Newsgroup sci.stat.math 12167

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I am looking for the formula for semi-variance.  I have been
told that the semi-deviation (I am assuming is the squre root
of the semi variance) is a good measure of the volatility 
below the mean and, hence, a way to determine the skewedness
of a distribution.  So, does anyone have the formula for
semivariance and can anyone comment further on its ability
to measure skewedness?  Thanks.
 - Dave

SEMATECH Statistical Methods Symposium: Call for Papers
Topic:  Categorical/Discrete Data Analysis
This is a yearly symposium held by the SEMATECH Statistical
Methods Group.  SEMATECH is a consortium of major US semiconductor
manufacturers.  The audience typically numbers about 100 engineers 
and statisticians from the SEMATECH community and the 
semiconductor equipment supplier community.  While the call for
papers is open to everyone, preference will be 
given to applied examples in the semiconductor industry.  
Attendance to the symposium is restricted to speakers, SEMATECH
and SEMI/SEMATECH community members.  
Date:  May 5-7, 1997
Location:  San Antonio, Texas
Submit abstract by:  	January 6, 1997  (extended deadline)
Rev 0 of paper due by:  March 10, 1997
Final paper due by:	April 6, 1997
Submit abstract to:
Diana Ballard
2706 Montopolis Drive
Austin, Texas 78741
or 
Don McCormack
2706 Montopolis Drive
Austin, Texas 78741
or
email:	Diana.Ballard@SEMATECH.Org
	Don.McCormack@SEMATECH.Org

"Less Wright"  wrote:
>Tonight on the way home from dinner I told some of my programming
>colleagues a funny story I got in email, but after the laughs it turned
>into a discussion of probability theory, followed by a bet on who's answer
>was right...please enjoy the short story below if you haven't already heard
>it, but if you can help settle our bet based on the resulting problem it
>brought up, I would appreciate it!
>The story as I recall it: (if you've already heard, please skip to 'the
>bet')...
>Two college students, supposedly from UVA, were doing quite well in their
>chemistry class and ended up having several free days during finals before
>their last exam in chemistry.  Given that they only needed to get a D on
>their chem final to pass the course, they decided to enjoy the free days
>before the exam by partying at a neighboring college...in their ensuing
>drunkeness, they ended up oversleeping on the day they were supposed to
>return to take their 'easy' chem exam...technically, this meant they got a
>0 and were now going to fail the course.  They drove back to campus, and on
>the way concocted a story about how they had returned on time, but were
>seriously delayed due to a flat tire, and the ensuing towing and repair
>time.  
>They told their professor the story, and begged for a make up exam - he
>consented, provided that they could not leave the classroom during the exam
>for any reason until they were done with their exam, and said they could
>take the exam the next day...they stayed up all night reviewing all of
>their chemistry knowledge.  They met the professor the next day, and he
>then placed them into seperate rooms on each side of the hall, gave them
>their exams, shut the door, and he then sat in a desk in the middle of the
>hallway.  The students then opened the exam book to see:
>1)Question - worth 100 points:
>Which tire ?
>The bet:
>Hopefully you enjoyed the story, but now for the statistical bet which
>evolved - what is the probability of both students guessing the correct
>tire, given that they hadn't agreed beforehand on which tire had gone flat
>in their concocted story?  Everyone but me said 25%, because their are four
>tires and only one correct choice.  I disagreed, because my vague
>recollection of my probability course says that the real  fact is that you
>are evaluating the probability of *two* people choosing the same tire, each
>of which has a 25% choice, so your odds of them both randomly picking the
>same tire should be less than 25%....i.e. potential outcomes are 16
>different combinations of tires (i.e. Student A chooses Right Front,
>Student B could choose RF, LF, RB, LB.  A could also choose Left Front, and
>B could again choose RF,LF,RB,LB...)  So is it still a 25% chance they both
>pick the same tire, or is it 1/16 probability or ??  Actually though, now
>that I have written this out, I am starting to think I am wrong - b/c A
>could pick any given tire, and B could then pick any tire with a 25%
>probability of B's pick matching A...anyway, if you know the definitive
>answer and could help us resolve this gentlemen's bet, I would greatly
>appreciate it....
>Thanks in Advance,
>Less
Amusing story. OK, number the tires 1 to 4 (cause its easier to talk
about than NearsideRear and so forth). Student A chooses one of the
tires and has 4 choices to do so. For each choice that student A
makes, student B has 4 choices and these are all the combinations of
choices - i.e. 16 possible combined choices. This is your total "state
space". Now, in how many cases do they pick the same tire? Write them
down and count them up, there are four cases of course. OK, the
probability is therefore:
p = (number of desired outcomes) / (total number of outcomes)
  = 4 / 16
  = 1 / 4    or 25%.
QED as they say.

