sweiner
☆

2020-01-22 19:22
(200 d 12:59 ago)

Posting: # 21081
Views: 1,724

## Higher-order cross­overs [Power / Sample Size]

Hello,
I would like to obtain an advice on BE study sample size estimation.
Let’s say, we are planning a 2x2 crossover study for BE, and the intra-subject coefficient of variation for our test compound, from a prior study, is 23%. The assumed geometric mean ratio is 1.00, alpha is 0.05, equivalence limits are 80%-125%, and we desire power of 80%. These criteria result in an overall sample size of 20 subjects, or 10 per sequence.
Subsequently, we want to add 2 extra dose levels of our test drug, resulting in a 4x4 crossover trial. Let’s assume that %CV and GMR are not changing, as no further data is available. For this new scenario, is it appropriate to: 1) use the original sample size of N=20 and simply divide it over 4 sequences (5 per sequence), or 2) take the original per sequence sample size of 10 and multiply by 4 to get 40 subjects overall needed? In either case 1) or 2), is there a supporting reference you can provide?
Thanks,
Sveta

Edit: Category and subject line changed; see also this post #1 and #2[Helmut]
Helmut
★★★

Vienna, Austria,
2020-01-23 12:48
(199 d 19:34 ago)

@ sweiner
Posting: # 21083
Views: 995

## Purpose of the study?

Hi Sveta,

» Let’s say, we are planning a 2x2 crossover study for BE, and the intra-subject coefficient of variation for our test compound, from a prior study, is 23%. The assumed geometric mean ratio is 1.00, …

Which PE did you find in the prior study? Its sample size would be helpful as well.

» … alpha is 0.05, equivalence limits are 80%-125%, and we desire power of 80%. These criteria result in an overall sample size of 20 subjects, or 10 per sequence.

Correct – if you are a believer of the ‘carved in stone’ approach (i.e., that in the planned study the CV will be exactly 23% and the PE exactly 1). I suggest to have a look at Example 3 of the ABE-Vignette of the R package PowerTOST to reconsider your assumptions. See also there (slide 8 and followings).

» Subsequently, …

What do you mean by ‘subsequently’?

» … we want to add 2 extra dose levels of our test drug, resulting in a 4x4 crossover trial.

What is the purpose of the study – dose proportionality? If yes, that’s another pot of tea.

» Let’s assume that %CV and GMR are not changing, as no further data is available.

Rather strong assumptions, right?

» For this new scenario, is it appropriate to: 1) use the original sample size of N=20 and simply divide it over 4 sequences (5 per sequence), or 2) take the original per sequence sample size of 10 and multiply by 4 to get 40 subjects overall needed?

It depends whether you want to show dose-normalized equivalence (i.e., strict dose normality) or dose proportionality by the power model $$E[Y]=\alpha \cdot D^{\; \beta}$$.
Say we have four formulations (A, the reference R, B, and C) and three dose levels (x, y, z), where A = R = x, B = y, C = z.
You randomize subjects either to the Latin Square ARBC|RBCA|BCAR|CARB or any one of the six Williams’ designs ACBR|RBCA|BARC|CRAB, ARBC|RCAB|BACR|CBRA, ACRB|RABC|BRCA|CBAR, ABRC|RACB|BCAR|CRBA, ABCR|RCBA|BRAC|CARB, ARCB|RBAC|BCRA|CABR.
• If you want to assess dose-normalized equivalence, follow the ‘Two‐at‐a‐Time Principle’, i.e., perform pairwise comparisons whilst excluding the others.1 Do not use a pooled ANOVA because it may give biased estimates and/or inflate the Type I Error.2,3
That means your first scenario is correct (you estimate the sample size like a 2×2×2 crossover) and get three incomplete block designs to evaluate: AR, xB/yR, and xC/zR.
If you insist in a pooled ANOVA for any reasons, you end up with 20 subjects as well (though with slightly higher power due to the higher degrees of freedom 3n–6 compared to the n–2 in the 2×2×2):
library(PowerTOST) x         <- data.frame(design = c("2x2x2", "4x4"), n = NA, power.pct = NA) x[1, 2:3] <- sampleN.TOST(CV = 0.23, theta0 = 1, design = "2x2x2",                           print = FALSE)[7:8] x[2, 2:3] <- sampleN.TOST(CV = 0.23, theta0 = 1, design = "4x4",                           print = FALSE)[7:8] x[3]      <- round(100*x[3], 2) print(x, row.names = FALSE) design  n power.pct  2x2x2 20     82.08    4x4 20     84.55

• If you want to assess dose proportionality, the acceptance range depends on the dose-range (the wider the dose-range the narrower it will be). For the sample size estimation you need the dose-range and an assumed slope of the power model β. Furthermore, it depends whether you plan a confirmatory4 or exploratory5 study.

