-
8/3/2019 Design and Analysis of Single Factor Experiments
1/59
Designandanalysisofsinglefactori
experiments
Objectives
of
the
topic: es gn ngan con uc ngeng neer ngexper men s
involvingasinglefactor.
Anal zin
ex erimentsusin
the
anal sis
of
variance.
Assessingmodeladequacy.
Identifyingdifferencesbetweenmeans.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
2/59
Introductiontoengineeringexperiments
Experimentsare
one
way
of
engineering
design.
Forexample,inaspecificproductionline,fourmodelsofa
machinearebeingconsideredinthefillingstage.
Theindustrial
engineer
will
take
a
sample
from
each
machine
.
Themachinemodelwhosemeanfillingisclosesttothetarget
200ml
will
be
chosen.
Thisexperimenthasonefactor(machine)andfourlevels
(model).
ereare
on y
wo
mac ne
mo e s
e ng
cons ere ,
en
thisexperimentwillbedesignedandanalyzedusingmethods
fortwopopulationsamples.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
3/59
Theanalysis
of
variance
(ANOVA)
is
used
for
comparing
means
whentherearemorethantwolevelsofasinglefactor.
Statisticallydesignedexperimentsareimportantinindustrial
engineeringto
improve
the
performance
of
a
manufacturing
rocess.
Resultsfromsuchexperimentscanleadto
Improvedprocess
output
Reducedvariability
Reduceddesignanddevelopmenttime
Reducedcostofoperation.
Attheproductdesignstage,theindustrialengineercan
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
,
selectmachinesettings,choosematerial,etc.
-
8/3/2019 Design and Analysis of Single Factor Experiments
4/59
Everyexperiment
involves
a
sequence
of
activities:
1. Conjecture theoriginalhypothesisthatmotivatesthe
experiment.
2. Experiment the
test
performed
to
investigate
the
.
3. Analysis thestatisticalanalysisofthedatafromthe
experiment.
4. Conclusion whathasbeenlearnedabouttheoriginal
conjecturefromtheexperiment. frequentlythe
,
experiment,andsoforth.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
5/59
Completelyrandomizedsinglefactorexperiment
Considera
company
manufacturing
plastic
bags
for
Panda
supermarket.
Thestrengthoftheplasticbagisafunctionofthe
concentrationof
a
chemical
added
to
the
mix
that
make
up
thematerial.
Theconcentrationcanrangefrom5to20%.
Theindustrial
engineer
at
the
plant
has
decided
to
investigate
fourlevelsofchemicalconcentration:5%,10%,15%,20%.
Hemadesixtestspecimensateachconcentrationlevel. The
.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
6/59
Theresult
of
the
experiment
is
shown
below:
Observations
Concentration 1 2 3 4 5 6
5 7 8 15 11 9 1010 12 17 13 18 19 15
15 14 18 19 17 16 18
20 19 25 22 23 18 20
experimentwithfourlevels(treatments)ofthefactor.
Inthis
example,
each
treatment
has
six
observations
(replicates).
Randomizationisextremelyimportantinengineering
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
.
-
8/3/2019 Design and Analysis of Single Factor Experiments
7/59
Analysisofvariance
Ingeneral,
there
are
a
treatments
of
a
single
factor.
Theresponseforeachoftheatreatmentsisarandom
variable.
Theobserved
data
can
be
organized
in
the
following
table:
Treatment Observations Totals Averages
1 y11 y12 y1n y1.2 y21 y22 y2n y2.
.2
.1
y
y
a ya1 ya2 Yan ya.
y.. ..
.
y
y a
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
8/59
Thejth observation
taken
under
treatment
i is
denoted
by
yij.
Theobservationsinthetablecanbedescribedbylinear
statisticalmodel
ijiijY ++=
where (overallaverage)isaparametercommontoalltreatments treatmenteffect isa arameterassociatedwiththeith treatment,ij isarandomerror. Theerrorisassumedtobenormalwithmeanzeroand
var ance .
