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I am writing a report on a 2 x 3 mixed study. There is ONE within IV with TWO levels and there is ONE between subjects IV THREE levels. There are two DVs.

Sample sizes for each group is 25, 25, 27.

I tried Mauchly's test but it does not produce any results. I read somewhere that you can only use that for studies with 3 or more within IVs.

So my question is do I use the Box's test and refer to the Greenhouse Geisser output?

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The short answer is that problems of sphericity of variance can only arise when you have more than two levels on your within-subjects term. (Details are on the Wikipedia page about Mauchly's sphericity test)

Sphericity is a measure of variance between pairs of levels (e.g. for three levels the variance of pair difference A - B is equivalent to variance of pair difference A - C).

Hence, when there are only two levels (one pair), sphericity is not a concern as you are just estimating one difference on those two levels (and hence why SPSS doesn't calculate Greenhouse-Geisser corrected degrees of freedom).

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  • $\begingroup$ Another quick question. When reporting results which output should I refer to, the 'Greenhouse-Geisser' or the 'Sphericity Assumed'? $\endgroup$
    – buket dogan
    Commented Mar 17, 2013 at 21:49
  • $\begingroup$ In this case, don't they give the same information? (i.e. the Greenhouse-Geisser row in SPSS can have no correction applied when only two levels on within-factor, and so degrees of freedom will be the same, and so the G-G line is actually just a reprint of the sphericity assumed line.) (to actually answer your question, you can report the sphericity assumed line here.) $\endgroup$
    – James Stanley
    Commented Mar 17, 2013 at 21:52
  • $\begingroup$ OK I see where I got confused now. Thank you. Does this sound right? A mixed design ANOVA with type of scene (static and dynamic) as a within subjects factor and condition type (CI, MI, and BOTH) as a between subjects factor was conducted. The analysis revealed an effect in type of scene for accuracy f(1,74) = 6.21, p < .05 and reaction time f(1,74) = 20.97, p < .001. $\endgroup$
    – buket dogan
    Commented Mar 17, 2013 at 21:56
  • $\begingroup$ Presuming that interaction terms were non-significant, that's fine. Small points: should be capital F instead of lower-case f for F statistic, and I'd report the actual p-value rather than just "< .05" or "< .001" -- and I presume you're reporting difference in means by condition elsewhere (e.g table/figure). $\endgroup$
    – James Stanley
    Commented Mar 17, 2013 at 22:00

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