Unfortunately, for the purposes of checking your math, there is no separate way to compute Instead, you simply add the for the two main effects and for error and then SSInteraction. SS subtract that sum from SSTotal. The degrees of freedom are fairly easy to compute, because they follow closely what you’ve learned for the one-way ANOVA. That is: = number of scores 1 = 20 1 = 19 dfTotal – – = number of levels of strategy 1 = 2 1 = 1 dfStrategy – – = Total number of levels of gender 1 = 2 1 = 1 dfGender – – = * = 1 * 1 = 1 dfSxG dfStrategy dfGender = (Number of scores per condition 1) * Number of conditions = 4 * 4 = 16 dfError – The computation of MS is straightforward as well: = / MSStrategy SSStrategy dfStrategy = / MSGender SSGender dfGender = / MSSxG SSSxG dfSxG = / MSError SSError dfError With three null hypotheses, you’ll be computing three In each case, the denominator F-ratios. will be MSError: H0 What’s being compared F = / and FStrategy MSStrategy = MRepetition MImagery Repetition MSError Imagery = / and FGender MSGender Male = Female MMale MFemale MSError = / No interaction in Cell means FSxG MSSxG MSError population The source table would look like this: Source SS df MS F Strategy 405 1 405 73.64 Gender 5 1 5 .91 Strategy x Gender 0 1 0 0 Error 88 16 5.5 498 19 FMax Because this is an independent groups design, we once again be interested in determining whether or not we had violated the homogeneity of variance assumption. That is, we need to compute and compare that value to When we have some FMax FMax Critical.
Unfortunately, for the purposes of checking your math, there is no separate way to compute SSInteraction. Instead, you simply add the SS for the two main effects and for error and then subtract that sum from SSTotal. The degrees of freedom are fairly easy to compute, because they follow closely what you’ve learned for the one-way ANOVA. That is: dfTotal = number of scores – 1 = 20 – 1 = 19 dfStrategy = number of levels of strategy – 1 = 2 – 1 = 1 dfGender = Total number of levels of gender – 1 = 2 – 1 = 1 dfSxG = dfStrategy * dfGender = 1 * 1 = 1 dfError = (Number of scores per condition – 1) * Number of conditions = 4 * 4 = 16 The computation of MS is straightforward as well: MSStrategy = SSStrategy / dfStrategy MSGender = SSGender / dfGender MSSxG = SSSxG / dfSxG MSError = SSError / dfError With three null hypotheses, you’ll be computing three F-ratios. In each case, the denominator will be MSError: F H0 What’s being compared FStrategy = MSStrategy / MSError Repetition = Imagery MRepetition and MImagery FGender = MSGender / MSError Male = Female MMale and MFemale FSxG = MSSxG / MSError No interaction in population Cell means The source table would look like this: Source SS df MS F Strategy 405 1 405 73.64 Gender 5 1 5 .91 Strategy x Gender 0 1 0 0 Error 88 16 5.5 498 19 FMax Because this is an independent groups design, we once again be interested in determining whether or not we had violated the homogeneity of variance assumption. That is, we need to compute FMax and compare that value to FMax Critical. When we have some
S.Ch14.docx