- On 10 May 2004 at 11:23:04, "Federica Vacondio" (fvacond6.aaa.ipruniv.cce.unipr.it) sent the message
Dear all,

Back to the Top

imagine that I have found in three indipendent titrations these three

values

of Normality for a titrant

0.1119

0.1018

0.1015

What is the most correct theoretical justification I could give to

exclude

the first from my mean?

I know it is basic, but please if you have time and will, answer me...

Thanks

Claudia - On 10 May 2004 at 10:21:11, Bernard.Murray.aaa.abbott.com sent the message
Federica,

Back to the Top

I am not a statitician but I understand that it is suitable to apply

the Q-test under these circumstances. The rejection coefficient

would be;

Q = ( 0.1119 - 0.1018 ) / ( 0.1119 - 0.1015 ) = 0.971

For n = 3 this just exceeds the threshold for 95% confidence (0.97)

but is less than that for 99% (0.994) so the high data point can be

legitimately eliminated by the Q test with 0.05 > p > 0.01

I hope that this helps.

All the very best,

Bernard

Bernard Murray, Ph.D.

Senior Research Investigator

Drug Metabolism, PCS, PPD, GPRD, Abbott Laboratories, Chicago, USA

Bernard.Murray.at.abbott.com - On 12 May 2004 at 15:18:30, "Porzio, Stefano" (Stefano.Porzio.-at-.ZambonGroup.com) sent the message
Statistics with n=3 are really statistics?

Back to the Top

Regards

Stefano - On 12 May 2004 at 15:02:26, Nick Holford (n.holford.at.auckland.ac.nz) sent the message
There are several philosophies you can choose from but the usual split is

Back to the Top

between the Bayesians and the Frequentists.

In this case where the data is so sparse and the reason for wanting to pick

one as an outlier is so tenuous I would choose to be an informed Bayesian

and want to know if there was any previous experience of this titration

process? What was the skill level of the person doing it? What is the

typical distribution of values obtained from a much larger sample (e.g. 20

or more replicates)? All of these might help make a decision about whether

to discard the 0.1119 or even to discard the two values of 0.1018 and 0.1015.

A naive frequentist would try to make a decision based only on the 3

numbers you supply without considering the real world context and also by

assuming some specific kind of distribution (e.g. normal) for the variability.

--

Nick Holford, Dept Pharmacology & Clinical Pharmacology

University of Auckland, 85 Park Rd, Private Bag 92019, Auckland, New Zealand

email:n.holford.-at-.auckland.ac.nz tel:+64(9)373-7599x86730 fax:373-7556

http://www.health.auckland.ac.nz/pharmacology/staff/nholford/ - On 12 May 2004 at 15:21:06, Dave Vowles (d.vowles.-at-.ncrl.co.uk) sent the message
Personally speaking I would look to see if there was any issue associated with

Back to the Top

the way the measurement was made. If not, then I am cautious of any judgements

that would allow me to exclude data. Maybe I am too defensive and others might

have different approaches

Dave - On 12 May 2004 at 15:19:35, "J.H.Proost" (J.H.Proost.-at-.farm.rug.nl) sent the message
Dear Federica,

Back to the Top

The 'outlier question' is tricky, whatever method one uses. I am not a

statistician, and so I cannot give a real answer to your question. If the

only information you have are the three values, any solution seems

questionable. But probably you have more information: for example, the

standard deviation of the assay, as determined during the validation. This

information may be useful for answering your question. E.g. if the standard

deviation is 0.0005, the value of 0.1119 is likely to be an outlier.

However, if the standard deviation is 0.0050 (probably unlikely in the case

of a simple titration, but not in case of bioanalytical assays), the value

of 0.1119 is about 2 sd's from the others, so there is nothing strange to

this value.

This is a kind of Bayesian inference that is often overlooked, but very

useful in such situations where there are only a few data.

Any contribution to the discussion is welcome!

