Design and Analysis of Single Factor Experiments

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


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.



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


,

selectmachinesettings,choosematerial,etc.


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.



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

.



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


.


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



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.



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.



10/59

Theanalysis

of

variance

is

used

to

test

for

equality

of

treatmenteffectsi.

Treatmenteffectsi aredefinedasdeviationsfromtheoverallaverage

:

a

01

==i

i



11/59

Define:

entith treatmunder thensobservatiotheoftotal==n

entith treatmunder thensobservatiotheofaverage.

1

.

==

=

yy ii.

j

nsobservatiothealloftotalgrand== yy

na n

ij..

nsobservatiotheallofmeangrand.. ===

= =

N

y

an

yy ....



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 .



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



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



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=


isan

unbiased

estimator

of

.


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 = =


TreatmentsSSSSSS TE =


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



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



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



20/59

Fromthe

F

tables,

f0.05,3,20 =

3.10.

Sincef0 >f0.05,3,20,H0 shouldberejectedanditisconcluded

thatthechemicalconcentrationsignificantlyaffectsthemean

strengthof

the

plastic

bags.



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 /



22/59

A100(1

)%confidenceintervalonthemeanoftheithtreatmentis

n

S

tyn

S

ty

E

naii

E

nai )1(,2/.)1(,2/. +



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



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



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.



26/59

A100(1

)%confidenceintervalonthedifferenceintwotreatmentmeansi jis

n

MStyy

n

MStyy EnajijiEnaji

2

2)1(,2/..)1(,2/.. +



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

.


67.15=

51.6

00.17.3

.

=

=

EMS

y



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



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 = =


TreatmentsSSSSSS TE =


30/59

HypothesisTestondifferencesintreatmentmeans

IfH0:

1 =2 ==a =0isrejected,someofthetreatmentmeansaredifferent.

Tofindifthereisanysignificantdifferencebetweentwo

treatmentmeans

i j,thefollowinghypothesiscanbetested:

H0:i =jH1:

i Theteststatisticis

YY ji ..

= nMSE /20


.


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.



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

.


00.10=

51.6

17.21.4

.

=

=

EMS

y



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


.


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



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.



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.



37/59


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.



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 ++=



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.



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


2

)1()( = = naEMSE EE


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



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



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.



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.



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.



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



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.



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



50/59

Theexpected

means

squares

are

2a

12

Treatments 1)( +==i

i

aMSE

1

2

2

Blocks

)( +=

=j

ja

MSE

2)( =EMSE



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

.



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

==



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


l


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?



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


SSBlocks =6.69 SSE =

0.96


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.



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

.


E l


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



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

.

Design and Analysis of Single Factor Experiments

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