New Contributor
Posts: 2

Sample Size Calculation for Log-Rank Test: Technical details

Dear all,

I hope this could the right place to raise a question regarding the technical details for sample size determination using proc power for comparing two survival curves:
http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_power_a00000...

My question is to understand the calculation of N(i+1) specified as
N_{j}(i+1) = N_{j}(i) * [1-h_{j}(t_{i})*(1/b) - \phi_{j}(t_{i})*(1/b) - (1/b)*(1/(T-\tao-t_{i}))* I_{t_{i}>\tao})]

It looks as if the very last term 1/(T+\tao-t_{i}) is for administrative censoring (i.e., patients who enrolled late in the accrual phase and survival longer than \tao could be censored due to study closure). For example, suppose that subject live forever and will never be censored, (i.e., h(t)=0 and \phi(t)=0 for all t), suppose that N(0)=100, T=4, \tao=0 and b=1, then:
N(1) = 100*(1-1/3).

However, assuming subject entry is uniformed distributed between [0,4] then only subject enters from [3,4] could be censored and should the right answer for N(1) is
75 = 100*(1-1/4)?

I really appreciate the insights from you .

New Contributor
Posts: 2

Re: Sample Size Calculation for Log-Rank Test: Technical details

Problem solved based on Cantor 1997 pp 83-92 and Joanna H Shih 1995 (Sample Size Calculation for Complex Clinical Trials with Survival Endpoints).

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