Bioequivalence and Bioavailability Forum • Sequential Sample Size and Power Estimate in R

xpresso
☆

United States,
2023-09-19 15:40
(360 d 03:40 ago)

Posting: # 23720
Views: 3,224

Sequential Sample Size and Power Estimate in R [Power / Sample Size]

Hi there! I am looking for insight into how to calculate power and sample size estimation for separate PK endpoints (Auclast, Aucinf, etc.). In particular, I am looking to randomly sample different sample sizes and estimate the power to pass for bioequivalence for each given sample size. For example, with 30 subjects (15 per arm) the power for this subset of samples to pass for bioequivalence would be ___ %.

I have already simulated the 1:1 parallel trial of 1,000 individuals and repeated the trial 10,000 times (denoted by the rep column below). I have been using the Rxode2 package in R to simulate, with wanting to calculate the power and sample size estimation based on their "Weight-based dosing example", seen here : https://nlmixrdevelopment.github.io/RxODE-manual/examples.html.

Below is what the dataset currently looks like:
[image]

Helmut
★★★

Vienna, Austria,
2023-09-19 16:04
(360 d 03:17 ago)

@ xpresso
Posting: # 23721
Views: 2,678

IUT

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Hi xpresso & welcome to the club!

I think that you are overdoing it.
If you have multiple PK metrics and all must pass (that’s the regulatory requirement) at level \(\small{\alpha}\) – in general 0.05 – you are protected against inflation of the type I error by the intersection-union principle.¹ No need for simulations,² simply calculate power for given T/R-ratio, CV, and the sample size. If any of the metrics shows power <50%, the study will fail (see this article).

Berger RL, Hsu JC. Bioequivalence Trials, Intersection-Union Tests and Equivalence Confidence Sets. Stat Sci. 1996; 11(4): 283–302. JSTOR:2246021.
Zeng A. The TOST confidence intervals and the coverage probabilities with R simulation. March 14, 2014. Online.

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

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xpresso
☆

United States,
2023-09-19 16:18
(360 d 03:02 ago)

@ Helmut
Posting: # 23722
Views: 2,659

IUT

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❝ Hi xpresso & welcome to the club!

Hi! Thank you for the warm welcome and response! I have been privately reading the materials in the forum over the past couple months and finally made an account.

❝ I think that you are overdoing it. […]

This is good to know! I have considered the intersection-union principle after reading previous posts in the forum. I was hoping to compute the predictive power for different numbers of subjects for each trial and was using bootstrapping to generate this value. Does this make more sense? Would you happen to have any resources/examples? I am new to this topic and working on this project out of pure interest (forgive me if this is not the place for these questions). Thank you for your help :-D

Edit: Full quote removed. Please delete everything from the text of the original poster which is not necessary in understanding your answer; see also this post #5! [Helmut]

Helmut
★★★

Vienna, Austria,
2023-09-19 17:46
(360 d 01:35 ago)

@ xpresso
Posting: # 23723
Views: 2,697

power is analytically accessible

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Hi xpresso

❝ […] I have considered the intersection-union principle after reading previous posts in the forum. I was hoping to compute the predictive power for different numbers of subjects for each trial and was using bootstrapping to generate this value. Does this make more sense? Would you happen to have any resources/examples?

Of course, you could do bootstrapping but I still think that it’s over the top. For given T/R-ratio, CV, n, and design power can be directly calculated (contrary to the sample size, where you need an iterative procedure). If your are interested in some combinations, see these examples for a parallel design and a 2×2×2 crossover.
In short, whatever you want, simply set up nested loops and assign the result to a data frame (or matrix if more than two dimensions).
As a single point metric C_max is in general the most variable one. Hence, assessing power for it should cover the other metrics as well. I like IUT.

❝ I am new to this topic and working on this project out of pure interest (forgive me if this is not the place for these questions).

That’s the best reason. It’s the right place.

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

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BEQool
★

2024-03-01 15:26
(196 d 02:54 ago)

(edited by on 2024-03-01 15:36)
@ Helmut
Posting: # 23887
Views: 1,606

Power<50%

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❝ If any of the metrics shows power <50%, the study will fail (see this article).

Hello, is the statement above always true?

For example, in a hypothetical (and very unrealistic :-D

) scenario for Cmax with PE=95%, CV=110%, 36 subjects in a 2x2x3 design we get the following 90% CI:

CI.BE(pe=0.95,n=36,design="2x2x3",CV=1.1)

   lower    upper 

0.701628 1.286294

Widened acceptance limits are (the maximum ones):

scABEL(CV=1.1)

    lower     upper 

0.6983678 1.4319102

So we pass the study although the power is <50%:

power.scABEL(theta0=0.95,n=36,design="2x2x3",CV=1.1)

  0.23718

Why do we pass the study if the power is below 50%?

