90% confidence interval for R_dnm [Study As­sess­ment]

posted by d_labes  – Berlin, Germany, 2019-01-05 15:01 (1994 d 10:23 ago) – Posting: # 19731
Views: 18,679

Dear Shuanghe,

First: Happy New Year to You and to All.

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

:cool: Late but hopefully not too late insight. As I sometimes stated: All answers (of asked or not asked questions) are here. You only have to dig out what you are interested in.

❝ 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).


Can you please elaborate where your difficulties arose? Able to obtain a point estimate of Rdnm but no 90% CI thereof?

❝ ...

❝ In his article (pp.1282, 2nd paragraph), Smith wrote that "[i]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 ...


For me this is a dubious description (the whole paragraph) I don't understand at all. Difference in log-transformed means of what?

I would go with the formula of Rdnm

R_dnm = r^(beta1 - 1)
with r= ratio of highest to lowest dose and beta1=slope


Use this with the point estimate of beta1 from your model and its 90% CI limits and you obtain the 90% CI of Rdnm if I'm correct.
Example Cmax in Table 2 of the Smith et al. paper:
beta1 = 0.7615 (0.679, 0.844)
R console:
c(10^(0.7615-1), 10^(0.679-1), 10^(0.844-1))
[1] 0.5774309 0.4775293 0.6982324

Smith reported in Table 2:
    0.577    (0.477,    0.698)
Good enough?

Hope this helps.

Regards,

Detlew

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