Michael (OP),
Not that it's necessary, but you might also find
confint(aa)
to be instructive.
Peter Ehlers
On 2012-02-29 07:20, R. Michael Weylandt wrote:
Formally, look at Pr(>|z|). Informally, look at the null and residual
deviances from print(aa).
Michael
On Wed, Feb 29, 2012 at 10:14 AM, Michael<comtech....@gmail.com> wrote:
How did you see it's non-significant?
Thanks!
On Wed, Feb 29, 2012 at 9:10 AM, Sarah Goslee<sarah.gos...@gmail.com>wrote:
On Wed, Feb 29, 2012 at 10:02 AM, Michael<comtech....@gmail.com> wrote:
Hi all,
As you can see from below, the result is strange...
Not really.
I would imagined that the bb result should be much higher and close to 1,
any way to improve the fit?
Any other classification methods?
Thank you!
data=data.frame(y=rep(c(0, 1), times=100), x=1:200)
aa=glm(y~x, data=data, family=binomial(link="logit"))
newdata=data.frame(x=6, y=100)
bb=predict(aa, newdata=newdata, type="response")
bb
bb
1
0.4929125
What did you expect? Your model is completely nonsignificant; there's no
way to predict y from x, and that's what your predicted value tells you.
summary(aa)
Call:
glm(formula = y ~ x, family = binomial(link = "logit"), data = data)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.190 -1.177 0.000 1.177 1.190
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.030152 0.283924 -0.106 0.915
x 0.000300 0.002450 0.122 0.903
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 277.26 on 199 degrees of freedom
Residual deviance: 277.24 on 198 degrees of freedom
AIC: 281.24
Number of Fisher Scoring iterations: 3
I can only assume that you didn't construct the data frame that
you intended to test.
--
Sarah Goslee
http://www.functionaldiversity.org
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