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90% confidence Intervals for IVRT [Nonparametrics]

posted by GM - India, 2017-10-30 06:22  - Posting: # 17944
Views: 2,261

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

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