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Back to the forum  Query: 2018-05-21 13:11 CEST (UTC+2h)
 
Louis52
Junior

2018-02-01 19:57

Posting: # 18328
Views: 1,415
 

 Finding the CI limits [Power / Sample Size]

Hello,

I'm new at this subject matter. I'm trying to learn more about BEQ studies, especially about how to get the CV(intra) when little or no (or not reliable) prior info available. I went over the "Biostatistics Sample Size Estimation for BE Studies" presentation (thank you) and at one point in one of the "Hints" slide it says that "All you need is the 90% geometric confidence interval and the sample size." My question is if I don't have no prior pilot study from which to get a CV intra, how would I use all this information? Am I understanding correctly that the CLlow and CLhigh would be the 90% confidence limits for a prior CV intra, or they refer to something else? I'm asking that because of the "...geometric confidence interval...". Why are we talking here about that?

Thanks!
Helmut
Hero
Homepage
Vienna, Austria,
2018-02-02 00:56

@ Louis52
Posting: # 18330
Views: 1,192
 

 CV from CI

Hi Louis,

» […] how to get the CV(intra) when little or no (or not reliable) prior info available. I went over the "Biostatistics Sample Size Estimation for BE Studies" presentation […] and at one point in one of the "Hints" slide it says that "All you need is the 90% geometric confidence interval and the sample size."

I guess you are referring to this presentation (slide 24), right?

» Am I understanding correctly that the CLlow and CLhigh would be the 90% confidence limits for a prior CV intra…

Essentially yes.

<nitpick>

The 100(1–2α) CI is calculated from

  1. the geometric mean ratio (GMR),
  2. the number of subjects / sequence,
  3. the t-value (for given α and degrees of freedom specific for the design), and
  4. the mean square error (which is an estimate of the variance).

It’s not rocket science to perform the calculation “backwards” (get the unknown #4 from the known #1–#3 and the CI) and then CVintra=√MSE–1. If the GMR is not known, use √CLlower×CLupper.


</nitpick>

» I'm asking that because of the "...geometric confidence interval...". Why are we talking here about that?

I wanted to make clear that we need for the calculations outlined in slides 25–28 the confidence interval derived from log-transformed data (which is required for most PK metrics like AUC, Cmax,…).
Note that the calculation in the presentation is outlined for 2×2 crossover designs. It can be a little bit tricky to perform the calculations “manually” if you are unsure how the degrees of freedom in other designs are defined. I recommend the package PowerTOST for R (open source & free of costs). :thumb up:

Examples:
  • 2×2 crossover, subjects in the two sequence (n1, n2) 11 and 10 (planned for 22 but one dropout), α 0.05 (90% CI), lower CL 0.91, upper CL 1.15: Calculate the CVintra and then the CI in order whether the input was correct (should give the original CI).
    library(PowerTOST)
    alpha  <- 0.05
    lower  <- 0.91
    upper  <- 1.15
    n      <- c(11, 10)
    design <- "2x2"
    CV     <- CVfromCI(alpha=alpha, lower=lower, upper=upper,
                       n=n, design=design)
    CI     <- CI.BE(alpha=alpha, pe=sqrt(lower*upper), CV=CV,
                    n=n, design=design)
    cat("\n CV =", sprintf("%.2f%%", 100*CV),
        "\n", sprintf("%g%%", 100*(1-2*alpha)), "CI =",
        paste(CI, collapse = " to "), "\n")

    Should give
     CV = 22.17%
     90% CI = 0.91 to 1.15

  • As above but the study was a parallel design in 2×20 subjects. Use n <- c(20, 20) and design <- "parallel" to get
     CV = 22.22%
     90% CI = 0.91 to 1.15

    Note that this is the total (or pooled) CV and >2 groups are not supported by the function.
  • If the study was a replicate design for reference-scaling evaluated for Average Bioequivalenc with Expanding Limits (ABEL: EMA, Health Canada, Russia, and many others): You need only the upper expanded limit (neither the CI, the design, or the sample size).
    U  <- 1.38
    CV <- CVwRfromU(U=U)
    cat("\n CVwR =", sprintf("%.2f%%", 100*CV), "\n")

    Should give
     CVwR = 44.35%
    Note that CVwR is the within-subject CV of the reference product.

[image]Cheers,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
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Louis52
Junior

2018-02-03 17:12

@ Helmut
Posting: # 18337
Views: 1,027
 

 CV from CI

I guess I'm still a bit confused. Yes, it is about what is on the slide 25 'Hints' in the last bullet point: "All you need is the 90% geometric confidence interval and the sample size." If I am to look on slide 26 when the PE is derived it seems that the CLlo and CLhi are for the ratio for AUCs or Cmaxs as Test vs Ref from previous study. So maybe this is case when there is no other information about a previous study other than that. Because otherwise if we know the MSE for AUC or Cmax then it it easy to go get directly Cvintra. Am I correct on this? Or PE is something else? I'm asking because when delta(CL) is derived it seems that we take the logs. Not sure why...
Thanks!
Helmut
Hero
Homepage
Vienna, Austria,
2018-02-03 21:48

@ Louis52
Posting: # 18338
Views: 1,028
 

 CV from CI

Hi Louis,

I don’t know which presentation you are referring to. Link?
I’m referring to the one held in Moscow 2014 (slides 24–28).

