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Is using a 2 way MANOVA to check whether height and weight differ across age groups and gender, the same as using a 2 way ANOVA to check whether BMI (wt in kg/ht in m$^2$) differ across age groups and gender?

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A side-remark. You asked 14 questions in the last several days, and several (though admittedly only few) of them seem to have received good answers. As you never "accepted" any answers on StackExchange, I guess that you might be unfamiliar with the system; so let me point out that if you received a satisfying answer you should consider "accepting" it by clicking on a green tick sign nearby. Having zero acceptance rate will likely prevent people from answering your questions. – amoeba Jul 28 '14 at 17:05
Adding on to what @amoeba said: you should also consider upvoting answers that you like (whether or not you accept them, whether or not you asked the original question) by clicking the up arrow to the top left of the answer. – Alexis Jul 28 '14 at 17:34

No. The manova tests the bivariate effects, and is based only on linear functions thereof; whereas the BMI measure is a nonlinear function of those two responses. Even if BMI were linear, you'd be restricting the comparison to a 1-dimensional subspace of the two dimensions in your data.

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Which sounds sensible? – user52672 Jul 28 '14 at 17:11
Both, in their own context. BMI is well understood and is an easier "sell" in discussing your analysis. But if the groups differ in a way substantially different than the BMI, the manova will pick up on that (provided it is linear). Another idea is to take logs. Note that log BMI = log wt - 2*log ht, so if you do the multivariate analysis on the logged responses, and see that the coefficients are more-or-less proportional to 1 and -2, that'd indicate that BMI is the big player and you could simplify. The log scale may make the diagnostics look better too. – rvl Jul 28 '14 at 17:22
Important caveat is that statistical inferences (frequentist hypothesis testing, or confidence intervals) on transformed variables do not apply to statistical inferences on untransformed variables, because $f(\sigma^{2}_{x}) \ne \sigma^{2}_{f(x)}$. – Alexis Jul 28 '14 at 17:32
@Alexis: I understand your example, but this is mainly because the assumption of normality made by t-test cannot be valid for both x and ln(x). And non-parametric tests such as ranksum will yield the same outcome in both cases. However, this does relate to the current discussion, because MANOVA obviously does rely on normality as well... – amoeba Jul 28 '14 at 22:18
People transform data all the time. Yes, it's true that the inference made isn't exactly equivalent to one made on the original scale. But that's the whole point. You're making an inference on a parameter, not on the measurements. An inference comparing the mean log BMI for two groups is in fact an inference on the ratio of the geometric means of the two groups. That's a different framework, but there's nothing invalid about making inferences about geometric means. – rvl Jul 28 '14 at 22:51

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