## ‘Two-at-a-Time’ approach with >3 treat­ments [Design Issues]

Dear all,

the EMA1–3 recommends for higher-order crossovers the ‘Two-at-a-Time’ approach, i.e., to exclude all but two treatments, leading to Incomplete Block Designs.4 That’s fine, since the ‘All at Once’ approach (an ANOVA of all data) may lead to biased estimates and an inflated Type I Error.5,6

So far I had to deal only with three treatments and planned pairwise comparisons $$\small{\text{A}\;vs\;\text{C}}$$ and $$\small{\text{B}\;vs\;\text{C}}$$. For a six-sequence Williams’ design we get$$\small{\begin{array}{c|ccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & \text{A} & \text{B} & \text{C}\\ 2 & \text{A} & \text{C} & \text{B}\\ 3 & \text{B} & \text{A} & \text{C}\\ 4 & \text{B} & \text{C} & \text{A}\\ 5 & \text{C} & \text{A} & \text{B}\\ 6 & \text{C} & \text{B} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{c|ccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & \text{A} & {\color{Red}\bullet} & \text{C}\\ 2 & \text{A} & \text{C} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{A} & \text{C}\\ 4 & {\color{Red}\bullet} & \text{C} & \text{A}\\ 5 & \text{C} & \text{A} & {\color{Red}\bullet}\\ 6 & \text{C} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|ccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & {\color{Red}\bullet} & \text{B} & \text{C}\\ 2 & {\color{Red}\bullet} & \text{C} & \text{B}\\ 3 & \text{B} & {\color{Red}\bullet} & \text{C}\\ 4 & \text{B} & \text{C} & {\color{Red}\bullet}\\ 5 & \text{C} & {\color{Red}\bullet} & \text{B}\\ 6 & \text{C} & \text{B} & {\color{Red}\bullet}\\ \end{array}}$$ where $$\small{{\color{Red}\bullet}}$$ denotes an excluded treatment. Both IBDs are balanced. Excellent.

What about four treatments in a Williams’ design with pairwise comparisons $$\small{\text{A}\;vs\;\text{D}}$$, $$\small{\text{B}\;vs\;\text{D}}$$, and $$\small{\text{C}\;vs\;\text{D}}$$? $$\small{\begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & \text{A} & \text{B} & \text{C} & \text{D}\\ 2 & \text{B} & \text{D} & \text{A} & \text{C}\\ 3 & \text{C} & \text{A} & \text{B} & \text{D}\\ 4 & \text{D} & \text{C} & \text{B} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV} \\\hline 1 & \text{A} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{D} & \text{A} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{A} & {\color{Red}\bullet} & \text{D}\\ 4 & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet} & \text{D}\\ 2 & \text{B} & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{B} & \text{D}\\ 4 & \text{D} & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{C} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet} & \text{C}\\ 3 & \text{C} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D}\\ 4 & \text{D} & \text{C} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ \end{array}}$$ Now the IBDs are imbalanced and hence, lacking period effects have to be assumed. Nasty.
Solution: Use one of the 24 arrangements given by Senn.6 Here the first as an example:
$$\small{\begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & \text{A} & \text{B} & \text{C} & \text{D}\\ 2 & \text{B} & \text{A} & \text{D} & \text{C}\\ 3 & \text{C} & \text{D} & \text{A} & \text{B}\\ 4 & \text{D} & \text{C} & \text{B} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV} \\\hline 1 & \text{A} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{A} & \text{D} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{D} & \text{A} & {\color{Red}\bullet}\\ 4 & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet} & \text{D}\\ 2 & \text{B} & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet} & \text{B}\\ 4 & \text{D} & {\color{Red}\bullet} & \text{B} & {\color{Red}\bullet}\\ \end{array}{\color{Blue}\wedge} \begin{array}{c|cccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{C} & \text{D}\\ 2 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D} & \text{C}\\ 3 & \text{C} & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ 4 & \text{D} & \text{C} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ \end{array}}$$ All is good. An -script at the end.
BTW, the common Latin Square ($$\small{\text{ABCD}\,|\,\text{BCDA}\,|\,\text{CDAB}\,|\,\text{DABC}}$$) would do as well.

