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Back to the forum  Query: 2017-11-22 10:25 CET (UTC+1h)
 
GM
Junior

India,
2017-10-30 06:22

Posting: # 17944
Views: 588
 

 90% confidence Intervals for IVRT [Nonparametrics]

Hello all,

I need help regarding In Vitro Release Testing of semisolid dosage forms.

Here is the guidance given by the FDA in page 22 of 40.
In Vitro Release Testing - FDA

My doubt is, How the 90% confidence Intervals are calculated..?

I was tried with "Mann-Whitney Median Confidence Interval with Hodges-Lehmann estimation" using below formula.

90% CI= mn/2 ± Zcrit*(sqrt((mn*(m+n+1))/12)

Lower limits are matching with guidance but not upper limits with only one digit difference. Here is my calculation for 6Runs, 2Runs and 2Runs with one cell missing.

m               18            6            5
n               18            6            6
mn             324           36           30
mn/2           162           18           15
(mn(m+n+1))/12 999           39           30
sqrt(above)     31.60696126   6.244997998  5.477225575
zcrit            1.645        1.645        1.645
zcrit*sqrt(D14) 51.99345127  10.27302171   9.010036071

L              110            8            6
U              214           28           24



Edit: Tabulators changed to spaces and BBcoded; see also this post #6. URL corrected. Please don’t link to a Google-India search term. [Helmut]
GM
Junior

India,
2017-11-01 06:51

@ GM
Posting: # 17960
Views: 486
 

 90% confidence Intervals for IVRT

Hello all,

Any idea for the above topic.Kindly let me know.

How to calculate the 90% CI for the median using Wilcoxon Rank Sum/Mann-Whitney rank test...?

Thanks in advance,
GM.


Edit: Relax; see also this post #8. [Helmut]
zizou
Junior

Plzeň, Czech Republic,
2017-11-01 19:49
(edited by zizou on 2017-11-01 20:03)

@ GM
Posting: # 17961
Views: 434
 

 90% confidence Intervals for IVRT

Dear GM.

First of all, I figured out your link to the reference:
https://www.fda.gov/downloads/drugs/guidances/ucm070930.pdf

Example from page 23 (PDF page 26 of 40).

Lower limit is the 8th value of sorted values and upper limit is 29th value as mentioned in the text (PDF page 27 of 40).

8 is 5th quantile calculated from the equation as you already mentioned:
w(0.05) = mn/2 - zcrit*sqrt(mn*(m+n+1)/12) ... approximation for m+n large enough.

And 29 is 95th quantile + 1. Here you probably forgot "+ 1" to get the right position of upper limit.
In this case there is even count of ordered data.
Lower limit is 8th from the start, upper limit is 8th from the end, i.e. 29th from the start.

So you get the positions of ordered data and then you have lower and upper limit directly, you know.

Best regards,
zizou
Helmut
Hero
Homepage
Vienna, Austria,
2017-11-01 23:07

@ zizou
Posting: # 17962
Views: 387
 

 critical values

Hi Václav and GM,

see also this rather old post with tables of the positions of the lower critical values and exact probabilities for m=n=3–32. Generally the normal approximation should not be used if m <8 or n <8.

[image]Regards,
Helmut Schütz 
[image]

The quality of responses received is directly proportional to the quality of the question asked. ☼
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GM
Junior

India,
2017-11-02 06:21

@ Helmut
Posting: # 17963
Views: 352
 

 critical values

Dear Zizou and Helmut,

Thank you so much for the response.

This is really helped me alot.

Thank you,
GM.
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