Emrah Soner Özdeş
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Turkey,
2014-12-24 08:38
(3768 d 16:31 ago)

Posting: # 14157
Views: 8,846
 

 Interim Analysis and Power [Two-Stage / GS Designs]

Dear All;

I recently performed an interim analysis of a two-stage study with Potvin Method C and here are the initial results (power evaluated with alpha level of 0.05)

LN AUCT: 77.81 - 96.71 Power: 91.55%
LN CMAX: 77.24 - 100.97 Power: 73.51

Given below the flowchart of Potvin method C, I believe this is a fail case where second stage is not required since power of only end-point is below 80%

Do you think is this the right way interpreting the results ?

Any help will be appreciated.


Thanks

C. Evaluate the power at stage 1 using the variance estimate from stage 1 and an a level of 0.05. If the power is greater than or equal to 80%, evaluate BE at stage 1 using an a level of 0.05 and stop whether BE is met or not. If the power is less than 80%, evaluate BE using an a of 0.0294. If the BE criterion is met, stop. If the BE criterion is not met, calculate the sample size based on the variance estimated at stage 1 and an a level of 0.0294 and continue to stage 2. Evaluate BE at stage 2 using data from both stages and an a level of 0.0294. Stop here whether BE is met or not and regardless of the power achieved.
Helmut
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Vienna, Austria,
2014-12-24 17:22
(3768 d 07:46 ago)

@ Emrah Soner Özdeş
Posting: # 14166
Views: 7,480
 

 Interim Analysis and Power

Merhaba Emrah,

❝ Do you think is this the right way interpreting the results ?


Maybe. Maybe not. We need more information: Can you please give us the sample size of the first stage?

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Emrah Soner Özdeş
☆    

Turkey,
2014-12-25 09:36
(3767 d 15:33 ago)

@ Helmut
Posting: # 14172
Views: 7,504
 

 Interim Analysis and Power

❝ Maybe. Maybe not. We need more information: Can you please give us the sample size of the first stage?


Hi Helmut;

First stage was performed with 24 subjects.
ElMaestro
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Denmark,
2014-12-25 13:32
(3767 d 11:37 ago)

@ Emrah Soner Özdeş
Posting: # 14173
Views: 7,468
 

 Interim Analysis and Power

Hi Emrah,

❝ I recently performed an interim analysis of a two-stage study with Potvin Method C and here are the initial results (power evaluated with alpha level of 0.05)


❝ LN AUCT: 77.81 - 96.71 Power: 91.55%

❝ LN CMAX: 77.24 - 100.97 Power: 73.51


❝ Given below the flowchart of Potvin method C, I believe this is a fail case where second stage is not required since power of only end-point is below 80%


❝ Do you think is this the right way interpreting the results ?


I assume you wrote in your protocol that your way through the decision tree would be dictated by Cmax and not AUCt? You'll be going into stage 2 before declaring success or failure.

Please help me understand the details. I get on basis of Cmax:
PE=88.31% and CV=27.5% with N1=24, and assuming balance you'll have a power of about 65% (alpha 5%, GMR 95% as per Potvin and not 88.31%). Where did I go wrong here?

Pass or fail!
ElMaestro
Emrah Soner Özdeş
☆    

Turkey,
2014-12-26 09:55
(3766 d 15:14 ago)

@ ElMaestro
Posting: # 14178
Views: 7,426
 

 Interim Analysis and Power

Herr Schütz;

❝ Please help me understand the details. I get on basis of Cmax:

❝ PE=88.31% and CV=27.5% with N1=24, and assuming balance you'll have a power of about 65% (alpha 5%, GMR 95% as per Potvin and not 88.31%). Where did I go wrong here?


It seems that I need to check my calculations for power value since point estimator and CV values are same with the one you calculated. Could you advice a reliable formula to check my calculations ?

❝ I assume you wrote in your protocol that your way through the decision tree would be dictated by Cmax and not AUCt? You'll be going into stage 2 before declaring success or failure.


