Pat, you wrote [in part]: > to zero. However, the values returned by SAS do not match those obtained > from binomial tables and excel, so it must be something else.

First of all, never trust stat results from Excel. If a stat package and Excel disagree about a stat result, trust the stat package. NOT MS Excel. Please. I think there's a recent paper in Journal of the ACM about the flaws in Excel results on test datasets.

> For example, with N=175 observations and M(Sign)=4.5 (implying 92 > "sucesses" out of 175 if values greater than zero are defined as > sucesses), SAS returns a Pr>=[M] of .5455. The p-value for the two > tailed sign test should be about .2249. It would be nice to have SAS

Funny, but I don't get .2249 either. I assume you normalized, and looked the value up in the normal tables [although you didn't say so]. A few problems: [1] the normal approximation is just that - an approximation {a tiny error}; [2] it looks like you failed to use the typical 1/2 continuity correction {still a small error}; and you forgot to multiply the tail probability by 2 in order to compute the 2-tailed p-value.

> automatically calculate the sign test for me. We have SAS version 6.12.

I got the same answer as SAS version 6.12 . Here's what I did.

P{S > |M|} = 2 * sum{j=0 to min(p,n-p)} nCj * 0.5**n

where n = number of non-zero values p = number of values greater than 0 nCj = binomial coefficient

You have to truncate the sum at min(p,n-p) in order to get the right 2-tailed test, and you need to remember to multiply by 2 afterward. Now either compute this directly [icky for n=175], or use the binomial approximation with continuity correction:

2 * Phi( (83+0.5 - 175*0.5)/sqrt(175*0.5*0.5) ) = 2 * Phi ( -.6047 ) which is about .546 .

Mathematica can do this much cleaner than I can.. as did SAS.

David -- David Cassell, OAO cassell@mail.cor.epa.gov Senior computing specialist mathematical statistician