Bioequivalence and Bioavailability Forum

Brus
★

Spain,
2023-06-14 18:34
(173 d 16:00 ago)

Posting: # 23593
Views: 1,088

Calculation of PE and 90% CI [General Statistics]

Dear all,

In a partial replicated design (3-periods, 3 sequences), only reference is replicated, with a sample size of 45, CRO sent to me the following with proc GLM for ANOVA for the R/R comparison:

Source   / DF /       SS      /     MS       / P

Seq      /  2 /  0.20523638   / 0.10261819   / 0.7686

Sbj(seq) / 40 / 15.497377771  / 0.3874344443 / <0.0001

Per      /  2 /  0.0922514826 / 0.0461257413 / 0.3597

Treat    /  1 /  0.0231184863 / 0.0231184863 / 0.4725

Error    / 37 /  1.6238230813 / 0.0438871103 / -



Least Squares Means:

R1: 2.522

R2: 2.419

Also they sent to me the following 90% CI for the comparative R1/R2:

Ratio = 110.79

90% Confidence Interval = 87.31 - 140.57

Intra-subject CV% = 21.18

The number of observations used for the calculation were 40 due to drop outs.

But I have several doubts:
- It surprises me a 90% CI so wide for a relatively low, at least not high, ISCV.
- If I back-calculated the CV from this 90% CI with “CVfromCI” of PowerTOST, I obtain a CV of around 70%. Why?
- In addition, I have tried to calculate the ratio and 90% CI with MSE and LSM according to the Helmut example from Pamplona lecture on 2018 "Basic Statistic on BE", slide 16 and 17:

SE = √(MSE/nps), where nps = (n1 +n2)/2
∆ = LSM (T) – LSM (R)
PE (GMR = e∆)
90% CI = ∆ ± t(α = 0.05, df) × SE

I used 0.043 from first table as MSE and I used the formula from Helmut presentation to convert it in SE and then to calculate the ratio and 90% CI.

I obtained a ratio of 110,8 but a 90% CI of 102,44 – 119,94. Same ratio as CRO but different 90 % CI. This is strange although it is in line with the ISCV presented by the CRO.

Furthermore, doing various tests, I have noticed that if you add the mean square obtained in the Sbj(seq) from first table to the MSE and you use the formula from Helmut presentation to convert this sum in SE and then to calculate the ratio and 90% CI, you obtain the same 90% CI as CRO. But, If the subj(seq) is significant, should it be calculated in this way?, why?, and should it be taken into account and impact the 90% CI range?

So, what is the correct 90% CI?

Thank you so much

Best regards,

Helmut
★★★

Vienna, Austria,
2023-06-15 00:26
(173 d 10:08 ago)

@ Brus
Posting: # 23594
Views: 930

No ‘R1’ and ‘R2’ in replicate designs

Post reply

Hi Brus,

❝ In a partial replicated design (3-periods, 3 sequences), only reference is replicated, with a sample size of 45, CRO sent to me the following with proc GLM for ANOVA for the R/R comparison:

❝ …

❝ Least Squares Means:

❝ R1: 2.522

❝ R2: 2.419

❝

❝ Also they sent to me the following 90% CI for the comparative R1/R2:

❝

❝ Ratio = 110.79

❝ 90% Confidence Interval = 87.31 - 140.57

❝ Intra-subject CV% = 21.18

I guess, your problems started already at the beginning.

There can be only one.

There is no ‘R₁’ and ‘R₂’, only one R repeatedly administered in the sequences in different periods. Say, the sequences are TRR | RTR | RRT. Did the CRO call the first administration in each sequence R₁ and the second R₂ (while dropping T)?
This gives after recoding: ░R₁R₂ | R₁░R₂ | R₁R₂░

❝ But I have several doubts:

❝ - It surprises me a 90% CI so wide for a relatively low, at least not high, ISCV.

As said above, such a ‘comparison’ will not work.

❝ - If I back-calculated the CV from this 90% CI with “CVfromCI” of PowerTOST, I obtain a CV of around 70%. Why?

You tried CI2CV(lower=0.8731, upper=1.4057, design="2x2x2", n=40), right? That’s not what you have. After recoding (wild guess) you have an Incomplete Block Design. Other degrees of freedom, :blahblah:

❝ - In addition, I have tried to calculate the ratio and 90% CI with MSE and LSM according to the Helmut example from Pamplona lecture on 2018 "Basic Statistic on BE", slide 16 and 17:

❝ …

❝ I obtained a ratio of 110,8 but a 90% CI of 102,44 – 119,94. Same ratio as CRO but different 90 % CI. This is strange although it is in line with the ISCV presented by the CRO.

These formulas are for a 2×2×2 crossover and are also implemented in CI2CV(). But again, there is only one R. A ‘PE’ and ‘CI’ doesn’t make sense.

❝ Furthermore, doing various tests, … why?

No idea.

❝ So, what is the correct 90% CI?

There is none. You can only calculate the CV_wR according to the EMA’s or the FDA’s models. The results will be similar, though quite often the one of the FDA is a bit smaller.

