Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 35ANalysis Of VArianceHow do changes in one variablecorrespond to changes inanother variable?TESTING4 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.2SUM OF SQUAREScalculating the sum of squares (within,between, total) can be messycan make it a bit easierfor each group, k, letTk=nk!iXikand letT =!kTkthenSSB=!kT2knk−T2NSSW=!knk!iX2ik−!kT2knkandSST=!knk!iX2ik−T2N3EXAMPLEA college professor wants to determine the best way topresent an important lecture topic to his class. He hasthe following three choices: (1) he can lecture, (2) hecan lecture plus assign supplementary reading, or (3) hecan show a film and assign supplementary reading. Hedecides to do an experiment to evaluate the threeoptions. He solicits 27 volunteers from his class andrandomly assigns 9 to each of three conditions. Incondition 1, he lectures to the students. In condition 2,he lectures plus assigns supplementary reading. Incondition 3, the students see a film on the topic plusreceive the same supplementary reading as the studentsin condition 2. The students are subsequently tested onthe material. The following scores (percentage correct)were obtained.4EXAMPLELecture Lecture + Reading Film + ReadingCondition 1 Condition 2 Condition 392 86 8186 93 8087 97 7276 81 8280 94 8387 89 8992 98 7683 90 8884 91 83T1=767 T2=819 T3=734!knk!iX2ik= 200, 428T = 23205(1) HYPOTHESESfor one-way ANOVA thehypotheses areH0: µ1= µ2= µ3Ha: µi!= µkfor some i, kTest with α =0.056(2) CRITICAL VALUEbetween degrees of freedom(numerator) [number of groupsminus 1]df = K − 1=3− 1=2within degrees of freedom(denominator)[number of subjectsminus number of groups]df = N − K = 27 − 3 = 24Total degrees of freedom [numberof subjects minus 1]df = N − 1 = 27 − 1 = 26From F distribution table we findFcv=3.407(3) STATISTICfrom the dataSSB=!kT2knk−T2NSSB=(767)29+(819)29+(734)29−(2320)227= 408.074SSW=!knk!iX2ik−!kT2knkSSW= 200428−(767)29+(819)29+(734)29= 671.778andSST=!knk!iX2ik−T2NSST= 200428 −(2320)227= 1079.852checkSST= SSB+SSW= 408.074+671.778 = 1079.8528(3) STATISTICMSB=SSBK − 1=408.0742= 204.037MSW=SSWN − K=671.77824= 27.991soF =MSBMSW=204.03727.991=7.299(4) INTERPRETsince7.29 = F > Fcv=3.40we reject H0. The methods ofpresentation are not equallyeffective. (Note, does not tell uswhich pair of means are different!)SUMMARY TABLESource of Sum of Degrees of Variancevariation squares freedom estimate F ratioBetween 408.074 2 204.037 7.29∗Within 671.778 24 27.991Total 1079.852 26With α =0.05, Fcv=3.40. Therefore H0is rejected.10ASSUMPTIONSto use ANOVA validly, the data mustmeet some restrictions1. The observations are random and in-dependent samples from the popula-tions.2. The distributions of the p opulations fromwhich samples are selected are nor-mal.3. The variances of the distributions inthe populations are equal. Homogene-ity of variance.How do we check these assumptions?1. Sample properly.2. χ2-test.3. Bartlett’s test.11ANOVA TESTING4 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.12ASSUMPTIONSit turns out thatviolations of normality have smalleffects on Type I error ratesviolations of homogeneity ofvariance have a big effect if thepopulation sizes are differentmeans that ANOVA is robust aslong as the sample sizes are thesame across populationsNote, though, the restrictions toapply ANOVA are more stringentthan for t-tests (no restrictions onpopulation distributions)13SIGNIFICANCEANOVA tells us if an effect(difference of means) isstatistically significantit does not tell us if it is importantone measure that helps usdetermine importance is thestrength of association ω2measures association betweenindependent and dependentvariables(similar to correlation coefficientof determination, r2)14SIGNIFICANCEwhat percentage of the variationin the data is accounted for by theindependent variable? (changes ingroups)ω2=SSB− (K − 1)MSWSST+ MSWin our example problem (lasttime) we plug in the numbersω2=408.074 − (2)27.9911079.852 + 27.991=0.3178the different lecture styles accountfor approximately 32% of thevariancesuggests that the results are notreally that strong15t testsif we have only two groupsH0: µ1= µ2Ha: µ1!= µ2we can use either ANOVA or thet-tests discussed previouslythey give identical results!16t testsit turns out that the Fdistribution for K − 1,N − K(1,N − 2) degrees of freedom issimply the t distribution forN − 2 df, squared.t2= Ft2cv= Fcvso using either technique producesthe same results (reject or notreject)17EXAMPLEA sociologist wants to determine whether sororityor dormitory women date more often. Herandomly samples 12 women who live insororities and 12 women who live in dormitoriesand determines the number of dates they eachhave during the ensuing month. The followingare the results.Sorority DormitoryWomen, X1Women, X28 95 76 34 412 47 89 710 55 83 67 35 5T1=81 T2=69X1=6.750 X2=5.75018t TESTtest with α =0.05, two-tailedH0: µ1− µ2=0Ha: µ1− µ2!=0we have equal numbers ofsubjects, so we do not need to testfor homogeneity of variance!df = n1+n2−2 = 12+12−2 = 22so, from the t distribution table,we findtcv= ±2.074from data we calculate the pooledestimate of population variances2=5.57019t TESTso standard error issX1−X2=())))))*s21n1+1n2sX1−X2=0.963andt =(X1− X2) − (µ1− µ2)sX1−X2t =1.00.963=1.038 < 2.074 = tcvwhich is not in the region ofrejection, so do not reject H0no evidence for a difference innumber of dates20ANOVAH0: µ1= µ2Ha: µ1!= µ2we have 1 df in the numerator and22 df in the denominator, so, fromthe F distribution tableFcv=4.30we can calculateSSB=6.00SSW= 122.500SST= 128.500check128.50 = SST= SSB+SSW=6.00+122.50021ANOVAMSB=SSBK − 1=6.001=6.00MSW=SSWN − K=122.50022=5.568F =MSBMSW=6.005.568=1.078sinceF =1.078 < 4.30 = Fcvwe do not reject H0note:F =1.078 ≈ 1.077 = (1.038)2= t2Fcv=4.30 = 4.30 = (2.074)2= t2cv22CONCLUSIONSrestrictions on use of one-wayANOVAstrength of association23NEXT TIMEdependent samplesrepeated measures ANOVAThe last