No it will be dictated by both Cmax and AUCt. Assuming that both endpoints are outside the 90% CI and power is below 80% for only one endpoint ? What would be the next step ?

Thanks in advance

Emrah
Helmut
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Vienna, Austria,
2014-12-26 18:51
(3766 d 06:18 ago)

@ Emrah Soner Özdeş
Posting: # 14179
Views: 7,531
 

 R-code & some thoughts

Hi Emrah,

❝ Herr Schütz;


First: You replied to ElMaestro and second: I’m fine to be called with my prename Helmut. ;-)

❝ Could you advice a reliable formula to check my calculations ?


Software for Power/sample size: Get R and the package PowerTOST.
I guess you used a ratio of 1 (should be 0.95 for Potvin’s & Montague’s methods) and the shifted (central) t-distribution. I got almost your results:

library(PowerTOST)
n <- 24
cat(sprintf("%.2f%%", 100*power.TOST(CV=CI2CV(lower=0.7781, upper=0.9671, n=n),
  theta0=1, n=n, method="shifted")), "\n")
91.56%
cat(sprintf("%.2f%%", 100*power.TOST(CV=CI2CV(lower=0.7724, upper=1.0097, n=n),
  theta0=1, n=n, method="shifted")), "\n")
73.48%

It’s better to use the exact method or at least the approximation by the noncentral t-dis­tri­bu­tion.

❝ ❝ I assume you wrote in your protocol that your way through the decision tree would be dictated by Cmax and not AUCt? You'll be going into stage 2 before declaring success or failure.


❝ No it will be dictated by both Cmax and AUCt. Assuming that both endpoints are outside the 90% CI and power is below 80% for only one endpoint ? What would be the next step ?


This is what I write in my protocols as well. The procedure for your study (note that you should not necessarily calculate the 90% CIs), Method C: I generally start with Cmax, which likely is higher variable:
  • Cmax
    PE 88.31%, CV 27.52%, power for T/R 0.95 is 64.87%. Since <80% you try the 94.12% CI (not 90%!), which is 75.60–103.17%. You fail BE, but due to the low power you may initiate a second stage. The total sample size (T/R 0.95, CV 27.52%, power 80%, α 0.0294) would be 40. You could perform the second stage in 16 subjects.
  • AUCt
    PE 86.75%, CV 27.20%, power for T/R 0.95 is 83.22%. Since ≥80% you calculate the (unadjusted) 90% CI, which is 77.81–96.71. Study failed in the first stage, end of story.
R-code:

