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Back to the forum  Query: 2017-08-21 12:11 CEST (UTC+2h)
 
mahmoud-teaima
Regular

Cairo, Egypt,
2016-07-20 11:03
(edited by mahmoud-teaima on 2016-07-20 19:27)

Posting: # 16496
Views: 3,155
 

 ANOVA for lambdaz, half life and tmax in bear 2.7.7 [R for BE/BA]

[/sub]Good morning all,
I want to know if bear 2.7.7 in R 3.3.1 running on macos 10.11.6 Elcaptin can perform ANOVA for lambda[sub]z
, t1/2 and tmax for test and reference products; also can it perform Kruskal Wallis analysis for tmax as these are non official requirements issued by the some members of the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt.
If not, can these be considered in the coming updates of bear!!!!.

Greetings for all.

Mahmoud Teaima, PhD.
Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy.
Vice-director at Center of Applied Research and Advanced Studies (CARAS).
Faculty of Pharmacy, Cairo University, Cairo, Egypt.
Helmut
Hero
Homepage
Vienna, Austria,
2016-07-20 12:25

@ mahmoud-teaima
Posting: # 16498
Views: 2,749
 

 Egypt: Strange requirements

Hi Mahmoud,

we have to wait for Yung-jin passing by when it comes to bear.

I have some questions / remarks.

» […] new requirements issued by the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt.

Are these requirements available on the net? The latest version I know is of January 2010, which states:

For tmax descriptive statistics should be given. If tmax is to be subjected to a statistical analysis this should be based on non-parametric methods and should be applied to untransformed data.
For parameters describing the elimination phase (t1/2) only descriptive statistics should be given.



» […] ANOVA for K (1st order elimination rate constant), t1/2 and tmax for test and reference products;
  • Identical clearance after T and R is one of the assumptions in bioequivalence. Think about the basic equation of PK: AUC = F × D / CL.
    For two treatments we have a set of two equations with six unknown variables (FT, FR, DT, DR, CLT, CLR) and two constants (AUCT, AUCR). Only if we assume [sic] that DT = DR and CLT = CLR we can eliminate these variables from the equations and end up with a comparison of Fs based on the ratio of observed AUCs. To ask for a comparison of clearances (expressed as k or t½) is strange. Furthermore, CL is a property of the drug not the formulation. In BE we are not interested in that.
    BTW, asking for a comparison of both k and t½ demonstrates a lack of understanding of statistics. They differ only by a factor of ln(2). Any comparison will give identical variance, CI, etc.
  • ANOVA for tmax doesn’t make sense. Contradicts the next point.

» […] also can it perform Kruskal Wallis analysis for tmax…

Kruskal Wallis makes only sense for more than two treatments and will give only a global comparison. IMHO, not very helpful since post-hoc tests for pairwise comparisons with a Bonferroni-adjustment are needed.

[image]All the best,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
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mahmoud-teaima
Regular

Cairo, Egypt,
2016-07-20 14:12

@ Helmut
Posting: # 16499
Views: 2,728
 

 Egypt: Strange requirements

Thanks Helmut for the prompt detailed fully scientific and statistically logic answers which i totally agree with you.
Unfortunately things don't run fully according to book here.
The official Egyptian guidelines to conduct BE studies totally agree with your words, but some members of the committee for evaluation of BE studies may insist on requests like the ones formentioned in my first post. that is why i see it will be helpful if bear can do such extra calculations.

Herein below is an example of data submitted before:

t-half
SOURCE       D.F     SS        MS        F         p
Period         1   8.53E-03  8.53E-03  7.04E-03  0.9339
Subject(Seq)  22  24.9975    1.13625   0.937355  0.5596
Formulation    1   1.48403   1.48403   1.22426   0.2805
Sequence       1   7.68      7.68      6.33565   0.01962
Error         22  26.6681    1.21219
Total         47  60.8382

k
SOURCE       D.F    SS        MS        F         p
Period         1  3.74E-04  3.74E-04  0.025692  0.8741
Subject(Seq)  22  0.78587   0.035721  2.45336   0.02028
Formulation    1  1.29E-02  1.29E-02  0.888474  0.3561
Sequence       1  0.117216  0.117216  8.05048   0.009586
Error         22  0.320324  1.46E-02
Total         47  1.23672


  Kruskal Wallis  DESIGN for tmax

N = 48

Si[1] = 595.5 ; ni[1] = 24
Si[2] = 580.5 ; ni[2] = 24

  H =             0.026032
  Mean Rank[1] = 24.81
  Mean Rank[2] = 24.19
  Chi Square ( 0.95 -   1 ) = 3.841

The difference between means is not significant.


I really don't know how these calculations were done, my guess it may be done by winnonlin!!!!!!