Therefore,eachtreatmentisnormalwithmean +i andvariance2.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
9/59
Twoquestions
that
can
asked
in
industrial
engineering
design:
1. Whatisthelevelofthefactorhavingthelargesteffecton
theresponse?
2. Doesthe
factor
have
any
effect
on
the
response?
en e rea men sarec osenspec ca yan e
hypothesisisaboutthetreatmentmeansandtreatment
effects,the
model
is
called
a
fixed
effects
model.
Ifthetreatmentsarearandomsampleforalargerpopulation
oftreatments,anditisdesiredtomakeconclusionstoall
,
effectsmodel.
Weconcentratefirstonfixedeffectsmodels.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
10/59
Theanalysis
of
variance
is
used
to
test
for
equality
of
treatmenteffectsi.
Treatmenteffectsi aredefinedasdeviationsfromtheoverallaverage
:
a
01
==i
i
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
11/59
Define:
entith treatmunder thensobservatiotheoftotal==n
entith treatmunder thensobservatiotheofaverage.
1
.
==
=
yy ii.
j
nsobservatiothealloftotalgrand== yy
na n
ij..
nsobservatiotheallofmeangrand.. ===
= =
N
y
an
yy ....
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
12/59
Ifit
is
desired
to
test
the
equality
of
the
treatment
means,
the
followinghypothesiscanbetested:
H0:1 =2 ==a =0
H1:i 0 foratleastonei 0 s rue, eneac o serva oncons s so eovera
meanplusarandomerrorij. E uivalentl ,
if
H is
true,
it
can
be
said
that
chan in
the
levelsofthefactorhasnoeffectonthemeanresponse.
Theobservationsinthiscasearenormalwithmeanand
var ance .
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
13/59
ANOVAandsignificanceoftreatments
ANOVApartitions
the
total
variability
in
the
sample
data
into
twocomponents,givenbytheANOVAsumofsquares
ent ty:
a a n
iiji.
a n
ij yyyynyy += 2.2..2.. )()()(
ET
i i ji j
SSSSSS +== = == =
Treatments
1 1 11 1
SST = totalvariabilityintheobservations
SStreatments
=sumofsquaresofdifferencesbetweentreatment
meansandthegrandmean
SSE =sumofsquaresofdifferencesofobservationswithina
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
14/59
Differencesbetween
observed
treatment
means
and
the
grandmeanmeasuresthedifferencesbetweentreatments.
Differencesofobservationswithinatreatmentfromthe
treatmentmean
is
attributed
to
random
error.
by
=a 22
Theexpectedvalueoftheerrorsumofsquaresis
=i 1iTreatments
2)1()( = naSSE E
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
15/59
SST has(an
1)
degrees
of
freedom,
SStreatments has
(a
1),
and
SSE hasa(n1);thepartitionofdegreesoffreedomis
an 1=a 1+a(n 1)
Themean
scare
for
treatments
is
given
by
= = = =
1-MS TreatmentsTreatments
a=
a ,
i =0andMSTreatments isanunbiasedestimatorof2. Theerrormeansquare
1)-(MS EE
na=
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
isan
unbiased
estimator
of
.
-
8/3/2019 Design and Analysis of Single Factor Experiments
16/59
Therandom
variables
MSTreatment and
MSE are
independent.
IfthenullhypothesisH0:1 =2 ==a =0istrue,theratio
EE MS
MS
naSS
aSSF
TreatmentsTreatments0
)]1(/[
)1/(
=
=
hasanFdistributionwitha1anda(n1)degreesoffreedom.
Hence,
H0 is
rejected
if
f0 >
f,a1,a(n1).
2
..2
N
yySS
a n
ijT = 2
..
1
2
.Treatments
N
y
n
ySS
a
i
i = =
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
TreatmentsSSSSSS TE =
-
8/3/2019 Design and Analysis of Single Factor Experiments
17/59
Thecomputations
are
organized
in
the
ANOVA
table:
Sourceof Degreesof Mean
Variation Sumofsquares freedom square F0
Treatments SSTreatments a 1 MSTreatments MSTreatments/MSEError SSE a(n 1) MSE
T
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
18/59
Example
Considerthe
plastic
bag
production.
