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I am currently conducting an experiment looking at the effect of visual experience on the development of neural network firing freq. One thing I am interested in knowing is when in development visual experience has an effect.

Since I have 2 independent variables (Age and Rearing condition) I conducted a two-way Anova (type III SS - due to unbalanced groups) in R but i find no interaction between Rearing condition and Age despite there being appearing to be a difference between Rearing condition at 7 dpf (see plot). Could some please explain why I am seeing no effect? Is it due to the variance at 5 dpf being larger than at 3 or 7?

enter image description here

Is a two way Anova the appropriate test to understand when in development visual experience has an effect? or would it be more appropriate to test between rearing conditions at each developmental age and within rearing condition at each age and correct for the multiple comparisons using a bonferroni correction since i am not interested in the comparisons between groups at different ages.

Anova code:

enter image description here

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  • $\begingroup$ Probably because you have unequal variance between groups (i.e. your equality of variance assumption is being violated). $\endgroup$
    – André.B
    Nov 27 '19 at 20:34
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A few notes:

It's best to keep everything in one model, as you have done...

I'm also surprised you don't see a significant interaction effect, but I suspect that has to do with the high variability in the data for age 5. Have you checked residuals for homoscedasticity? You might look in to using tests that are robust to heteroscedasticity, just as white.adjust in car:Anova ...

When using type III anova in R, be sure to change the type of contrasts used in the model. Alternatively, you might read up on type II SS, and see if that convinces you that you'd rather use type II ...

If your main interest is really just comparing within each age, you might go ahead and use emmeans to make these comparisons. The correct grammar might be marginal = emmeans(model, ~ Rearing_condition | age); pairs(marginal).

Also check the residuals for normality. I have a suspicion this kind of data may be modeled better with a Gamma distribution. This is easy in R with glm. But if the residuals pass visual checks for a Gaussian (OLS) model, there's no reason to make things more complicated with a glm.

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    $\begingroup$ Thank you this was super helpful. Visually inspecting the residuals suggests that there is heteroscedasticity and levene test revealed significant differences in variance between the conditions. As suggested white.adjust in car:Anova resulted in significant interaction between the factors. $\endgroup$
    – neurosoup
    Nov 28 '19 at 11:48

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