Edit: Tabulators changed to spaces and BBcoded; see also this post #6. [Helmut]

Mahmoud Teaima, PhD.
Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy.
Vice-director at Center of Applied Research and Advanced Studies (CARAS).
Faculty of Pharmacy, Cairo University, Cairo, Egypt.
mittyri
Senior

Russia,
2016-07-20 17:00

@ mahmoud-teaima
Posting: # 16503
Views: 2,684
 

 Output is not from Phoenix/Winnonlin

Hi Mahmoud,

this doesn't look like Winnonlin/Phoenix output:
1. there is no Total row in Phoenix SS tables. Of course you can add it manually
2. As Helmut mentioned Kruskal Wallis test extends the Mann–Whitney U test when there are more than two groups. You won't find this test in Phoenix/Winnonlin, Mann–Whitney U test only

By the way I don't see any difficulties to get these results in R using BEAR output as source data

Kind regards,
Mittyri
d_labes
Hero

Berlin, Germany,
2016-07-20 16:05

@ Helmut
Posting: # 16501
Views: 2,710
 

 ke versus t1/2

Dear Helmut,

» BTW, asking for a comparison of both k and t½ demonstrates a lack of understanding of statistics. They differ only by a factor of ln(2). Any comparison will give identical variance, CI, etc.

I think here you err.
t½ = ln(2)/k (some sort of 'inverse transformation').

Lets consider for simplicity the GMR's (I consider the metric names are the geometric means here):
t½T/t½R = (ln(2)/kT)/(ln(2)/kR) = kR/kT
That's not identical to the GMR for k IMHO :no:.

BTW: @Mahmoud.
k or ke will be better called lambdaZ - terminal rate constant.
» » 1st order elimination rate constant
is only correct within an one-compartment model.

Regards,

Detlew
Helmut
Hero
Homepage
Vienna, Austria,
2016-07-20 16:31

@ d_labes
Posting: # 16502
Views: 2,690
 

 ke versus t1/2

Dear Detlew,

» » BTW, asking for a comparison of both k and t½ demonstrates a lack of understanding of statistics. They differ only by a factor of ln(2). Any comparison will give identical variance, CI, etc.
»
» I think here you err.

I stand partly corrected. Just checked one of my studies (never compared this stuff before).
λz:   99.66% (90% CI: 92.23% – 107.69%)
t½: 100.34% (90% CI: 92.86% – 108.42%)

Within- and between-subject variances are identical. Hence, any conclusions (pass/fail) will be the same.

Note the PEs close to 100% (“proving” the assumption?) and 1/1.0034 ≈ 0.9966 (and so are the CIs: 1/CLlower of one metric = CLupper of the other). Hence, testing both is nuts. :-D

[image]All the best,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
Science Quotes
d_labes
Hero

Berlin, Germany,
2016-07-21 14:42

@ Helmut
Posting: # 16510
Views: 2,549
 

 ke versus t1/2

Dear Helmut,

» ... Hence, testing both is nuts. :-D

Totally correct IMHO. That'also my opinion. My concern was [nitpicking] "Any comparison will give identical variance, CI, etc." [/nitpicking].

But I must confess that I couldn't convince all sponsors I worked with of this opinion. Especially one from Upper-Bavaria:waving: was resistant.

Regards,

Detlew
Helmut
Hero
Homepage
Vienna, Austria,
2016-07-21 14:54

@ d_labes
Posting: # 16511
Views: 2,553
 

 “double“ testing

Dear Detlew,

» But I must confess that I couldn't convince all sponsors I worked with of this opinion. Especially one from Upper-Bavaria:waving: was resistant.

Ha-ha! They once forced me to test AUC0–τ and Caverage:crying:

„Doppelt genäht hält besser”

[image]All the best,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
Science Quotes
yjlee168
Senior
Homepage
Kaohsiung, Taiwan,
2016-07-20 21:10

@ mahmoud-teaima
Posting: # 16505
Views: 2,670
 

 tmax in bear 2.7.7

Dear Mahmoud,

No Kruskal Wallis analysis for tmax with bear at this moment. Sorry about that. Probably you can do it manually or using other statistical package for now.

» ...these are non official requirements issued by the some members of the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt.

I see. sound interesting.

» If not, can these be considered in the coming updates of bear!!!!.

I will consider your suggestion. BTW, thanks for letting me know that bear works well on the new MacOS.

All the best,
---Yung-jin Lee
[image]bear v2.8.3:- created by Hsin-ya Lee & Yung-jin Lee
Kaohsiung, Taiwan http://pkpd.kmu.edu.tw/bear
Download link (updated) -> here
mahmoud-teaima
Regular

Cairo, Egypt,
2016-07-20 22:40

@ yjlee168
Posting: # 16506
Views: 2,637
 

 tmax in bear 2.7.7

thanks Yung-jin Lee for your reply,
here is the results of kruskal wallis test of tmax data performed in R 3.3.1 on macos 10.11.6 Elcaptin.