Hi,
I have a general question about research methodology. I have a 
particular problem (that I won't trouble you with) and I wish 
to know for exactly which parameters a stopping time T of a given 
nonnegative martingale  has a finite mean. To show that in 
some situations it has an infinite mean I use a theorem in 
Shiryayev's 1984 text that says 
If E(T) < infty, and (*) E(|x_{n+1})-x_n|x_0,...,x_n) < C < infty,
then E(x_T)=E(x_0).
So what I do is show that the conclusion E(x_T)=E(x_0) is false always,
and argue that for parameters a certain way, (*) holds. Then by way
of contradiction, I know that the premise E(T) < infty is false. 
But I have been unable to come up with a proof methodology for deducing
that under the complementary parameter set (ar any, for that matter), 
E(T) < infty.  I am missing a useful trick here, I think. 
Any inspired or experienced help would be greatly appreciated.
Thanks,
Lones
  .-.     .-.     .-.     .-.     .-.     .-.    
 / L \ O / N \ E / S \   / S \ M / I \ T / H \   
/     `-'     `-'     `-'     `-'     `-'     ` 
 Lones Smith, Economics Department, M.I.T., E52-252C, Cambridge MA 02139
 (617)-253-0914 (work)  253-6915 (fax)   lones@lones.mit.edu 

I have some data to bring to bear on the problem of 
the probability that four students will correctly select
the same tire when asked which one caused them to be late.
While the conventional, a priori answer must of course be
0.25, I can report from experience that the probability
is 1.0.  This is an old scam (I first tried it in
1968 while skipping out from high school), and can
verify that even then, students were well prepared to
be asked this question by school administrators, so
that by prior agreement, the tire that went flat was
designated by fiat to be the right front.:-}
Regards,
-- 
=-=-=-=-=-=-=-=-=-==-=-=-=
Mike Lacy, Sociology Dept., Colo. State Univ. FT COLLINS CO 80523
voice (970) 491-6721        fax   (970) 491-2191

Hi Less,
	Thanks for the funny story.  I'll remember that the next time a 
student comes along with a dubious sob-story.
	As for the probability of both students coming up with the same 
tire, I think you were right to doubt your initial answer.  The sample space 
is indeed a 4x4 grid.  IF there had indeed been a flat tire, and IF the 
professor knew which one it was, and IF neither of the students knew, 
then there would only be one correct cell in the grid.  That is to say, 
under those (nonsensical) conditions, the probablility of both students 
getting it right would be 1/16.
	But in the situation you describe, the only requirement is that 
BOTH students pick the same tire.  (For convenience, I will number them 
1-4.)  Note that there are 4 ways this could happen:  1-1, 2-2, 3-3, and 
4-4.  So the probability that both students pick the same tire is 4/16, 
or 1/4.
Cheers,
Bruce Weaver
UWB, Psychology
pss091@bangor.ac.uk
On 12 Dec 1996, Less Wright wrote:
> Tonight on the way home from dinner I told some of my programming
> colleagues a funny story I got in email, but after the laughs it turned
> into a discussion of probability theory, followed by a bet on who's answer
> was right...please enjoy the short story below if you haven't already heard
> it, but if you can help settle our bet based on the resulting problem it
> brought up, I would appreciate it!
> 
> The story as I recall it: (if you've already heard, please skip to 'the
> bet')...
> Two college students, supposedly from UVA, were doing quite well in their
> chemistry class and ended up having several free days during finals before
> their last exam in chemistry.  Given that they only needed to get a D on
> their chem final to pass the course, they decided to enjoy the free days
> before the exam by partying at a neighboring college...in their ensuing
> drunkeness, they ended up oversleeping on the day they were supposed to
> return to take their 'easy' chem exam...technically, this meant they got a
> 0 and were now going to fail the course.  They drove back to campus, and on
> the way concocted a story about how they had returned on time, but were
> seriously delayed due to a flat tire, and the ensuing towing and repair
> time.  
> They told their professor the story, and begged for a make up exam - he
> consented, provided that they could not leave the classroom during the exam
> for any reason until they were done with their exam, and said they could
> take the exam the next day...they stayed up all night reviewing all of
> their chemistry knowledge.  They met the professor the next day, and he
> then placed them into seperate rooms on each side of the hall, gave them
> their exams, shut the door, and he then sat in a desk in the middle of the
> hallway.  The students then opened the exam book to see:
> 1)Question - worth 100 points:
> Which tire ?
> 
> The bet:
> Hopefully you enjoyed the story, but now for the statistical bet which
> evolved - what is the probability of both students guessing the correct
> tire, given that they hadn't agreed beforehand on which tire had gone flat
> in their concocted story?  Everyone but me said 25%, because their are four
> tires and only one correct choice.  I disagreed, because my vague
> recollection of my probability course says that the real  fact is that you
> are evaluating the probability of *two* people choosing the same tire, each
> of which has a 25% choice, so your odds of them both randomly picking the
> same tire should be less than 25%....i.e. potential outcomes are 16
> different combinations of tires (i.e. Student A chooses Right Front,
> Student B could choose RF, LF, RB, LB.  A could also choose Left Front, and
> B could again choose RF,LF,RB,LB...)  So is it still a 25% chance they both
> pick the same tire, or is it 1/16 probability or ??  Actually though, now
> that I have written this out, I am starting to think I am wrong - b/c A
> could pick any given tire, and B could then pick any tire with a 25%
> probability of B's pick matching A...anyway, if you know the definitive
> answer and could help us resolve this gentlemen's bet, I would greatly
> appreciate it....
> 
> Thanks in Advance,
> Less
> 
> 

