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From |
"Rosy Reynolds" <rr@dandr.demon.co.uk> |

To |
<statalist@hsphsun2.harvard.edu> |

Subject |
Re: st: power or sample size by survival vs. comparison of proportions |

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" <ymarchenko@stata.com>

To: "statalist" <statalist@hsphsun2.harvard.edu>

Sent: Sunday, June 22, 2008 1:29 AM

Subject: Re: st: power or sample size by survival vs. comparison of proportions

"Rosy Reynolds" <rr@dandr.demon.co.uk> 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.Rosy used p1 = 0.6 and p2 = 0.4 as the group-specific proportions ofIf 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. ...

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

ymarchenko@stata.com

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**References**:**Re: st: power or sample size by survival vs. comparison of proportions***From:*ymarchenko@stata.com (Yulia Marchenko, StataCorp LP)

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