Kruskal-Wallis rank sum test of tmax

ref = c(1, 0.5, 2, 0.75, 1, 1.5, 2, 2, 1, 1.5, 0.75, 1.5, 0.75, 0.75, 1, 1, 0.75, 1.5, 1, 3, 0.75, 0.75, 1, 0.75)
test = c(1, 0.5, 2, 0.8, 0.8, 1.5, 1, 1.5, 1, 2, 0.5, 1.5, 0.8, 1, 1, 0.8, 1, 1, 1, 1.5, 0.8, 0.8, 1, 0.8)
dati = list(g1=ref, g2=test)
kruskal.test(dati)

   Kruskal-Wallis rank sum test

data:  dati
Kruskal-Wallis chi-squared = 0.0054846, df = 1,
p-value = 0.941


so no significant difference in tmax values between reference and test products.

thanks for your advise in that.

waiting your advise on how to make the remaining statistical analysis of lambdaz and t1/2?!!!!!.

Greetings.

Mahmoud Teaima, PhD.
Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy.
Vice-director at Center of Applied Research and Advanced Studies (CARAS).
Faculty of Pharmacy, Cairo University, Cairo, Egypt.
yjlee168
Senior
Homepage
Kaohsiung, Taiwan,
2016-07-21 11:11
(edited by yjlee168 on 2016-07-21 11:44)

@ mahmoud-teaima
Posting: # 16509
Views: 2,616
 

 tmax in bear 2.7.7

Dear Mahmoud,

» ...
» Kruskal-Wallis rank sum test of tmax ...
» Kruskal-Wallis chi-squared = 0.0054846, df = 1,
» p-value = 0.941

Great. much better than SPSS, right?

» waiting your advise on how to make the remaining statistical analysis of lambdaz and t1/2?

I guess the descriptive statistics for λz and t1/2 should be good enough based on Egypt's BE GL, as mentioned previously by Helmut. I am not sure if the challenge was due to that you inappropriately submitted ANOVA for λz and t1/2 to the regulatory before. If so, you may consider to remove ANOVA stuffs and just present descriptive statistics for λz and t1/2.

All the best,
---Yung-jin Lee
[image]bear v2.8.3:- created by Hsin-ya Lee & Yung-jin Lee
Kaohsiung, Taiwan http://pkpd.kmu.edu.tw/bear
Download link (updated) -> here
mahmoud-teaima
Regular

Cairo, Egypt,
2016-07-27 12:13

@ yjlee168
Posting: # 16520
Views: 2,384
 

 tmax results in R 3.3.1 and spss 20 on macos 10.11.6

This is the results of tmax for my BE project from both R3.3.3 and spss20 on macos 10.11.6

from spss:
NPAR TESTS
/K-W=Tmax BY formulation(1 2)
/STATISTICS DESCRIPTIVES
/MISSING ANALYSIS.
NPar Tests
Syntax NPAR TESTS
/K-W=Tmax BY formulation(1 2)
/STATISTICS DESCRIPTIVES
/MISSING ANALYSIS.

Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
Tmax 48 1.1271 .50010 .50 3.00
treatment 48 1.50 .505 1 2

Kruskal-Wallis Test
Ranks
treatment N Mean Rank
Tmax ref 24 24.35
test 24 24.65
Total 48

Test Statisticsa,b
Tmax
Chi-Square .005
df 1
Asymp. Sig. .941
a Kruskal Wallis Test
b Grouping Variable: treatment


From R:
Kruskal-Wallis rank sum test of tmax

> ref = c(1, 0.5, 2, 0.75, 1, 1.5, 2, 2, 1, 1.5, 0.75, 1.5, 0.75, 0.75, 1, 1, 0.75, 1.5, 1, 3, 0.75, 0.75, 1, 0.75)
> test = c(1, 0.5, 2, 0.8, 0.8, 1.5, 1, 1.5, 1, 2, 0.5, 1.5, 0.8, 1, 1, 0.8, 1, 1, 1, 1.5, 0.8, 0.8, 1, 0.8)
> dati = list(g1=ref, g2=test)
> kruskal.test(dati)

Kruskal-Wallis rank sum test

data: dati
Kruskal-Wallis chi-squared = 0.0054846, df = 1,
p-value = 0.941

Can anyone see a difference?

Mahmoud Teaima, PhD.
Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy.
Vice-director at Center of Applied Research and Advanced Studies (CARAS).
Faculty of Pharmacy, Cairo University, Cairo, Egypt.
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