PROC MIXED pvalue calculation UPPER option

[atavenard]

Hi,

I would need help to understand how the pvalue is computed when using UPPER option in a PROC MIXED.

Here is my mixed procedure:

PROC MIXED data=dummyall plots=residualpanel method=reml;

     CLASS   diet timec id timeplan;

     MODEL pulse= diet timec diet*timec pulsebase/ ddfm=kr;

     REPEATED timec / type=sp(pow)(timeplan) subject=id;

     LSMEANS diet*timec/diff=all cl;

     LSMESTIMATE diet*timec 'Change at Timec=2 - Diet 1 vs diet 2'  1 0 0 -1 0 0 / upper alpha=0.05 cl ;

RUN;

I have the impression the pvalue of the one sided test is 0.5*(pvalue two tailed) in case the pvalue is significant and the pvalue of the one sided test is 1-0.5*(pvalue two tailed) in case the pvalue is not significant. Though I thought the pvalue calculation was based on which alternative hypothesis is considered (diff>0 or diff<0). Why does the way the pvalue is computed change according to the results?

Can anyone help me to understand?

Thanks!

[SteveDenham]

I believe your interpretation is correct.  The directionality of one-tailed tests leads to the definition of the p value.  Let's suppose the alternative is that d>0, but you obtain a negative estimate for d.  The p value is then everything under the pdf curve to the right of d.  That is, 0.5 + (1 - the amount to the LEFT of the value of d).  A similar case comes out of an alternative hypothesis of d<0, and you obtain a positive estimate for d.  A pictorial representation of this should make this clearer, so just draw a Gaussian curve centered on 0, and put in some values for d, and then look at the part under the curve.

SteveDenham