BEQool

Helmut
★★★

Vienna, Austria,
2024-03-01 15:44
(196 d 02:36 ago)

@ BEQool
Posting: # 23888
Views: 1,634

Power<50%

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Hi BEQool,

❝ Why do we pass the study if the power is below 50%?

Let’s dive deeper into you example:

library(PowerTOST) n <- sampleN.scABEL(CV = 1.1, theta0 = 0.95, design = "2x2x3", details = FALSE, print = FALSE)[["Sample size"]] n <- seq(n, 36, -2) res <- data.frame(n = n, lower = NA_real_, upper = NA_real_, power = NA_real_) for (j in seq_along(n)) { res[j, 2:3] <- CI.BE(CV = 1.1, pe = 0.95, n = n[j], design = "2x2x3") res[j, 4] <- power.scABEL(CV = 1.1, theta0 = 0.95, n = n[j], design = "2x2x3") } print(signif(res, 4), row.names = FALSE) n lower upper power 88 0.7838 1.151 0.8080 86 0.7821 1.154 0.7974 84 0.7803 1.157 0.7866 82 0.7784 1.159 0.7750 80 0.7764 1.162 0.7626 78 0.7744 1.165 0.7503 76 0.7723 1.169 0.7360 74 0.7701 1.172 0.7226 72 0.7679 1.175 0.7070 70 0.7655 1.179 0.6911 68 0.7631 1.183 0.6743 66 0.7606 1.187 0.6567 64 0.7579 1.191 0.6373 62 0.7551 1.195 0.6180 60 0.7522 1.200 0.5972 58 0.7492 1.205 0.5745 56 0.7460 1.210 0.5504 54 0.7426 1.215 0.5267 52 0.7391 1.221 0.5008 50 0.7353 1.227 0.4733 48 0.7314 1.234 0.4453 46 0.7272 1.241 0.4134 44 0.7227 1.249 0.3816 42 0.7180 1.257 0.3490 40 0.7129 1.266 0.3122 38 0.7075 1.276 0.2751 36 0.7016 1.286 0.2372

Strange. I have to think about it… :ponder:

CI.BE() gives us the realized results (i.e., the values observed in a particular study), whereas power.scABEL() the results of 10⁵ simulations. Both the CV and PE are skewed to the right and, therefore, I would expect that simulated power is lower than assessing whether a particular study passes. However, such a large discrepancy is surprising for me.

—
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz

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Science Quotes

BEQool
★

2024-03-02 00:02
(195 d 18:18 ago)

(edited by on 2024-03-02 08:47)
@ Helmut
Posting: # 23889
Views: 1,668

Power<50%

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❝ CI.BE() gives us the realized results (i.e., the values observed in a particular study), whereas power.scABEL() the results of 10⁵ simulations. Both the CV and PE are skewed to the right and, therefore, I would expect that simulated power is lower than assessing whether a particular study passes. However, such a large discrepancy is surprising for me.

What about function power.TOST()? I think that it doesnt give us the result based on simulations? Am I right? And this calculated power is also way below 50% (similar to the power calculated above):

power.TOST(theta0=0.95,n=36,design="2x2x3",CV=1.1, theta1=0.6983678, theta2=1.4319102)

 0.2355959

I can find more such examples :-)

All are as expected close to the limit:
a)

CI.BE(pe=0.95,n=36,design="2x2x3",CV=0.5)

    lower     upper 

0.8089197 1.1156855 



power.TOST(theta0=0.95,n=36,design="2x2x3",CV=0.5)

 0.4273464

CI.BE(pe=0.95,n=36,design="2x2",CV=0.44)

    lower     upper 

0.8033552 1.1234134 



power.TOST(theta0=0.95,n=36,design="2x2",CV=0.44)

0.3786147

CI.BE(pe=0.94,n=18,design="2x2",CV=0.28)

    lower     upper 

0.8011079 1.1029725 



power.TOST(theta0=0.94,n=18,design="2x2",CV=0.28)

0.4259032

...

❝ If any of the metrics shows power <50%, the study will fail (see this article).

Is it possible that the opposite is always true and this isnt?
So that if the study fails, the power will always be <50%; but if the power is <50%, it doesnt always mean that the study failed (will fail)?