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From: Adam on 24 Feb 2010 08:39
In retrospect, it wouldn't need to be as complicated as I had originally
thought. If I could just get a macro to create a data table like this:

table4:
LINE1 LINE2 LINE3 LINE4 dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
A p1 5.7 1 1 0 0 0 0
A p2 5.6 1 1 0 0 0 0
B p3 5.4 0 0 1 1 0 0
B C p4 5.3 0 0 1 1 1 0
C p5 5.1 0 0 0 1 1 0
D p6 4.7 0 0 0 0 0 1

Using this as a starting point:

table5:
dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
p1 5.7 1 1 0 0 0 0
p2 5.6 1 1 0 0 0 0
p3 5.4 0 0 1 1 0 0
p4 5.3 0 0 1 1 1 0
p5 5.1 0 0 0 1 1 0
p6 4.7 0 0 0 0 0 1
From: "Data _null_;" on 24 Feb 2010 10:56
Maybe this will help.

A Macro for Converting Mean Separation Output to Letter Groupings in PROC MIXED
http://www2.sas.com/proceedings/sugi23/Stats/p230.pdf

I tried a method using GLM on summary data. Supply your N's and
STDERRs for the lsmeans, and I will test the program.

this is the program for anova from summary data.

http://support.sas.com/kb/25/020.html

On 2/24/10, Adam <clown_rt(a)hotmail.com> wrote:
> In retrospect, it wouldn't need to be as complicated as I had originally
> thought. If I could just get a macro to create a data table like this:
>
> table4:
> LINE1 LINE2 LINE3 LINE4 dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
> A p1 5.7 1 1 0 0 0 0
> A p2 5.6 1 1 0 0 0 0
> B p3 5.4 0 0 1 1 0 0
> B C p4 5.3 0 0 1 1 1 0
> C p5 5.1 0 0 0 1 1 0
> D p6 4.7 0 0 0 0 0 1
>
> Using this as a starting point:
>
> table5:
> dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
> p1 5.7 1 1 0 0 0 0
> p2 5.6 1 1 0 0 0 0
> p3 5.4 0 0 1 1 0 0
> p4 5.3 0 0 1 1 1 0
> p5 5.1 0 0 0 1 1 0
> p6 4.7 0 0 0 0 0 1
>
From: Jerry Davis on 24 Feb 2010 13:13
Adam wrote:

[Adam wants letter groupings assigned to his mean means]

> In retrospect, it wouldn't need to be as complicated as I had originally
> thought. If I could just get a macro to create a data table like this:
>
> table4:
> LINE1 LINE2 LINE3 LINE4 dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
> A p1 5.7 1 1 0 0 0 0
> A p2 5.6 1 1 0 0 0 0
> B p3 5.4 0 0 1 1 0 0
> B C p4 5.3 0 0 1 1 1 0
> C p5 5.1 0 0 0 1 1 0
> D p6 4.7 0 0 0 0 0 1

You could use PROC GLIMMIX to fit your model. It has a lines option for
the lsmeans statement and assigns letter groupings based on the results
of pairwise t-tests.

Jerry
--
Jerry Davis
Experimental Statistics
UGA, CAES, Griffin Campus
From: "Data _null_;" on 24 Feb 2010 16:48
Google: PDMIX800.SAS

On 2/24/10, Data _null_; <iebupdte(a)gmail.com> wrote:
> Maybe this will help.
>
> A Macro for Converting Mean Separation Output to Letter Groupings in PROC MIXED
> http://www2.sas.com/proceedings/sugi23/Stats/p230.pdf
>
> I tried a method using GLM on summary data. Supply your N's and
> STDERRs for the lsmeans, and I will test the program.
>
> this is the program for anova from summary data.
>
> http://support.sas.com/kb/25/020.html
>
> On 2/24/10, Adam <clown_rt(a)hotmail.com> wrote:
> > In retrospect, it wouldn't need to be as complicated as I had originally
> > thought. If I could just get a macro to create a data table like this:
> >
> > table4:
> > LINE1 LINE2 LINE3 LINE4 dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
> > A p1 5.7 1 1 0 0 0 0
> > A p2 5.6 1 1 0 0 0 0
> > B p3 5.4 0 0 1 1 0 0
> > B C p4 5.3 0 0 1 1 1 0
> > C p5 5.1 0 0 0 1 1 0
> > D p6 4.7 0 0 0 0 0 1
> >
> > Using this as a starting point:
> >
> > table5:
> > dependent Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
> > p1 5.7 1 1 0 0 0 0
> > p2 5.6 1 1 0 0 0 0
> > p3 5.4 0 0 1 1 0 0
> > p4 5.3 0 0 1 1 1 0
> > p5 5.1 0 0 0 1 1 0
> > p6 4.7 0 0 0 0 0 1
> >
>
From: Adam on 25 Feb 2010 09:23
Thank you Jerry and "Data _null_;" for your suggestions. I've had an
opportunity to investigate both of your solutions/hints and this mostly
handles my issue. I now realize, however, that I may have been a little too
specific in my original posting -- to generalize my problem, suppose that
PROC MIXED is not necessarily my starting point. Suppose that I began with
a table of p-values for multiple comparisons (perhaps proportions would be a
better example) and that from that table of p-values I have created the table:

table6:
dependent Prop Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
p1 0.57 1 1 0 0 0 0
p2 0.56 1 1 0 0 0 0
p3 0.54 0 0 1 1 0 0
p4 0.53 0 0 1 1 1 0
p5 0.51 0 0 0 1 1 0
p6 0.47 0 0 0 0 0 1

with the goal of producing this table:

table7:
LINE1 LINE2 LINE3 LINE4 prop Mean Eqp1 Eqp2 Eqp3 Eqp4 Eqp5 Eqp6
A p1 0.57 1 1 0 0 0 0
A p2 0.56 1 1 0 0 0 0
B p3 0.54 0 0 1 1 0 0
B C p4 0.53 0 0 1 1 1 0
C p5 0.51 0 0 0 1 1 0
D p6 0.47 0 0 0 0 0 1
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