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Re: st: power or sample size by survival vs. comparison of proportions
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From |
"Rosy Reynolds" <[email protected]> |
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To |
<[email protected]> |
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Subject |
Re: st: power or sample size by survival vs. comparison of proportions |
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Date |
Sun, 22 Jun 2008 09:41:31 +0100 |
Yulia,
Thank you so much for the detailed explanation. That's just what I needed.
Rosy
----- Original Message -----
From: "Yulia Marchenko, StataCorp LP" <[email protected]>
To: "statalist" <[email protected]>
Sent: Sunday, June 22, 2008 1:29 AM
Subject: Re: st: power or sample size by survival vs. comparison of
proportions
"Rosy Reynolds" <[email protected]> asks about differences in estimated
sample sizes obtained using the binomial distribution (-sampsi-) and Cox
proportional hazards model (-stpower cox-) for survival data:
Suppose there are two groups of equal size in a (proportional hazards)
survival study. I follow them up for such a time that overall 50% of
participants die, and I am looking for a hazard ratio of 1.5. By the
end of follow-up, therefore, 60% would die in the higher risk group
and 40% in the lower risk group.
If I was going to analyse simply by comparing the proportions who died
in the two groups, I could estimate the number needed for 80% power,
5% significance, with
. sampsi 0.6 0.4, power(0.8) alpha(0.05)
If I was going to analyse by using Cox regression, I could estimate
the number with
. stpower cox, hratio(1.5) power(0.8) alpha(0.05) failprob(0.5)
-sampsi- estimates that I need 214 participants (107 in each group),
while -stpower- estimates a need for 382 participants to observe
191 deaths.
...
Rosy used p1 = 0.6 and p2 = 0.4 as the group-specific proportions of
participants who are expected to die by the end of the study in
the -sampsi-
command. However, I believe that Rosy obtained these values using the
formula
for a hazard ratio based on hazard rates (hr = h1/h2, 1.5 = 0.6/0.4)
rather
than based on proportions surviving by the end of the study
(hr = ln(1-p1)/ln(1-p2)). This produced unexpected results from
the -sampsi-
and -stpower- commands.
Rosy either has a hazard ratio of 1.5 (in which case the proportions are
not
0.6 and 0.4), or she has proportions of 0.6 and 0.4 (in which case the
hazard
ratio is not 1.5).
In the first case, assuming a hazard ratio of 1.5 and an average death
rate of
50%, Rosy would need to solve a nonlinear equation (1-p1) = (1-p2)^1.5
subject
to the constraint that (p1+p2)/2 = 0.5 to obtain the correct p1 = 0.57 and
p2
= 0.43 for use in the -sampsi- command. Using these values, the required
total sample size is 428 (214 per group) which is comparable to the sample
size of 382 Rosy obtained from -stpower cox-.
. sampsi 0.57 0.43, p(0.8)
Estimated sample size for two-sample comparison of proportions
Test Ho: p1 = p2, where p1 is the proportion in population 1
and p2 is the proportion in population 2
Assumptions:
alpha = 0.0500 (two-sided)
power = 0.8000
p1 = 0.5700
p2 = 0.4300
n2/n1 = 1.00
Estimated required sample sizes:
n1 = 214
n2 = 214
In the second case, Rosy can use a hazard ratio of 1.79 =
ln(1-0.6)/ln(1-0.4)
with -stpower cox-. For example, if we specify -hratio(1.79)- instead of
-hratio(1.5)- with -stpower cox-, we obtain a required sample size of 186
which
is comparable to 214 obtained by Rosy from -sampsi-.
. stpower cox, hratio(1.79) power(0.8) alpha(0.05) failprob(0.5)
Estimated sample size for Cox PH regression
Wald test, log-hazard metric
Ho: [b1, b2, ..., bp] = [0, b2, ..., bp]
Input parameters:
alpha = 0.0500 (two sided)
b1 = 0.5822
sd = 0.5000
power = 0.8000
Pr(event) = 0.5000
Estimated number of events and sample size:
E = 93
N = 186
-- Yulia
[email protected]
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