This question makes me laugh and let me tell you why. The fact of the matter stems from the scale being made my humans. Humans tried to make 100 degrees Fahrenheit what our body temperature would be but didn’t quite scale it properly. Guess what 0 degrees Fahrenheit is? The temperature at which sea water freezes was scaled to be at zero degrees Fahrenheit. Pog fact indeed…

This is an app to convert CSV to XPT, XPT to CSV, CSV to Mat, Mat to CSV, CSV to xlsx or xls, and xlsx to CSV.

Online CSV Viewer

I asked this question to an apiarist. Her answer was so clear. There is not enough water in honey to sustain life. That is why it has antibacterial-like properties. It is not that honey kills. It is that it is not life sustaining.

Dear all,

I used to live in South Korea. I love the Korean culture and people. I still immerse myself in the language and culture although I have not lived there for nearly two decades. I weep for North Koreans. I cannot understand the atrocities of North Korea’s totalitarian regime. He is such an ass. Seriously, what the fetching hell is wrong with him? Pompous ass hole! Why does the world do nothing for them? Why not help North Korea? So so sad. Too bad no one from North Korea will read this. If they did they would be executed. What a sick, sick mind.

There are 2 answers to you wanting to look up what the mystery liquid is on top of your yogurt. 1) You may actually be interested in the answer; and 2) You may actually need to seek a therapist for you to get over your concern.

The way you make numbers repeat themselves in fraction form is as follows:

If you want 0.123123123… then you just use 123/999

If you want 123.123123… then use 123/0.999

If you want 123456789 then divide it by the same number of 9s: 123456789/999999999 =0.123456789123456789

If you want: 0.102102102 then simply use 102/999

With that said, my favorite number is 1/0.98

The 0.98 above is not a typo:
1.020408163265

That five is not a typo at the end.

Why was it not 64 at the end? 😉

The answer may surprise you. The answer is that zero times anything is not always zero. I have 2 examples of zero being multiplied by something else and not mathematically resulting in zero.

1) 0 * (1/0) = undefined
2) 0! = 1

You may think the second one doesn’t count but it does. Factorial or the exclamation point means you take the number and multiply it by every number below it all the way down to 1. Zero is different though. 0! is equal to 1.

Also, for fun, you should ask Siri what zero divided by zero is. You’ll get a good laugh.

Best wishes,

Pharmacoengineering.com

This article summarizes: Journal of Clinical Pharamacology, 2012; 52:1601-1606

Introduction
Paragraph 1:
Most development programs have one chance to obtain an informative set of trials; generally, thereafter, companies lose financial incentive.
PK information is useful because you can use it to help select a dose range, assess drug exposure for efficacy and safety purposes (via matching adult exposures), and support dosing approval.
Guidances have been published to elucidate the role of pediatric studies in the context of ADME.
Although provided, sample size selection for pediatric PK and safety studies have been very different and have not had clear justification.
A uniform definition of study quality for pediatric PK studies is needed.

Paragraph 2:

An important goal is to ensure precise estimates of PK parameters, such as CL and Vd, which are use to justify the safe and effective dose.

Regulatory guidance has been as follows:

The study must be prospectively powered to target a
95% CI [confidence interval] within 60% and 140%
of the geometric mean estimates of clearance and
volume of distribution for DRUG NAME in each
pediatric sub-group with at least 80% power.

Paragraph 3:

The article of interest will report how to do this based on an NCA or popPK approach.

Note that in terms of the NCA analysis discussion, they assume the PK parameters have been robustly evaluated (appropriate blood sampling; on a patient level).

In terms of safety objectives, there should be a minimum number of participants required and this number may be more than that which is considered required for the PK analysis objectives. The sponsor will still report to the regulatory authority via the same language if the number required for the safety analysis is greater than that required for the PK objectives.

3 pediatric clinical trials will be presented to demonstrate what is being proposed, while complying with the regulatory request.

Potential impacts of the quality standard on pediatric drug development is discussed as well.

Method

Sample Size Calculation for Rich PK Sampling Design Intended for NCA Analysis

Step 1: Derive a reasonable estimate of variability.

the standard deviation (SD) of log-transformed individual clearance for adults or any prior and related study can be used. This would also be done for WT. 

Also, the CL SD can be predicted within the pediatric patients using allometry.

If you know the %CV for the untransformed CLi is M% then use the following:

SD = sqrt(log(%CV^2+1)) =sqrt(log((M/100)^2+1)) 

Use this for each of the pediatric patient age subgroups.

