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Course: MATH 333, Spring 2011School: NJIT
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Ifasampleofsizen 11/17/2011 8 Estimatingthevariance StatisticalIntervals foraSingleSample isdrawnfromanormal population withvariance2 andsamplevariances2 calculated,s2 willbethepointestimatefor2. Confidenceintervalof2canbeobtainedusingthe statisticX2. Requirement:NormalPopulation OUTLINE 84ConfidenceIntervalon2 & ofa NormalDistribution 85LargeSampleConfidenceInterval foraPopulationProportion...
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Ifasampleofsizen 11/17/2011
8
Estimatingthevariance
StatisticalIntervals
foraSingleSample
isdrawnfromanormal
population withvariance2 andsamplevariances2
calculated,s2 willbethepointestimatefor2.
Confidenceintervalof2canbeobtainedusingthe
statisticX2.
Requirement:NormalPopulation
OUTLINE
84ConfidenceIntervalon2 & ofa
NormalDistribution
85LargeSampleConfidenceInterval
foraPopulationProportion
Lecture 21
11/16/11
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Chi squaredistribution
Chi square(2 )distribution
Theareaundereachcurveofthechisquare
distributionequalsone.
Themeanandvarianceof2distributionarekand
2krespectively,wherekisthenumberofdegreesof
freedom.
Chisquarerandomvariableisnonnegative
Chisquaredistributionisskewedtotheright
Askincreases,thedistributionbecomesmore
symmetric
Ask,thelimi ngformofchisquaredistribution
isthenormaldistribution
Thepercentagepointsofthe2 distributionaregiven
inTableIVoftheappendix.
Theareas arethecolumnheadingsandthe
degreesoffreedomk aregivenintheleftcolumn.
Several 2 distributions
are shown in the figure.
4
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Chi square(2 )distribution
Thechisquarevaluewith10degreesoffreedom
havinganarea(probability)of0.05totherightof
20.05,10 =18.31
Thisvalueisoftencalledanupper5%ofchisquare
with10degreesoffreedom.
Converselyalower5%pointofchisquarewith10
degreesoffreedomwouldbe20.95,10 =3.94(from
thetableintheappendix).(areshowninthefigure)
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11/17/2011
Chi square(2 )distribution
Define2,kasthepercentagepointorvalueofthechi
squarerandomvariablewithk degreesoffreedomsuch
thattheprobabilityofX2exceedsthisvalueis.
V
ThisprobabilityisshownastheshadedareainFigure
Percentage points of the chi square distribution
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
8
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
100(1)%CIfor
100(1)%CIfor2
Ifs2 isthesamplevariancefromarandomsampleofn
observationsfromanormaldistributionwithunknown
variance2,thena100(1)%confidenceintervalon 2 is
Aconfidenceintervalfor haslowerandupper
limitsthatarethesquarerootsofthecorresponding
limitsinthe100(1)%CIfor2.
wherearetheupperandlower
100/2percentagepointsofthechisquaredistribution
withn1degreesoffreedom,respectively.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
10
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Example1
ConfidenceIntervalontheVarianceandStandard
DeviationofaNormalDistribution
OneSidedConfidenceBounds
10.12
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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Example2
Example1
Arivetistobeinsertedintoahole.Arandomsample
ofn=15partsisselected,andtheholediameteris
measured.
Thesamplestandarddeviationoftheholediameter
measurementsiss=0.008millimeters.
Constructa99%lowerconfidenceboundfor
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Solution2
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Solution2
Example3
Thesugarcontentofthesyrupiscannedpeaches
normallydistributed.
Arandomsampleofn=10cansyieldsasample
standarddeviationofs=4.8milligrams.
