Impact of minimum stage 2 sample size on the TIE: example [Two-Stage / GS Designs]

posted by Helmut Homepage – Vienna, Austria, 2016-12-30 14:01  – Posting: # 16911
Views: 4,149

Hi ElMaestro,

» So let me ask the forbidden question: "Can you reformulate?"
»
» » Higher sample size ⇒ more degrees of freedom ⇒ narrower CI ⇒ higher probability to pass BE.
» » In other words, the TIE will also increase and one would have to use a lower adjusted α.
»
» This is one thing I did not get.

I’ll give two examples. Both at the location (n1 12, CV 20%) of the maximum TIE.
Simulating for power (at 0.95):
  1. No lower limit of n2
    library(Power2Stage)
    power.2stage(method="B", alpha=rep(0.0294, 2), CV=0.2,
                 n1=12, GMR=0.95, targetpower=0.8, min.n2=0)

    TSD with 2x2 crossover
    Method B: alpha (s1/s2) = 0.0294 0.0294
    Target power in power monitoring and sample size est. = 0.8
    Power calculation via non-central t approx.
    CV1 and GMR = 0.95 in sample size est. used
    No futility criterion
    BE acceptance range = 0.8 ... 1.25

    CV = 0.2; n(stage 1) = 12; GMR= 0.95

    1e+05 sims at theta0 = 0.95 (p(BE)='power').
    p(BE)    = 0.84174
    p(BE) s1 = 0.41333
    Studies in stage 2 = 56.34%

    Distribution of n(total)
    - mean (range) = 20.6 (12 ... 82)
    - percentiles
     5% 50% 95%
     12  18  40


  2. Lower limit of n2 = 1.5 × n1
    power.2stage(method="B", alpha=rep(0.0294, 2), CV=0.2,
                 n1=12, GMR=0.95, targetpower=0.8, min.n2=18)

    TSD with 2x2 crossover
    Method B: alpha (s1/s2) = 0.0294 0.0294
    Target power in power monitoring and sample size est. = 0.8
    Power calculation via non-central t approx.
    CV1 and GMR = 0.95 in sample size est. used
    No futility criterion
    Minimum sample size in stage 2 = 18
    BE acceptance range = 0.8 ... 1.25

    CV = 0.2; n(stage 1) = 12; GMR= 0.95

    1e+05 sims at theta0 = 0.95 (p(BE)='power').
    p(BE)    = 0.91564
    p(BE) s1 = 0.41333
    Studies in stage 2 = 56.34%

    Distribution of n(total)
    - mean (range) = 23.5 (12 ... 82)
    - percentiles
     5% 50% 95%
     12  30  40
If we require that the minimum sample size in the second stage = 1.5 × n1, naturally the same percent of studies will proceed to the second stage. However, the expected total sample sizes will be larger (E[N] 23.5 vs. 20.6, median 30 vs. 18). The sponsor gains power (91.6% vs. 84.2%).

» Does that logic also work when we simulate true GMR 0.8 or 1.25 for type I error? I find it hard to convince myself.

Yes, it does – and this was my point. This time simulating for the TIE (at 1.25):
  1. No lower limit of n2
    library(Power2Stage)
    power.2stage(method="B", alpha=rep(0.0294, 2), CV=0.2,
                 n1=12, GMR=0.95, targetpower=0.8, min.n2=0, theta0=1.25)

    TSD with 2x2 crossover
    Method B: alpha (s1/s2) = 0.0294 0.0294
    Target power in power monitoring and sample size est. = 0.8
    Power calculation via non-central t approx.
    CV1 and GMR = 0.95 in sample size est. used
    No futility criterion
    BE acceptance range = 0.8 ... 1.25

    CV = 0.2; n(stage 1) = 12; GMR= 0.95

    1e+06 sims at theta0 = 1.25 (p(BE)='alpha').
    p(BE)    = 0.046262
    p(BE) s1 = 0.028849
    Studies in stage 2 = 87.86%

    Distribution of n(total)
    - mean (range) = 23.1 (12 ... 98)
    - percentiles
     5% 50% 95%
     12  22  40


  2. Lower limit of n2 = 1.5 × n1
    power.2stage(method="B", alpha=rep(0.0294, 2), CV=0.2,
                 n1=12, GMR=0.95, targetpower=0.8, min.n2=18, theta0=1.25)

    TSD with 2x2 crossover
    Method B: alpha (s1/s2) = 0.0294 0.0294
    Target power in power monitoring and sample size est. = 0.8
    Power calculation via non-central t approx.
    CV1 and GMR = 0.95 in sample size est. used
    No futility criterion
    Minimum sample size in stage 2 = 18
    BE acceptance range = 0.8 ... 1.25

    CV = 0.2; n(stage 1) = 12; GMR= 0.95

    1e+06 sims at theta0 = 1.25 (p(BE)='alpha').
    p(BE)    = 0.048816
    p(BE) s1 = 0.028849
    Studies in stage 2 = 87.86%

    Distribution of n(total)
    - mean (range) = 29.2 (12 ... 98)
    - percentiles
     5% 50% 95%
     12  30  40
The Type I Error increases from 0.046262 (no minimum n2) to 0.048816 (minimum n2 = 1.5 × n1). No problem with ‘Method B’ but gets nasty with ‘Method C’ (see the plot in my OP).

» Somehow I guess regulators just wanted to say that inclusion of a single subject in stage 2 would not be ok. They are right and that is not rocket science.

I think not to perform the second stage with one subject is a no-brainer. I guess that two was a compromise. AFAIK, Alfredo suggested 12 subjects to the BSWP.*



Cheers,
Helmut Schütz
[image]

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes

Complete thread:

Activity
 Admin contact
20,343 posts in 4,272 threads, 1,401 registered users;
online 9 (2 registered, 7 guests [including 5 identified bots]).
Forum time (Europe/Vienna): 04:15 CET

That which is static and repetitive is boring.
That which is dynamic and random is confusing.
In between lies art.    John Locke

The Bioequivalence and Bioavailability Forum is hosted by
BEBAC Ing. Helmut Schütz
HTML5