R help - Aug 2005 - calculate likelihood based on logit regression

Hi,

I just ran the following logit regression. But can
anyone tell me how to calculate how much more likely
males (Male=1) could show such symptom than
females(Male=0)? I know it must be simple to get once
I have the coefficients, but I just don't recall.
Thank you very much!

Call:
glm(formula = Symptoms ~ 1 + Male, family binomial(link = logit), 
    data = HA)
Deviance Residuals: 
       Min          1Q      Median          3Q        
Max  
-6.038e+03  -2.067e-06   0.000e+00   0.000e+00  
6.720e+03  
Coefficients:
                      Estimate Std. Error    z value
Pr(>|z|)    
(Intercept)          5.711e+00  8.466e-02  6.746e+01  
<2e-16 ***
Male        -6.493e+00  8.540e-02 -7.603e+01   <2e-16
***


Best,
Ed

I think males are z times more likely to show such
symptom, where:

z=(1+exp(6.493-5.711))/(1+exp(-5.711))

Is this right? Thanks!
Ed.

--- Haibo Huang <edhuang00 at yahoo.com> wrote:
> Hi,
> 
> I just ran the following logit regression. But can
> anyone tell me how to calculate how much more likely
> males (Male=1) could show such symptom than
> females(Male=0)? I know it must be simple to get
> once
> I have the coefficients, but I just don't recall.
> Thank you very much!
> 
> Call:
> glm(formula = Symptoms ~ 1 + Male, family > binomial(link = logit), 
>     data = HA)
> Deviance Residuals: 
>        Min          1Q      Median          3Q      
>  
> Max  
> -6.038e+03  -2.067e-06   0.000e+00   0.000e+00  
> 6.720e+03  
> Coefficients:
>                       Estimate Std. Error    z value
> Pr(>|z|)    
> (Intercept)          5.711e+00  8.466e-02  6.746e+01
>  
> <2e-16 ***
> Male        -6.493e+00  8.540e-02 -7.603e+01  
> <2e-16
> ***
> 
> 
> Best,
> Ed
> 
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