Calculatea95%twosidedconfidenceintervalfor.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
3
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Solution3
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Example4
Solution4
Determinethe2percentilethatisrequiredto
constructeachofthefollowingCIs:
A)Confidencelevel=95%,degreesoffreedom=24,
onesided(upper)
B)Confidencelevel=99%,degreesoffreedom=9,one
sided(lower)
C)Confidencelevel=90%,degreesoffreedom=19,
twosided
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
LargeSampleConfidenceintervalforp
Supposearandomsampleofsizen hasbeentaken
fromalarge(possiblyinfinite)populationandthat
Xobservationsinthissamplebelongtoaclassof
interest
Largesampleconfidenceinterval
forpopulationproportion
andp Notethatn aretheparametersofabinomial
distribution
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
4
11/17/2011
Estimatingpopulationproportion
Estimatingpopulationproportion
Apointestimatoroftheproportion p is,
whereXrepresentsthenumberofsuccessesin
ntrials
Bythecentrallimittheorem,ifn issufficiently
large,isapproximately normally
distributedwithmeanp andvariance
p(1p)/n.
Hence,sampleproportionwillbethe
pointestimateofthepopulation proportion p.
Iftheunknownp isnottoocloseto0or1,
confidenceintervalforp canbeconstructed
consideringthesamplingdistributionof
25
Estimatingpopulationproportion
26
85ALargeSampleConfidenceIntervalFora
PopulationProportion
Wemakeuseofnormalapproximationfora
binomialproportiontoobtainconfidenceintervals
forp.
NormalApproximationforBinomialProportion
Typically,toapplynormalapproximation
np andn(1p)shouldbegreaterthanorequal
to5
Thequantity iscalledthestandarderrorofthe
pointestimator.
27
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
LargeSampleConfidenceintervalforp
Example1
Ifistheproportionofobservationsinarandom
sampleofsizenthatbelongstoaclassofinterest,an
approximate100(1)%confidenceintervalonthe
proportionpof thepopulationthatbelongstothis
classis:
wherez/2istheupper/2percentagepointofthe
standardnormaldistribution.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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Example1
Choiceofsamplesize
Sinceisthepointestimatorofp
ErrorinestimatingpbywouldbeE=|p
|
Weareapproximately100(1)%confidentthatthis
errorislessthan
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Choiceofsamplesize
SampleSizeforaSpecifiedErroronaBinomial
Proportion
Insituationswheresamplesizecanbeselectedwe
maychoosen tobe100(1)%confidentthatthe
errorislessthansomespecifiedvalueE.
ChoiceofSampleSize
ThesamplesizeforaspecifiedvalueE isgivenby
Ifweset
andsolveforn,wecanobtainapprox.samplesize.
33
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Example2
SampleSizeforaSpecifiedErroronaBinomial
Proportion
1.
Anupperboundonn isgivenby
Sample size will be maximum for p=0.5. Substituting
p=0.5 in the previous equation, we would obtain an
upper bound on n . Hence p(1-p)=0.5(1-0.5)=0.25
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11/17/2011
OneSidedConfidenceBounds
Example2
Theapproximate100(1)%lowerandupperconfidence
boundsare
respectively.
37
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Problem1
Solution1
Thefractionofdefectiveintegratedcircuitsproduced
inaphotolithographyprocessisbeingstudied.A
randomsampleof300 circuitsistested,revealing13
defectives.
A)Calculatea95%twosidedCIonthefraction of
defectivecircuitsproducedbythisparticulartool.
B)Calculatea95%upperconfidenceboundonthe
fraction ofdefectivecircuits.
39
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Solution1
Problem2
AnarticleinJournaloftheAmericanStatistical
Associationmeasuredweightof30 ratsunder
experimentcontrols.Supposethatthereare12
underweightrats.
A)Calculatea95%twosidedconfidenceintervalonthe
trueproportionofratsthatwouldshowunderweight
fromtheexperiment
B)Usingthepointestimateofp obtainedfromthe
preliminarysample,whatsamplesizeisneededtobe
95%confidentthattheerrorinestimatingthetruevalue
ofpis lessthan0.02?
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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Problem2
Solution2
C)Howlargemustthesamplebeifwewishtobeatleast
95% confidentthattheerror inestimatingp islessthan
0.02,regardlessofthetruevalueofp
,
,
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
Solution2
Solution2
B)Usingthepointestimateofp obtainedfromthe
preliminarysample,whatsamplesizeisneededtobe
95%confidentthattheerrorinestimatingthetrue
valueofp islessthan0.02?
C)Howlargemustthesamplebeifwewishtobeatleast
95%confidentthattheerrorinestimatingp islessthan
0.02,regardlessofthetruevalueofp
45
JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger.
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