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Dear PharmPKers,
Could someone explain why the SD for a cross-over study would be
SD=sqrt(2*MSE);
why it have to times 2 of MSE (sigma^2)?
Xia Luo
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Hi Xia Luo,
SD=sqrt(2*MSE) could be the correct SD for a difference.
For example, if the variances for cmax under 2 different treatments are equal, you have
var(diff_cmax) = 2*var(cmax)
Peter
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Thank you Peter.
I am using nQuery to calculate sample size. For a bioequivalence study in a cross-over study design,
given the PK is log normally distributed and the Xik1, Xik2 are log transformed PK parameters from
ith subject in kth sequence in during periods 1 or 2, when we fill known MSE (in log scale) from
experience, the nQuery also gives out SD in log scale: sqrtn (2*MSE) (in log scale).
Wondering where I can find details about how these relation of SD and MSE is derived. Or in another
word, I would like to understand more theory behind it. I know SD(iXk1)=sqrt((Xik1-Xk1-bar)^2/n-1),
so does SD(Xik2). Perhaps MSE is not a simple expression and it depends on the model. Still at last,
how is the equation of SD =sprt(2*MSE) derived?
From your answer below Var(diff_cmax)=2*var(cmax), does it say: Var(Yi-Xi)=Var(Y) + Var(X) assuming
Var are identical for the two treatment periods for Cmax? We can not assume the two values (Xik1 and
Xik2) from two treatment period are independent, they are correlated. So what else, derived that
equation?
Any recommendation for a reference book?
Thank you
Xia
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Hi Xia,
For BE test, the purpose is to estimate (X1hat-X2hat), X1hat: the estimate for test drug; X2hat: the
estimate for reference drug.
For Cross-over BE studies, the statistical test for BE is through constructing a new variable from
crossover data, as it is not as simple as parallel study where you can use mean of each group to
estimate X1hat and X2hat.
For example, if you have 2 treatment, 2 period, 2 way cross-over design, you would do ANOVA test on
the data, and you need to analyze the fixed effects of "period","treatment" and "sequence", random
effects of "intra-individual" and "interindividual" variability.
For individual m,
Yijkm=Ai(sequence 1/2)+ Bj(period 1/2)+ Xk(treatment 1/2) + mu(m,i) (interindividual)+ epsi(i,j,m)
(intraindividual)
Using ANOVA, you assume that groups in comparison are independent and the variance within groups are
uniform because you have defined the fixed and random effects in the model. Another assumption is
normal distribution, that is why log transformation is recommended.
In this design, there are three factors(period, sequence and treatment), each with two levels(1 or
2).
In summary, you can assume that It is a process to extract only those random parts for comparison-
variance between groups comparing to variance within groups.
And within group (intraindividual) variance is considered to be uniform among all groups- sigma^2.
And the (X1hat-X2hat) is constructed through Y11,Y12,Y21,Y22 (means):
Y11(mean): sequence1, period 1,treatment 1, --------A(1)+B(1)+X(1)
Y12(mean): sequence1, period 2,treatment 2, --------A(1)+B(2)+X(2)
Y21(mean): sequence2, period 1, treatment 2, -------A(2)+B(1)+X(2)
Y22(mean): sequence2, period 2,treatment 1,-------- A(2)+B(2)+X(1)
X(2)-X(1) = 1/2*((Y12-Y11)+(Y21-Y22)) is constructed to estimate (X2hat-X1hat)
VAR(X2-X1)=1/4*(Var(Y12)+VAR(Y11)+VAR(Y21)+VAR(Y22))
1/4*(sigma^2/n12+sigma^2/n11+sigma^2/n21+sigma^2/n22) Note: Since Y12, Y11, Y21, Y22 is the mean of
(n12, n11, n21, n22) numbers of Yijkm.
Assume equal numbers n12=n11=n21=n22=N/2 (N is the total pts number for the study), You would get:
VAR(X2-X1)=1/4*(4*2*Sigma^2/N)=2*Sigma^2/N
SD(X2-X1)=Sqrt(2*MSE). MSE: mean squared error
I would recommend ANOVA related chapters in statistics for you to further understand this part. Hope
this helps!
Jie Zhou
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Hi Xia,
you may look into:
Shein-Chung Chow, Jen-Pei Liu:
Design and Analysis of Bioavailability and Bioequivalence Studies.
Peter
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