This works well for well-divided age groups (but not for a group spanning the entire pediatric age group).

Note the reported BSV is usually a combination of BSV+IOC (intra-occasion variability; within subject variability).

If the true BSV and IOC were separated, the total BSV should be used to avoid underestimating this SD value.

If different SDs can be used for different age sub-groups, it would increase the probability of success.

Step 2: Calculate sample size to achieve the target. The 95% CI for the geometric mean CL (or WT) in one subgroup can be constructed as follows:

The 95% CI can be calculated as follows:

CLbar_geo is the sample geometric mean of CLis.

The S is the sample standard deviation of logCLi.

N is the number of patients in that age subgroup

t_0.975, N-1 is the t value corresponding to the 97.5th percentile of a Student t distribution with N-1 degrees of freedom (df).

To fulfill the requirement, the above equation should result within the following bounds: (0.6, 1.4).

Given the sampling distribution of S, a required N is needed to ensure that:

or

with a certain level of confidence. This is referred to as “power” in the requirement. Note that no hypothesis testing is involved here.

R Code for this is as follows:

f <- function(s, n, sigma){2*((n-1)/2/sigma**2)((n-1)/2)/gamma(0.5*(n-1))*s**(n-2)*exp(-(n-1)***s2/2/sigma**2)}

nmin <- 4

nmax <- 25

nsub <- nmax – nmin + 1

result <- rep(0, nsub)

sd <- 0.4 # SD of logX from adult data

for (i in 1:nsub) { n <- nmin + i – 1

tv <- qt(0.975, n – 1)

sup <- log(1.4)*sqrt(n)/tv
g <- function(x){f(x,n,sd)}
poweri <- integrate(g,0, sup, subdivisions = 100)
result[i] <- poweri[[1]]
}
power <- data.frame(nsub=c(nmin:nmax), power=result)
power

For further convenience, you could also change the “sd <- 0.4” line above to:

percCV <- 35 # percent CV
sd <- sqrt(log((percCV/100)**2+1))

And that would allow you to directly enter the % CV values.

You can also add the following at the end of the script for convenience (also add library(dplyr) at the top if you do):

powerCutoff <- 0.8

powerConcise <- (power %>% filter(power > powerCutoff))[1,]
sprintf(‘The number of subjects that should be used for %f is %i’, powerCutoff, powerConcise$nsub)

The answers are the numbers required per age group, FYI.

Sample Size Calculation for Sparse/Rich PK Sampling Design Intended for popPK Analysis

The script written in the article was not clear so I re-wrote it here.

#Assume covariate model CL=theta1(wt/70)* theta2*age/(age+theta3) and

#input parameter estimates

thetap1=3.7421
theta2=1.0078
theta3=4.8422

#input variance-covariance matrix for the 3 parameters above

covp3=matrix(
c(
0.29810, 0.05782, 1.27120,
0.05782, 0.02921, 0.02073,
1.27120, 0.02073, 8.42210
),nrow=3, byrow=T
)

#define the weight and age combination for SE estimation

wt=14
age=3

#define covariate model in R

f2 <- function(x,y,z) x + ylog(wt/70) + log(age/(age+z))

df<- deriv(body(f2), c(“x”,”y”,”z”))

x=thetap1

y=theta2

z=theta3

out=eval(df)

dfout=attr(out,”gradient”)

varlcl=dfout%%covp3%*%t(dfout)

#SE of LCL 0.09436884

SElcl=sqrt(varlcl)
se <- as.numeric(SElcl)
se

library(dplyr)
f <- function(s, n, sigma){2*((n-1)/2/sigma**2)**((n-1)/2)/gamma(0.5*(n-1))*s**(n-2)*exp(-(n-1)*s**2/2/sigma**2)}

nmin <- 2 # > 1
nmax <- 25
nsub <- nmax – nmin + 1
result <- rep(0, nsub)

for (i in 1:nsub) {
n <- nmin + i – 1
tv <- qt(0.975, n – 1)
sup <- log(1.4)*sqrt(n)/tv

g <- function(x){f(x,n,se*sqrt(n))}
poweri <- integrate(g,0, sup, subdivisions = 100)
result[i] <- poweri[[1]]
}
power <- data.frame(nsub=c(nmin:nmax), power=result)
power

powerCutoff <- 0.8

powerConcise <- (power %>% filter(power > powerCutoff))[1,]
sprintf(‘The number of subjects that should be used for %f is %i’, powerCutoff, powerConcise$nsub)

#End

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