Bioequivalence and Bioavailability Forum • Sample size calculation

BE-proff
●

2016-12-14 09:56
(3072 d 12:49 ago)

Posting: # 16843
Views: 6,917

Sample size calculation [RSABE / ABEL]

Hi All,

Let's say I want to calculate sample size for a BE-study using the following data:

CVintra of Cmax= 39.8%
Power= at least 80%
GMR = 1.15

I want to utilize 2x2x4 design with reference scaling and PowerTOST gives me:

library(PowerTOST)

sampleN.scABEL.ad(CV=0.398, theta0=1.15, targetpower=0.80, design="2x2x4")



+++++++++++ scaled (widened) ABEL +++++++++++

            Sample size estimation

        for iteratively adjusted alpha

---------------------------------------------

Study design: 2x2x4 (RTRT|TRTR)

log-transformed data (multiplicative model)

1,000,000 studies in each iteration simulated.



Expected CVwR 0.398

Nominal alpha      : 0.05 

True ratio         : 1.1500 

Target power       : 0.8 

Regulatory settings: EMA (ABEL)

Switching CVwR     : 0.3 

 Regulatory constant: 0.76 

Expanded limits    : 0.7472...1.3383

Upper scaling cap  : CVwR > 0.5 

PE constraints     : 0.8000 ... 1.2500

n  48,   adj. alpha: 0.04444 (power 0.8100), TIE: 0.05000

Is it correct that:

- at least 48 subjects to be randomized
- CI for Cmax is 0.74 - 1.34
- CIs for AUCs are 0.80-1.25

Helmut
★★★

Vienna, Austria,
2016-12-14 17:24
(3072 d 05:21 ago)

@ BE-proff
Posting: # 16845
Views: 6,208

Sample size estimation for iteratively adjusted α

Post reply

Hi BE-proff,

❝ library(PowerTOST)

❝ sampleN.scABEL.ad(CV=0.398, theta0=1.15, targetpower=0.80, design="2x2x4")

THX for being concerned about the consumer risk! With a CV_wR of 39.8% we can expect a slight inflation of the Type I Error (TIE).

❝ Is it correct that:

❝ - at least 48 subjects to be randomized

Yes.

❝ - CI for Cmax is 0.74 - 1.34

No. For C_max the expanded limits are 0.7472...1.3383. The confidence interval has to lie entirely within these limits and additionally the PE must lie within 0.8–1.25.
Note that you have to use an adjusted α. Only if the CV_wR in the study would be exactly as assumed and there will be no dropouts, you could used the adjusted α from the sample size estimation above and calculate the 100(1–2×0.04444)=91.112% CI.

If you observe a different CV_wR in the study (likely) and there were dropouts you have to check whether the original iteratively adjusted α still suits and – if not – get a new one based on the study’s data. Examples for one dropout in the first sequence and two in the other:

CV_wR 38% (lower than expected):

library(PowerTOST) scABEL.ad(CV=0.38, theta0=1.15, design="2x2x4", n=c(23, 22), alpha.pre=0.0444) +++++++++++ scaled (widened) ABEL +++++++++++ iteratively adjusted alpha --------------------------------------------- Study design: 2x2x4 (RTRT|TRTR) log-transformed data (multiplicative model) 1,000,000 studies in each iteration simulated. CVwR 0.38, n(i) 23|22 (N 45) Nominal alpha : 0.05, pre-specified alpha 0.0444 True ratio : 1.1500 Regulatory settings : EMA (ABEL) Empiric TIE for alpha 0.0444 : 0.05485 Power for theta0 1.1500 : 0.761 Iteratively adjusted alpha : 0.03910 Empiric TIE for adjusted alpha: 0.05000 Power for theta0 1.1500 : 0.745

Since the CV_wR is closer to the critical range (maximum TIE at CV_wR 30%) you have to use the adjusted α 0.03910 (which is lower than the planned one). Hence you loose some power. Assess ABEL based on the 100(1–2×0.03910)=92.18% CI.
CV_wR 42% (higher than expected):

scABEL.ad(CV=0.42, theta0=1.15, design="2x2x4", n=c(23, 22), alpha.pre=0.0444) +++++++++++ scaled (widened) ABEL +++++++++++ iteratively adjusted alpha --------------------------------------------- Study design: 2x2x4 (RTRT|TRTR) log-transformed data (multiplicative model) 1,000,000 studies in each iteration simulated. CVwR 0.42, n(i) 23|22 (N 45) Nominal alpha : 0.05, pre-specified alpha 0.0444 True ratio : 1.1500 Regulatory settings : EMA (ABEL) Empiric TIE for alpha 0.0444 : 0.04491 Power for theta0 1.1500 : 0.811 TIE not > nominal alpha; the chosen pre-specified alpha is justified.

