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DESAINEKSPERIMENSIMPLECOMPARATIVE
EXPERIMENTS
SemesterGenap2017/2018JurusanTeknikIndustriUniversitasBrawijaya
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Outline• Introduction• BasicStatisticalConcepts• TheHypothesisTestingFramework
– Two-Samplet-Test– TheTwo-Sample(Pooled)t-Test– Two-samplet-test:Choiceofsamplesize
• AnIntroductiontoExperimentalDesign-Howtosample?– CompletelyRandomizedDesign–Howtosample
• Importanceofthet-Test• Two-samplet-test—ConfidenceIntervals• Othertopics
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Introduction(1/3)
• Formulationofacementmortar• Originalformulationandmodifiedformulation• 10samplesforeachformulation• Onefactoràformulation• Twoformulations:àtwotreatmentsàtwolevelsofthefactorformulation
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Introduction(2/3)
4
Results:
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Introduction(3/3)
5
Dot diagram
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BasicStatisticalConcepts(1/27)
• Experiencesfromaboveexample– Run–eachofaboveobservations– Noise,experimentalerror,error–theindividualrunsdifference
– Statisticalerror–arisesfromvariationthatisuncontrolledandgenerallyunavoidable
– Thepresenceoferrormeansthattheresponsevariableisarandomvariable
– Randomvariablecouldbediscreteorcontinuous
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BasicStatisticalConcepts(2/27)
• Describingsampledata– Graphicaldescriptions
• Dotdiagram—centraltendency,spread• Boxplot–• Histogram
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8
BasicStatisticalConcepts(3/27)
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BasicStatisticalConcepts(4/27)
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BasicStatisticalConcepts(5/27)
10
• Discrete vs continuous
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BasicStatisticalConcepts(6/27)
• Probabilitydistribution– Discrete
– Continuous
11
∑ =
==
≤≤
jy
j
jjj
jj
yp
yypyyPyyp
of valuesall
1)(
of valuesall )()( of valuesall 1)(0
1)(
)()(
)(0
=
=≤≤
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∫
∫∞
∞−dyyf
dyyfbyap
yfb
a
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BasicStatisticalConcepts(7/27)
• Probabilitydistribution– Mean—measureofitscentraltendency
– Expectedvalue–long-runaveragevalue
12
⎪⎩
⎪⎨
⎧= ∑∫∞
∞−
yallyyypydyyyf
discrete )(continuous )(
µ
⎪⎩
⎪⎨
⎧== ∑∫∞
∞−
yallyyypydyyyf
yE
discrete )(continuous )(
)(µ
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BasicStatisticalConcepts(8/27)
• Probabilitydistribution– Variance—variabilityordispersionofadistribution
13
2
22
(
2
2
)(
])[(
discrete )()(continuous )()(
σ
µσ
µ
µσ
=
−=
⎪⎩
⎪⎨
⎧
−
−= ∑∫∞
∞−
yVor
yEor
yypyydyyfy
yall
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BasicStatisticalConcepts(9/27)
• Probabilitydistribution– Properties:cisaconstant
• E(c)=c• E(y)=μ• E(cy)=cE(y)=cμ• V(c)=0• V(y)=σ2• V(cy)=c2σ2• E(y1+y2)=μ1+μ2
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BasicStatisticalConcepts(10/27)
• Probabilitydistribution– Properties:cisaconstant
• V(y1+y2)=V(y1)+V(y2)+2Cov(y1,y2)• V(y1-y2)=V(y1)+V(y2)-2Cov(y1,y2)• Ify1andy2areindependent,Cov(y1,y2)=0• E(y1*y2)=E(y1)*V(y2)=μ1*μ2• E(y1/y2)isnotnecessaryequaltoE(y1)/V(y2)
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BasicStatisticalConcepts(11/27)
• Samplingandsamplingdistribution– Randomsamples--ifthepopulationcontainsNelementsandasampleofnofthemistobeselected,andifeachofN!/[(N-n)!n!]possiblesampleshasequalprobabilitybeingchosen
– Randomsampling–aboveprocedure– Statistic–anyfunctionoftheobservationsinasamplethatdoesnotcontainunknownparameters
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BasicStatisticalConcepts(12/27)
• Samplingandsamplingdistribution– Samplemean
– Samplevariance
17
n
yy
n
ii∑
== 1
1
)(1
2
2
−
−=∑=
n
yys
n
ii
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BasicStatisticalConcepts(13/27)
• Samplingandsamplingdistribution– Estimator–astatisticthatcorrespondtoanunknownparameter
