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: 4,031

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
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