AngusMcLean ★★ USA, 2010-11-07 15:55 (5303 d 16:00 ago) Posting: # 6111 Views: 15,799 |
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Sunday November 7, 201: I discovered this slide from a presentation from Pharsight From: 80-125% to ± 1.25σ(Ref)/0.25
![]() The observed ratio must be within 80-125%; the study design must be partially or fully replicated. What do you think of it? Angus Edit: Category changed. [Helmut] |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-07 17:08 (5303 d 14:48 ago) @ AngusMcLean Posting: # 6112 Views: 14,194 |
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Hi Angus! ❝ Sunday November 7, 201: It’s not necessary to date your posts. But if you do so – do you have a time-machine? ![]() ❝ I discovered this slide from a presentation from Pharsight Well, do you have a link? — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
AngusMcLean ★★ USA, 2010-11-08 23:22 (5302 d 08:34 ago) @ Helmut Posting: # 6115 Views: 14,105 |
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Helmut: there is a link; here it is. I am interested in what folks think of this expansion of the confidence intervals. |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-09 03:13 (5302 d 04:43 ago) @ AngusMcLean Posting: # 6116 Views: 15,138 |
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Hi Angus! ❝ I am interested in what folks think of this … Well, I have no idea what Pharsight calculated here – I get other values. ![]() CV% Pharsight slide 15 FDA’s limits EMA’s limits Values in red: The calculated limits at CV=30% are not applicable with FDA's method; 80%-125% are used instead. Expanded limits for CV>50% are not allowed by the EMA; the limits for CV=50% are used instead. ❝ … expansion of the confidence intervals. <nitpicking> The confidence interval is the same – the acceptance range (bioequivalence limits) is expanded. </nitpicking> It’s important to note that FDA’s and EMA’s approaches are different; FDA’s leads to a discontinuity of the acceptance range at CV=30%, because the scaling CV is 25% – but only applied at CV >30%: See the following two plots (lin and log scale): ![]() ![]() The dotted red lines demonstrate FDA’s procedure: from their formula scaling would start at CV 25%, but is only allowed for >30% – with the scaling factor of 25%. Therefore this nice step. So far about global harmonization. Orange dots are Pharsight’s values. A vendor-specific ‘compromise’ across the Atlantic? ![]() For some background see
If you have to deal with the FDA, simply follow the progesterone guidance from April 2010. The restriction on the point estimate (GMR) of 80–125% in both regulations is statistically not justified, but entirely political. P.S.: Search the forum for RSABE, reference scaled bioequivalence, or HVD – a lot of stuff is waiting for you. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
d_labes ★★★ Berlin, Germany, 2010-11-09 11:42 (5301 d 20:13 ago) @ Helmut Posting: # 6118 Views: 13,872 |
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Dear Helmut, ❝ Well, I have no idea what Pharsight calculated here ... Me too. Their formula given on the slide +1.25(sigmaRef/0.25) is clearly nonsense. It would give negative lower bounds! The correct one is exp(+log(1.25)*sigmaRef/0.25) I think. Taking the numbers for the widened acceptance ranges given, they seems to have a label problem. (In German: Wo 35% draufsteht sollte auch 35% drin sein!) Label their CV=35% with 30%, their CV=40% with 35% and so on and you have perfect coincidence with your values within rounding error ![]() Eventually one should remind the author of the slides that labelling falls under the policy of GMP ![]() Full ACK to all other of your points. BTW: <nitpicking> ------------------ I get for the FDA's widened limits:
</nitpicking> ------------------ ![]() — Regards, Detlew |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-09 16:10 (5301 d 15:46 ago) @ d_labes Posting: # 6121 Views: 13,867 |
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Dear D. Labes! ❝ ❝ Well, I have no idea what Pharsight calculated here … ❝ ❝ Me too. Their formula given on the slide ❝ ❝ is clearly nonsense. It would give negative lower bounds! Sure. But note that the upper bound is OK, maybe the have calculated L=1/U ?❝ The correct one is ❝ ❝ I think. Right – as always. ![]() ❝ Taking the numbers for the widened acceptance ranges given, they seems to have a label problem. Oh no! ![]() ❝ Eventually one should remind the author of the slides that labelling falls under the policy of GMP I have linked this thread to Pharsight’s Extranet ten hours ago… ❝ BTW: ❝ <nitpicking> ------------------ ❝ I get for the FDA's widened limits: [...] ❝ </nitpicking> ------------------ Me too – even with bloody M$-Excel (excuse me, Martin). Don’t know what László has done here. Was too late to produce a nice looking plot in R… — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
d_labes ★★★ Berlin, Germany, 2010-11-09 17:46 (5301 d 14:10 ago) @ Helmut Posting: # 6122 Views: 13,857 |
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Dear Helmut! ❝ Me too - even with bloody M$-Excel (excuse me, Martin). Don’t know what László has done here … He has taken log(1.25)/0.25 ~ 0.89 exact is 0.892574205... ![]() — Regards, Detlew |
AngusMcLean ★★ USA, 2010-11-09 18:12 (5301 d 13:44 ago) @ d_labes Posting: # 6123 Views: 13,821 |