Observations
oncen ra on
5 7 8 15 11 9 10
10 12 17 13 18 19 15
15 14 18 19 17 16 18
We
want
to
test
the
hypothesis
that
different
chemical
20 19 25 22 23 18 20
.
Thehypothesisweshouldtestis
H :
=
=
=
=
0H1:i 0foratleastonei
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
19/59
Thesummations
are
96.512383
20...872
222 =+++=
79.382383127...9460
24
2222
Treatments =+++=SS
17.13079.38296.512 ==ESS
Sourceof Degreesof Mean
0
Treatments 382.79 3 127.60 19.60
Error 130.17 20 6.51
Total 512.96 23
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
20/59
Fromthe
F
tables,
f0.05,3,20 =
3.10.
Sincef0 >f0.05,3,20,H0 shouldberejectedanditisconcluded
thatthechemicalconcentrationsignificantlyaffectsthemean
strengthof
the
plastic
bags.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
21/59
Confidenceintervalontreatmentmean
Themean
of
the
ith treatment
is
i = +i Apointestimatorofi is
. ii Y= Sincetheerrorisassumedtobenormalwithmeanzeroand
variance2,eachtreatmentaverageisnormalwithmeaniand
variance
2 /n.
UsingMSE asanestimatorof2,thentherandomvariableY
Tii .
=
hasatdistributionwitha(n 1)degreesoffreedom.
nSE /
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
22/59
A100(1
)%confidenceintervalonthemeanoftheithtreatmentis
n
S
tyn
S
ty
E
naii
E
nai )1(,2/.)1(,2/. +
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
23/59
Example
Considerthe
plastic
bag
production.
Observations
oncen ra on
5 7 8 15 11 9 10
10 12 17 13 18 19 15
15 14 18 19 17 16 18
We
want
to
construct
a
95%
confidence
interval
on
the
mean
20 19 25 22 23 18 20
.
Thefollowingquantitieshavebeencalculated
51.6
167.21.4
=
=
EMS
y
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
24/59
A95%
confidence
interval
on
the
mean
of
the
20%
treatment
is
6
51.6
086.2167.216
51.6
086.2167.21 4 +.. 4
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
25/59
Confidenceintervalonthedifferenceoftwotreatment
means
Apoint
estimator
of
the
difference
between
two
treatment
meansi j is ... ji YY ThevarianceV(i j)canbeestimatedby
YYV222 2
=+=
assumingtreatmentmeansareindependentrandomvariables.
2
nnn..
,
YYT
jiji )(.. =
hastdistributionwitha(n 1)degreesoffreedom.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
26/59
A100(1
)%confidenceintervalonthedifferenceintwotreatmentmeansi jis
n
MStyy
n
MStyy EnajijiEnaji
2
2)1(,2/..)1(,2/.. +
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
27/59
Example
Considerthe
plastic
bag
production.
Observations
oncen ra on
5 7 8 15 11 9 10
10 12 17 13 18 19 1515 14 18 19 17 16 18
Wewanttoconstructa95%confidenceintervalonthe
20 19 25 22 23 18 20
.
Thefollowingquantitieshavebeencalculated
67.15=
51.6
00.17.3
.
=
=
EMS
y
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
28/59
a95%
confidence
interval
on
the
difference
in
treatment
means3 2 is
6)51.6(2086.267.1500.17
6)51.6(2086.267.1500.17 23 +
40.474.1 23
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
29/59
Unbalanceddesign
Whenthe
number
of
observations
under
each
treatment
are
different,thedesigniscalledunbalanced.
Letni thenumberofobservationstakenundertreatmenti.
Thetotal
number
of
observations
is
a
=
=i
inN1
2
..2
N
yySS
a n
ijT
i
= = =
2
..
