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reedwinter
Calcite | Level 5

Hi,

I have a question about the model statement for binomial distribution, one way to model it is:

model count/n=trt/dist=binomial;

While for binomial distribution count and n should be integers.

However, I noticed that event count and n are not integers (such as the average of counts or n over samples), the model can still work. Is the result still trustable? How does the model work in this case?

 

Thanks,

4 REPLIES 4
Ksharp
Super User
Can you post some data ?
From your code, I see you are doing Logistic Regression?
But you said it is average of count, it lead to more like ANOVA ?

PGStats
Opal | Level 21

Inference about binomial data requires the number of events and trials to be known. It also assumes that trials are independent. Your counts (number of events) and n (number of trials) should therefore be the total numbers, not averages.

 

It is also likely that a certain level of dependence occurs for units belonging to the same sample, i.e. that unidentified effects, other than trt, are at play for each sample. These should be modelled as random effects.

PG
reedwinter
Calcite | Level 5

Yes, there is random effect,  I did not list it just to simplify my quesiton.

 

I always used the total numbers as count and n, and I did not expect the non-integer count and n can work. But someone in my group showed this model and used the average for some special reason, and it worked without generating any error message. So I am just wondering why it could work and if the result is trustable or not.

 

FYI, I found this paper:http://support.sas.com/resources/papers/proceedings11/349-2011.pdf also used non-integer count on page 11.

 

SteveDenham
Jade | Level 19

The ratio of two random variables that is bounded on the set (0, 1) is more likely to be beta distributed than binomially distributed.  In this case, you have two means (events and counts), which results in a ratio of two RVs.  Whatever procedure you are using can use non-integers for a binomial, but whether you should is another matter altogether, as @Ksharp and @PGStats have pointed out.

 

Steve Denham

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