– Estimate–aparticularnumericalvalueofanestimator
– Pointestimator:toμands2toσ2– Propertiesonsamplemeanandvariance:
• Thepointestimatorshouldbeunbiased• Anunbiasedestimatorshouldhaveminimumvariance
18
y
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BasicStatisticalConcepts(14/27)
• Samplingandsamplingdistribution– Sumofsquares,SSin
19
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n
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n
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Sumofsquares,SS,canbedefinedas
∑=
−=n
ii yySS
1
2)(
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BasicStatisticalConcepts(15/27)
• Samplingandsamplingdistribution– Degreeoffreedom,v,numberofindependentelementsinasumofsquarein
20
⎥⎥⎥⎥
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⎤
⎢⎢⎢⎢
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−
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∑=
1
)()( 1
2
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n
yyESE
n
ii
Degreeoffreedom,v,canbedefinedas1−= nv
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BasicStatisticalConcepts(16/27)
• Samplingandsamplingdistribution– Normaldistribution,N
21
∞<<∞= −− y- 2
1)(2]/))[(2/1( σµ
πσyeyf
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BasicStatisticalConcepts(17/27)
• Samplingandsamplingdistribution– StandardNormaldistribution,z,anormaldistributionwithμ=0andσ2=1
22
),z~N(ei
yz
10.,.
σµ−
=
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BasicStatisticalConcepts(18/27)
• Samplingandsamplingdistribution– CentralLimitTheorem–Ify1,y2,…,ynisasequenceofnindependentandidenticallydistributedrandomvariableswithE(yi)=μandV(yi)=σ2andx=y1+y2+…+yn,thenthelimitingformofthedistributionof
asnà∞,isthestandardnormaldistribution
23
2σ
µ
nnxzn
−=
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BasicStatisticalConcepts(19/27)
• Samplingandsamplingdistribution– Chi-square,χ2,distribution–Ifz1,z2,…,zkarenormallyandindependentlydistributedrandomvariableswithmean0andvariance1,NID(0,1),therandomvariable
followsthechi-squaredistributionwithkdegreeoffreedom.
24
222
21 ... kzzzx +++=
2/1)2/(2/ )2/(21)( xk
k exk
xf −−
Γ=
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BasicStatisticalConcepts(20/27)
• Samplingandsamplingdistribution– Chi-squaredistribution–exampleIfy1,y2,…,ynarerandomsamplesfromN(μ,σ2),distribution,
– SamplevariancefromNID(μ,σ2),
25
212
1
2
2 ~)(
−=∑ −
= n
n
ii yy
SSχ
σσ
21
222 )]1/([~ .,. 1 −−
−= nnSeinSSS χσ
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BasicStatisticalConcepts(21/27)
• Samplingandsamplingdistribution– tdistribution–Ifzandareindependentstandardnormalandchi-squarerandomvariables,respectively,therandomvariable
followstdistributionwithkdegreesoffreedom
26
2kχ
/2 kztk
kχ
=
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BasicStatisticalConcepts(22/27)
• Samplingandsamplingdistribution– pdfoftdistribution–
μ=0,σ2=k/(k-2)fork>2
27
∞<<∞−+Γ
+Γ= + t
ktkkktf k
]1)/[(1
)2/(]2/)1[()( 2/)1(2π
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BasicStatisticalConcepts(23/27)
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BasicStatisticalConcepts(24/27)
• Samplingandsamplingdistribution– Ify1,y2,…,ynarerandomsamplesfromN(μ,σ2),thequantity
isdistributedastwithn-1degreesoffreedom
29
nSyt/µ−
=
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BasicStatisticalConcepts(25/27)
• Samplingandsamplingdistribution– Fdistribution—Ifandaretwoindependentchi-squarerandomvariableswithuandvdegreesoffreedom,respectively
followsFdistributionwithunumeratordegreesoffreedomandvdenominatordegreesoffreedom
30
2uχ
2vχ
vuF
v
uvu /
/2
2
, χχ
=
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BasicStatisticalConcepts(26/27)
• Samplingandsamplingdistribution– pdfofFdistribution–
31
∞<<+ΓΓ
+Γ= +
−
xxvuvxuxvuvuxh vu
uu
0 ]1)/)[(2/()/(
)/](2/)[()( 2/)(
1)2/(2/
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BasicStatisticalConcepts(27/27)
