Power of equality [Software]

posted by d_labes  – Berlin, Germany, 2011-11-10 17:42 (4543 d 01:25 ago) – Posting: # 7652
Views: 8,788

Dear Helmut,

❝ was really nice meeting you yesterday!


Meeting You, yes it was, really :yes: :smoke:!

❝ BTW, fog at both Berlin’s and Vienna’s airports; arrived at home at midnight… :sleeping:


For that I feel sorry for you. Eventually we should had smoked to a lesser degree? :-D

Regarding the WLN power I must correct myself to a certain degree.
Chow, Shao and Wang1) call a test with the hypotheses pair
 H0 µT = µR
 HA µT ≠ µR

a test of equality to distinguish it from a superiority test with the hypothesis pair
 H0 µT-µR ≤ delta
 HA µT-µR > delta

with delta the superiority margin. Ok, I think this is semantics.

These authors give for the cross-over design the following formula for the power of the 'equality test' (have translated it in R notation):
  power = 1 - pt(tcrit,df,ncp=eps*sqrt(2*n)/sigma)
            + pt(-tcrit,df,ncp=eps*sqrt(2*n)/sigma)

where eps is the true difference for which the power shall calculated and tcrit is the quantile of the central t-distribution to the confidence level 1-alpha/2 with df = n-2 degrees of freedom.
Setting eps=0.2*LSMref and sigma/sqrt(2*n)=SEdiff and let b=0.2*LSMref/SEdiff
we get
  power = 1 - pt(tcrit,df,ncp=b) + pt(-tcrit,df,ncp=b)

With the approximation of the non-central t-distri via 'shifted' central t-distri according to
  pt(x,df,ncp) ~ pt(x-ncp,df)

we obtain yicaotings formulas!

Lets calculate via non-central t-distribution using your given data:
Full dataset, untransformed
b  = 4.42291
df = 22
tcrit = qt(0.95,22) = 1.717144...
power = 1 - 0.004184843 + 1.415275e-09 = 0.9958152

Seems to agree within the approximation used in WNL. At least the right order of magnitude.

Full dataset, logtransformed
b  = 30.57179
df = 22
tcrit = qt(0.95,22) = 1.717144...
power = 1-2.395435e-180+0 = 1

Seem not to function! No agreement to the WNL results. Due to insufficient degree of approximation for large non-centrality or due to false formula?

b  = ln(0.8)/SEdiff = -7.788037 (yicaotings formula)
df = 22
tcrit = qt(0.95,22) = 1.717144...
power = 1-1+1 = 1

Seems also not to function! This is surprising! :confused:
yicaoting: Which SEdiff did you use?

Reduced dataset is the homework for all :-D.
Sorry for all that numbers with insufficient decimals.


1) Chow, Shao and Wang
"Sample size calculations in clinical research"
Marcel Dekker, New York, NY 2003

Regards,

Detlew

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