Since a proportion is formally a mean (of a binary variable whose possible values are 0 and 1), and the usual two-sample test for proportions is formally identical to the two-sample test for means, I should think that if the latter is based on a LRT, so should be the former. There may be some complication induced by the use of a large-sample normal approximation, for which the LRT argument above would hold, versus the binomial distribution procedure, for which I am less certain.
On 10 Mar 2003, Jason Owen wrote: > Is the two-sample test for comparing proportions based on > a likelihood ratio test? > > The one-sample test is, and both the one and two-sample > test for means are. However, I'm not seeing that the > pivotal quantity in the two-sample test for proportions > is borne out of a LRT. Is this correct? ----------------------------------------------------------------------- Donald F. Burrill [EMAIL PROTECTED] 56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816 . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