#####################################
# Interim analysis, Potvin Method C #
#####################################
library(PowerTOST)
PK   <- c("Cmax", "AUCt")
n1   <- c(24, 24)
PE   <- c(0.883115, 0.867468)
CV   <- c(0.275241, 0.221999)
TR   <- 0.95
al   <- c(0.05, 0.0294)
CIad <- sprintf("%.2f%%", 100*(1-(2*al[2])))
fail <- 0
n2   <- vector()
for(j in seq_along(PK)) {
  Pw <- power.TOST(alpha=al[1], CV=CV[j], theta0=TR, n=n1[j])
  cat("\nInterim analysis of", PK[j], "\n————————————————————————",
      "\nPower:", paste0(round(100*Pw, 2), "%\n"))
  if(Pw >= 0.8) {
    CI <- CI.BE(alpha=al[1], pe=PE[j], CV=CV[j], n=n1[j])
    if(CI["lower"] < 0.8 | CI["upper"] > 1.25) {
      cat("First stage 90% CI (unadjusted, since power \u226580%):",
        sprintf("%.2f \u2013 %.2f%%", 100*CI[1], 100*CI[2]),
        "\nFailed BE; stop the study.\n")
      fail <- fail + 1
    } else {
    cat("First stage 90% CI (unadjusted, since power \u226580%):",
        sprintf("%.2f \u2013 %.2f%%", 100*CI[1], 100*CI[2]),
        "\nPassed BE; stop the study.\n")
    }
  } else {
    CI <- CI.BE(alpha=al[2], pe=PE[j], CV=CV[j], n=n1[j])
    if(CI["lower"] < 0.8 | CI["upper"] > 1.25) {
      cat("First stage", CIad, "CI (adjusted, since power <80%):",
          sprintf("%.2f \u2013 %.2f%%", 100*CI[1], 100*CI[2]),
          "\nFailed BE; initiate the second stage.\n")
      n <- sampleN.TOST(alpha=al[2], CV=CV[j], theta0=TR,
                        print=FALSE)["Sample size"]
      n2[j] <- as.numeric(n-n1[j])
      cat("Second stage sample size:", n2[j], "subjects.\n")
    } else {
      cat("First stage", CIad, ", CI (power <80%):\n",
          sprintf("%.2f \u2013 %.2f%%", 100*CI[1], 100*CI[2]),
          "\nPassed BE; stop the study.\n")
    }
  }
  if(j == length(PK)) {
    if(fail == 0) {
      cat("\nInitiate the second stage with", max(n2), "subjects.\n\n")
    } else {
      cat("\nSince", fail,
          "PK metric(s) failed in the first stage with \u226580% power,",
          "stop the study.\n")
    }
  }
}


Gives:

Interim analysis of Cmax
————————————————————————
Power: 64.87%
First stage 94.12% CI (adjusted, since power <80%): 75.60 – 103.17%
Failed BE; initiate the second stage.
Second stage sample size: 16 subjects.

Interim analysis of AUCt
————————————————————————
Power: 83.22%
First stage 90% CI (unadjusted, since power ≥80%): 77.81 – 96.71%
Failed BE; stop the study.

Since 1 PK metric(s) failed in the first stage with ≥80% power, stop the study.


I think it is worthwhile to look at the PEs, which were <90%. Did you have some clues which let you assuming them to be 95%? If yes, OK. Shit happens. A conventional fixed-sample design – in up to 60 (!) subjects – would have failed as well.
Let’s play the devil’s advocate: You had some hints that the T/R will be 90%. Then you should better have used Montague’s Method D (α 0.0280) – instead of being overly optimistic:
  • Cmax
    Power for T/R 0.90 is 42.05%. Since <80% you try the 94.40% CI, which is 75.45–103.36%. You fail. The total sample size (T/R 0.90, CV 27.52%, power 80%, α 0.0280) would be 82. You could perform the second stage in 58 subjects.
  • AUCt
    Power for T/R 0.90 is 56.23%. Since <80% you try the 94.40% CI, which is 76.34–98.57%. You fail. The total sample size (T/R 0.90, CV 22.20%, power 80%, α 0.0280) would be 56. You could perform the second stage in 32 subjects.
Whether it makes sense to run a study in 82 subjects is yet another story… If you decide to repeat the study (since you failed according to the protocol) I would try to tweak the test formulation or find a closer reference.

No Two-Stage Method is published yet which would allow to stop a study if the PE is outside an acceptable range. Charles Bon suggested at the AAPS Annual Meeting 2007 a futility criterion of <85.00 and >117.65%. Before you apply such a method, you would have to find and validate a suitable αadj. Since your PEs were pretty low there would be some chances that the study stops in the first stage (contrary to Method D which would lead to high sample sizes). Try the function power.2stage.fC() in the package Power2Stage.

Dif-tor heh smusma 🖖🏼 Довге життя Україна! [image]
Helmut Schütz
[image]

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

Turkey,
2014-12-29 08:49
(3763 d 16:20 ago)

@ Helmut
Posting: # 14195
Views: 7,104
 

 R-code & some thoughts

Dear Helmut;

Thanks for this satisfying answer !!!. It is kind of a lecture on interim analysis and two-stage desgins. It really helped a lot.

Regards
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