• Samplingandsamplingdistribution– Fdistribution–exampleSupposewehavetwoindependentnormaldistributionswithcommonvarianceσ2,ify11,y12,…,y1n1isarandomsampleofn1observationsfromthefirstpopulationandy21,y22,…,y2n2isarandomsampleofn2observationsfromthesecondpopulation
32
1 ,122
21
21~ −− nnFS
S
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TheHypothesisTestingFramework
• Statisticalhypothesistestingisausefulframeworkformanyexperimentalsituations
• Originsofthemethodologydatefromtheearly1900s
• Wewilluseaprocedureknownasthetwo-samplet-test
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Two-Samplet-Test(1/8)
– Supposewehavetwoindependentnormal,ify11,y12,…,y1n1isarandomsampleofn1observationsfromthefirstpopulationandy21,y22,…,y2n2isarandomsampleofn2observationsfromthesecondpopulation
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Two-Samplet-Test(2/8)
– Amodelfordata
εisarandomerror
35
),0(~,,...,2,12,1
{ 2iij
jijiij NID
nji
y σεεµ=
=++
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Two-Samplet-Test(3/8)
• Samplingfromanormaldistribution• Statisticalhypotheses:
36
0 1 2
1 1 2
::
HH
µ µ
µ µ
=
≠12/02/18 Desain Eksperimen - Semester Genap
2017/2018
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Two-Samplet-Test(4/8)
• H0iscalledthenullhypothesisandH1iscallalternativehypothesis.
• One-sidedvstwo-sidedhypothesis• TypeIerror,α:thenullhypothesisisrejectedwhenitistrue
• TypeIIerror,β:thenullhypothesisisnotrejectedwhenitisfalse
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false) is |reject tofail()error II type() trueis |reject ()error I type(
00
00
HHPPHHPP
==
==
β
α
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Two-Samplet-Test(5/8)
• Powerofthetest:
• TypeIerroràsignificancelevel• 1-α=confidencelevel
38
false) is |reject (1 00 HHPPower =−= β
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Two-Samplet-Test(6/8)
• Two-sample-t-test– Hypothesis:
– Teststatistic:
where
39
11
210 11
nnS
yyt
p +
−=
0 1 2
1 1 2
::
HH
µ µ
µ µ
=
≠
)2()1()1(
21
222
2112
−+−+−
=nn
SnSnSp
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Two-Samplet-Test(7/8)
40
1
2 2 2
1
1 estimates the population mean
1 ( ) estimates the variance 1
n
ii
n
ii
y yn
S y yn
µ
σ
=
=
=
= −−
∑
∑
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Two-Samplet-Test(8/8)
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Example–Two-Samplet-Test(1/6)
121
1
1
16.76
0.1000.31610
ySSn
=
=
=
=
222
2
2
17.04
0.0610.24810
ySSn
=
=
=
=
42
Formulation 1
“New recipe”
Formulation 2
“Original recipe”
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Example–Two-Samplet-Test(2/6)
43
1 2
22y
Use the sample means to draw inferences about the population means16.76 17.04 0.28
Difference in sample meansStandard deviation of the difference in sample means
This suggests a statistic:
y y
nσ
σ
− = − = −
=
1 20 2 2
1 2
1 2
Z y y
n nσ σ
−=
+
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Example–Two-Samplet-Test(3/6)
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2 2 2 21 2 1 2
1 22 2
1 2
1 2
2 2 21 2
2 22 1 1 2 2
1 2
Use and to estimate and
The previous ratio becomes
However, we have the case where Pool the individual sample variances:
( 1) ( 1)2p
S Sy yS Sn n
n S n SSn n
σ σ
σ σ σ
−
+
= =
− + −=
+ −
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Example–Two-Samplet-Test(4/6)
• Valuesoft0thatarenearzeroareconsistentwiththenullhypothesis• Valuesoft0thatareverydifferentfromzeroareconsistentwiththe
alternativehypothesis• t0isa“distance”measure-howfaraparttheaveragesareexpressedin
standarddeviationunits• Noticetheinterpretationoft0asasignal-to-noiseratio
45
1 20
1 2
The test statistic is
1 1
p
y ytS
n n
−=
+
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Example–Two-Samplet-Test(5/6)
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2 22 1 1 2 2
1 2
1 20
1 2
( 1) ( 1) 9(0.100) 9(0.061) 0.0812 10 10 2
0.284
16.76 17.04 2.201 1 1 10.284
10 10
The two sample means are a little over two standard deviations apartIs t
p
p
p
n S n SSn n
S
y ytS
n n
− + − += = =
+ − + −
=
− −= = = −
+ +
his a "large" difference?
