Quantiles of the F-distribution [General Sta­tis­tics]

posted by d_labes  – Berlin, Germany, 2015-09-28 09:45  – Posting: # 15475
Views: 18,980

Dear Shuanghe,

» However, there's no sas code for the variability comparison ...

Wow! Seems correct. Was not aware of this fact.

At moment I don't have answers to all of your other questions but only to your points 1., 2. and 4.

Lets start with 2. and 4, the simpler ones:

» 2. From the sas code for sWR (from Dij=R1-R2 then running with PROC MIXED, output CovParms, and calculate sWR = SQRT(estimate/2)), is it correct to assume that we need to modify the code to have something like DijT = T1 - T2, then run the same procedure as for sWR to calculate sWT?


» 4. Since FDA request that only subject who complete all 4 periods should be included in the analysis, for obtaining F-distribution value, it seems the degree of freedom vT and vR will always be equal, which is always Ntotal - 2 for full replicate. Right?

Correct for full replicate 4-periods with 2 sequences with mentioned precondition.

Now to the mysterious lower/upper (upper/lower?) CI of the variabilities (your point 1.):
The answer lies hidden in the text of the Warfarin guidance.
"... Fα/2,ν1,ν2 is the value of the F-distribution with ν1 (numerator) and ν2 (denominator) degrees of freedom that has probability of α/2 to its right." (Emphasis be me. Similar text for F1-α/2,ν1,ν2).

Example in R speak with df1=12, df2=12, alpha/2=0.05:
F1 <- qf(0.05, df1=12, df2=12, lower.tail=FALSE)
gives F1 = 2.686637. This is the same as
F1 <- qf(1-0.05, df1=12, df2=12, lower.tail=TRUE)

Usually (and in SAS) the probability to the left is given back (integral over F-density from 0 to F).
# lower.tail=TRUE is default
F2 <- qf(0.05, df1=12, df2=12)

gives F2 = 0.3722125.
Data Finv;
  df1=12; df2=12; alpha=0.05;
  F1=Finv(1-alpha, df1, df2);
  F2=Finv(alpha, df1, df2);

gives also F1 = 2.686637, F2 = 0.3722125.

Hope this helps.



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