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Interpretation of Bayes Repeated Mesures Anova

edited November 2015 in JASP & BayesFactor

Hi,

I am new in JASP. I have conducted an Bayesian Repeated Mesures Anova and I am having problems to interpret correctly the results. Besides, and in contrast to the Bayesian t-test analysis, i do not find the Cauchy prior width plot in this analysis..Could someone give me some hints?

I attached the image with the output.

Thanks in advance,

G.

Comments

  • edited 9:36PM

    Hi,

    That's an excellent question! Good to baptise the JAPS & BayesFactor subforum with. I'll leave it to the experts to answer, but for now I'd like to point out that you haven't uploaded the image. And could explain in more detail about which aspects of the interpretation you are unsure?

    Cheers,
    Sebastiaan

  • EJEJ
    edited 9:36PM

    Hi Guillon,
    Thanks for your question! With respect to the interpretation, it would indeed be good to add the image. With respect to the Cauchy prior, your are right, we did not implement this for the ANOVA. We discussed it recently and then decided that it would perhaps not be all to helpful -- the current default works well. But I'll bring this up again. In the mean time, if you really want to tweak this parameter, you can do so in the BayesFactor package.
    Cheers,
    E.J.

  • edited November 2015

    Thanks for your answers. I was also wondering where my image was..

    You can check it here:

    image

    G.

  • EJEJ
    edited 9:36PM

    Hi Guillon,

    OK. An entire discussion of all of the entries is perhaps going too far;
    Maarten Marsman and I will write a paper doing just that (for a special issue --
    we hope to be done in a few weeks). Anyway, let me tell you what I would conclude.

    First I look at the "BF_10" column of the top table. This compares all the models against the null model. We see that "Condition" is a huge improvement. Not so for "Time" -- there is actually a little evidence against adding "Time" to the null model (BF_10 = 0.284, which means BF_01 = 1/.284 = 3.5 in favor of the null). To judge the value of adding the interaction, we can compare the model with two main effect (Condition + Time, with BF_10 = 2.772e+9) to the model that adds the interaction term (Condition + Time + ConditionTime, with BF_10 = 1.663e+8). As you see, the evidence against the null model is more compelling *without the interaction term. If we want to know the BF against the interaction term we exploit the fact that Bayes factors are transitive: denote the null model by 0, the model with main effects by 1, and the model that adds the interactions by 2. By transitivity we get BF_12 = BF_10 * BF_02 = BF_10 / BF_20 = 2.772e+9/1.663e+8 = 16.66867. That's pretty good evidence against the interaction.

    A similar conclusion can be drawn from the analysis of effects shown in the second table. This analysis model-averages over the models shown in the top table. However, with few models you can get what you need from the top table (this is a little different when there are many models).

    Perhaps this is also useful: http://www.ejwagenmakers.com/inpress/RouderEtAlinpressANOVAPM.pdf

    Cheers,
    E.J.

  • edited 9:36PM

    Tks a lot for this exahustive answer EJ,
    Best,

    G.

  • edited 9:36PM

    Hi All,

    I'm also new to JASP and wanted to ask a quick follow-up question on this topic.
    I have a slightly larger design with two within vars (task-type, time) and one between (group). Im attaching a screen shot. Theoretically, I'm only interested in the Group x Time interaction. NHST for the Group x Time interaction is not significant, and I want to use JASP to indicate the amount of evidence in favor of H0.

    1. What is the accurate way to indicate the evidence in favor of H0 regarding Group x Time ? Can I use the same calculation EJ showed above with B01?

    2. Given that I am not interested in other effects, should I check "is Nuisance"?

    Thank you !
    Nitzan.

    image

  • EJEJ
    edited 9:36PM

    Hi Nitzan,

    Yes, you can use the transitivity trick to compare the model with the interaction to the model that has the other terms but lacks the interaction. However, as you suggest, you can also designate those other terms as nuisance. These nuisance terms will then be part of the null model, and then you can simply inspect the BF of the interaction model versus the null model directly. Both the transitivity trick and the nuisance method should give you the same answer of course.

    Cheers,
    E.J.

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