Dear all,Back to the Top
imagine that I have found in three indipendent titrations these three
of Normality for a titrant
What is the most correct theoretical justification I could give to
the first from my mean?
I know it is basic, but please if you have time and will, answer me...
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
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 Murray, Ph.D.
Senior Research Investigator
Drug Metabolism, PCS, PPD, GPRD, Abbott Laboratories, Chicago, USA
Statistics with n=3 are really statistics?Back to the Top
There are several philosophies you can choose from but the usual split isBack 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
Personally speaking I would look to see if there was any issue associated withBack 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
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 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!
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
HiBack 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
Federica,Back to the Top
There is no way to exclude an outlier from a chemical test that has already
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 Bales, Ph.D.
Senior Regulatory Consultant
Worldwide Regulatory Affairs
2520 Meridian Pkwy, Suite 200
Durham, NC 27713
(919) 294-5297 Phone
(919) 294-5573 Fax
I'm perplexed... truly people think it's possible to define the normalBack 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
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!
Dr. Stefano Porzio
Pharmacokin & Tox. Dept.
Inpharzam Ricerche SA -ZambonGroup
Taverne - Switzerland
If indeed only 3 measurements are available then it would be difficult toBack 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
Use Dixon Outlier test for small sample size (such as n=3 in your case) toBack to the Top
judge outliers. Just type Dixon outlier test in Yahoo and you will get lot
of information on how to use it.
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