Bioequivalence and Bioavailability Forum • Parallel Power Problems

ElMaestro
★★★

Denmark,
2013-02-14 15:46
(4874 d 23:34 ago)

Posting: # 10026
Views: 8,629

Parallel Power Problems [Two-Stage / GS Designs]

Hi all,

Potvin's method were developed for 2,2,2-BE crossover trials but I am interested in extending the methodology to parallel designs. In this regard I have two major areas of doubt:

I want to use the power formula that Potvin used, only with modification for the parallel situation. This is the formula that approximates power on basis of the central t-distribution. For technical reason (that means speed!) I am not going to us Owen's Q or shifted t. So what would the formula look like for the parallel case¹? In particular please describe your interpretation of n and df for your answer, if possible. Let's assume equal variances for simplicity, so that s is a pooled st.dev. estimate.
If we assume non-equal variances, what would the (or a) formula looks like then²?

Best regards and muchas gracias for any input you can give,

EM.

Notes:

I tried to read Julious, C&L, and even the documentation for power.TOST but stil have my doubts.
A less stringent solution at this step would be to simplify and assume equal variances for the powering step alone. This would then imply derivation of a pooled estimate of s which goes into the formula under question 1.

d_labes
★★★

Berlin, Germany,
2013-02-15 10:16
(4874 d 05:04 ago)

@ ElMaestro
Posting: # 10028
Views: 7,284

Parallel Power shifted

Post reply

Dear ElMaestro,

❝ Notes:

❝ 1: I tried to read Julious, C&L, and even the documentation for power.TOST but stil have my doubts.

Since I have done the documentation of PowerTOST I'm not able to answer until I know where your doubts are.

I will try with the fundamental relationship of approximating the non-central t-distribution with the 'shifted' central t-distribution:

pt(t, df, delta) ~ pt(t-delta, df)

Using this relationship the strategy to follow is: Use the power formulas given by Julious for the parallel group design, based on non-central t-distribution, and apply the approximation via 'shifted' central t-distribution.

Hope this helps.

BTW: Do you really need that last "Quäntchen" for speed? I can't believe that the usage of non-central t-distribution versus central t-distribution will make much a difference in a compiler environment.

—
Regards,

Detlew

ElMaestro
★★★

Denmark,
2013-02-15 12:37
(4874 d 02:44 ago)

@ d_labes
Posting: # 10029
Views: 7,282

Doubts explained

Post reply

Dear d_labes,

thanks for answering.

I will try to describe. The power given by Potvin is
P= Ft(a/(s*sqrt(2/n)) - t, DF) - Ft(b/(s*sqrt(2/n)) - t, DF)

where a and b are simple constants, s is derived from the CV and n is the total sample size, t is the critical value given some alpha and the df (=n-2). In a 2,2,2BE crossover we have n*2 observations. DF loss:
Treatment 2 (or intercept 1 + treatment 1)
Sequence 1
Subject n-1
Period 1
and then we gain 1 due to redundancy/crossover:
So
DF₂₂₂=2n-2-1-(n-1)-1 + 1 = n-2

In a parallel study with n subjects we have n observations just lose 2 df's due to the two treatments (or one treatment plus intercept).
DF_par=n-2

Ach so....
So if power formulae (whcihever Owen's Q, shifted t, central t ...) can be ported without to Parallel studies then a parallel study and a 2,2,2-be study should have the same power for a given n, theta and s?
No. We probably need to take some kind of design constant into consideration. In this post the design constants and n were defined slightly different from the current version of power.TOST I think, at least it looks like that when I go
known.designs() in your brilliant package (installed last week).

About here here I have lost orientation. I do not know how to take the design into consideration when using Potvin's equation to calculate power for the parallel situation.

(Post changed here due to slow-working brain!)

❝ BTW: Do you really need that last "Quäntchen" for speed? I can't believe that the usage of non-central t-distribution versus central t-distribution will make much a difference in a compiler environment.

Yes I believe I need it. I can explain the trick later.

—
Pass or fail!
ElMaestro

d_labes
★★★

Berlin, Germany,
2013-02-15 15:31
(4873 d 23:49 ago)

@ ElMaestro
Posting: # 10033
Views: 7,358

Formulas

Post reply

Dear Ol'Simulant,

according to Julious¹⁾, log-transformed data, parallel group design with common variance, non-central t-approximation:

power = pt(-tcrit, df, delta2) - pt(tcrit, df, delta1)

with

df     = nT+nR-2           # unequal sizes of both groups allowed

tcrit  = qt(1-alpha, df)   # t-quantil, R function nomenclature       

delta1 = (log(GMR) - log(0.8))/SE

delta2 = (log(GMR) - log(1.25))/SE

and

SE  = sqrt((1/nT+1/nR)*mse)

mse = log(1.0 + CV^2).

If you prefer to think in total number n=nT+nR and have a balanced design with nT=nR=n/2 some of the formula reduce to:

df = n-2

SE = sqrt((4/n)*mse)

Note the design constant=4 in the last formulas (in terms of total n). Different from the 2 in your formulas for the classical 2x2x2 crossover.

If it comes to unequal CV's, variances: Duno exactly.

One option would be simple pooling. mse = ((nT-1)*sT^2 + (nR-1)*sR^2)/(nT+nR-2).
Another option is thinking in direction of Welch-t-test. But I'm absolutely not sure if the t-distribution does apply in this case. And non-integer df's are horror for me.

Hope I have complied all these stuff without typos.
The application of the 'shifted' central t-approximation is left to you :cool:

¹⁾ S.A. Julious
"TUTORIAL IN BIOSTATISTICS
Sample sizes for clinical trials with Normalal data"
Statistics in Medicine 2004; 23: 1921-1986
page 1970, formula 68

—
Regards,

Detlew

ElMaestro ★★★ Denmark, 2013-02-15 17:16 (4873 d 22:05 ago) @ d_labes Posting: # 10035 Views: 7,103	Formulas Post reply
	Thanks a lot, d_labes, [rest of post deleted due to low IQ] — Pass or fail! ElMaestro