» If I am to look on slide 26 when the PE is derived it seems that the CLlo and CLhi are for the ratio for AUCs or Cmaxs as Test vs Ref from previous study.

No. See slide 27: The confidence interval (CI) is given with 0.91–1.15. Hence, the lower confidence limit (CLlo) was 0.91 and the upper one (CLhi) 1.15. The point estimate (PE) aka geometric means ratio (GMR) was not given. It can be calculated by √CLlo×CLhi, i.e., √0.91×1.15=1.023.

» So maybe this is case when there is no other information about a previous study other than that. Because otherwise if we know the MSE for AUC or Cmax then it it easy to go get directly Cvintra. Am I correct on this?

Yes. This entire game is only necessary if the MSE is not given.

» I'm asking because when delta(CL) is derived it seems that we take the logs. Not sure why...

As I already wrote, all calculations are done in the logarithmic* domain. Did you ever wonder why the acceptance range of 80–125% is not symmetric around 100%? Most biologic variables follow a log-normal distribution. Makes sense because negative and even zero values are not possible (what’s a negative concentration?). It means also that log-transformed values follow a normal distribution which is one of the assumptions in the ANOVA. In BE we accept a deviation (Δ) of 20% as to be clinically not relevant. Consequently the lower limit of the acceptance range is 100(1–Δ)=80% and the upper one its reciprocal 100(1–Δ)–1=125%. Transformed to logs we get –0.2231 and +0.2231 which are symmetric around zero. Bingo. We could apply the ANOVA.
Remember some useful rules for working with logs:
  • log(x·y)=log(x)+log(y), log(x∕y)=log(x)–log(y)
  • log(xy)=y·log(x), log(yx)=log(x)∕y
  • The back-transformed arithmetic mean of logs is the geometric mean of original values: ℯ(log(x1),…,log(xi))∕n=n∏(x1,…,xi). Before the age of computers the first formula was used. With the latter even pocket calculators reach their limit soon. Not to speak of paper/pencil/brain.
In the formula for the CI we need the Δ (i.e., log(PE)), its standard error (SE), and the t-value:
CI=ℯlog(PE)∓t·SEΔ. SEΔ is calculated from the mean square error MSE and the total sample size (N) as 2·MSE∕N (slide 28). See slide 27 how the MSE is calculated from the difference between one of the CLs and PE, the subjects in both sequences (n1, n2), and the t-value. Note that log(CLhi)–log(PE), log(PE)–log(CLlo), or the logs of ratios give the same result: log(1.15)–log(1.023)=log(1.023)–log(0.91)=log(1.15∕1.023)=log(1.023∕0.91)=0.11702.
Once we have the MSE, calculation of the CV is easy. If you do it by hand (I hope not), I suggest to calculate the CI afterwards. If you don’t get the original values, something went wrong.


  • log(x), the natural of Napierian logarithm (base ℯ≈2.71828…). If ever possible, avoid the common logarithm log10(x).
    If your analysis was done on log10-transformed data CVintra=√10log(10)·MSE–1.

[image]Cheers,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
Science Quotes
Louis52
Junior

2018-02-04 14:38

@ Helmut
Posting: # 18341
Views: 974
 

 CV from CI

Hi Helmut,
Thank you for the explanation :-)
Probably, most of the times we would actually know the PE from a previous study together with the CI. That's why I would not see the need the recalculate the PE (as the GMR for either AUCs or Cmaxs). But I completely agree that if we get from a published article only the information about the GMR limits for AUC or Cmax, then this is a very clever way of deriving the CVintra.
Couple of more questions:
  1. Presumably, n1 and n2 are form the previous study, right?
  2. Where 0.91 and 1.15 are coming from? Did you select them just as an example or there is a particular reason?
Thanks!
Helmut
Hero
Homepage
Vienna, Austria,
2018-02-05 11:39

@ Louis52
Posting: # 18346
Views: 916
 

 CV from CI

Hi Louis,

» 1. Presumably, n1 and n2 are form the previous study, right?

As already explained these are the number of subjects / sequence (TR or RT). If known, fine. The calculated CV will be exact (to the numeric precision of the input). If unknown (rarely given in the literature) use the total number of subjects N and split it in such a way that the study is as balanced as possible. If the study was more unbalanced than assumed you will get a CV which is higher than the true one. Since this is conservative you will be one the safe side basing your sample size estimation on it (see there for an example).

» 2. Where 0.91 and 1.15 are coming from? Did you select them just as an example or there is a particular reason?

What do you think the word example in

■ Example: 90% CI [0.91 – 1.15], N 21 (n1 = 11, n2 = 10)

in this presentation means?

[image]Cheers,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
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