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Example–Two-Samplet-Test(6/6)
• P-value–Thesmallestlevelofsignificancethatwouldleadtorejectionofthenullhypothesis.
• Computerapplication
Two-SampleT-TestandCISample NMeanStDevSEMean1 1016.7600.3160.102 1017.0400.2480.078Difference=mu(1)-mu(2)Estimatefordifference:-0.28095%CIfordifference:(-0.547,-0.013)T-Testofdifference=0(vsnot=):T-Value=-2.20P-Value=0.041DF=18BothusePooledStDev=0.2840
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WilliamSealyGosset(1876,1937)
Gosset'sinterestinbarleycultivationledhimtospeculatethatdesignofexperimentsshouldaim,notonlyatimprovingtheaverageyield,butalsoatbreedingvarietieswhoseyieldwasinsensitive(robust)tovariationinsoilandclimate.Developedthet-test(1908)GossetwasafriendofbothKarlPearsonandR.A.Fisher,anachievement,foreachhadamonumentalegoandaloathingfortheother.Gossetwasamodestmanwhocutshortanadmirerwiththecommentthat“Fisherwouldhavediscovereditallanyway.”
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TheTwo-Sample(Pooled)t-Test(1/3)
• Sofar,wehaven’treallydoneany“statistics”
• Weneedanobjectivebasisfordecidinghowlargetheteststatistict0reallyis
• In1908,W.S.Gossetderivedthereferencedistributionfort0…calledthetdistribution
• Tablesofthetdistribution–seetextbookappendix
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t0 = -2.20
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TheTwo-Sample(Pooled)t-Test(2/3)
• Avalueoft0between–2.101and2.101isconsistentwithequalityofmeans
• Itispossibleforthemeanstobeequalandt0toexceedeither2.101or–2.101,butitwouldbea“rareevent”…leadstotheconclusionthatthemeansaredifferent
• CouldalsousetheP-valueapproach
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t0 = -2.20
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TheTwo-Sample(Pooled)t-Test(3/3)
• TheP-valueisthearea(probability)inthetailsofthet-distributionbeyond-2.20+theprobabilitybeyond+2.20(it’satwo-sidedtest)
• TheP-valueisameasureofhowunusualthevalueoftheteststatisticisgiventhatthenullhypothesisistrue
• TheP-valuetheriskofwronglyrejectingthenullhypothesisofequalmeans(itmeasuresrarenessoftheevent)
• TheP-valueinourproblemisP=0.042
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t0 = -2.20
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CheckingAssumptions–TheNormalProbabilityPlot
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Two-samplet-testChoiceofsamplesize(1/4)
• ThechoiceofsamplesizeandtheprobabilityoftypeIIerrorβarecloselyrelatedconnected
• Supposethatwearetestingthehypothesis
• AndThemeanarenotequalsothatδ=μ1-μ2
• BecauseH0isnottrueàwecareabouttheprobabilityofwronglyfailingtorejectH0àtypeIIerror
53/72
0 1 2
1 1 2
::
HH
µ µ
µ µ
=
≠
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Two-samplet-testChoiceofsamplesize(2/4)
• Define
• Onecanfindthesamplesizebyvaryingpower(1-β)andδ
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σδ
σµµ
2221 =
−=d
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Two-samplet-testChoiceofsamplesize(3/4)
Testingmean1=mean2(versusnot=)Calculatingpowerformean1=mean2+differenceAlpha=0.05Assumedstandarddeviation=0.25 SampleTargetDifferenceSize PowerActualPower0.25 27 0.950.9500770.25 23 0.900.9124980.25 10 0.550.5620070.50 8 0.950.9602210.50 7 0.900.9290700.50 4 0.550.656876Thesamplesizeisforeachgroup.