Since no inflation of the TIE is expected with α 0.0444, drop it from the arguments (i.e., try with the nominal α 0.05 only).

scABEL.ad(CV=0.42, theta0=1.15, design="2x2x4", n=c(23, 22)) +++++++++++ scaled (widened) ABEL +++++++++++ iteratively adjusted alpha --------------------------------------------- Study design: 2x2x4 (RTRT|TRTR) log-transformed data (multiplicative model) 1,000,000 studies in each iteration simulated. CVwR 0.42, n(i) 23|22 (N 45) Nominal alpha : 0.05 True ratio : 1.1500 Regulatory settings : EMA (ABEL) Empiric TIE for alpha 0.0500 : 0.04803 Power for theta0 1.1500 : 0.823 TIE not > nominal alpha; no adjustment of alpha is required.

Perfect. Use the nominal α 0.05 and assess ABEL based on the 100(1–2×0.05)=90% CI. You gain power.

❝ - CIs for AUCs are 0.80-1.25

The acceptance limits.

—
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

BE-proff ● 2016-12-14 20:43 (3072 d 02:03 ago) @ Helmut Posting: # 16846 Views: 5,863	Sample size estimation for iteratively adjusted α Post reply
	Hi Helmut, Many thanks for your very detailed explanation! It really helps!

BE-proff ● 2016-12-20 22:47 (3065 d 23:58 ago) @ Helmut Posting: # 16868 Views: 5,717	Sample size estimation for iteratively adjusted α Post reply
	Hi Helmut, Which R script should be used to assess BE in 92.18% CI? Afaik, "classic" BE is analysed in 90% CI. Correct?

d_labes
★★★

Berlin, Germany,
2016-12-21 12:15
(3065 d 10:30 ago)

@ BE-proff
Posting: # 16870
Views: 5,691

(1-2*alpha) CI

Post reply

Dear BE-prof!

❝ Which R script should be used to assess BE in 92.18% CI? :confused:

Which script do you use until now?
A serious script should IMHO not have a hard coded alpha=0.05 or the other way round a hard coded 90% CI. But rather allow the specification of the alpha to use.

Have a look into the function CI.BE() of the package PowerTOST.
For a 92.18% CI specify alpha=(1-0.9218)/2=0.0391, give the point estimate, CV and number of subjects in your study and voila you get the 1-2*alpha CI.
Example:

CI.BE(alpha=(1-0.9218)/2, pe=1.1557, CV=0.3567, n=24, design="2x2")

    lower     upper 

0.9609407 1.3899323

Now compare this to the (widened) acceptance range to judge BE or not BE.

❝ Afaik, "classic" BE is analysed in 90% CI. Correct? ;-)

Correct. alpha=0.05.

—
Regards,

Detlew

Helmut
★★★

Vienna, Austria,
2016-12-21 19:09
(3065 d 03:37 ago)

@ d_labes
Posting: # 16871
Views: 5,859

Script kiddy

Post reply

Hi Detlew!

I guess BE-proff was referring to this post.

This one should do it all:

library(PowerTOST)

regulator <- "EMA" # "EMA" or "HC"

CV        <- 0.25

theta0    <- 1.15

design    <- "2x2x4"

n         <- c(23, 22)

alpha.pre <- 0.0444

BE.H0 <- CI.H0 <- PE.H0 <- TRUE

res       <- scABEL.ad(CV=CV, theta0=theta0, design=design,

                       n=n, alpha.pre=alpha.pre, print=FALSE)

reg       <- reg_const(regulator=regulator)

AR        <- scABEL(CV=CV, regulator=regulator)

if (is.na(res$alpha.adj)) {

  CI  <- CI.BE(alpha=res$alpha, pe=theta0, CV=CV, n=n, design=design)

} else {

  CI  <- CI.BE(alpha=res$alpha.adj, pe=theta0, CV=CV, n=n, design=design)

}

if (CI[["lower"]] > AR[["lower"]] & CI[["upper"]] < AR[["upper"]]) CI.H0 <- FALSE

if (theta0 > 0.8 & theta0 < 1.25) PE.H0 <- FALSE

if (sum(PE.H0, PE.H0) == 0) BE.H0 <- FALSE

txt <- paste("\nDesign             :", design,

    "\nSubjects / sequence:", paste0(n, collapse=" / "),

    sprintf("\nCVwR               : %.2f%%", 100*CV))

if (CV > 0.3) {

  txt <- paste(txt, "(reference-scaling applicable)",

         "\nRegulatory settings:", regulator,

         sprintf("\nCVswitch           : %.2f%%", 100*reg$CVswitch),

         sprintf("\nCVcap              : %.2f%%", 100*reg$CVcap),

         sprintf("\nRegulatory constant: %.3f", reg$r_const))

}

txt <- paste(txt, sprintf("\nNominal \u03B1          : %.4f", res$alpha),

       sprintf("(TIE %.4f)", res$TIE.unadj))

if (!is.na(res$alpha.adj)) {

  txt <- paste(txt, sprintf("\nIteratively adj. \u03B1 : %.4f", res$alpha.adj),

              sprintf("(TIE %.4f)", res$TIE.adj))

}

  txt <- paste(txt, sprintf("\nPoint estimate     : %6.2f%%", 100*theta0),

         sprintf("\nAcceptance range   : %6.2f %s %6.2f%%",

                                    100*AR[["lower"]], "\u2013",

                                    100*AR[["upper"]]))

if (AR[["lower"]] < 0.8) {

  txt <- paste(txt, "(expanded = \'ABEL\')")

} else

  txt <- paste(txt, "(conventional)")

}

if (is.na(res$alpha.adj)) {

  txt <- paste(txt, sprintf("\n%5.2f%% CI          :", 100*(1-2*res$alpha)))

} else {

  txt <- paste(txt, sprintf("\n%5.2f%% CI          :", 100*(1-2*res$alpha.adj)))

}

txt <- paste(txt, sprintf("%6.2f %s %6.2f%%",

                          100*CI[["lower"]], "\u2013",

                          100*CI[["upper"]]),

    "\nAssessment:")

if (!CI.H0) {

  txt <- paste(txt, "\n  Confidence interval entirely within acceptance range.")

} else {

  txt <- paste(txt, "\n  Confidence interval not entirely within acceptance range.")

}

if (CV < 0.3) {

  if (!CI.H0) {

    txt <- paste(txt, "\n  Passed (ABE shown).\n")

  } else {

    txt <- paste(txt, "\n  Failed (ABE not shown).\n")

  }

} else {

  if (!PE.H0) {

    txt <- paste(txt, "\n  Point estimate within 80.00 \u2013 125.00%.")

  } else {

    txt <- paste(txt, "\n  Point estimate not within 80.00 \u2013 125.00%.")

  }

  if (sum(CI.H0, BE.H0) == 0) {

    txt <- paste(txt, "\n  Passed both criteria (ABEL shown).\n")

  } else {

    txt <- paste(txt, "\n  Failed at least one criterion (ABEL not shown).\n")

  }

}

cat(txt)

Voilà!

Design : 2x2x4 Subjects / sequence: 23 / 22 CVwR : 38.00% (reference-scaling applicable) Regulatory settings: EMA CVswitch : 30.00% CVcap : 50.00% Regulatory constant: 0.760 Nominal α : 0.0500 (TIE 0.0549) Iteratively adj. α : 0.0391 (TIE 0.0500) Point estimate : 115.00% Acceptance range : 75.65 – 132.20% (expanded = 'ABEL') 92.18% CI : 104.35 – 126.74% Assessment: Confidence interval entirely within acceptance range. Point estimate within 80.00 – 125.00%. Passed both criteria (ABEL shown).

Higher CV_wR 41% when the inflation of the Type I Error will hit no more:

Design : 2x2x4 Subjects / sequence: 23 / 22 CVwR : 41.00% (reference-scaling applicable) Regulatory settings: EMA CVswitch : 30.00% CVcap : 50.00% Regulatory constant: 0.760 Nominal α : 0.0500 (TIE 0.0478) Point estimate : 115.00% Acceptance range : 74.11 – 134.93% (expanded = 'ABEL') 90.00% CI : 104.33 – 126.76% Assessment: Confidence interval entirely within acceptance range. Point estimate within 80.00 – 125.00%. Passed both criteria (ABEL shown).

CV_wR 28%; no reference-scaling but still some adjustment required:

Design : 2x2x4 Subjects / sequence: 23 / 22 CVwR : 28.00% Nominal α : 0.0500 (TIE 0.0547) Iteratively adj. α : 0.0404 (TIE 0.0500) Point estimate : 115.00% Acceptance range : 80.00 – 125.00% (conventional) 91.93% CI : 107.00 – 123.60% Assessment: Confidence interval entirely within acceptance range. Passed (ABE shown).

Even lower CV_wR 27%:

Design : 2x2x4 Subjects / sequence: 23 / 22 CVwR : 27.00% Nominal α : 0.0500 (TIE 0.0494) Point estimate : 115.00% Acceptance range : 80.00 – 125.00% (conventional) 90.00% CI : 107.71 – 122.79% Assessment: Confidence interval entirely within acceptance range. Passed (ABE shown).

CV_wR 80% (!) and point estimate 124% (‼) for Health Canada’s conditions. Amazing.

Design : 2x2x4 Subjects / sequence: 23 / 22 CVwR : 80.00% (reference-scaling applicable) Regulatory settings: HC CVswitch : 30.00% CVcap : 57.38% Regulatory constant: 0.760 Nominal α : 0.0500 (TIE 0.0440) Point estimate : 124.00% Acceptance range : 66.67 – 150.00% (expanded = 'ABEL') 90.00% CI : 104.22 – 147.53% Assessment: Confidence interval entirely within acceptance range. Point estimate within 80.00 – 125.00%. Passed both criteria (ABEL shown).

No way for the EMA, 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. 🚮
Science Quotes