If you want, send me the raw data off-list.

—
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

Brus
★

Spain,
2023-06-15 16:07
(172 d 18:27 ago)

@ Helmut
Posting: # 23598
Views: 874

No ‘R1’ and ‘R2’ in replicate designs

Post reply

Hi Helmut,

Sorry, because I am lego and I can´t understand all correctly.

❝ There is no ‘R₁’ and ‘R₂’, only one R repeatedly administered in the sequences in different periods. Say, the sequences are TRR | RTR | RRT. Did the CRO call the first administration in each sequence R₁ and the second R₂ (while dropping T)?

Yes. But, although this is not the usual approach, if you set the R treatment as R1 and R2, you can make an R1 / R2 comparison, as if they were different treatments, right? And in this way, Is the 90% CI calculated by the CRO well calculated? It seems to me to be a very large interval for a CV of 20%.

❝ As said above, such a ‘comparison’ will not work.

Why? Can you explain?

❝ You tried CI2CV(lower=0.8731, upper=1.4057, design="2x2x2", n=40), right? That’s not what you have. After recoding (wild guess) you have an Incomplete Block Design. Other degrees of freedom, :blahblah:

Yes, I know, but in any case, the difference should be decimals, not changes from 20% to 70%, right? I can't understand how they get a CV of 20% and I get 70%. What's wrong? Besides, with the CI being so wide, it seems to be much more than 20% CV.

❝ These formulas are for a 2×2×2 crossover and are also implemented in CI2CV(). But again, there is only one R. A ‘PE’ and ‘CI’ doesn’t make sense.

❝ You can only calculate the CV_wR according to the EMA’s or the FDA’s models. The results will be similar, though quite often the one of the FDA is a bit smaller.

But, although this is not the usual approach, if you set the R treatment as R1 and R2, you can make an R1 / R2 comparison, if you want to do it just for curiosity, considering them as different treatments, it will be feasible, right? And in this way, Is the 90% CI calculated by the CRO well calculated? It seems to me to be a very large interval for a CV of 20%. Or mine?

Edit: Standard quotes restored; see also this post #8. [Helmut]

Helmut
★★★

Vienna, Austria,
2023-06-15 16:49
(172 d 17:45 ago)

@ Brus
Posting: # 23599
Views: 863

Rose is a rose is a rose is a rose

Post reply

Hi Brus,

❝ Sorry, because I am lego and I can´t understand all correctly.

❝ ❝ There is no ‘R₁’ and ‘R₂’, only one R repeatedly administered in the sequences in different periods. Say, the sequences are TRR | RTR | RRT. Did the CRO call the first administration in each sequence R₁ and the second R₂ (while dropping T)?

❝ Yes. But, although this is not the usual approach, if you set the R treatment as R1 and R2, you can make an R1 / R2 comparison, as if they were different treatments, right?

Nope. Though you call them different treatments, they aren’t. You have one and only one treatment – administered in different sequences / periods.

❝ And in this way, Is the 90% CI calculated by the CRO well calculated? It seems to me to be a very large interval for a CV of 20%.

There is no right or wrong. Calculating a PE and CI doesn’t make sense.

❝ ❝ As said above, such a ‘comparison’ will not work.

❝ Why? Can you explain?

I’ll try… In a 2×2×2 crossover you have two different but similar things, say, apples and oranges and compare their weights (© Stephen Senn).
In your partial replicate you ignored the apple and tried to compare oranges with oranges. What? Heck, they are the same. Only conditions differed.

❝ ❝ You can only calculate the CV_wR according to the EMA’s or the FDA’s models. The results will be similar, though quite often the one of the FDA is a bit smaller.

❝ But, although this is not the usual approach, if you set the R treatment as R1 and R2, you can make an R1 / R2 comparison, if you want to do it just for curiosity, considering them as different treatments, it will be feasible, right? And in this way, Is the 90% CI calculated by the CRO well calculated?

No, that’s nonsense. I think we are going in circles. What do you want achieve?

Rose is a rose is a rose is a rose.

Let’s explore the EMA’s reference data set II, both with the approach given in the Q&A as well as with yours. See the [image]

-script at the end.

Type III AOV (CVwR: EMA approach) Response: log(PK) Df Sum Sq Mean Sq F value Pr(>F) sequence 2 0.023936 0.01197 0.08523 0.91862 period 2 0.039640 0.01982 1.43328 0.24898 sequence:subject 21 2.948968 0.14043 10.15485 4.6897e-11 Residuals 46 0.636114 0.01383 CVwR (%) 11.80027 Type III AOV (pseudo 'R2' vs 'R1') Response: log(PK) Df Sum Sq Mean Sq F value Pr(>F) sequence 2 0.049616 0.02481 0.28973 0.75141 period 2 0.018554 0.00928 0.71416 0.50112 treatment 1 0.000044 0.00004 0.00336 0.95435 sequence:subject 21 1.798142 0.08563 6.59175 3.0663e-05 Residuals 21 0.272787 0.01299 CVw (%) 11.43441 'PE': 99.43% 'CI': 83.90%, 117.84%

