cochran-mantel-haenszel and breslow-day test output in proc freq

[MaryA_Marion]

I am doing a cochran-mantel-haenszel test using proc freq. The output is very confusing. Is there an order to the options included with the tables statement ?  For example, the Breslow-Day statistic is not always output and there are two cmh tables with different values. What are those supposed to be? I expected one cmh table with cmh1 cmh2 and cmh3.

SAS CODE:

data b;
input confounder $ treatment $ response $ count;
cards;
1 drug1    yes 19
1 drug1     no 3
1 drug2    yes 14
1 drug2     no 11
2 drug1    yes 27
2 drug1     no 18
2 drug2    yes 21
2 drug2     no 15
;

proc freq data=b order=data;
table treatment*response / nopercent norow nocol chisq relrisk;
weight count;
run cancel;

proc freq data=b order=data;
tables treatment*response / nopercent norow nocol  relrisk chisq cmh(mf);
by confounder;
weight count;
run;

Thanks.

MM

[Ksharp · Accepted solution]

CMH test is usually for three dimension contingency table .

I think you should involve one more variable in your model.

proc freq data=b order=data;
tables confounder*treatment*response / nopercent norow nocol  relrisk chisq cmh bdt;
weight count;
run;

[MaryA_Marion]

I know how to do it right but I had to keep removing table options ending up with only chisq and cmh(mf). This has happened to me multiple times with different datasets. These options generate code which sometimes conflict with each other. I think that's what is happening here. The order is important but I don't know the order. Thank you for the reply.

[MaryA_Marion]

Thanks. I had put a "by confounder" statement in one of my tries. I won't make that mistake again. Your approach gives good output. I am verifyng doing it by hand. I like using cmh(mf). MM