MSE calculation from CI of 3x3x3 study [General Sta­tis­tics]

posted by d_labes  – Berlin, Germany, 2012-09-14 11:43 (4035 d 05:48 ago) – Posting: # 9215
Views: 10,664

Dear John

❝ Hi Helmut,

Not interested in the opinion of other members? :-D

The common formula for the 90% CIs of cross-over studies is (in the log domain):
  [lower, upper] = point ± tval(1-alpha/2, df)*sqrt(MSE)*sqrt(bk(ni)*sum(1/ni))
ni are the number of subjects in the sequence groups, bk(ni) is the so called design constant (here if ni are given) and df the degrees of freedom from the cross-over ANOVA.

Under the assumption of balanced designs the above formula can be given in terms of total number of subjects (N=sum(ni)):
  [lower, upper] = point + tval(1-alpha/2, df)*sqrt(MSE)*sqrt(bk/N)
with bk the design constant in terms of total number if subjects.

Here the design characteristics for your question (tmt x seq x period):
design   df                       bk(ni)    bk
2x2x2   n1+n2-2                   1/2      2
3x3x3   2*(n1+n2+n3)-4            2/9      2
3x6x3   2*(n1+n2+n3+n4+n5+n6)-4   1/18     2

The latter design is 6-sequence williams design for 3 treatments and 3 periods.

Insertion of the 2x2x2 characteristics will give the formulas in Helmut's lectures.

With a little algebra you should be able to derive the MSE formula for your case :cool: . But you could save the effort for better things. Use the function CVfromCI() from the R-package PowerTOST which has some more designs you eventually need.
But in the moment it works only for balanced designs i.e. same number of subjects in sequence groups. Therefore it needs only the total number of subjects under study. This is in most cases accessible from literature while the number of subjects in sequences is usually not given.

BTW: The red emphasis above shows where the great admin of this forum and itinerant preacher in the matter of BE :-D errs if we talk about alpha=0.1 i.e 90% CIs.



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