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Two-samplet-testChoiceofsamplesize(4/4)
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AnIntroductiontoExperimentalDesign-Howtosample?
n Acompletelyrandomizeddesignisanexperimentaldesigninwhichthetreatmentsarerandomlyassignedtotheexperimentalunits.
n Iftheexperimentalunitsareheterogeneous,blockingcanbeusedtoformhomogeneousgroups,resultinginarandomizedblockdesign.
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CompletelyRandomizedDesign–Howtosample?(1/8)
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n Recall Simple Random Sampling n Finite populations are often defined by lists
such as: n Organization membership roster n Credit card account numbers n Inventory product numbers
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CompletelyRandomizedDesign–Howtosample?(2/8)
n A simple random sample of size n from a finite population of size N is a sample selected such that each possible sample of size n has the same probability of being selected.
n Replacing each sampled element before selecting subsequent elements is called sampling with replacement.
n Sampling without replacement is the procedure used most often.
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CompletelyRandomizedDesign–Howtosample?(3/8)
n Inlargesamplingprojects,computer-generatedrandomnumbersareoftenusedtoautomatethesampleselectionprocess.
n Excelprovidesafunctionforgeneratingrandomnumbersinitsworksheets.
n Infinitepopulationsareoftendefinedbyanongoingprocesswherebytheelementsofthepopulationconsistofitemsgeneratedasthoughtheprocesswouldoperateindefinitely.
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CompletelyRandomizedDesign–Howtosample?(4/8)
n A simple random sample from an infinite population is a sample selected such that the following conditions are satisfied. n Each element selected comes from the same population. n Each element is selected independently.
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CompletelyRandomizedDesign–Howtosample?(5/8)
n Random Numbers: the numbers in the table are random, these four-digit numbers are equally likely.
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CompletelyRandomizedDesign–Howtosample?(6/8)
n Most experiments have critical error on random sampling.
n Ex: sampling 8 samples from a production line in one day n Wrong method:
n Get one sample every 3 hoursà not random!
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CompletelyRandomizedDesign–Howtosample?(7/8)
n Ex: sampling 8 samples from a production line n Correct method:
n You can get one sample at each 3 hours interval but not every 3 hours à correct but not a simple random sampling
n Get 8 samples in 24 hours n Maximum population is 24, getting 8 samplesà two digits n 63, 27, 15, 99, 86, 71, 74, 45, 10, 21, 51, … n Larger than 24 is discarded n So eight samples are collected at:
15, 10, 21, … hour
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CompletelyRandomizedDesign–Howtosample?(8/8)
n In Completely Randomized Design, samples are randomly collected by simple random sampling method.
n Only one factor is concerned in Completely Randomized Design, and k levels in this factor.
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Importanceofthet-Test
• Providesanobjectiveframeworkforsimplecomparativeexperiments
• Couldbeusedtotestallrelevanthypothesesinatwo-levelfactorialdesign,becauseallofthesehypothesesinvolvethemeanresponseatone“side”ofthecubeversusthemeanresponseattheopposite“side”ofthecube
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Two-samplet-test—ConfidenceIntervals(1/2)
• Hypothesistestinggivesanobjectivestatementconcerningthedifferenceinmeans,butitdoesn’tspecify“howdifferent”theyare
• Generalformofaconfidenceinterval
• The100(1-α)%confidenceintervalonthedifferenceintwomeans:
67
where ( ) 1 L U P L Uθ θ α≤ ≤ ≤ ≤ = −
1 2
1 2
1 2 / 2, 2 1 2 1 2
1 2 / 2, 2 1 2
(1/ ) (1/ )
(1/ ) (1/ )n n p
n n p
y y t S n n
y y t S n nα
α
µ µ+ −
+ −
− − + ≤ − ≤
− + +
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Two-samplet-test—ConfidenceIntervals(1/2)
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OtherTopics
• Hypothesistestingwhenthevariancesareknown—two-sample-z-test
• Onesampleinference—one-sample-zorone-sample-ttests
• Hypothesistestsonvariances–chi-squaretest• Pairedexperiments
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