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The reason for the confusion is we do not know the calculation that is being made by Pharsight. Consequently we do not know conceptually it is incorrect or if there is an error in it. I had difficulty getting the same number you got so I must not be interpreting your equation correctly. exp(+log(1.25)*sigmaRef/0.25) For sigma = 0.30 I get 1.3221 for above calculation (when exp(+log(1.25) is used. Am I misunderstanding the parenthesis used? First I evaluated e^log1.25= 1.101761 That is (1.101761)*0.3/0.25 =1.3221 Please can you advise as to my error, Angus Edit: Full quote removed. Please delete anything from the text of the original poster which is not necessary in understanding your answer; see also this post! [Helmut] |
d_labes ★★★ Berlin, Germany, 2010-11-09 20:48 (5301 d 11:08 ago) @ AngusMcLean Posting: # 6124 Views: 13,803 |
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Dear Angus, ❝ Please can you advise as to my error, Lets try it step by step: First evaluate log(1.25)/0.25 = 0.892574205... .Second: Convert the CV to the within-subject standard error for Reference: sigmaRef=sqrt(log((CV/100)2+1)) gives for CV=30% sigmaRef=0.293560379.... Third: Calculate exp(+0.892574205*sigmaRef Should give you for the lower bound exp(-0.892574205*0.293560379)=0.769492231 and for the upper bound exp(+0.892574205*0.293560379)=1.29955828 .Fouth: Change the numbers to percent is left to you. ![]() BTW: I think you could figure out this all by your self in a couple of minutes if you remember your high school math. Or didn't you sympathize with your math teacher? ![]() Hope this helps. — Regards, Detlew |
AngusMcLean ★★ USA, 2010-11-09 21:46 (5301 d 10:10 ago) @ d_labes Posting: # 6127 Views: 13,905 |
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Many Thanks! At highschool in the UK log(1.25)/0.25 =0.096910013/0.25 =0.38764. In other words the the term "log" pertained to a logarithm to the base 10. Now for a log to the base e (otherwise known as a natural logarithm or Napierian logarithm) we were taught to use the term "ln". They were first used by John Napier, who was a Scottish mathematician (late 17th century). So ln(1.25)/0.25=0.892574205, which is what you have. On the calculators here in the US there are log and ln functions to use as appropriate. So now I realize that in future when I read an equation that I have to look out for this difference in notation. ![]() Angus |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-09 23:07 (5301 d 08:49 ago) @ AngusMcLean Posting: # 6130 Views: 14,101 |
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![]() ❝ At highschool in the UK [...] the term "log" pertained to a logarithm to the base 10. ❝ Now for a log to the base e (otherwise known as a natural logarithm or Napierian logarithm) we were taught to use the term "ln". They were first used by John Napier, who was a Scottish mathematician (late 17th century). Yeah! I guess you’re still a patriot – even after your move to the former colonies! BTW, in my posts I’ve used ln rather than log in order to avoid any ambiguities (see also the label of the y-axis in my second plot above). I guess Detlew used log assuming to be evident from the following exp which kind is meant. In SAS and R/S+ log() calls the natural logarithm – so we might be excused.Since I was trained as a chemical engineer, I’m used to switch my brain. In mechanical engineering and (in-)organic technology we used solely log10, because it’s easy to guess a value from the part before the decimal point: a log10 of 3.69897 means something between 103 and 104. Only an idiot savant can tell 103.69897=5000… In physical chemistry we used mainly ln; like in PK solving systems of differential equations is much easier in the ‘world’ of e. Well known exceptions in chemistry, where the decadic log is used are pK-values. Neutral pH means –log10([H3O+])=7 or 10-7 molar H3O+. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
d_labes ★★★ Berlin, Germany, 2010-11-10 09:31 (5300 d 22:25 ago) @ AngusMcLean Posting: # 6134 Views: 13,881 |
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Dear Angus! ❝ On the calculators here in the US there are log and ln functions to use as appropriate. Thanks for pointing at this. Since I do not own an US calculator I was not aware of this distinction. My use of log() as synonym of the natural logarithm's aka ln() came from the fact that all programming languages, spreedsheets and statistical systems I came across so far (and this are more than 2 ![]() log() for that and eventually log10() for the logs to base 10.That leads me to the opinion that english-like speaking people use log() always in that context. So I must abandon this opinion. "Man wird alt wie eine Kuh und lernt immer noch dazu" was a slogan of my grandma (One gets old as an cow and learns still thereto) ![]() BTW: For the Napierian logarithm you may find this interesting. — Regards, Detlew |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-10 14:42 (5300 d 17:14 ago) @ d_labes Posting: # 6137 Views: 13,821 |
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Dear D. Labes! ❝ spreedsheets […] use the function Not quite. M$-Excel, OpenOffice-Calc: LN(EXP(1)) natural (Napierian)LOG10(10^1)) decadicLOG(10^1)) no base given, fallback to decadicLOG(10^1,10)) base 10, same as LOG10() LOG(2^1,2) base 2 (logarithmus dualis)LOG(PI()^1,PI()) crazy, but working— Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-09 21:11 (5301 d 10:44 ago) @ AngusMcLean Posting: # 6125 Views: 14,119 |
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Dear Angus! ❝ The reason for the confusion is we do not know the calculation that is being made by Pharsight. Consequently we do not know conceptually it is incorrect or if there is an error in it.
