatM25C.MMMAMrando
mMsampleMofM16MbatchesMofMdetergentMisMcollected,MandMtheMaverageMviscosityMisM812.MSupposeMweMkno
wMthatMtheMstandardMdeviationM ofMviscosityMisMM=M25Mcentistokes.

(a) StateMtheMhypothesesMthatM shouldMbeMtested.

H0:M M=M800 H1:M MM800

(b) TestMtheseMhypothesesMusingMM=M0.05.M WhatMareMyourM conclusions?




2-1

, SolutionsMfromMMontgomery,M D.MC.M (2001)MDesignMandMAnalysisMofMExperiments,MWiley
,M NY

yMMoM 812MM800M 12M SinceMz/2M =Mz0.025M =M1.96,MdoMnotMreject.
M
zMM1.92 M M M M
o
 25 25
n 16 4

(c) WhatMisMtheMP-valueMforM theMtest? PMM2(0.0274)MM0.0549

(d) FindMaM 95MpercentM confidenceMintervalM onM theMmean.

TheM95%MconfidenceMintervalM is
 
yMMz MM M yMMz
2
n 2
n
812MM1.9625M 4MMM M812MM1.9625M 4
812M12.25MM M M812MM12.25
799.75MMM M824.25


2-3 TheMdiametersMofMsteelMshaftsMproducedMbyMaMcertainMmanufacturingMprocessMshouldMhaveMaMmea
nMdiameterMofM0.255Minches.MTheMdiameterMisMknownMtoMhaveMaMstandardMdeviationMofMM=M0.0001Minch
.M AMrandomM sampleMofM10MshaftsMhasManM averageMdiameterM ofM0.2545Minches.

(a) SetMupMtheMappropriateMhypothesesMonM theMmeanM .

H0:M M=M0.255 H1:M MM0.255

(b) TestMtheseMhypothesesMusingMM=M0.05.M WhatMareMyourM conclusions?

nM=M 10,M M=M 0.0001,M yM=M0.2545

yMMoM M 0.2545MM0.255M
zM M M M15.81

o
 0.0001
n 10
SinceMz0.025M =M1.96,MrejectMH0.

(c) FindMtheMP-valueMforM thisMtest.M P=2.6547x10-56

(d) ConstructMaM 95MpercentMconfidenceMintervalM onM theMmeanM shaftM diameter.

TheM95%M confidenceMinterval
M is  
yMMz MM M yMMz
2
n 2
n
M 0.0001MM M 0.0001M
0.2545MM1.96 M M M 0.2545MM1.96 
   
 10   10 

0.254438MMM M0.254562


2-4 AMnormallyMdistributedMrandomMvariableMhasManMunknownMmeanMMandMaMknownMvarianceM2M=M
9.MFindMtheMsampleMsizeMrequiredMtoMconstructMaM95MpercentMconfidenceMintervalMonMtheMmean,MthatMhas
MtotalMwidthMofM1.0.




2-2

, SolutionsMfromMMontgomery,M D.MC.M (2001)MDesignMandMAnalysisMofMExperiments,MWiley
,M NY

SinceMyMMN(,9),MaM95%Mtwo-sidedMconfidenceMintervalMonM Mis

MMM 
yMMz MM M yMMz 
2 2
n n
3M 3
yMM(1.96 MM M yMM(1.96)M
) n
n

IfMtheMtotalM intervalM isMtoMhaveMwidthM 1.0,M thenM theMhalf-intervalM isM0.5.M SinceMzM /2M =M z0.025M =M 1.96,


1.963 n   0.5 M M


n  1.963 0.5  11.76
M M M M


n  11.76  138.30  139
2 M
M M M M M




2-5 TheM shelfM lifeM ofM aM carbonatedM beverageMisMofMinterest.M TenM bottlesMareMrandomlyMselectedM andM
tested,MandMtheMfollowingMresultsMareMobtained:

Days
108 138
124 163
124 159
106 134
115 139

(a) WeMwouldMlikeMtoMdemonstrateMthatM theMmeanM shelfMlifeMexceedsM120Mdays.M SetM upMappr
opriateMhypothesesMforM investigatingMthisMclaim.

H0:M M=M120 H1:M M>M120

(b) TestMtheseMhypothesesMusingMM=M0.01.M WhatMareMyourM conclusions?

yM=M131
s2M =M[M(108M-M131)2M +M(124M-M131)2M +M(124M-M131)2M +M(106M-M131)2M +M(115M-M131)2M +M (138M-M131)2
+M(163M-M131)2M +M (159M-M131)2M +M (134M-M131)2M +M(M139M-M131)2M ]M/M(10M-M1)

s2M =M3438M/M9M=M382
sM 382 M19.54

yMMo 131M120
tMMMM 
M
M1.78
o n 19.54 10
s

sinceMt0.01,9M =M2.821;MdoMnotMrejectMH0

MinitabMOutput
T-TestMofMtheMMean

TestMofMmuM=M120.00MvsMmuM>M120.00

Variable N Mean StDev SEMMean T P
ShelfMLifeM 10 131.00 19.54 6.18 1.78 0.054

TMConfidenceMInterval
s

2-3

, SolutionsMfromMMontgomery,M D.MC.M (2001)MDesignMandMAnalysisMofMExperiments,MWiley
,M NY

Variable Mean StDevM SEMMean 99.0M%MCI
NM M131.0 19.54 6.18M (M 110.91,M 151.09)
ShelfMLifeM 10 0


(c) FindMtheMP-valueMforM theMtestM inM partM (b).M P=0.054

(d) ConstructMaM99MpercentMconfidenceMintervalMonM theMmeanM shelfMlife.
s s
TheM95%M confidenceMinterval yM M2M,n MM M yM M2M,nM1
M is Mt M1 n Mt n


M1954M M M1954M 
131M3.250 M M M131M3.250
    
 10   10 

110.91MM M M151.09



2-6 ConsiderMtheMshelfMlifeMdataMinMProblemM2-
5.MCanMshelfMlifeMbeMdescribedMorMmodeledMadequatelyMbyMaMnormalMdistribution?MWhatMeffectMwouldMvi
olationMofMthisMassumptionMhaveMonMtheMtestMprocedureMyouMusedMinMsolvingMProblemM2-5?

AMnormalMprobabilityMplot,MobtainedMfromMMinitab,MisMshown.M ThereMisMnoMreasonMtoMdoubtMtheMadequa
cyMofMtheMnormalityMassumption.MIfMshelfMlifeMisMnotMnormallyMdistributed,MthenMtheMimpactMofMthisMonMt
heMt-testMinMproblemM2-5MisMnotMtooMseriousMunlessMtheMdepartureMfromMnormalityMisMsevere.

NormalMProbabilityMPlotMforMShelfMLife
MLMEstimates



99 MLMEstimates
Mean 131
95
StDev 18.5418
90
GoodnessMofMFit
80
70 AD* 1.292
Percent




60
50
40
30
20

10
5


1


86 96 106 116 126 136 146 156 166 176
Data




2-7 TheM timeM toM repairM anM electronicM instrumentM isM aM normallyM distributedM randomM variableM measu
redM inMhours.M TheMrepairM timeMforM 16MsuchM instrumentsMchosenM atMrandomMareMasMfollows:

Hours
159 280 101 212
224 379 179 264
222 362 168 250
149 260 485 170



2-4