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.-at-.farm.rug.nl - On 12 May 2004 at 15:22:10, Prashant Musmade (pmusmade.-at-.yahoo.co.in) sent the message
Hi

Back to the Top

in these three values you take a mean and calculate the normality.

But for three titration the +/- 0.1ml error is consider. For that you refer

the United state pharmacopoeia 24

I think it will help you - On 12 May 2004 at 15:24:27, "Bales, Frank" (Frank.Bales.-a-.parexel.com) sent the message
Federica,

Back to the Top

There is no way to exclude an outlier from a chemical test that has already

happened.

But the important question is, why were three titrations made? Does your

method call for taking three determinations? If your method just calls for

doing the titration, then the first determination is the result to report.

If that first determination was outside spec, then this is now an Out Of

Specification (OOS) situation. What does your OOS SOP call for? Usually,

additional determinations, like the second and third determinations, are

only allowed to investigate the method. Usually the initial result stands

unless there is an identifiable error such as using the wrong stock standard.

It is possible to develop a method that allows for a specified number of

outliers to be excluded from specific calculations as long as certain

conditions are met. Unit Content Uniformity tests can be like that: if one

out of ten determinations is outside a range, but all are inside another

range, then an additional twenty determinations can be made and all but two

out of all thirty must be within range.

Similarly, it is possible to develop a method that requires three

determinations to be made, and then those determinations averaged to yield

a single result.

But this brings us back to the fundamental OOS question, why is there so

much variability in the titrations?

Frank

Frank Bales, Ph.D.

Senior Regulatory Consultant

Worldwide Regulatory Affairs

PAREXEL Intl.

2520 Meridian Pkwy, Suite 200

Durham, NC 27713

(919) 294-5297 Phone

(919) 294-5573 Fax

Email: frank.bales.aaa.parexel.com - On 13 May 2004 at 14:49:28, "Porzio, Stefano" (Stefano.Porzio.-a-.ZambonGroup.com) sent the message
I'm perplexed... truly people think it's possible to define the normal

Back to the Top

distribution of a sample with n=3?

The only way probably is to know the "population" distribution (i.e. the

distribution of all possible points measured with your method, in your

conditions and in your experimental outline conditions) and to check what

is the probability that your "sample" (i.e. your experimental points)

belong to this population.

The definition of outlier comes from the knowledge of the population

distribution.

The easier way to define the population distribution is to obtain a

statistically representative sample from the population, to calculate

population parameters and to use, after easy extrapolations, the sample

parameters to estimate the population parameters: OBVIOUSLY A

REPRESENTATIVE SAMPLE OF POPULATION CONTAINS MORE THAN 3 SUBJECTS!

Best regards

Dr. Stefano Porzio

Pharmacokin & Tox. Dept.

Inpharzam Ricerche SA -ZambonGroup

Taverne - Switzerland - On 14 May 2004 at 13:57:04, Laszlo Endrenyi (l.endrenyi.at.utoronto.ca) sent the message
If indeed only 3 measurements are available then it would be difficult to

Back to the Top

make a statement and unreasonable to omit any reading. However, the

measurements could be part of a larger, for instance, standardization in

which triplicate measurements are recorded at various concentrations. In

that case, it would be reasonable, and not unusual, to take the median of

each triplicate.

Laszlo Endrenyi - On 16 May 2004 at 20:31:52, Garcia PK (garcia_pk.at.yahoo.com) sent the message
Use Dixon Outlier test for small sample size (such as n=3 in your case) to

Back to the Top

judge outliers. Just type Dixon outlier test in Yahoo and you will get lot

of information on how to use it.

Garcia.

Want to post a follow-up message on this topic? If this link does not work with your browser send a follow-up message to PharmPK@boomer.org with "Outlier question" as the subject

PharmPK Discussion List Archive Index page

Copyright 1995-2010 David W. A. Bourne (david@boomer.org)