❝ I had difficulty getting the same number you got so I must not be interpreting your equation correctly. ❝ ❝ ❝ For sigma = 0.30 I get 1.3221 for above calculation (when exp(+log(1.25) is used. Am I misunderstanding the parenthesis used? [...] Don’t know, but let's see how to get this right. I didn’t check your parentheses business, but mainly you mixed up CVWR with sigmaWR , in other words you plugged in CVWR , whereas a conversion to sigmaWR is required first. See one of my lectures (slides 34pp): sigmaWR=sqrt(ln(CVWR²+1)) .For example (including all intermediate steps): CVWR : 0.30CVWR² : 0.09CVWR²+1 : 1.09ln(CVWR²+1) : 0.86178sqrt(ln(CVWR²+1)) : 0.29356 (Hurrah, that’s sigmaWR!)ln(1.25) : 0.22314 (Aha – the usual suspect: upper limit in log-scale.)ln(1.25)*sigmaWR/0.25 : 0.26202 (Scaled limit in log scale; note FDA’s magic 0.25…)exp(±ln(1.25)*sigmaWR/0.25) : 0.7695-1.2996 (Welcome back to the linear world.)A word about the magic 0.25 in FDA’s method. Note that this value does not correspond to a theoretical start of switching of CVWR 25%, but corresponds to a CVWR=sqrt(exp(0.25²)-1)) 25.396% (sigmaWR 0.25). That’s the CV where the dashed lines in my plots intersect the conventional limits. If I recall it right, this value was chosen for convenience (if you are masochistic, follow the links here). Haidar et al. suggested 0.25 in their paper without any justification.![]() — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
AngusMcLean ★★ USA, 2010-11-09 21:57 (5301 d 09:58 ago) @ Helmut Posting: # 6128 Views: 13,856 |
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Helmut: Thank you for your elegant outline. Please see my new post. It seems that Log X can mean Lnx (otherwise know as logeX. You see I was taught that Log X means Log10X and for Log eX we differentiate and use Ln X. Angus |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-09 22:31 (5301 d 09:24 ago) @ AngusMcLean Posting: # 6129 Views: 14,093 |
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Hi Angus! Yes, I saw your post already. I guessed that also, but concentrated on CV vs. sigma – which may make one’s life miserable. In PK (with a few exceptions) ln is used, because it comes handy dealing with the calculus. It’s much easier with base e than with base 10. Of course any base can be used, even the logarithmus dualis (ld – base 2); not uncommon in the IT world. As long as you are consistent, everything is fine. Some anecdotes: I once saw an application for Australia, where log10 was used. So far, so good. But next they fiddled around with a posteriori power (sigh!). In calculating the CVintra they used the common formula, ignoring that their MSE was in log10 rather than loge scale and got stuck. Gotcha! I also saw an enquiry from the Malaysian authority, asking whether log10 or loge was used in the analysis. Obviously they had a good reason to ask. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
SDavis ★★ ![]() UK, 2010-11-19 19:02 (5291 d 12:54 ago) @ Helmut Posting: # 6178 Views: 13,828 |
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Angus, Helmut, et al. Apologies for the delay, been a busy few weeks; the authors are going to look at that slide in the presentation next week and hopefully I/they can post some clarification shortly; thanks for bringing it to our attention. Have a good week-end, Simon. — Simon Senior Scientific Trainer, Certara™ [link=https://www.youtube.com/watch?v=xX-yCO5Rzag[/link] https://www.certarauniversity.com/dashboard https://support.certara.com/forums/ |
yjlee168 ★★★ ![]() ![]() Kaohsiung, Taiwan, 2010-11-30 08:16 (5280 d 23:40 ago) @ Helmut Posting: # 6251 Views: 13,795 |
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Dear Helmut, I would like to add one more links for your RSABE ref. list. BASS XVII Presentations. And Endrenyi & Tothfalusi's presentation slides (Title: SCALED AVERAGE BIOEQUIVALENCE: AN APPROACH TO RESOLVE A DIFFICULT PROGRAM, pdf) can be found there. BASS website also has many nice presentations. — All the best, -- Yung-jin Lee bear v2.9.2:- created by Hsin-ya Lee & Yung-jin Lee Kaohsiung, Taiwan https://www.pkpd168.com/bear Download link (updated) -> here |
Helmut ★★★ ![]() ![]() Vienna, Austria, 2010-11-30 16:44 (5280 d 15:12 ago) @ yjlee168 Posting: # 6257 Views: 13,567 |
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Dear Yung-jin, THX, nice presentations! — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![]() Helmut Schütz ![]() The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |