Imph
☆

Algeria,
2022-09-25 15:52
(129 d 04:27 ago)

Posting: # 23318
Views: 1,326

## Washout based on mean ± SD of t½ [Design Issues]

hello,

I want to use a pharmacokinetic data from the literature and this data is provided in the form: mean ±SD, is it correct to use only the mean without taking into account the standard deviation?
For example I want to calculate a washout duration of seven times t1/2. This t1/2 is provided in the form: mean ±SD. Is it right to do "7 x t1/2 mean" without taking into consideration the SD?.

Thank you in advance

Edit: Category and subject line changed; see also this post #1[Helmut]
Helmut
★★★

Vienna, Austria,
2022-09-25 18:08
(129 d 02:11 ago)

@ Imph
Posting: # 23319
Views: 1,120

## Washout based on mean ± SD of t½

Hi Imph,

❝ Is it right to do "7 x t1/2 mean" without taking into consideration the SD?

That’s extremely risky. You may see residual concentrations in subjects with a half life longer than the reported mean and have to exclude them to avoid carryover (i.e., if the predose concentration is >5% Cmax).

See this presentation (slides 64/65).

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

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Imph
☆

Algeria,
2022-09-26 17:24
(128 d 02:54 ago)

@ Helmut
Posting: # 23320
Views: 1,064

## Washout based on mean ± SD of t½

Hi,

Thank you very much for your valuable informations.

I would like to know if this method can be applied to each pharmacokinetic measurement, for example to determine the duration of the sampling (5 x t1/2) or to calculate the predose limit of 5% Cmax when the Cmax is provided in the literature under mean+/-SD.

Why plan for 95% coverage (z0.95 = 1.96), is it related to alpha =5%.

In the reference you gave, a normal distribution is assumed, but pharmacokinetic measurements do not generally follow a normal distribution but a log-normal one. Does this method still apply?

Thank you verry much.
Shuanghe
★★

Spain,
2022-09-26 19:34
(128 d 00:44 ago)

@ Imph
Posting: # 23321
Views: 1,039

## Washout based on mean ± SD of t½

Hi Imph

❝ I would like to know if this method can be applied to each pharmacokinetic measurement, for example to determine the duration of the sampling (5 x t1/2) or to calculate the predose limit of 5% Cmax when the Cmax is provided in the literature under mean+/-SD.

I don't know about "every" PK parameter, but FDA use this approach to plan partial AUC as well.

For example, for methylphenidate HCl ER tablet, among other parameters, FDA requests partial AUC0--3h and AUC0--4h for fasting and fed studies, respectively. The reason was that the Tmax was 2h under fasting and 3h under fed. the standard deviation (SD) of Tmax is about 0.5h, so 2 x SD is 1h, according to FDA:

> "For Tmax, two standard deviations = 1.0; Generally, approximately 95% of observations fall within two standard deviations of the mean; Thus, since the Tmax from the immediate-release portion of this formulation is about 2 hours under fasting conditions and 3 hours under fed conditions, pAUCs calculated to 0-3 hours in a fasting BE study and 0-4 hours in a fed BE study should capture the responses of 95% of the subjects. This should provide assurance that a test and reference product will be therapeutically equivalent over the early part of the daily dosing interval, corresponding to onset of response." (see ref linked here)

This is the same as what Helmut did for t1/2 in his PPT cited in previous post.

❝ In the reference you gave, a normal distribution is assumed, but pharmacokinetic measurements do not generally follow a normal distribution but a log-normal one. Does this method still apply?

Well, Tmax is discrete so it is not normal distribution either. FDA still use it for the approximation for pAUC.

I'd use this 5 x (mean(PK) + 2xSD(PK)) instead of the empirical "7 x mean(PK)".

All the best,
Shuanghe
Helmut
★★★

Vienna, Austria,
2022-09-26 19:35
(128 d 00:43 ago)

@ Imph
Posting: # 23322
Views: 1,051

## Washout based on mean ± SD of t½

Hi Imph,

❝ I would like to know if this method can be applied to each pharmacokinetic measurement, for example to determine the duration of the sampling (5 x t1/2) or to calculate the predose limit of 5% Cmax when the Cmax is provided in the literature under mean+/-SD.

