Subject: std error of sum of normals of different variances?
From: rweir@cybercom.net (Robert Weir)
Date: Mon, 16 Dec 1996 05:21:50 GMT
I've forgotten almost everything I've learned about statistics and
sold the textbooks to boot! Could I get a little help on this
application?
I have a project that involves N tasks, to be completed sequentially,
each one with a predicted length C[i] and variance on that length
V[i]. The questions is what is the expected range of end dates for the
projects?
If my intuition serves me, the mean of the sum is the sum of all the
estimated lengths, and if all the lengths had the same variance, we
could say that the error of the sum went as the stdev/sqrt(N). But
what if each estimate has a different variance? For example:
step 1 10 days +/- 2 days
step 2 15 days +/- 3 days
step 3 8 days +/- 1 day
step 4 3 days +/- .5 days
step 5 20 days +/- 5 days
Perhaps we could use a similar formula but use a weighted average of
the variances?
Thanks in advance!
-Rob
Subject: Improve the Calendar ?
From: BillBecker@msn.com (William H. Becker)
Date: 16 Dec 96 06:48:37 -0800
Wouldn't an improved international civil calendar
be a great boon in many sectors; scheduling, communications,
better statistical comparisons, budgeting, reduced confusion,
fixed day-date relationships, etc. etc.?
With the upcoming start of a new year, new century, and new
millennium, isn't this a good time to give this issue some
attention ? I sent U.S. Vice Pres. Gore info similar to that
covered in URL listed below and in Nov. 1993 he wrote me that the
idea of an improved calendar "deserves serious consideration".
ISN'T IT ABOUT TIME ? ?
Suggest you look at ideas on Home Page for Calendar Reform at URL:
http://ecuvax.cis.ecu.edu/~pymccart/calendar-reform.html
billbecker@msn.com
Subject: Early Stopping and Unblinded Assessment
From: david.hadorn@vuw.ac.nz (David Hadorn)
Date: 16 Dec 1996 10:02:43 GMT
I am currently working on a paper concerning the effect of early stopping on
randomized controlled trials (RCTs). My earlier post to this newsgroup
concerning sequential analysis produced some helpful responses, so I'd like
to put my line in the water again.
The particular angle I am interested in pursuing here concerns the extent to
which studies conducted without observer blinding (and without
inter-observer reliability testing) might be more likely than properly
blinded studies to meet early-stopping criteria, particularly if the outcome
assessors believe the treatment is effective. This situation has occurred
in some important settings recently.
I believe the epidemiological literature is clear that RCTs using "soft"
endpoints (anything other than death or, perhaps, things like "major
strokes"), should employ blinded assessment techniques, or, where this
absolutely isn't possible, that some attempt should be made in a subsample
of patients to compare the assessments of unblinded observers with blinded
ones. (This latter task should probably be done in any case because of the
generally poor inter-observer reliability in the assessment of most soft
clinical endpoints.)
Where neither blinding nor reliability testing is done, and where observers
believe the treatment is effective, the study is obviously a set-up for
"proving" the treatment "works". Moreover, I think, and this is where my
question really lies, such studies are probably more susceptible to early
stopping than properly conducted studies--which aggravates the bias. The
reason for this seems obvious, intuitively, but it would be good to develop
a somewhat more rigorous analysis of the situation, if possible.
I'm about to begin a review of a series of recent RCTs to see if my theory
is supported empirically, but first I thought I'd describe the situation to
this group. I realize this might be more of an epidemiological than a
statistical question, but it doesn't hurt to ask.
Any thoughts? (Please respond [also] via private e-mail due to server
unreliability.)
David Hadorn
david.hadorn@vuw.ac.nz
Subject: Books forsale
From: wong@leconte.seas.ucla.edu (Ling S. Wong)
Date: Mon, 16 Dec 1996 14:54:07 GMT
Hi everyone,
I can't find the *.forsale group corresponding to this newsgroup.
So hopefully nobody is offended by my posting the forsale list here. If
someone is interested in several volumes then we could negotiate the
price. Thanks very much for your kind attention.
Lance Wong
301-613-0829
or reply to wong@enh.nist.gov
All books are in excellent to pristine conditions. Prices excludes postage.
