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
calculate likelihood based on logit regression
2 messages · Haibo Huang
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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