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I was wondering what to do in this situation.

I wanted to check for interaction between two IV's on the DV. First IV is gender, second musicianship (Yes/No). Only one (musicianship) main effect is significant, and the two IV's are correlated r(295)=0.33, p<0.001. Design is unbalanced, but that can be corrected with type III sum of squares, if I am not mistaken. Looking at the graph I would assume significant interaction, but the term is not significant, p=0.19

My research question includes testing for differences in DV based on gender and musicianship, but the interaction term between the two is not hypothesized, but still needs to be reported.

The big question is this: for testing group differences (gender/musicianship) should I use results from two way anova (main effect) to say that there is a difference in DV based on musicianship, or use the t-test which says there is no difference? Since t2=F and there are only two groups per variable, I presume one way anova and independent samples t-test would say the same thing. I know in two-way anova SPSS uses estimated marginal means, which is different from the mean used by t-test. Or should I do a two way anova without the interaction term, so i could control for the correlation between gender and musicianship, thus getting the gender unbiased estimate of musicianship effect on DV?

edit: what if I analyzed effect of musicianship on DV separately by gender? How would that relate to the interaction term?

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  • $\begingroup$ I suggest the latter. $\endgroup$
    – David Lane
    Commented Aug 5, 2017 at 18:38
  • $\begingroup$ @DavidLane Do you mean t-test or anova without interaction? Could you please explain why because I have to defend my logic later on. Many thanks! $\endgroup$ Commented Aug 5, 2017 at 18:49
  • $\begingroup$ This: " Or should I do a two way anova without the interaction term, so i could control for the correlation between gender and musicianship, thus getting the gender unbiased estimate of musicianship effect on DV?" Since your variables are correlated, they would be confounded if one were ignored and a t test done on the other. As you noted, only unconfounded variance is attributed to a variable with Type III sums of squares, so that is a conservative choice. $\endgroup$
    – David Lane
    Commented Aug 6, 2017 at 18:46
  • $\begingroup$ @DavidLane I notice that some sources reccommend against doing a new analysis without interaction and we have never done it in class. If I understand correctly, type III SS controls for A in B and vice versa, but it leaves the other factor in the interaction term? Therefore, when interaction is in the model p(gender)=0.04 and p(musicianship)=0.01, but their p increases when interaction is omitted. I am sceptic about this approach, since, I repeat, this was never done in class. $\endgroup$ Commented Aug 7, 2017 at 9:57

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