1. EMA, CHMP. Guideline on the Investigation of Bioequivalence. CPMP/EWP/QWP/1401/98 Rev. 1/Corr **. London. 20 January 2010. Online.
2. Brown D. Presentation at the 3rd EGA Symposium on Bioequivalence. London. June 2010. Slides.
3. EGA. Revised EMA Bioequivalence Guideline. Questions & Answers. Online.
4. Cochran WG, Cox GM. Experimental Designs. New York: Wiley; 2nd ed. 1957. Chapter 9, p. 376–95.
5. Schuirmann DJ. Two at a Time? Or All at Once? International Biometric Society, Eastern North American Region, Spring Meeting. Pittsburgh, PA. March 28–31, 2004. Abstract.
6. D’Angelo P. Testing for Bioequivalence in Higher‐Order Crossover Designs: Two‐at‐a‐Time Principle Versus Pooled ANOVA. 2nd Conference of the Global Bioequivalence Harmonisation Initiative. Rockville, MD. 15–16 Sep­tem­ber, 2016.
7. Senn S. Cross-over Trials in Clinical Research. Chichester: Wiley; 2nd ed. 2002. Table 5.1. p. 163.

library(PowerTOST) library(randomizeBE) # sample size for evaluation by ‘Two at a Time’ # note that the design has to specified as "2x2" n    <- sampleN.TOST(CV = 0.20, theta0 = 0.95, targetpower = 0.8,                      design = "2x2", print = FALSE)[["Sample size"]] # specify the sequences and get the randomization list seqs <- c("ABCD", "BADC", "CDAB", "DCBA") rl   <- RL4(nsubj = n, seqs = seqs)$rl rl <- cbind(rl, IBD.1 = NA, IBD.2 = NA, IBD.3 = NA) # extract pairwise comparisons (A vs D, B vs D, C vs D) and build the IBDs for (j in 1:nrow(rl)) { rl$IBD.1[j] <- gsub("[B, C]", "•", rl$sequence[j]) # keep A and D rl$IBD.2[j] <- gsub("[C, A]", "•", rl$sequence[j]) # keep B and D rl$IBD.3[j] <- gsub("[A, B]", "•", rl$sequence[j]) # keep C and D } rl <- rl[, -2] # remove seqno # check IBDs for balance c.1 <- c.2 <- c.3 <- NA for (j in 1:4) { c.1[j] <- sum(substr(rl$IBD.1, j, j) == "A") == sum(substr(rl$IBD.1, j, j) == "D") c.2[j] <- sum(substr(rl$IBD.2, j, j) == "B") == sum(substr(rl$IBD.2, j, j) == "D") c.3[j] <- sum(substr(rl$IBD.3, j, j) == "C") == sum(substr(rl\$IBD.3, j, j) == "D") } chk  <- setNames(c(sum(c.1), sum(c.2), sum(c.3)), c("IBD.1", "IBD.2", "IBD.3")) txt  <- paste("Specified sequences:", paste(seqs, collapse = ", ")) if (length(unique(chk)) == 1) {   txt <- paste(txt,                "\nPassed check       : All IBDs are balanced.\n")   } else {   txt <- paste(txt,               "\nFailed check       : IBDs are not balanced:",               "Examine the specified sequences.\n") } print(rl, row.names = FALSE) cat(txt)

Gives
 subject sequence IBD.1 IBD.2 IBD.3        1     DCBA  D••A  D•B•  DC••        2     CDAB  •DA•  •D•B  CD••        3     CDAB  •DA•  •D•B  CD••        4     BADC  •AD•  B•D•  ••DC        5     BADC  •AD•  B•D•  ••DC        6     ABCD  A••D  •B•D  ••CD        7     DCBA  D••A  D•B•  DC••        8     ABCD  A••D  •B•D  ••CD        9     CDAB  •DA•  •D•B  CD••       10     BADC  •AD•  B•D•  ••DC       11     BADC  •AD•  B•D•  ••DC       12     ABCD  A••D  •B•D  ••CD       13     DCBA  D••A  D•B•  DC••       14     DCBA  D••A  D•B•  DC••       15     CDAB  •DA•  •D•B  CD••       16     ABCD  A••D  •B•D  ••CD       17     DCBA  D••A  D•B•  DC••       18     CDAB  •DA•  •D•B  CD••       19     ABCD  A••D  •B•D  ••CD       20     BADC  •AD•  B•D•  ••DC Specified sequences: ABCD, BADC, CDAB, DCBA Passed check       : All IBDs are balanced.

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