[R] Need help interpreting the logit regression function

[cberry at tajo.ucsd.edu]

Rohit Pandey <rohitpandey576 at gmail.com> writes:

> Hello R community,
>
> I have a question about the logistic regression function.
> Specifically, when the predictor variable has not just 0's and 1's,
> but also fractional values (between zero and one). I get a warning
> when I use the "glm(formula = ... , family = binomial(link =
> "logit"))" which says:
>
> "In eval(expr, envir, enclos) : non-integer #successes in a binomial glm!"
>
> I looked this up on a few forums and got the impression that I could
> go ahead and ignore this warning. If that is not so, please suggest
> another function that can be used (and no need to read further).
>
> But if this is usable, then the numbers are not making sense for me.
>
> What I am seeing is best explained through an example:
>
> #Setting up the data:
>>c(1,1,0,0,1,1,0,1,1,0,0,0,0,1,1,0,0,0,0,1)->resp
>>indep<-rep(2:3,10)
> #Running the logistic regression:
>>glm(formula = resp~as.factor(indep) , family = binomial(link = "logit"),na.action = na.pass)
> #Gives me:
> Coefficients:
>       (Intercept)  as.factor(indep)3
>        -1.692e-16         -4.055e-01
>>table(indep,resp)
>      resp
> indep 0 1
>     2   5 5
>     3   6 4
>
> Meaning that the odds for '2' to have a favorable outcome are 5/5 =1;
> and the odds for 3 are 4/6 = 0.6666
>
> And indeed, exp( -1.692e-16) = 1 and exp( -1.692e-16+-4.055e-01) = 0.66.
>
> But when I change all the 1's of the independent variable to 0.5, the
> odds should both be half of what they were before (right?).

No. Reducing the number of successes (but not the number of trials) by
1/2 reduces the relative frequencies by 1/2. And the fitted
probabilities reflect that.

Try 

    predict( glm( ... ), type = "response" )


under each setup to see.

HTH,

Chuck
>
> However, when I run the logistic function after doing this,
>
>>resp[resp==1]=0.5
>>glm(formula = resp~as.factor(indep) , family = binomial(link = "logit"),na.action = na.pass)
> (Intercept)  as.factor(indep)3
>           -1.0986            -0.2877
>
> Now, exp(-1.0986) = 0.3333 and exp(-1.0986-0.2877) = 0.25.
>
> These are nothing like the previous odds halved. So, either I am using
> the function wrong or interpreting it wrong. Can some one please point
> me in the right direction..

-- 
Charles C. Berry                            Dept of Family/Preventive Medicine
cberry at ucsd edu			    UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901