Bioequivalence and Bioavailability Forum • 90% confidence Intervals for IVRT

GM
★

India,
2017-10-30 07:22
(2703 d 09:05 ago)

Posting: # 17944
Views: 9,150

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]

—
Best Regards,
GM

GM ★ India, 2017-11-01 07:51 (2701 d 08:36 ago) @ GM Posting: # 17960 Views: 7,827	90% confidence Intervals for IVRT Post reply
	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] — Best Regards, GM

zizou
★

Plzeň, Czech Republic,
2017-11-01 20:49
(2700 d 19:39 ago)

@ GM
Posting: # 17961
Views: 7,750

90% confidence Intervals for IVRT

Post reply

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

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

And 29 is 95^th 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 8^th from the start, upper limit is 8^th from the end, i.e. 29^th 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
★★★

Vienna, Austria,
2017-11-02 00:07
(2700 d 16:20 ago)

@ zizou
Posting: # 17962
Views: 7,713

critical values

Post reply

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.

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Helmut Schütz

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GM ★ India, 2017-11-02 07:21 (2700 d 09:06 ago) @ Helmut Posting: # 17963 Views: 7,600	critical values Post reply
	Dear Zizou and Helmut, Thank you so much for the response. This is really helped me alot. Thank you, GM. — Best Regards, GM