90% confidence interval for R_dnm [Study As­sess­ment]

posted by Shuanghe  – Spain, 2019-01-07 11:53 (1603 d 07:26 ago) – Posting: # 19739
Views: 16,445
(edited by Shuanghe on 2019-01-07 12:51)

Dear Detlew and all,

Happy New Year!

❝ :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.


Reproduce Smith's results in SAS is easy enough (except the 90% CI of Rdnm, which is not realy need for judging dose proportionality), but I struggled for a loooooong loooooooooong time with R as I was playing lm and glm and to make it worse, random effect was coded wrong.... Obviously, my R skill needs to be improved much more :crying:
Anyway, not too late for me. I'll check the R code mentioned by Zizou and Helmut et al later.

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


yes.

❝ 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.

❝ 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?


This must be it!
According to his article, Smith use the formula EXP(beta0 + beta1*ln(dose)) to calculate each mean to get ratio Rdnm. so the value is Rdnm = 0.577402 (red digit from my sas). But this is equivalent to what you wrote so if you use the full precision figure we would have obtained the same values (90% CI of (0.477381, 0.698378)). I checked with the AUC data as well, they matches what's reported in table 2.


❝ Hope this helps.

Definitely helps! Thanks.
—
All the best,
Shuanghe

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