1
2
.Treatments
N
y
n
ySS
a
i
i = =
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
TreatmentsSSSSSS TE =
-
8/3/2019 Design and Analysis of Single Factor Experiments
30/59
HypothesisTestondifferencesintreatmentmeans
IfH0:
1 =2 ==a =0isrejected,someofthetreatmentmeansaredifferent.
Tofindifthereisanysignificantdifferencebetweentwo
treatmentmeans
i j,thefollowinghypothesiscanbetested:
H0:i =jH1:
i Theteststatisticis
YY ji ..
= nMSE /20
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
.
-
8/3/2019 Design and Analysis of Single Factor Experiments
31/59
Thenull
hypothesis
H0 should
be
rejected
if
|t0|>
t/2,a(n1).
Whenthesamplesizesaredifferent,theteststatistic
becomes
= ji YYT ..0
+ji
Enn
MS11
withN adegreesoffreedom.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
32/59
Example
Considerthe
plastic
bag
production.
Observations
oncen ra on
5 7 8 15 11 9 10
10 12 17 13 18 19 1515 14 18 19 17 16 18
Wewanttotestthehypothesisthatthetreatmentmeansof
20 19 25 22 23 18 20
.
Thefollowingquantitieshavebeencalculated
00.10=
51.6
17.21.4
.
=
=
EMS
y
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
33/59
Thehypothesis
that
is
going
to
be
tested
is
H0:1 =4H1:1 4
Thevalue
of
the
test
statistic
is
60.747.1
.
6/)51.6(2
..0 =
=
=t
, 0.025,20 . .
Since|t0|=7.60>t0.025,20,H0 shouldberejected.
themeansoftreatments5%and20%concentrations,and
thatthe5%and20%concentrationshaveaneffectonthe
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
.
-
8/3/2019 Design and Analysis of Single Factor Experiments
34/59
Samplesizeand OC
curves
are
available
to
combine
the
probability
of
type
II
error,samplesize,andtheminimumdeviationintreatment
means.
Thedeviation
that
needs
to
be
detected
is
defined
by
the
standardizeddeviation
2
a
i
2
1
== id
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
35/59
Thehorizontal
lines
in
the
OC
curves
are
the
values
computedas
da
n=
w wo egreeso ree omv1 =a an v2 =a n .
Tofindn,severalvaluesofnshouldbeevaluatedandthebest
nis
selected.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
36/59
Residualanalysis
Theresidual
is
the
difference
between
the
observation
and
thetreatmentmean:
Itis
assumed
that
the
residuals
are
normal
with
mean
zero
2
.iijij yye =
.
Thenormalityoftheresidualscanbeassessedusingthe
normalprobability
plot.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
37/59
-
8/3/2019 Design and Analysis of Single Factor Experiments
38/59
Tocheck
the
assumption
of
equal
variances
at
each
factor
level,plottheresidualsagainstthefactorlevelsandcompare
t esprea nt eres ua s.
Theindependence
assumption
can
be
checked
by
plotting
the
residuals a ainstthetimeorrunorderinwhichthe
experimentwasperformed.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
39/59
Randomeffectsmodel
Whena
factor
levels
are
randomly
selected
from
the
populationoffactorlevels,thefactorissaidtobearandom
actor.
Becausethe
levels
of
the
factor
used
in
the
experiment
are
chosenrandoml theconclusionsreachedwillbevalidforthe
entirepopulationoffactorlevels.
Itwill
be
assumed
that
the
population
of
the
factor
levels
are
n n te.
Similartothefixedeffectscase,thestatisticalmodelinthe
ijiijY ++=
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
40/59
Thetreatment
effects
i andtheerrorsi areindependentrandomvariables.
Ifthevarianceofthetreatmenteffecti is2,thevarianceof
theresponse
is
whichis
called
the
random
effects
model.
