Bioequivalence and Bioavailability Forum • Finding the CI limits

Louis52
☆

2018-02-01 20:57
(2604 d 04:03 ago)

Posting: # 18328
Views: 7,997

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
★★★

Vienna, Austria,
2018-02-02 01:56
(2603 d 23:04 ago)

@ Louis52
Posting: # 18330
Views: 7,231

CV from CI

Post reply

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

the geometric mean ratio (GMR),
the number of subjects / sequence,
the t-value (for given α and degrees of freedom specific for the design), and
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 \(CV_{intra} = \sqrt{e^{MSE} - 1}\). If the GMR is not known, use \(\sqrt{CL_{lower} \times CL_{upper}}\).

</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, C_max,…).
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).

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 CV_intra 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 CV_wR is the within-subject CV of the reference product.

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

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

Louis52
☆

2018-02-03 18:12
(2602 d 06:48 ago)

@ Helmut
Posting: # 18337
Views: 6,789

CV from CI

Post reply

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
★★★

Vienna, Austria,
2018-02-03 22:48
(2602 d 02:13 ago)

@ Louis52
Posting: # 18338
Views: 6,748

CV from CI

Post reply

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 (CL_lo) was 0.91 and the upper one (CL_hi) 1.15. The point estimate (PE) aka geometric means ratio (GMR) was not given. It can be calculated by \(\sqrt{CL_{lo} \times CL_{hi}}\), i.e., \(\sqrt{0.91 \times 1.15} = 1.023\).

❝ 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:

\(\log {(x \cdot y)} = \log {(x)} + \log {(y)}, \log {(x / y)} = \log {(x)} - \log {(y)}\)
\(\log {(x^y)} = y \cdot \log {(x)}, \sqrt[y]{x} = \log {(x)} / y\)
The back-transformed arithmetic mean of logs is the geometric mean of original values:
\(e^{1/n \cdot \sum {\log {(x_1)}, \ldots, \log {(x_i)}})} =\sqrt[n]{\prod {x_1, \ldots, x_n}}\)
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:
\(CI=e^{\log(PE) \mp t \cdot SE_\Delta )}\). SE_Δ is calculated from the mean square error MSE and the total sample size (N) as \(\sqrt{2 \cdot MSE / N}\) (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 \(\log(CL_{hi}) - \log(PE), \log(PE) - \log(CL_{lo})\), or the logs of ratios give the same result: \(\log(1.15) - \log(1.024) = \log(1.023) - \log(0.91) = \log((1.15/1.023)) = \log((1.023/0.91)) = 0.11702\)
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_ℯ(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: \(CV_{intra} = \sqrt{10^{\log_e(10) \cdot MSE} - 1}\).

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

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

Louis52
☆

2018-02-04 15:38
(2601 d 09:22 ago)

@ Helmut
Posting: # 18341
Views: 6,712

CV from CI

Post reply

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:

Presumably, n1 and n2 are form the previous study, right?
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
★★★

Vienna, Austria,
2018-02-05 12:39
(2600 d 12:22 ago)

@ Louis52
Posting: # 18346
Views: 6,715

CV from CI

Post reply

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?

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

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