Meaning of "likelihood ratio criterion is distribution free" This is in ref to pp. 54-55 in McDonald,R.P (1985) [1] in the context of exploratory factor analysis (EFA) estimation: I am confused as to the meaning of: 

"(the likelihood ratio criterion) is a distribution free measure of
  "misfit" of the estimated parameters to sample correlation".

Is it the LRC that is distribution free (isn't it chi-square distributed?) or is it no distribution assumption is being made about the variables being modeled (isn't supposed be multivariate normal). 
I am also not clear about the subsequent sentence: 

...it is a direct measure of the departure of correlation matrix of
  the residuals from an identity matrix...

I find the book full of insights explained to statistically unsophisticated like me, which is why I am keen to find out what it is that I am missing.I am relatively new to inferential statistics and my grasp on all this is bit tenuous, so apologies if the question is badly worded.     
[1] Roderick P. McDonald (1985),
Factor Analysis and Related Methods,
Psychology Press
 A: What is meant by a procedure/criterion being distribution-free is that - whatever the distribution from the data were drawn* - the result (in this case, the LRC criterion) has the same distribution under the null.
* (there can be some conditions; often many of the more traditional distribution free procedures require continuity and iid)
"Distribution-free" doesn't mean that the test statistic or criterion can't have a given distribution (in fact, it typically means that it does). 
The claim being made in the text there is that -- irrespective of what distribution you started with  -- the likelihood ratio criterion (presumably the thing it's calling $Q*$ on the previous page) has (at least asymptotically) the same chi-square distribution.
So yes, normally you have a multivariate normal assumption for EFA -- but the text is saying that if it had some other distribution, the result would still have a chi-square distribution (in the null case).
The point is that in sufficiently large samples the distribution of the calculated result doesn't depend on what distributional model that was. In that sense, it could be said to be (asymptotically) distribution free.
[However, I'm not sure that the earlier claim (p54) about least squares requiring the normal assumption is accurate; unless I mistake something about the situation - which I might, since I don't have the whole document -  I think asymptotically it is also distribution free in a similar sense -- but note that "least squares" isn't a statistic, so it may depend on what the actual statistic is being taken to be.]
