CVfromCI for higher order crossover designs [Power / Sample Size]

posted by d_labes  – Berlin, Germany, 2012-11-07 09:45 (3981 d 07:08 ago) – Posting: # 9519
Views: 3,058

Dear lovemysoul,

❝ When we calculate CV% of 2x2 design, MSE has to be calculated.

❝ MSE = 2 x [(CI_high - PE) / {Root(1/nA + 1/ nB) x TINV(1-0.05,Df)}]^2

❝ In case of 2x2 design, Df = nA + nB - 2

(Little correction by me. Note also that CI_high and PE are the values in the log domain, i.e. ln(CI_high), ln(PE) if you have the confidence interval as ratio!)

What you describe here is obtaining the MSE from the 90% confidence intervals if it is not directly given in the literature. Normally we obtain MSE directly from the corresponding ANOVA table.

❝ but Df of 6x3 williams design is 2n-4...

Correct. But note that n is the total number of subjects.

❝ in case of 6x3 williams design, how can i get CV%? just change Df formula in MSE formula is enough?

No. For a generalisation you have to adapt the square-root(...) also to account for the six sequence groups and adapt the constant 2 to another value appropriate to the design (sometimes called design constant).

In the moment I'm not in the mood to fuzzle out the resulting formula. Consider using R and the package PowerTOST. It contains a function CVfromCI() that will do the calculations for you, however assuming a sequence balanced design (all sequence groups with the same number of subjects). But in case of an unbalanced design you usually don't have informations about the number of subjects in the sequence groups and have to stick with the total number as implemented in CVfromCI().

❝ and can i get a sample size of 2x2 crossover design using the data that i got from 6x3 williams design study?

Yes. Why not?



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