Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 34ANalysis Of VArianceMeasure twice, cut once.MULTIPLE t- TESTSif we have K = 5 populationmeans, we might want to compareeach mean to all the othersrequiresc =K(K − 1)2= 10different t-testssuppose each test is with α =0.05The probability of making a TypeI error over the whole set is muchlarger2ANOVA VARIABLESindependent variables: variable thatforms groupingsone-way ANOVA: one indep endentvariablelevels: number of groups, number ofpopulationse.gMethod of teaching is an independentvariableyou may teach in 17 different ways(levels) and have 17 different samplegroups with sample meansX1, X2, ........X16, X17,so that for your hypothesis test youwould want to test whether all thepopulation means of the different levelsare the same3ANOVA VARIABLESwe need additional subscripts tokeep track of variablesxikis the score for the ith subject inthe kth level (group)nkis the number of scores in the kthlevel!ixikis the sum of scores in the kthlevel!knk!ixikis the sum of all scores4HYPOTHESESfor one-way ANOVA thehypotheses areH0: µ1= µ2= ... = µKHa: µi"= µkfor some i, kthe null hypothesis is that allpopulation means are the samethe alternative hypothesis is thatat least one mean is different fromanother5INTUITIONthe basic approach of ANOVA isto make two calculatio ns ofvariance(1) We can calculate variance of eachgroup separately and combine them toestimate the variance of all scores.(within variance, s2W)(2) We can also calculate the varianceamong all the group means, relative to agrand mean. (between variance, s2B)these estimates will be the same ifH0is true!these estimates will be different ifH0is not true!6INTUITIONwe compare the estimates usingthe F ratioF =s2Bs2Wif F ≈ 1, do not reject H0if F>1, reject H0how big depends on the samplesizes, significance, ...7SCORESwhat contributes to a particularscore?assume a linear modelXik= µ + αk+ eik• Xikis the ith score in the kthgroup• µ is the grand mean for the pop-ulation, across all groups• αk= µk− µ is the effect of be-longing to group k• eikis random error associated withthe scoreeikchanges b ecause of randomsampling (normally distributed,mean of zero, σ2e)8SUM OF SQUARESwe want to estimate σ2e(varianceof population if H0is true)need sum of squaresΣkΣi(Xik− X)2consider one score(Xik−X)=(Xik−Xk)+(Xk−X)so(Xik−X)2= [(Xik−Xk)+(Xk−X)]2or(Xik−X)2=(Xik−Xk)2+2(Xk−X)(Xik−Xk)+(Xk−X)29SUM OF SQUARESif we sum across all subjects incategory knk!i(Xik−X)2=nk!i(Xik−Xk)2+2(Xk−X)nk!i(Xik−Xk)+nk!i(Xk−X)2since deviations from a meanequal zero, this reduces tonk!i(Xik−X)2=nk!i(Xik−Xk)2+nk!i(Xk−X)2in addition,nk!i(Xk− X)2= nk(Xk− X)2so we getnk!i(Xik−X)2=nk!i(Xik−Xk)2+nk(Xk−X)210SUM OF SQUARESnow, we sum across the k groups to getthe total sum of squares!k!i(Xik−X)2=!knk!i(Xik− Xk)2+ nk(Xk− X)2which becomes!k!i(Xik−X)2=!knk!i(Xik−Xk)2+!knk(Xk−X)2orSST= SSW+ SSBwhere• SSTis the total sum of squares.• SSWis the within sum of squares.Deviation of scores from the groupmean.• SSBis the between sum of squares.Deviation of group means fromthe grand mean.11DEVIATIONSSSW=!knk!i(Xik− Xk)2what causes this to be greaterthan zero?sinceXik= µ + αk+ eikµ + αkis fixed as i variesthus, deviations from Xkmust b edue to the eikterm (randomerror)12DEVIATIONSsinceXik= µ + αk+ eikthe mean of group k isXk=ΣiXiknk= µ + αk+ΣieiknksoXik− Xk= µ +αk+ eik−µ + αk+Σieiknk= eik−Σieiknkso only the random variationacross subjects is considered here13ESTIMATE OF σ2ewithin each group, deviationsfrom the mean are due to theerror terms eik, sos2k=Σi(Xik−Xk)2nk− 1→ σ2eto get a better estimate, poolacross all groupsSSWN − K= MSW→ σ2ehere MSWstands for meansquares withinN − K is the degrees of freedom14DEVIATIONSSSB=!knk(Xk− X)2what causes this to be greaterthan zero?sinceXik= µ + αk+ eikthe mean of group k isXk=ΣiXiknk= µ + αk+Σieiknkas k changes, X ≈ µ stays thesameso any deviations from X are dueto changes in αk(changes betweengroups) or to changes inΣieiknk(random error)15ESTIMATE OF σ2eif H0is true, then all αk= 0 andany deviations must be due onlyto the random error terms(Σieik/nk)so we can agai n estimate σ2easMSB=SSBK − 1=Σknk(Xk− X)2K − 1→ σ2ehere K − 1 is degrees of freedomon the other hand, if H0is nottrue, then MSBincludesdeviations due to positive αk, soMSB> σ2e16F statisticso, we do not know what σ2eis,but we have two estimates1. MSW: always estimates σ2e2. MSB: estimates σ2eif H0is true.Larger than σ2eif H0is false.compare the estimates bycomputingF =MSBMSWif H0is true, should get F = 1, ifH0is not true, should get F>117F criticalas always for inferential statistics,we need to know if F issignificantly greater than 1.0depends on degrees of freedomlook up Fcvfrom the Fdistribution table using K − 1 d.f.in the numerator and N − K d.f.in the denominatoralso written asF(K−1,N−K)18TESTING4 STEPS1. State the hypothesis. : H0:µ1= µ2= ... = µK, Ha: µi"=µjfor some i, j.2. Set the criterion. F(K−1,N−K)3. Compute the test statistic. F =MSB/M SW4. Interpret results.19CONCLUSIONStesting multiple meanstwo estimates of populationvarianceone estimate always estimatesvarianceother estimate is true only if H0istruelets us test H020NEXT TIMEan exampleassumptions underlying ANOVAmeasure of associationrepeated measures ANOVAThe last