TeX in Practice by von Bechtolsheim (4 volumes) $125
LaTeX (for 2e) by L. Lamport $ 16
NeXTSTEP Programming by Garfinkel & Mahoney $ 20
The TeXbook by D. Knuth (Hardcover) $30
Commands A-L, M-Z for SVR4.2; Unix Press $50
The Macintosh Bible 5/e by DiNucci, et al. $15
Principles & Applications of Organotransition Metal Chemistry by Collman, et
al. $25
Main Group Chemistry by Massey $25
Reacton mechanisms of inortganic and organometallic systems by Jordan $25
An introduction to ultrathin organic films (from L-B to Self-assembly) by
Ulman $35
Stereochemistry for Organic Compands by Eliel, et al. $35
Principles of Polymer Chemistry by Flory $45
Physical Chemistry by Adamson (5/e) $30
NMR of proteins & nucleic acids by Wuthrich $30
Electrode kinetics for chemists, chemical engineers, and material scientists
by Gileadi $35
Electrochemistry by Brett & Brett $20
Modern Electrochemistry 1 & 2 by Bockris & Reddy $30
Atomic & Molecular Spectroscopy by Svanberg $30
Writing the Lab. Notebook by Kanare (ACS) $8.
Subject: Re: std error of sum of normals of different variances?
From: mcohen@cpcug.org (Michael Cohen)
Date: 16 Dec 1996 18:05:55 GMT
Robert Weir (rweir@cybercom.net) wrote:
:
: I have a project that involves N tasks, to be completed sequentially,
: each one with a predicted length C[i] and variance on that length
: V[i]. The questions is what is the expected range of end dates for the
: projects?
:
: If my intuition serves me, the mean of the sum is the sum of all the
: estimated lengths, and if all the lengths had the same variance, we
: could say that the error of the sum went as the stdev/sqrt(N).
That's the error of the mean. For the sum, it is sqrt(N) stdev.
: But
: what if each estimate has a different variance? For example:
:
:
: step 1 10 days +/- 2 days
: step 2 15 days +/- 3 days
: step 3 8 days +/- 1 day
: step 4 3 days +/- .5 days
: step 5 20 days +/- 5 days
:
: Perhaps we could use a similar formula but use a weighted average of
: the variances?
If the steps are independent, the variance of the sum will be the sum of
the variances. More generally, it's the sum of the variances plus twice
all the covariance terms. The variances will be proportional to the
squares of the lengths of your confidence intervals. Then take the square
root to get the standard error of the sum.
--
Michael P. Cohen home phone 202-232-4651
1615 Q Street NW #T-1 office phone 202-219-1917
Washington, DC 20009-6331 office fax 202-219-2061
mcohen@cpcug.org
Subject: Re: Multivariate bernoulli distribution
From: mcohen@cpcug.org (Michael Cohen)
Date: 16 Dec 1996 19:21:11 GMT
Theodore Sternberg (strnbrg@rahul.net) wrote:
: This problem arose in an insurance application... Consider three Bernoulli
: random variables A, B and C, i.e. each one equals either success or
: failure. Suppose we know the three marginal probabilities, as well as all
: the two-way joint probabilities p(A,B), p(B,C) and p(A,C).
:
: But we don't know the three-way joint probabilities. What is the most
: "natural" guess at the three-way probabilities? OK, that sounds vague,
: but the background is that the only "data" are results from a survey that
: asked people about their subjective probabilities, and it only asked about
: 2-way joint probabilities.
:
: Had the survey asked about only the marginals p(A), p(B) and p(C), I'd say
: the most natural guess at p(A,B) would be to assume independence and take
: p(A)p(B). Is there an analogous "independence" condition available here?
: Or failing that, some kind of "neutral" assumption?
:
This is an interesting question. I vote for:
If p(A)p(B)p(C)=0 then p(A,B,C)=0; else
p(A,B,C)=[p(A,B)p(B,C)p(A,C)]/[p(A)p(B)p(C)].
My justification is that if, say, C is independent of A and
B, then the above reduces to p(A,B)=p(A,B), not unnecessarily
constraining the equation.
I considered p(A,B,C)=sqrt[p(A,B)p(B,C)p(A,C)] but then
if C is independent of A and B, we get p(A,B)=p(A)p(B),
forcing independence of A and B.
It would be nice to derive a justification based on minimum
entropy or some similar criterion.
--
Michael P. Cohen home phone 202-232-4651
1615 Q Street NW #T-1 office phone 202-219-1917
Washington, DC 20009-6331 office fax 202-219-2061
mcohen@cpcug.org
Subject: Re: Semivariance
From: middleto@mcmail.cis.McMaster.CA (Gerard Middleton)
Date: 16 Dec 1996 14:29:02 -0500
I would not have thought the semivariance was a good way to approach
skewness. It is essentially tied to spatial statistics:
The best, but not the easiest, reference is the book by Cressie on
spatial statistics. An easier one is by Isaaks and Srivastava. Both
will give you information on how to actually compute the semivariance.
