[R] Survival Analysis with an Historical Control
Paul Miller
pjmiller_57 at yahoo.com
Tue Jul 22 19:06:51 CEST 2014
Hi Chris,
Thanks for your reply. Certainly no need to apologize for a delayed response. Appreciate your taking the time to answer my question.
My concern was about the value of "16". My understanding is it is based on one expert's opinion. I wondered if this is in keeping with whatever the standard practice might be. I feel uncomfortable with the idea of using this as the basis but perhaps I am alone in feeling this way. My preference would be to see citations from previous studies to justify the number. A meta analysis of results from previous studies might be even better.
Your code is very interesting. Especially the parametric part. Nice to get an actual p-value for the test.
Thanks,
Paul
--------------------------------------------
On Mon, 7/21/14, Andrews, Chris <chrisaa at med.umich.edu> wrote:
Subject: RE: [R] Survival Analysis with an Historical Control
Cc: "r-help at r-project.org" <r-help at r-project.org>
Received: Monday, July 21, 2014, 10:33 AM
Hi Paul,
Sorry for the delayed reply. I was away last
week. I'm not clear what you want confirmed about your
approach.
(a)
"20.57"- computing the rejection region of the
analysis. The formulas implemented at the addresses you
gave in your original post are from a reputable source -
Lawless (1982) - at least that is claimed in the help file
(http://www.swogstat.org/stat/Public/Help/survival1.html).
It seems that another person derived 20.57 from some
combination of input. I don't see how to back
calculate what the input was from the information
provided. Perhaps elsewhere in your study protocol there
is discussion of accrual time, et al. that can help you.
(b) "16" - the value
of the parameter in the null hypothesis is very important as
you noted in your response to Terry. But this is not
really a statistical question. It may be derived from
historical data, expert opinion, regulatory mandate, or some
combination of these and other factors. Presumably this
study was undertaken because 16 was an important number to
somebody.
(c) performing
the one-sample survival analysis itself. This is what I
did with the data you provided.
# Non-parametric
(km <-
survfit(Surv(WKS, 1-CENS) ~ 1, data=hsv,
type="kaplan-meier", conf.type="log",
conf.int=0.9))
summary(km)
# Compare to median survival =
16
# (Used 90% CI above to get 0.05 one
sided test here)
quantile(km,
prob=0.5)$lower > 16
#
Parametric
(paraExp <- survreg(Surv(WKS,
1-CENS) ~ 1, data=hsv, dist="exponential"))
summary(paraExp)
# Compare to median survival = 16
# That is, compare to beta0 = log(16/log(2)) =
3.1391...
# one sided test
pnorm(c((coef(paraExp) - log(16/log(2))) /
sqrt(vcov(paraExp))), lower.tail=FALSE)
Chris
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