In principle, yes. I prefer to start with a single dose profile (even if read from a published figure) and perform nonparametric superposition.

❝ Why plan for 95% coverage (z0.95 = 1.96), is it related to alpha =5%.

Convention. You can use a more conservative value.

❝ In the reference you gave, a normal distribution is assumed, but pharmacokinetic measurements do not generally follow a normal distribution but a log-normal one. Does this method still apply?

You are right. But if the arithmetic mean ± SD is all you have… If you have results for the geometric mean, better, of course.

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

The quality of responses received is directly proportional to the quality of the question asked. 🚮
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Helmut
★★★

Vienna, Austria,
2022-09-27 13:52
(127 d 06:27 ago)

@ Imph
Posting: # 23323
Views: 1,014

## If you have data…

Hi Imph,

if you have individual half lives, you can calculate the geometric mean / SD and use any confidence interval you like. For planning the washout use the upper confidence limit.
-script (results in blue):

g.fun <- function(x, alpha = 0.05, digits = 4, print.only = FALSE) {   # calculate geometric mean, SD, and CI of geometric mean   if (sum(is.na(x)) > 0) { # Remove NA(s)     message("NA(s) removed from the vector.")     x <- x[-is.na(x)]   }   if (sum(x <= 0) > 0) {   # Only positive numbers are allowed     message("Geometric mean applicable to",             "\npositive numbers only; others removed.")     x <- x[-which(x <= 0)]   }   mean.log <- mean(log(x))   SD.log   <- sd(log(x))   gMean    <- exp(mean.log)   gSD      <- exp(SD.log)   # confidence interval based on the t-distribution with n–1 degrees of freedom   CI       <- exp(mean.log + c(-1, +1) *                   qt(alpha / 2, length(x) - 1, lower.tail = FALSE) * SD.log)   loc.dis  <- setNames(c(gMean, gSD, CI),                        c("Geom. mean", "Geom. SD",                          paste0("lower ", 100 * (1 - alpha), "%"),                          paste0("upper ", 100 * (1 - alpha), "%")))   if (print.only) {     print(signif(loc.dis, digits))   } else {     return(loc.dis)   } } # give your data in a t12 vector, e.g., # t12 <- c(7.15, 5.44, 5.33, ...) and proceed with descriptive statistics # lognormal-distributed example data set.seed(123456) n   <- 24   # sample size mue <- 6    # geometric mean of half life CV  <- 0.25 # coefficient of variation t12 <- rlnorm(n = n,               meanlog = log(mue) - 0.5 * log(CV^2 + 1),               sdlog = sqrt(log(CV^2 + 1))) exp(summary(log(t12)))        # descriptive statistics  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 4.425   5.412   6.383   6.530   7.764  10.780 g.fun(t12, print.only = TRUE) # default: 95% CI Geom. mean   Geom. SD  lower 95%  upper 95%      6.530      1.275      3.951     10.790 alpha      <- 0.01            # conservative: 99% CI g.fun(t12, alpha = alpha, print.only = TRUE) Geom. mean   Geom. SD  lower 99%  upper 99%      6.530      1.275      3.302     12.910 half.lives <- c(5, 7, 10) col.name   <- paste0("upper ", 100 * (1 - alpha), "%") res        <- data.frame(half.lives = half.lives,                          washout = half.lives * g.fun(t12, alpha = alpha)[[col.name]]) res$days <- ceiling(res$washout / 24) # round up washout (h) to days print(round(res, 1), row.names = FALSE)  half.lives washout days           5    64.6    3           7    90.4    4          10   129.1    6

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
Imph
☆

Algeria,
2022-10-25 11:21
(99 d 08:57 ago)

@ Helmut
Posting: # 23348
Views: 732

## If you have data…

Hello,

Thank you very much it was a great help for us.

Best regards.
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