22
)( +=ijYV
Theerrorsareassumedtobenormalwithmeanzeroand
variance2. The
treatment
effects
are
assumed
to
be
normal
with
mean
zeroandvariance2.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
41/59
ANOVA
Toexamine
the
contribution
of
the
random
effects,
the
followinghypothesisistested:
H0:2 =0
H1:2 > 0
= ,a rea men sare en ca ; u or , ere savariabilitybetweentreatments.
UnderANOVA,
the
total
variabilit
is
SST =SSTreatments +SSE
Theexpectedtreatmentsanderrormeansquaresare2
2TreatmentsTreatments
1)( +=
= na
EMSE
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
2
)1()( = = naEMSE EE
-
8/3/2019 Design and Analysis of Single Factor Experiments
42/59
WhenH0 is
true,
both
MSE and
MSTreatments estimate
2. SinceMSE andMSTreatments areindependent,theratio
EMS
MSF
Treatments0 =
hasanFdistributionwitha1anda(n 1)degreesoffreedom
whenH0 istrue.
,a ,a n .
Theestimatesofthevariancesare
=2
n
MSMS E= Treatments2
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
43/59
Example
Ina
textile
manufacturing
company,
the
industrial
engineering
isinterestedtotesttheeffectofloomsonthetensilestrength
o a r c.
Hehas
picked
randomly
four
looms
out
of
10
looms:
Observations
Loom
number 1 2 3 4
3 91 90 93 92
7 96 95 97 95
Arethereanydifferencesinfabricstrengthsproducedby
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
44/59
Toanswer
this
question,
the
following
hypothesis
has
to
be
tested.
H0:2 =0
H1:2 > 0
esumso squaresare
SST =111.94 with3d.f.
=reatments . . .
SSE =22.75 with15d.f.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
45/59
TheANOVA
table
is
ourceo
Variation Sumofsquares
egreeso
freedom
ean
square f 0Treatments 89.19 3 29.73 15.65
Error 22.75 12 1.90
Total 111.94 15
, 0.05,3,12 . .
Sincef0 >f0.05,3,12,H0 shouldberejected.
Hence,theloomsproducedifferentfabricstrengths.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
46/59
Randomizedcompleteblockdesign
Supposein
the
plastic
bag
manufacturing,
there
could
be
a
nuisancefactorthatisnotofinterestinthisexperiment,such
as eattreatment.
Toeliminate
the
effect
of
a
nuisance
factor,
we
say
the
factor
shouldbeblockedout.
Blockingisdonebyholdingthenuisancefactorconstantand
allowingthe
factor
of
interest
to
vary.
Forexample,toblocktheeffectofheatontheexperiment,
severalchemicalconcentrationsareobservedunderthe
Theobservationyij isobtainedwhentreatmentleveli istested
underblockj.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
47/59
Thegeneral
procedure
for
a
randomized
complete
block
designistoselectbblocksandrunacompletereplicateof
t eexper ment neac oc .
Thereare
a
observations
in
each
block.
Blocks
Treatments 1 2 b Totals Averages .
2 y21 y22 y2b y2.
.
.2y
a a a a.Totals y.1 y.2 y.b y..
Averages
.a
1.y 2.y by. ..y
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
48/59
Theobservations
are
represented
by
the
linear
statistical
model:
ijjiijY +++=
n srepresen a on, j s ee ec o e oc .
Thetreatmentsandblockeffectsaredefinedasdeviations
fromthe
over
all
avera e
hence
=0and =0. Itisassumedthattreatmentsandblocksdonotinteract.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
49/59
Hypothesisontheequalityoftreatmenteffects
Thetreatment
effects
hypothesis
is
H0:1 =2 ==a =0
H1:i 0 foratleastonei ANOVA
can
be
used
to
test
the
hypothesis.
TherandomizedblockANOVAidentityis
)()()()(2
..
2
..
2
..
