## FDA's HVD SAS Code / Stan­dard Error / CV [Regulatives / Guidelines]

Hi j_kevin,

❝ Q1: May I ask why in 2*2*2 cross over design, SE = sqrt(MSE)*sqrt(2/N)?

Correct is $$SE=\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right),}\tag{1}$$ where $$\small{n_1}$$ and $$\small{n_2}$$ are the number of subjects in sequences $$\small{1}$$ and $$\small{2}$$, respectively.*
If and only if sequences are balanced ($$\small{n_1=n_2\to N=n_1+n_2}$$) you can use \eqalign{SE&=\sqrt{\frac{2\,MSE}{N}}\\ &=\sqrt{MSE}\,\cdot\,\sqrt{\frac{2}{N}}}\tag{2} instead. In order to avoid falling into a trap (i.e., using $$(2)$$ if $$\small{n_1\neq n_2}$$), I strongly recommend to use always $$(1)$$.

❝ Here is the SAS code for the analysis of ABE with fully replicate design data.

PROC MIXED;

❝ CLASSES SEQ SUBJ PER TRT;

❝ MODEL Y = SEQ PER TRT/ DDFM=SATTERTH;

❝ RANDOM TRT/TYPE=FA0(2) SUB=SUBJ G;

❝ REPEATED/GRP=TRT SUB=SUBJ;

❝ ESTIMATE 'T vs. R' TRT 1 -1/CL ALPHA=0.1;

❝ Q2: When should I write 1 -1 and when should I write -1 1 in estimate statements?

All software use lexical order. Hence, it depends how you coded test and reference, i.e., in the FDA’s SAS-code as T and R. By default the comparison will be done R - T because in lexical order T > R and last one will act as the reference. If you have other ones, you may have to code TRT in ESTIMATE differently.

❝ Here is the covariance output from SAS

 Covariance Parameter Estimates

❝ Cov Parm  Subject       Group         Estimate

❝ FA(1,1)   subject                     0.2001

❝ FA(2,1)   subject                     0.1263

❝ FA(2,2)   subject                     0.1507

❝ Residual  subject   formulation Ref   0.01495

❝ Residual  subject   formulation Test  0.03265

❝ G(1,1) = 0.04005, G(2,1)=G(1,2)=0.02527, G(2,2)=0.03866

❝ In replicate design, the intra-subject variabilities are still calculated from the residuals as in a 2-way crossover, using:

❝ IntraCV_T = 100%*sqrt(exp(sig_WT^2)-1)

❝ IntraCV_R = 100%*sqrt(exp(sig_WR^2)-1)

If you remove the %, correct.

❝ The data include 10 subjects and 5 in seq 1 and 5 in seq2. This is a 2*2*4, 2 sequence 2 treatment and 4 period design, 90% CI of T/R is (0.8450984199, 1.104975448) and point estimate is 0.9663400049.

❝ I tried to use CVfromCI function from R package PowerTOST to calculate CV, and here is my r code:

100*CVfromCI(alpha=0.05, lower=0.8450985339, upper=1.104975299, n=c(5,5), design="2x2x4")

❝ and I get CV = 25.24456.

❝ Q3: Is this CV a pooled CV?

Correct, since the 90% CI is based on it.

❝ If I want to get CVwt and CVwr based on confidence interval, how can I do it using R package

That’s not possible because there is an infinite number of combinations of CVwT and CVwR giving the same CVw. Example:

library(PowerTOST) CV         <- 0.2524456 # CVw (pooled from the – unknown – CVwT and CVwR) # (random) variance ratios ratios     <- unique(sort(c(1, exp(runif(n = 10, min = log(0.5), max = log(2)))))) CVs        <- data.frame(var.ratio = ratios,                          CVwT = NA_real_,                          CVwR = NA_real_,                          CVw = NA_real_) CVs[, 2:3] <- CVp2CV(CV = CV, ratio = ratios) # pool CVwT and CVwR CVs[, 4]   <- mse2CV((CV2mse(CVs$CVwT) + CV2mse(CVs$CVwR)) / 2) print(CVs, row.names = FALSE) # q.e.d.  var.ratio      CVwT      CVwR       CVw  0.5431852 0.2108360 0.2887240 0.2524456  0.5917929 0.2168147 0.2841055 0.2524456  0.5928429 0.2169383 0.2840082 0.2524456  0.6628259 0.2246914 0.2777378 0.2524456  0.8504489 0.2417246 0.2627799 0.2524456  1.0000000 0.2524456 0.2524456 0.2524456  1.1854322 0.2632860 0.2411789 0.2524456  1.3868117 0.2728234 0.2305029 0.2524456  1.6521422 0.2828633 0.2183837 0.2524456  1.7311022 0.2854248 0.2151306 0.2524456  1.8842596 0.2899458 0.2092142 0.2524456

❝ Q4: I try to use the formula 100%*sqrt(exp(sig_W^2)-1) where sig_w^2 = (sig_wt^2+sig_wr^2)/2 to get the same CV(25.24456) based on the covariance output above, but the value is different. Is my calculation wrong? If so, can anyone please tell me the right way to get same CV?

No idea, I don’s speak SAS. AFAIK, FA(2,2) is the subject-by-formulation interaction. 0.1507 is unusually large.

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