As has been answered several times on rhelp ... the baseline hazard is
for a case with the "mean value". It's not a meaningful case with all
factor variables. There can be no cases where fidelity>3 has a
fractional value. You should be using predict() and survfit() to
display estimates for particular meaningful cases. You might want to
use a different rhs formula as well because the one offered omits the
main effects. Choosing ~bucket*factor(fidelity > 3) would include
those terms.
--
David.
On May 1, 2012, at 8:37 PM, Georges Dupret wrote:
Hi,
In the following results I interpret exp(coef) as the factor that
multiplies
the base hazard rate if the corresponding variable is TRUE. For
example,
when the bucket is ks008 and fidelity <= 3, then the rate, compared
to the
base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to
what case
does the base hazard rate correspond to? I would expect the
reference to be
the first factor value, i.e. bucket jpc001 with fidelity <= 3, but its
exp(coef) is not one. I verified the contrasts, and the row
corresponding to
the first factor value is zero everywhere; moreover, I didn't change
the
default, so a bad setting of the contrasts doesn't seem to be the
issue.
Best,
ge
Call:
coxph(formula = Surv(time, event = (censored ==
FALSE)) ~ bucket:factor(fidelity > 3), data = week.15)
coef exp(coef) se(coef)
z p
bucketjpc001:factor(fidelity > 3)FALSE -1.606 0.201 0.00624
-257.5 0
bucketks006:factor(fidelity > 3)FALSE -1.613 0.199 0.00627
-257.5 0
bucketks007:factor(fidelity > 3)FALSE -1.620 0.198 0.00626
-258.8 0
bucketks008:factor(fidelity > 3)FALSE -1.611 0.200 0.00625
-257.6 0
bucketks009:factor(fidelity > 3)FALSE -1.620 0.198 0.00626
-258.9 0
bucketks010:factor(fidelity > 3)FALSE -1.619 0.198 0.00626
-258.6 0
bucketjpc001:factor(fidelity > 3)TRUE -0.156 0.856 0.00596
-26.2 0
bucketks006:factor(fidelity > 3)TRUE -0.171 0.842 0.00600
-28.6 0
bucketks007:factor(fidelity > 3)TRUE -0.168 0.845 0.00602
-28.0 0
bucketks008:factor(fidelity > 3)TRUE -0.167 0.846 0.00600
-27.8 0
bucketks009:factor(fidelity > 3)TRUE -0.170 0.844 0.00599
-28.4 0
bucketks010:factor(fidelity > 3)TRUE NA NA 0.00000
NA NA
Likelihood ratio test=294562 on 11 df, p=0 n= 1173838, number of
events=
629383
--
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