2
.. +++= yyyyyyayybyya b
i..jij
a b
.ji.
a b
ij
)1)(1(111
BlocksTreatments
1 11 11 1
++=
++== == == =
babaab
SSSSSSSS ET
i ji ji j
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
50/59
Theexpected
means
squares
are
2a
12
Treatments 1)( +==i
i
aMSE
1
2
2
Blocks
)( +=
=j
ja
MSE
2)( =EMSE
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
51/59
IfH0 is
true,
MSTreatments is
an
unbiased
estimator
of
2. IfH0 isfalse,MSTreatments overestimates2. Theteststatistic
MSF Treatments=
hasanFdistributionwitha 1and(a 1)(b 1)degreesof
E
.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
52/59
Thesummations
are
calculates
as
2
..2 ya b
2
..2
1 1
1 y
ab
a
i j
ijT
= =
2
..2
1
.Treatments
1 ySS
abb
b
i
i
=
=
BlocksTreatments
1
.
SSSSSSSS
aba
TE
j
==
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
53/59
Thecomputations
are
organized
in
the
ANOVA
table:
Sourceof Degreesof
Variation Sumofsquares freedom Meansquare f 0
Treatments SSTreatments a
1 SSTreatments/(a
1) MSTreatments/MSEBlocks SSBlocks b1 SSBlocks/(b1)
E E
Total SST ab1
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
l
-
8/3/2019 Design and Analysis of Single Factor Experiments
54/59
Example
Anindustrial
engineer
is
investigating
the
effect
of
four
chemicalsonthestrengthofafabric.
Inordertoremovetheeffectofthefabrictypeonthefabric
strength,five
fabric
samples
have
been
selected:
Blocks
Treatments 1 2 3 4 5
1 1.3 1.6 0.5 1.2 1.1. . . . .3 1.8 1.7 0.6 1.5 1.34 3.9 4.4 2.0 4.1 3.4
Aretheredifferencesinthetreatmentmeans?
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
55/59
Thesummations
and
the
averages
of
treatments
and
blocks
are
Blocks
Treatments 1 2 3 4 5 Totals Averages1 1.3 1.6 0.5 1.2 1.1 5.7 1.1
. . . . . . .
3 1.8 1.7 0.6 1.5 1.3 6.9 1.4
4 3.9 4.4 2.0 4.1 3.4 17.8 3.6o a s . . . . . .
Averages 2.3 2.5 0.9 2.2 1.9
Thesummationsare
SST =25.69 SSTreatments =18.04
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
SSBlocks =6.69 SSE =
0.96
-
8/3/2019 Design and Analysis of Single Factor Experiments
56/59
TheANOVA
table
is
Sourceof Degreesof
Variation Sumofsquares freedom Meansquare f 0
Treatments 18.04 3 6.01 75.13Blocks 6.69 4 1.67
. .
Total 25.69 19
, 0.01,3,12 . .
At =0.01,H0 shouldberejected.
strength.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
57/59
Confidenceintervalsonthedifferenceintreatmentmeans
Totest
for
a
significant
difference
between
treatment
means,
thefollowinghypothesisisconsidered:
H0:i =jH1:
i j e es s a s c s
YY
T
ji ..
0
=whichfollowstdistributionwith(a1)(b1) degreesoffreedom.
H shouldberejectedift >t
.
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
E l
-
8/3/2019 Design and Analysis of Single Factor Experiments
58/59
Example
Considerthe
data:
Blocks
1 1.3 1.6 0.5 1.2 1.1
2 2.2 2.4 0.4 2.0 1.83 1.8 1.7 0.6 1.5 1.34 3.9 4.4 2.0 4.1 3.4
treatments1and4?
Thehypothesisthatshouldbetestedis
H0:1 =4H1:1 4
DrMuhammadAlSalamah,IndustrialEngineering,KFUPM
-
8/3/2019 Design and Analysis of Single Factor Experiments
59/59
Thevalue
of
the
test
statistic
is
56.314.1-=
=
Fromthe
ttables,
t0.025,12 =2.18.
4/)08.0(20 .
H0 isrejected.
Thereisasignificantdifferencebetweenthemeansof
.