Not only the SEs are different but also the degrees of freedom and therefore, the MSEs / CVs.
CI2CV() from the ‘pseudo R comparison’ does not work with any design-argument.

library(PowerTOST) # EMA Dataset II (partial replicate, balanced, complete, 24 subjects) EMA.II <- read.csv("https://bebac.at/downloads/ds02.csv", colClasses = c(rep("factor", 4), "numeric")) options(digits = 12) # more digits for anova # The EMA’s approach for CVwR mod.CVwR <- lm(log(PK) ~ sequence + subject %in% sequence + period, data = EMA.II) typeIII.a <- anova(mod.CVwR) attr(typeIII.a, "heading")[1] <- "Type III AOV (CVwR: EMA approach)" MSdenom <- typeIII.a["sequence:subject", "Mean Sq"] df2 <- typeIII.a["sequence:subject", "Df"] fvalue <- typeIII.a["sequence", "Mean Sq"] / MSdenom df1 <- typeIII.a["sequence", "Df"] typeIII.a["sequence", 4] <- fvalue typeIII.a["sequence", 5] <- pf(fvalue, df1, df2, lower.tail = FALSE) CVwR <- 100 * mse2CV(typeIII.a["Residuals", "Mean Sq"]) typeIII.a <- rbind(typeIII.a, CVwR = c(NA, NA, CVwR, NA, NA)) row.names(typeIII.a)[5] <- "CVwR (%)" # Only R; clumsy but transparent # Keep the codes of sequence and period EMA.II.R <- EMA.II[EMA.II$treatment == "R", ] EMA.II.Rs <- cbind(EMA.II.R, dummy = NA_character_) EMA.II.Rs$dummy[EMA.II.Rs$sequence == "TRR" & EMA.II.Rs$period == 2] <- "R1" EMA.II.Rs$dummy[EMA.II.Rs$sequence == "TRR" & EMA.II.Rs$period == 3] <- "R2" EMA.II.Rs$dummy[EMA.II.Rs$sequence == "RTR" & EMA.II.Rs$period == 1] <- "R1" EMA.II.Rs$dummy[EMA.II.Rs$sequence == "RTR" & EMA.II.Rs$period == 3] <- "R2" EMA.II.Rs$dummy[EMA.II.Rs$sequence == "RRT" & EMA.II.Rs$period == 1] <- "R1" EMA.II.Rs$dummy[EMA.II.Rs$sequence == "RRT" & EMA.II.Rs$period == 2] <- "R2" EMA.II.Rs$dummy <- as.factor(EMA.II.Rs$dummy) EMA.II.Rs$treatment <- EMA.II.Rs$dummy # Compare second administration of R with the first mod.ABE <- lm(log(PK) ~ sequence + subject %in% sequence + period + treatment, data = EMA.II.Rs) typeIII.b <- anova(mod.ABE) attr(typeIII.b, "heading")[1] <- "Type III AOV (pseudo \'R2\' vs \'R1\')" MSdenom <- typeIII.b["sequence:subject", "Mean Sq"] df2 <- typeIII.b["sequence:subject", "Df"] fvalue <- typeIII.b["sequence", "Mean Sq"] / MSdenom df1 <- typeIII.b["sequence", "Df"] typeIII.b["sequence", 4] <- fvalue typeIII.b["sequence", 5] <- pf(fvalue, df1, df2, lower.tail = FALSE) CVw <- 100 * mse2CV(typeIII.b["Residuals", "Mean Sq"]) typeIII.b <- rbind(typeIII.b, CVw = c(NA, NA, CVwR, NA, NA)) row.names(typeIII.b)[6] <- "CVw (%)" PE <- 100 * exp(coef(mod.ABE)[["treatmentR2"]]) CI <- setNames(100 * as.numeric(exp(confint(mod.ABE, "treatmentR2", level= 1 - 2 * 0.05))), c("lower", "upper")) txt <- paste("\n\'PE\':", sprintf("%6.2f%%", PE), "\n\'CI\':", sprintf("%6.2f%%, %6.2f%%\n", CI["lower"], CI["upper"])) print(typeIII.a, digits = 6, signif.stars = FALSE) print(typeIII.b, digits = 6, signif.stars = FALSE); cat(txt)

—
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

Achievwin ★ US, 2023-06-19 21:42 (168 d 12:52 ago) @ Brus Posting: # 23604 Views: 818	No ‘R1’ and ‘R2’ in replicate designs Post reply
	Rose is a rose is a rose is a rose Not always... .... there are some blessing from hell formulations.....

Calculation of PE and 90% CI [General Sta­tis­tics]

No ‘R1’ and ‘R2’ in replicate designs

No ‘R1’ and ‘R2’ in replicate designs

Rose is a rose is a rose is a rose

No ‘R1’ and ‘R2’ in replicate designs

Calculation of PE and 90% CI [General Statistics]