Date: Sun, 7 Sep 2008 13:19:51 -0700
Reply-To: "globalreviewer@gmail.com" <globalreviewer@GMAIL.COM>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: "globalreviewer@gmail.com" <globalreviewer@GMAIL.COM>
Organization: http://groups.google.com
Subject: Re: interpreting SAS output
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Thank you for helping me with your inputs. I am not expecting
different SEs rather I am expecting 'similar' SEs once the values are
back-transformed (not after transformation).
My mixed code is given below. I used log transformation on my data
because they were not normally distributed:
proc mixed data=PlotObs_Trans ;
class Month Sex ;
Model LogVRate=Month Sex Month*Sex /solution ddfm=satterth
outp=OutOldTrans1;
Lsmeans month*Sex/slice=Month;
Lsmeans month*sex/slice=sex;
LSmeans Month Sex month*Sex/PDIFF CL ADJUST=TUKEY;
Run;
Data OutOldTrans1;
Merge PlotObs_Old_Trans OutOldTrans1;
aresid = ABS(resid);
Drop StdErrPred Lower Upper;
Run;
Proc plot data=OutOldTrans1 vpercent=50;
plot LogVRate*Month LogVRate*Sex;
Plot Resid*Month resid*Sex /vref=0;
plot Resid*Pred aresid*Pred/Vref=0;
Quit;
TITLE7 'Examination of variance homogeneity using rank correlation';
TITLE8 'between absolute residuals vs predicted values';
PROC CORR SPEARMAN DATA=OutOldTrans1;
VAR aresid Pred;
QUIT;
TITLE8 'Examination of normality of residuals';
PROC UNIVARIATE NORMAL PLOT DATA=OutOldTrans1;
/*tests for normality W shud be closer to 1 and p shud not be <.05*/
VAR Resid;
Quit;
My sample data looks like follows: (it is twice this size..so it is
still not very big by the standards of social studies)
Month Day Sex Modmonth VRate VFreq
1 1 1 1 26.13 0.77
1 2 1 1 21.13 0.33
1 3 1 1 20.63 0.34
1 2 2 2 10.63 0.17
1 3 2 2 12.5 0.21
2 1 1 3 21.38 0.36
2 2 1 3 19 0.28
2 3 1 3 21.38 0.33
2 1 2 4 12.5 0.21
2 2 2 4 8.5 0.12
2 3 2 4 9.25 0.14
2 1 2 4 9.75 0.13
2 2 2 4 3 0.07
2 3 2 4 8.63 0.15
3 1 1 5 19.38 0.25
3 2 1 5 21.5 0.25
3 3 1 5 12.38 0.22
3 1 2 6 6.25 0.12
3 2 2 6 4.13 0.05
3 3 2 6 6 0.11
4 1 1 7 4.38 0.36
4 2 1 7 0.71
4 3 1 7 0
4 1 1 7 27.5 0.28
4 2 1 7 21.25 0.29
4 3 1 7 15.25 0.23
4 1 2 8 0.25 0.02
4 2 2 8 0.29
4 3 2 8 0
5 1 1 9 0.67 0.67
5 2 1 9 0
5 3 1 9 0
5 1 2 10 2.5 0.83
5 2 2 10 0.83 0.28
5 3 2 10 0.67 0.17
6 1 1 11 0
6 2 1 11 0 0
6 3 1 11 0
6 1 2 12 0
6 2 2 12 0.75 0.38
6 3 2 12 0
Also, what is the best method to show the interaction graphically..
plotting the means from the raw data or the estimates from proc
mixed ? I thought one would back-transform the estimates and the SE
and plot them on the graph (???).
Thanks again for your help..and for directing me to the other websites
that have clear explanations..
G Ren
On Sep 7, 8:58 am, art...@NETSCAPE.NET (Arthur Tabachneck) wrote:
> G Ren,
>
> It would help the list if you showed your code and some sample data.
>
> I don't know why you would expect to get different SEs. You might want to
> look at UCLA's stats page for some explanations of the output (e.g.,http://www.ats.ucla.edu/stat/sas/library/SASSlice_os.htm).http://www.ats.ucla.edu/stat/sas/libraryhas a lot of good examples.
>
> I would expect that you would be plotting means in order to see the
> interactions.
>
> As for applying transformations, I would think you would only do that if
> you had a reason (e.g., the data don't appear to be normally distributed or
> theory suggests that they aren't normally distributed).
>
> Art
> --------
> On Sat, 6 Sep 2008 21:47:52 -0700, globalrevie...@gmail.com
>
> <globalrevie...@GMAIL.COM> wrote:
> >Hello,
> >I have some very basic questions on reading SAS output. I ran a two
> >way anova with the model :
> >[x = a b a x b ] as a Mixed analysis and got significant result for
> >the interaction term (a x b). Factor 'a' has 2 levels and 'b' has 6. I
> >plotted the estimates from this analysis as follows:
>
> >a1b1, a1b2, ...a1b6, a2b1, a2b2 .... a2b6. I also got the SE values
> >from the output as (estimate x SE) values and I plotted it as a line
> >graph. I am interested in each level of interaction so I also wrote
> >contrasts for the levels of interest. However, on plotting the above
> >estimate values with the SE values, factors that show significance in
> >my contrast statement (or even the pdiff statement) dont show
> >significance on the graph. I am not sure what I am doing wrong.. also
> >I cannot seem to recover 'similar' SE values if I use different
> >transformations (ie sqrt and log). I am sure I am doing some real
> >fundamental mistake because I have not encountered this problem in any
> >of my classes where I learnt SAS!
>
> >Please help...
> >G Ren
