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### lecture21prn

Course: MATH 333, Spring 2011
School: NJIT
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Word Count: 1664

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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 &amp; 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 2 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 3 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) 6 5 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 1 11/17/2011 Chi square(2 )distribution Define2,kasthepercentagepointorvalueofthechi squarerandomvariablewithk degreesoffreedomsuch thattheprobabilityofX2exceedsthisvalueis. V ThisprobabilityisshownastheshadedareainFigure Percentage points of the chi square distribution 7 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 8 8 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. 9 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 10 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. Example1 ConfidenceIntervalontheVarianceandStandard DeviationofaNormalDistribution OneSidedConfidenceBounds 10.12 11 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 12 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 2 11/17/2011 Example2 Example1 Arivetistobeinsertedintoahole.Arandomsample ofn=15partsisselected,andtheholediameteris measured. Thesamplestandarddeviationoftheholediameter measurementsiss=0.008millimeters. Constructa99%lowerconfidenceboundfor 14 13 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. Solution2 15 16 16 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. Solution2 Example3 Thesugarcontentofthesyrupiscannedpeaches normallydistributed. Arandomsampleofn=10cansyieldsasample standarddeviationofs=4.8milligrams. Calculatea95%twosidedconfidenceintervalfor. 17 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 18 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 3 11/17/2011 Solution3 19 19 20 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 21 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 22 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. LargeSampleConfidenceintervalforp Supposearandomsampleofsizen hasbeentaken fromalarge(possiblyinfinite)populationandthat Xobservationsinthissamplebelongtoaclassof interest Largesampleconfidenceinterval forpopulationproportion andp Notethatn aretheparametersofabinomial distribution 24 23 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 28 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. LargeSampleConfidenceintervalforp Example1 Ifistheproportionofobservationsinarandom sampleofsizenthatbelongstoaclassofinterest,an approximate100(1)%confidenceintervalonthe proportionpof thepopulationthatbelongstothis classis: wherez/2istheupper/2percentagepointofthe standardnormaldistribution. 29 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 30 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 5 11/17/2011 Example1 Choiceofsamplesize Sinceisthepointestimatorofp ErrorinestimatingpbywouldbeE=|p | Weareapproximately100(1)%confidentthatthis errorislessthan 31 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 32 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. Choiceofsamplesize SampleSizeforaSpecifiedErroronaBinomial Proportion Insituationswheresamplesizecanbeselectedwe maychoosen tobe100(1)%confidentthatthe errorislessthansomespecifiedvalueE. ChoiceofSampleSize ThesamplesizeforaspecifiedvalueE isgivenby Ifweset andsolveforn,wecanobtainapprox.samplesize. 33 34 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 36 35 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 6 11/17/2011 OneSidedConfidenceBounds Example2 Theapproximate100(1)%lowerandupperconfidence boundsare respectively. 37 38 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. 40 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? 41 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 42 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 7 11/17/2011 Problem2 Solution2 C)Howlargemustthesamplebeifwewishtobeatleast 95% confidentthattheerror inestimatingp islessthan 0.02,regardlessofthetruevalueofp , , 43 44 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. 46 JohnWiley&Sons, Inc. AppliedStatisticsandProbabilityforEngineers,byMontgomery andRunger. 8
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