[R-sig-ME] LMM reduction following marginality taking out "item" before "subject:item" grouping factor
Maarten Jung
M@@rten@Jung @ending from m@ilbox@tu-dre@den@de
Sat Dec 1 17:58:11 CET 2018
Hi Jake,
this clears things up for me - thank you.
Regards,
Maarten
On Thu, Nov 29, 2018 at 3:54 PM Jake Westfall <jake.a.westfall using gmail.com>
wrote:
> Maarten,
>
> Just to double-check if I get this right: the entries in each cell of the
>> table are the numbers by which the variance components are divided in the
>> equation of the noncentrality parameter. Is this correct?
>
>
> Almost. They multiply the variance components, not divide them.
> Essentially each row gives the weights of a weighted sum of variance
> components. Then to translate that to what appears in the denominator of
> the noncentrality parameter, the entire thing is divided by the total
> sample size *and we remove the variance component for the effect in
> question *(I forgot to mention that part in my last email).
>
> For example, consider the simple design with random participants (P)
> nested in fixed groups (G). So g is the number of groups, p is the number
> of participants per group, and # is the number of replicates. (This is
> design 2 in the dropdown menu of examples.) The EMS table shows that, for
> the between-group effect, the coefficients for the error, participant, and
> group variance components are, respectively, 1, #, and #p. So the expected
> mean square is var_error + # * var_participants + # * p * var_groups. The
> total sample size is pg#, so in the noncentrality parameter expression this
> becomes sqrt(var_error / pg# + var_participants / pg). Note that this only
> gives most of the denominator of the of the noncentrality parameter
> expression -- it ignores the variance of the contrast weights -- you can
> see more in the PANGEA working paper, linked in the app.
>
> Jake
>
> On Thu, Nov 29, 2018 at 7:36 AM Maarten Jung <
> Maarten.Jung using mailbox.tu-dresden.de> wrote:
>
>> Hi Jake,
>>
>> So, regarding this issue, there is no difference between taking out
>>>> variance components for main effects before interactions within the same
>>>> grouping factor, e.g. reducing (1 + A*B | subject) to (1 + A:B | subject),
>>>> and taking out the whole grouping factor "item" (i.e. all variance
>>>> components of it) before "subject:item"?
>>>
>>>
>>> I think that if you have strong evidence that this is the appropriate
>>> random effects structure, then it makes sense to modify your model
>>> accordingly, yes.
>>>
>>
>> This makes sense to me.
>>
>> Do all variances of the random slopes (for interactions and main effects)
>>>> of a single grouping factor contribute to the standard errors of the fixed
>>>> main effects and interactions in the same way?
>>>
>>>
>>> No -- in general, with unbalanced datasets and continuous predictors,
>>> it's hard to say much for sure other than "no." But it can be informative
>>> to think of simpler, approximately balanced ANOVA-like designs where it's
>>> much easier to say much more about which variance components enter which
>>> standard errors and how.
>>>
>>> The standard error for a particular fixed effect is proportional to the
>>> (square root of the) corresponding mean square divided by the total sample
>>> size, that is, by the product of all the factor sample sizes. So examining
>>> the mean square for an effect will tell you which variance components enter
>>> its standard error and which sample sizes they are divided by in the
>>> expression.
>>>
>>
>> Your app is very useful, too. Just to double-check if I get this right:
>> the entries in each cell of the table are the numbers by which the variance
>> components are divided in the equation of the noncentrality parameter. Is
>> this correct?
>>
>>
>> Regards,
>> Maarten
>>
>
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