90% confidence interval for R_dnm [Study As­sess­ment]

posted by Shuanghe  – Spain, 2019-01-04 18:45 (1932 d 20:40 ago) – Posting: # 19728
Views: 18,112

Dear all,

Man, I should checked here before I started my work. It could save me a lot of time...

Recently I was helping one of my colleagues for dose proportionality study and power model in Smith's article is the preferred method. While I did figured out the "correct" degree of freedom method and reproduce all reported results such as intercept, slope, and 90% CI of those values, ρ1, ρ2, the ratio of dose-normalised geometric mean value Rdnm,..., I could not figure out how Smith obtained the 90% confidence interval for Rdnm (0.477, 0.698).

According to Smith , testing θL < Rdnm < θU to draw conclusion of dose proportionality is equivalent to testing 1 + ln(θL)/ln(r) < β1 < 1 + ln(θH)/ln(r). The latter is what we do and obtaining 90% CI for slope is easy enough and we can judge of dose proportionality based on that. However, it's also interesting to be able to reproduce all Smith's results.

In his article (pp.1282, 2nd paragraph), Smith wrote that "The 90% CI for the difference in log-transformed means was calculated within the MIXED procedure. Exponentiation of each limit and division by r gave the 90% CI for Rdnm. This CI lay completely outside (0.80, 1.25), indicating a disproportionate increase in Cmax."


What limit was he talking about? My sas code is basically identical to detlew's above and there's no apparent "limit" in the output that's similar to what Smith mentioned. I tried also adding ESTIMATE or LSMEANS statement with various codes, couldn't get it at all. Any help?

Many thanks.
—
All the best,
Shuanghe

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