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Introduction to Statistics inPsychologyPSY 201Professor Greg FrancisLecture 33ANalysis Of VArianceError is sneaky.HYPOTHESIS TESTINGwe know how to test the differenceof two meansH0: µ1− µ2=0Ha: µ1− µ2"=0by using the t distribution andestimates of standard errorwhat if you have more populationsand want to know if they are allequal?2MULTIPLE 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 =0.05What is the Type I error?3MULTIPLE t- TESTSwe have a risk of making a type Ierror for each t testsince we have c = 10 differentt-tests, with α =0.05, the Type Ierror rate becomes1 − (1 − α)c=0.40bigger risk of error than you mightexpect!would need to set α much smallerto insure that Type I error rate isbelow 0.05! (kind of messy to do)4DEMONSTRATIONcount off by fours and split into fourgroupsthese are four populationsk = 1, k = 2, k = 3, k =4within each group do the following:1. Go through your backpacks and countthe total number of books (textbooks,novels, notebooks) in the group, nk.2. Across the entire group, compute theaverage number of pages across the books,Xk, and the standard deviation sk:(a) Xk=!nki=1Xkink(b) sk="##$!iX2ki−[(!iXki)2/nk]nk−15DEMONSTRATIONnow, go get Xk, skand nkfromthe other three groupsRun 3 hypothesis tests to test if any of the meansare different from the mean of your group’ssample:(we’ll assume homogeneity of variance)(1) State the hypothesis:H0: µk= µjHa: µk"= µjα =0.05(2) Find the critical value:df = nk+ nj− 2tcv=6DEMONSTRATION(3) Compute test statistic:s2=(nk− 1)s2k+(nj− 1)s2jnk+ nj− 2=sXk−Xj="#######$s21nk+1nj=t =(Xk− Xj) − (0)sXk−Xj(4) Interpret your results:t = < ? > = tcv7DEMONSTRATIONNow let’s see how often you rejectH0• R → H0was rejected• N → H0was not rejectedµ1µ2µ3µ4µ1×µ2×µ3×µ4×any rejections are probably Type Ierrors (people were randomlyassigned to samples, so thereshould be no difference acrosspopulations)8WHAT DO WE MAKE OFTHIS?Not only is it a pain to makemultiple comparisions of meansbut it tends to lead to more TypeI error than α indicateswe could simply decrease α to asmaller value so that the overallType I error is how we want itbut there is a better method9CONCLUSIONStesting multiple meansloss of control of Type I error10NEXT TIMEANOVAtwo measures of varianceMeasure twice, cut


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