Bioequivalence and Bioavailability Forum • ANOVA for lambdaz, half life and tmax in bear 2.7.7

mahmoud-teaima
★

2016-07-20 13:03
(3163 d 13:16 ago)

(edited on 2016-07-20 19:27)
Posting: # 16496
Views: 11,442

ANOVA for lambdaz, half life and tmax in bear 2.7.7 [🇷 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.

Helmut
★★★

Vienna, Austria,
2016-07-20 14:25
(3163 d 11:54 ago)

@ mahmoud-teaima
Posting: # 16498
Views: 10,421

Egypt: Strange requirements

Post reply

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 t_max descriptive statistics should be given. If t_max 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 (t_1/2) 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 (F_T, F_R, D_T, D_R, CL_T, CL_R) and two constants (AUC_T, AUC_R). Only if we assume [sic] that D_T = D_R and CL_T = CL_R 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 t_½) 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 t_½ 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 t_max 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.

—
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

mahmoud-teaima
★

2016-07-20 16:12
(3163 d 10:07 ago)

@ Helmut
Posting: # 16499
Views: 10,407

Egypt: Strange requirements

Post reply

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.

mittyri
★★

Russia,
2016-07-20 19:00
(3163 d 07:19 ago)

@ mahmoud-teaima
Posting: # 16503
Views: 10,329

Output is not from Phoenix/Winnonlin

Post reply

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
★★★

Berlin, Germany,
2016-07-20 18:05
(3163 d 08:14 ago)

@ Helmut
Posting: # 16501
Views: 10,386

ke versus t1/2

Post reply

Dear Helmut,

❝ BTW, asking for a comparison of both k and t_½ 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.
t_½ = 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):
t_½T/t_½R = (ln(2)/k_T)/(ln(2)/k_R) = k_R/k_T
That's not identical to the GMR for k IMHO :no:

.

BTW: @Mahmoud.
k or k_e will be better called lambda_Z - terminal rate constant.

❝ ❝ 1st order elimination rate constant

is only correct within an one-compartment model.

—
Regards,

Detlew

Helmut
★★★

Vienna, Austria,
2016-07-20 18:31
(3163 d 07:48 ago)

@ d_labes
Posting: # 16502
Views: 10,281

ke versus t1/2

Post reply

Dear Detlew,

❝ ❝ BTW, asking for a comparison of both k and t_½ 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).
λ_z: 99.66% (90% CI: 92.23% – 107.69%)
t_½: 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/CL_lower of one metric = CL_upper of the other). Hence, testing both is nuts. :-D

—
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, 2016-07-21 16:42 (3162 d 09:37 ago) @ Helmut Posting: # 16510 Views: 10,169	ke versus t1/2 Post reply
	Dear Helmut, ❝ ... Hence, testing both is nuts. 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 was resistant. — Regards, Detlew

Helmut
★★★

Vienna, Austria,
2016-07-21 16:54
(3162 d 09:25 ago)

@ d_labes
Posting: # 16511
Views: 10,138

“double“ testing

Post reply

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 AUC_0–τ and C_average… :crying:

„Doppelt genäht hält besser”

—
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

yjlee168
★★★

Kaohsiung, Taiwan,
2016-07-20 23:10
(3163 d 03:09 ago)

@ mahmoud-teaima
Posting: # 16505
Views: 10,371

tmax in bear 2.7.7

Post reply

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
bear v2.9.2:- created by Hsin-ya Lee & Yung-jin Lee
Kaohsiung, Taiwan https://www.pkpd168.com/bear
Download link (updated) -> here

mahmoud-teaima
★

2016-07-21 00:40
(3163 d 01:39 ago)

@ yjlee168
Posting: # 16506
Views: 10,289

tmax in bear 2.7.7

Post reply

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.

yjlee168
★★★

Kaohsiung, Taiwan,
2016-07-21 13:11
(3162 d 13:08 ago)

@ mahmoud-teaima
Posting: # 16509
Views: 10,256

tmax in bear 2.7.7

Post reply

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 t_1/2 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 t_1/2 to the regulatory before. If so, you may consider to remove ANOVA stuffs and just present descriptive statistics for λz and t_1/2.

—
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

mahmoud-teaima
★

2016-07-27 14:13
(3156 d 12:06 ago)

@ yjlee168
Posting: # 16520
Views: 10,071

tmax results in R 3.3.1 and spss 20 on macos 10.11.6

Post reply

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.