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yoyong555
Obsidian | Level 7

Hi everyone.

 

The scenario is this:

 

A participant (par) needs to visit a center for three times to receive medication. The initial visit is the baseline. The second visit happens 6 months after and the last visit happens after a year from the first visit (twelvemonths). During each visit, the number of medications is counted (count). 

 

How do I determine if the number of medications  is significantly different across the three visits and participants?

 

Note: The initial visit date is different for each participant.

 

Thank you.

 

data setup;
input par $ 1-5 visit $ 8-19 count 21-22;
datalines;
par1   baseline     2
par1   sixmonths    0
par1   twelvemonths 2 
par2   baseline     0
par2   sixmonths    0
par2   twelvemonths 1 
par3   baseline     4
par3   sixmonths    2
par3   twelvemonths 5 
par5   baseline     4
par5   sixmonths    2
par5   twelvemonths 2 
par6   baseline     0
par6   sixmonths    1
par6   twelvemonths 1 
par7   baseline     7
par7   sixmonths    3
par7   twelvemonths 3 
par8   baseline     4
par8   sixmonths    5
par8   twelvemonths 5 
par9   baseline     14
par9   sixmonths    2
par9   twelvemonths 6 
par10  baseline     4
par10  sixmonths    4
par10  twelvemonths 8
;
run;
6 REPLIES 6
Jagadishkatam
Amethyst | Level 16

you can try the below code and we get the below results

 

The initial table in this listing is the Analysis of Variance Table. The most important line to observe in this table is the “Model.” At the right of this line is the p-value for the overall ANOVA test. It is listed as “Pr > F” and is p = 0.2892. This tests the overall model to determine if there is a difference in means between visits. In this case, since the p-value is large, you can conclude that there no significance difference between visit (medications) ie. visits mean are equal. we fail to reject the null hypothesis.

 

proc anova data=setup;
class visit;
model count=visit;
run;

image.png

 

Thanks,
Jag
Ksharp
Super User

I think you should post it at Stat forum since it is about statistical question.

And try RPOC GLIMMIX + Possion or Tweedie distribution .

ed_sas_member
Meteorite | Level 14

Hi @yoyong555 

 

In my opinion, you should use a Friedman test, as you have paired measures.

It involves ranking each row (=block represented by the variable 'par') together, then considering the values of ranks by columns (= variable 'visit'.

 

In SAS, the Friedman test can be done with PROC FREQ. The CMH2 SCORES=RANK option gives the Friedman test.

The p-value is given in the "Row Mean Scores Differs" of the output.

 

proc freq data=setup;
	table visit*par*count / cmh2 scores=rank noprint;
run;
PaigeMiller
Diamond | Level 26

@yoyong555 you have not answered my questions from your earlier thread ... what distribution do you want to assume for this data?? It probably isn't normal, but what do you think is the right distribution?

--
Paige Miller
PaigeMiller
Diamond | Level 26

Okay, Poisson (although I'm a little skeptical, let's go with that)

 

If there were only two different levels of VISIT, this could be analyzed using the traditional "paired t-test" or whatever the Poisson equivalent is. You can obtain the equivalent of a "paired t-test" in a generalized linear model by placing the participant into the model as a class variable.

 

proc glimmix data=setup;
    class par visit;
    model count=par visit/dist=poisson;
    lsmeans visit/ilink lines;    
run;

 

--
Paige Miller

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