Like everything else in statistics it is quite simple, except if you are
serious, when it becomes quite complicated!
--
Gerry Middleton
Department of Geology, McMaster University
Tel: (905) 525-9140 ext 24187 FAX 522-3141
Subject: Re: Multivariate bernoulli distribution
From: cberry@tajo.edu (Charles C. Berry)
Date: 16 Dec 1996 20:51:56 GMT
Theodore Sternberg (strnbrg@rahul.net) wrote:
: This problem arose in an insurance application... Consider three Bernoulli
: random variables A, B and C, i.e. each one equals either success or
: failure. Suppose we know the three marginal probabilities, as well as all
: the two-way joint probabilities p(A,B), p(B,C) and p(A,C).
: But we don't know the three-way joint probabilities. What is the most
: "natural" guess at the three-way probabilities? OK, that sounds vague,
: but the background is that the only "data" are results from a survey that
: asked people about their subjective probabilities, and it only asked about
: 2-way joint probabilities.
: Had the survey asked about only the marginals p(A), p(B) and p(C), I'd say
: the most natural guess at p(A,B) would be to assume independence and take
: p(A)p(B). Is there an analogous "independence" condition available here?
: Or failing that, some kind of "neutral" assumption?
: Ted Sternberg
: San Jose, California
The analogous condition is "lack of three-way interaction" in a
log-linear model of the joint probability.
A good source for discussion of this is
Bishop, Y.M.M., S.E. Fienberg, and P.W. Holland. (1975) Discrete
Multivariate Analysis. The MIT Press, Cambridge.
The actual calculation of p(A,B,C) can be carried out by a method
known as "raking" or "iterative proportional fitting". Software to
analyze tables of counts by log-linear models can usually perform this
procedure, but it is so simple that a spreadsheet is really all that
you would need to pull it off.
--
Charles C. Berry (619) 534-2098
Dept of Family/Preventive Medicine
E mailto:cberry@tajo.ucsd.edu UC San Diego
http://hacuna.ucsd.edu/members/ccb.html La Jolla, San Diego 92093-0622
Subject: Re: Multivariate bernoulli distribution
From: Theodore Sternberg
Date: 16 Dec 1996 22:27:22 GMT
In article <5947f7$rv5@news4.digex.net>,
Michael Cohen wrote:
>
>p(A,B,C)=[p(A,B)p(B,C)p(A,C)]/[p(A)p(B)p(C)].
>
>My justification is that if, say, C is independent of A and
>B, then the above reduces to p(A,B)=p(A,B), not unnecessarily
>constraining the equation.
This formula is indeed attractive, and has everything going for it except
that it doesn't guarantee
p(A,B,C) + p(A,B,~C) = p(A,B) .
In fact, I've found (experimentally) that the discrepancy can be quite
large (e.g. it's common for p(A,B,C) + p(A,B,~C) to exceed p(A,B) by a
factor of 3.
Does this have anything to do with "belief nets"?
-- TS
Subject: Re: Good Technical Books?
From: ch1grh@sunc.sheffield.ac.uk (G Harris)
Date: 17 Dec 1996 09:56:37 GMT
Chris Hecker (checker@netcom.com) wrote:
: blosskf@apci.com (Karl F. Bloss) writes:
: >* Numerical Recipes in C/FORTRAN
: Anyone thinking of using the algorithms NR should look at this page:
: http://math.jpl.nasa.gov/nr/
: The page starts with, "We have found Numerical Recipes to be generally
: unreliable," and then goes on to show why.
For the other side of the story, it is probably worthwhile
to look at a rebuttal from Numerical Receipes:
http://nr.harvard.edu/nr/bug-rebutt.html
and to visit their homepage:
http://cfata2.harvard.edu/nr/
In a nutshell, they say that many of the supposed bugs are
misunderstandings, and that in the case of genuine bugs,
these have pretty much been fixed by the second edition.
I am neutral in this debate. Anyone who accepts an algorithm
or piece of code without testing it or checking out alternatives
shouldn't be programming. My own experience with NR has been fine
(eg Runge-Kutta, root-finding etc), but I've found one generally
has to do a "driver routine" oneself, (but then the authors
recommend doing just that).
-------------------------------------------
Glen Harris, Chemistry, Sheffield Univ, GB.
tel.: 44-114-2824518, fax: 44-114-2738673
email: g.r.harris@sheffield.ac.uk
www: http://www.shef.ac.uk/~ch1grh/
-------------------------------------------
