0
$\begingroup$

I want to simulate a dataset that has a "grand mean", and then group means (with some deviation from the grand mean). Nevertheless, I have a bit of a problem conceptualising the problem: if I want (e.g.) the "grand mean" to be $\mu$ = 2, and then $\mu_{group1}$ = 1.5, and $\mu_{group2}$ = 2.5. Is it possible to somehow impose the constraint that $\mu$ still equals 2? What if the groups have unequal sizes?

Or am I thinking about it in a completely wrong way?

$\endgroup$
1
$\begingroup$

The problem here is, I think, the use of terminology and notation. Notation is defined within the framework of a model. You can have a model $$ Y_{ij}=\mu+\delta_j+\epsilon_{ij},\ \sum_{j=1}^J\delta_j=0, $$ where $\mu=2$ is what you call "grand mean", $J=2$ is the number of groups, $\delta_j$ the group effect for group $j$, these adding up to zero, i.e., -0.5 and 0.5 in your example, $\epsilon_{ij}$ the i.i.d. random error of observation $i=1,\ldots,N_j$ in group $j$ with $N_j$ observations.

There's nothing wrong with simulating from this model if the $N_j$ are not all equal, and the value $\mu$ is what it's defined to be ($\mu=2$ here) in any case.

However you are right observing that the expected value of the mean of the data sampled from such a model will not be $\mu$ unless $N_1=N_2$. So $\mu$ may not deserve to be called "grand mean" then (a terminology which I believe comes from designs where the group sizes are all the same). It's still $\mu$ though.

Ultimately you have to decide what you want: Data with $\mu=2$ in the model above, or data for which the expected value of the sample mean is 2, which requires different parameter choices in case the group sizes are not equal.

$\endgroup$
1
  • $\begingroup$ Thanks! This clears things up. Looking at the model definitely makes the conceptualisation easier. I should've done in in the first place :) $\endgroup$
    – Zlo
    Dec 2 '21 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.