[R-sig-ME] Proper analysis for the Machines dataset in lme4

[Douglas Bates]

On Tue, May 13, 2008 at 6:06 PM, Reinhold Kliegl
<reinhold.kliegl at gmail.com> wrote:
> Dear Michael,
>
> My following statement was not correct:
>
> On Mon, Apr 28, 2008 at 8:38 AM, Reinhold Kliegl
> <reinhold.kliegl at gmail.com> wrote:
>>  The comparison of m1 and m2 (or m1r and m2r) is conceptually
>>  questionable. m1 assumes there are 6 Workers; m2 assumes that there
>>  are 18 Workers, that is different groups of 6 persons worked on each
>>  of the 3 machines. Presumably, the experimental design decides whether
>>  m1 (m1r) or m2 (m2r) is the correct choice.
>
> After reading Douglas Bates's explanation of random effects for the
> Machines data today, I must add the following correction:
>
> mr1: score ~ Machine + (1 | Worker)
> mr2: score ~ Machine + (1 | Worker/Machine) ==  score ~ Machine + (1 |
> Worker) + (1 | Worker:Machine)
> mr3: score ~ Machine + (Machine | Worker)
>
> I wrongly assumed "(1 | Worker/Machine)" would force the six workers
> to be nested within Machine, that the program would treat them as 3
> groups of 6 workers although they are coded 1 to 6. Rather this syntax
> is shorthand for:  "(1 | Worker) + (1 | Worker:Machine)".
> Viewed this way, mr1 and mr2 can be compared, of course, as can mr1
> and mr3; these are nested models;  I am not sure about mr2 and mr3.

mr2 and mr3 are nested.  It is easiest to see this if mr3 is written
in the equivalent form

mr3: score ~ Machine + (0 + Machine | Worker)

If mr2 were re-expressed in this form the 3 by 3 variance-covariance
matrix for the levels of Machine given Worker would have a compound
symmetry form  in which all the diagonal elements are equal to
sigma^2_1 + sigma^2_2 and all the off-diagonal elements are equal to
sigma^2_2.