Weighting scheme [Power / Sample Size]

posted by Helmut Homepage – Vienna, Austria, 2009-05-26 21:52 (5822 d 06:47 ago) – Posting: # 3771
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Dear ElMaestro!

❝ ❝ α>0 and β≠0, weighted regression (w = 1/x)


❝ Could you explain where that weight comes from?


Chow & Liu (all editions), Chapter ‘Dose Proportionality Study’:

Let Y be AUC or Cmax and X be the dose level. Because, often, the standard deviation of Y increases as the dose increases, the primary assumption of dose proportionality is that the standard deviation of Y is proportional to X; that is,

Var(Y) = X2σ2,

where σ2 usually consists of inter- and intra-subject variability.

After describing the different models, they continue

It can be seen that model 1 can be used to evaluate dose proportionality by […] using a weighted linear regression with weights equal to X -1 based on the original data (X, Y).
[…] Model 2 […] can be tested using a weighted linear regression with weights equal to X -1 and with the original data (X, Y).
[…] similar to model 2, model 3 can be tested using a weighted linear regression with log-transformed data (logX, logY).


❝ I would intuitively say it should be non-weighted, as nonlinearity can easily be e.g. a phenomenon only visible/measurable in the upper dose region. In this case I would think this weight would work against the chance of finding it.


Hhm – maybe. Weighting comes from the quoted proportionality of the standard deviation of Y to X. I just checked it with one of my data sets (6 dose levels from 0.5–12, lower than dose proportional: b=0.59) and got an AIC of 286 for the unweighted model and an AIC of 263 for w=1/X. Residuals of the unweighted model showed the classical funnel shape. So I think it's not unreasonable.

Martin pointed me to a recent review*. Again, w=1/X is claimed for the linear model. They don’t give a weighting scheme for the power model, but derive a sample size estimation!
Interesting stuff.


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