No interpolation [NCA / SHAM]

posted by Ken Peh – Malaysia, 2013-05-30 21:55 (4402 d 14:51 ago) – Posting: # 10688
Views: 19,963

Dear Helmut,

❝ Interpolated values for imputation:

linear: C12*=33.17+|(12-8)/(16-8)|(7.86-33.17)=20.52


ok.

linlog: C12*=exp(log(33.17)+|(12-8)/(16-8)|log(7.86/33.17))=16.15


How did you get the readings of 16.15? I tried to follow your equation but I could not get the answer.:confused:

❝ Calculation based on imputed values:

linear/lin imp.: pAUC8-16=0.5(12- 8)(33.17+20.52)+

                          0.5(16-12)(20.52+ 7.86)=164.1

linear/log imp.: pAUC8-16=0.5(12- 8)(33.17+20.52)+

                          0.5(16-12)(20.52+ 7.86)=146.7


linear/lin imp and linear/log imp have the same data set. Why the output is different ?:confused:

linlog/log imp.: pAUC8-16=(12- 8)(16.15-33.17)/log(16.15/33.17)+

                          (16-12)( 7.86-16.15)/log( 7.86/16.15)=140.6


❝ Direct calculation without imputation:

linear: pAUC8-16=0.5(16-8)(33.17+7.86)=164.1


ok.

linlog: pAUC8-16=(16-8)(7.86-33.17)/log(7.86/33.17)=140.6


Similarly, I could not get the answer of 140.61. How did you calculate ?:confused:

Please kindly elaborate.

Thank you.

Ken

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