1. Schuirmann D. Two at a Time? Or All at Once? Pittsburgh: International Biometric Society, Eastern North American Region, Spring Meeting; March 28–31, 2005. Abstract.
2. European Medicines Agency, CHMP. Guideline on the Investigation of Bioequivalence. London; 20 January 2010. Doc. Ref. CPMP/EWP/QWP/1401/98 Rev. 1/ Corr **.
3. D’Angelo P. Testing for Bioequivalence in Higher‐Order Crossover Designs: Two‐at‐a‐Time Principle Versus Pooled ANOVA. Rockville: 2nd Workshop of the Global Bioequivalence Harmonisation Initiative; 15–16 September, 2016. Some of her slides in this post.
4. Smith BP, Vandenhende FR, DeSante KA, Farid NA, Welch PA, Callaghan JT, Forgue ST. Confidence Interval Criteria for Assessment of Dose Proportionality. Pharm Res. 2000; 17(19): 1278–83. doi:10.1023/a:1026451721686.
5. Hummel J, McKendrick S, Brindley C, French R. Exploratory assessment of dose proportionality: review of current approaches and proposal for a practical criterion. Pharm. Stat. 2009; 8(1): 38–49. doi:10.1002/pst.326.

Dif-tor heh smusma 🖖
Helmut Schütz

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

2020-01-26 15:50
(196 d 16:31 ago)

@ sweiner
Posting: # 21086
Views: 845

## Higher-order cross­overs

Helmut, thank you for the detailed response, examples in R, and references! I put the answers to the questions below (I tried to change color on responses, but not sure if that worked). Thanks. Sveta

» Which PE did you find in the prior study? Its sample size would be helpful as well.

This is more a hypothetical example, so really not known

» » … alpha is 0.05, equivalence limits are 80%-125%, and we desire power of 80%. These criteria result in an overall sample size of 20 subjects, or 10 per sequence.

» Correct – if you are a believer of the ‘carved in stone’ approach (i.e., that in the planned study the CV will be exactly 23% and the PE exactly 1). I suggest to have a look at Example 3 of the ABE-Vignette of the R package PowerTOST to reconsider your assumptions. See also there (slide 8 and followings).

» » Subsequently, …
» What do you mean by ‘subsequently’?

‘Subsequently’ here refers to the next study on somewhat different population (where we expect smaller value of CV), with higher order of crossover

» » … we want to add 2 extra dose levels of our test drug, resulting in a 4x4 crossover trial.
» What is the purpose of the study – dose proportionality? If yes, that’s another pot of tea.

The purpose is to show dose-normalized equivalence

» » Let’s assume that %CV and GMR are not changing, as no further data is available.
» Rather strong assumptions, right?

We actually expect a smaller value of %CV, not sure on GMR though

» » For this new scenario, is it appropriate to: 1) use the original sample size of N=20 and simply divide it over 4 sequences (5 per sequence), or 2) take the original per sequence sample size of 10 and multiply by 4 to get 40 subjects overall needed?
» It depends whether you want to show dose-normalized equivalence (i.e., strict dose normality) or dose proportionality by the power model $$E[Y]=\alpha \cdot D^{\; \beta}$$.
» Say we have four formulations (A, the reference R, B, and C) and three dose levels (x, y, z), where A = R = x, B = y, C = z.
» You randomize subjects either to the Latin Square ARBC|RBCA|BCAR|CARB or any one of the six Williams’ designs ACBR|RBCA|BARC|CRAB, ARBC|RCAB|BACR|CBRA, ACRB|RABC|BRCA|CBAR, ABRC|RACB|BCAR|CRBA, ABCR|RCBA|BRAC|CARB, ARCB|RBAC|BCRA|CABR.

» — If you want to assess dose-normalized equivalence, follow the ‘Two‐at‐a‐Time Principle’, i.e., perform pairwise comparisons whilst excluding the others.1 Do not use a pooled ANOVA because it may give biased estimates and/or inflate the Type I Error.2,3
» That means your first scenario is correct (you estimate the sample size like a 2×2×2 crossover) and get three incomplete block designs to evaluate: AR, xB/yR, and xC/zR.
» If you insist in a pooled ANOVA for any reasons, you end up with 20 subjects as well (though with slightly higher power due to the higher degrees of freedom 3n–6 compared to the n–2 in the 2×2×2):
» library(PowerTOST)
» x         <- data.frame(design = c("2x2x2", "4x4"), n = NA, power.pct = NA)
» x[1, 2:3] <- sampleN.TOST(CV = 0.23, theta0 = 1, design = "2x2x2",
»                           print = FALSE)[7:8]
» x[2, 2:3] <- sampleN.TOST(CV = 0.23, theta0 = 1, design = "4x4",
»                           print = FALSE)[7:8]
» x[3]      <- round(100*x[3], 2)
» print(x, row.names = FALSE)
»
» design  n power.pct
»  2x2x2 20     82.08

This is what we are trying to do, so this is exactly the response I was looking for.

I used the following example in R with CV=1.04 and got similar overall SS from either 2x2 or 4x4, so then the overall SS will just need to be split to higher order crossover – about 67 subjects per sequence for 2x2 and about 34 per sequence for 4x4. I assume that if we had to do even higher order crossover, then we would follow similar approach. R does not support 5x5 for example, but if we had to use 5x5 it would be about 26 subjects per sequence, it seems to be,is it correct?

CV=1.04, design = 2x2:
sampleN.TOST(CV = 1.04, theta1=0.67, theta2 = 1.5,  theta0 = 1.1, design = "2x2", print = FALSE, logscale=TRUE, targetpower=0.9)[7:8]   Sample size Achieved power 1         134      0.9029653

CV=1.04, design = 4x4
sampleN.TOST(CV = 1.04, theta1=0.67, theta2 = 1.5,  theta0 = 1.1, design = "4x4", print = FALSE, logscale=TRUE, targetpower=0.9)[7:8]   Sample size Achieved power 1           132      0.9008229