90% confidence interval for R_dnm [Study As­sess­ment]

posted by Shuanghe  – Spain, 2019-01-07 17:11  – Posting: # 19751
Views: 4,370

Dear Detlew,

» Could you please give a detailed example for what you did here?

Sorry. It seems I didn't explain it clearly. My SAS code is almost the same as yours:

PROC MIXED DATA = smith METHOD = ML;
  CLASS subj ;
  MODEL lcmax = ldose / DDFM=SATTERTHWAITE CL ALPHA = 0.1 ;
  RANDOM INTERCEPT / SUBJECT = subj;
  ODS OUTPUT solutionf = out1;
RUN;


I just take the output and calculated some of the numbers according to Smith's article to reproduce his result.

PROC SQL;
  CREATE TABLE result AS
    SELECT
      d2.estimate AS beta1,
      d2.lower AS beta1_l,
      d2.upper AS beta1_u,
      1 + LOG(0.8)/LOG(10) AS dpcrit_l,
      1 + LOG(1.25)/LOG(10) AS dpcrit_u,
      1.25**(1/MAX(1-d2.lower, d2.upper-1)) AS roh1,
      1.25**(1/MAX(d2.lower-1, 1-d2.upper)) AS roh2,
      EXP(d3.estimate + d2.estimate*LOG(250))/EXP(d3.estimate + d2.estimate*LOG(25)) * (25/250) AS R_dnm,     
      10**(d2.estimate-1) AS Rdnm,
      10**(d2.lower-1) AS Rdnm_l,
      10**(d2.upper-1) AS Rdnm_u,
      EXP(d3.estimate + d2.estimate*(LOG(25))) AS dosepred_l,
      EXP(d3.estimate + d2.estimate*(LOG(250))) AS dosepred_h
    FROM out1 AS d2, out1 AS d3
    WHERE UPCASE(d2.effect) EQ "LDOSE" AND
          UPCASE(d3.effect) EQ "INTERCEPT";
  SELECT * FROM result;
QUIT;

This gives (Smith's reported value in blue):

beta1:    0.7615 (0.7615)
beta1_l:  0.6789 (0.679)
beta1_u:  0.8441 (0.844)
dpcrit_l: 0.90309 (0.903)
dpcrit_u: 1.09691 (1.097)

Those are slope and its 90% CI, and the corresponding criteria calculated with dose ratio r = 10 and θL and θU of 0.8 and 1.25, respectively. The rest are for information purpose only.

roh1:        2.003428 (2.0)
roh2:        4.183885 (4.2)
R_dnm:       0.577402 (0.577)
Rdnm:        0.577402 (0.577)
Rdnm_l:      0.477381 (0.477)
Rdnm_u:      0.698378 (0.698)
dosepred_l:  80.92991 (80.9)
dosepred_h:  467.2906 (467)


where R_dnm is the one I used previously, since Smith mentioned in the article that each mean PK was calcuated as exp(beta_0 + beta1*ln(dose)). R_dnm is dose-normalised mean ratio, hence the long line in PROC SQL:
EXP(d3.estimate + d2.estimate*LOG(250))/EXP(d3.estimate + d2.estimate*LOG(25)) * (25/250).
Rdnm is the same thing but calculated with your code which is the one I use now since it's equivalent to previous one but much shorter. I guess that my explanation in previous post with this regard is not clear so I added both of them here. :-D

The last 2 are predicted geometric mean PK values at the dose levels of 25 and 250, as given in the 1st column in Table 2 in Smith's article. These 2nd part of the result are not really necessary to judge dose proportionality (though roh1 and roh2 are useful to know) but as I said, I prefer to reproduce all results as kind of "validation".

By the way, Helmut, I don't know how to make a table (e.g. with 1 row) with heading here so I manually entered all values above; also, I copy/paste greek letter from elsewhere. Is there any helper section with BBCode for special symble and greek letters and table making? I vaguely recall there used to be a section with BBcode examples but couldn't find it now.

All the best,
Shuanghe

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