so now I am thinking I should just export the table and then manually do stuff in excel to make it look like the example I was provided and import to SAS but that I am sure that is not what past programmers in my company have done. I am sure they figured out a way to do it all in SAS and my boss has even said I should do this all in SAS but I am left to figure it out on my own and as we can see by my number of posts I am not able too. Have read tons of books and just spinning my wheels and wasting time. I have my meeting Friday to show my data but I am no where near able to do that unless I figure this out.
I think I got it. I played and played and this is the code I used:
proc sql;
create table chi4 as
select * from
(select distinct who as source, who, svcd, svdesc, admission, count from chi)
union all
(select distinct "national", a.who, b.svcd, b.svdesc, b.admission, b.count from
chi as a inner join chi3 as b
on a.svcd = b.svcd and
a.svdesc = b.svdesc and
a.admission = b.admission)
order by who, svcd, source, admission);
run;
The output chi4 table looks like this:
who who svcd svdesc admission count
1111 1111 1 minor no 1
1111 1111 1 minor yes 0
national 1111 1 minor no 31645
national 1111 1 minor yes 52
2222 2222 2 moderate no 1
2222 2222 2 moderate yes 0
national 2222 2 moderate no 41864
national 2222 2 moderate yes 154
etc
Now I have to do a step to drop the 2nd who column or rename the label or something so when I do my chi it looks at the who with the numbers and national and then the remaining columns. Because if I use the 2nd column then it will add all the counts together. If I use first column it will chi it correctly from what I can tell.
I added a data step because even though the labels in the chi4 table both show who, the actual column name is source and who so I did this step:
data chi5 (drop = who);
set chi4;
run;
It drops the 2nd who column and leaves the other who column which is actually source and looks the way it should. So let me see how my chi works now.
Why on earth would you want to perform a Chi-Squared Test on a table with this many cells?
And with so many cells with virtually no representation (counts<2)? A sensible aggregation seems in order here.
Steve Denham
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