The 29th meeting of the
        Interface of Computer Science and Statistics
will be held in Houston, Texas, May 14-17, 1997,
sponsored by the Interface Foundation of North America.
Program chair is David W. Scott and host institutions include Rice
University and M.D.Anderson Cancer Institute.
The theme of Interface'97 is:
        "Mining and Modeling of Massive Data Sets
         in Science, Engineering, and Business"
The keynote speaker is Jerry Friedman, Stanford University.
Over 25 invited paper sessions have been organized,
and contributed papers are sought.  Partial funding for
young investigators and students may be available,
subject to final grant funding.
For further information, please go to
        http://www.ruf.rice.edu/~stat/interface97.html
or write to interface97@stat.rice.edu.
David W. Scott
-- 
David W. Scott
Dept. of Statistics, MS-138
Rice University
6100 Main St.
Houston, TX  77005-1953
Office Phone: (713) 527-6037
Dept.  Phone: (713) 527-6032
Fax: (713) 285-5476
Email: scottdw@rice.edu

Hi,
While reading about Wald statistic, 
I found that one of its criticism is
that it is not invariant under 
reparameterizations (for example, in nonlinear models
with or without nonlinear restrictions, in linear
models with nonlinear restrictions). 
I understand that invariance to reparametrizations
is desirable and maybe much needed to make 
coherent decisions. But I wonder if there are
any examples, where this parametrization 
dependence is desirable ? I couldn't find examples
in any of the statistics books I have.
Also, are there any alternatives to Wald statistic
which are invariant to parametrization, but 
like Wald statistic, require computation of unrestricted
maximum likelihood estimate only.
Thanks in advance.
-- 
Ashutosh Sabharwal

At http://www.stat.ucla.edu/journals/jss/v01/i03/ you will find
Thomas Lumley: 
XLISP-Stat Tools for Building Generalised Estimating Equation Models. 
browsable, downloadable, and (soon) executable.
-- 
Jan de Leeuw; UCLA Department of Statistics; UCLA Statistical Consulting
US mail: 8118 Math Sciences, 405 Hilgard Ave, Los Angeles, CA 90095-1554
phone (310)-825-9550;  fax (310)-206-5658;  email: deleeuw@stat.ucla.edu  
                 www: http://www.stat.ucla.edu/~deleeuw

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

Hello
      I am not able to find those books, about stochastic process,(I am 
interesting mainly in level crossing statistics):
        - H.Cramer and M.R. Leadbetter,Stationary and Related Processes,
          J.Wyley,N.Y.,1967
        - V.I. Tikhonov, Level-crossing in stochastic processes,(Ed. by 
          Science publishers, in Russia.
        - M.R.Leadbetter,G.Lindgren and H.Rootzen, Extremes and related   
        properties of Random sequences and Processes.Springer           
Verlag,.N.Y. 1983.
      Any help to find this books, of other about related fields suchs, 
reliability, etc., with strong emphasis in level crossing processes,will 
be very appreciated.
      Thanks a lot, in advance
      C.Antonio Fernandez Ameal
      caameal@lander.es

Hello
      I am not able to find those books, about stochastic process,(I am 
interesting mainly in level crossing statistics):
        - H.Cramer and M.R. Leadbetter,Stationary and Related Processes,
          J.Wyley,N.Y.,1967
        - V.I. Tikhonov, Level-crossing in stochastic processes,(Ed. by 
          Science publishers, in Russia.
        - M.R.Leadbetter,G.Lindgren and H.Rootzen, Extremes and related   
        properties of Random sequences and Processes.Springer           
Verlag,.N.Y. 1983.
      Any help to find this books, of other about related fields suchs, 
reliability, etc., with strong emphasis in level crossing processes,will 
be very appreciated.
      Thanks a lot, in advance
      C.Antonio Fernandez Ameal
      caameal@lander.es

Could someone please tell me the titles of books or papers that discuss	
simulation for multivariate distributions?
Thank you very much.
Cheng Li

~~~ VACANCY FOR AN ENVIRONMENTAL MODELLER (TWO YEAR RESEARCH POSITION) 
~~~
ZENECA Agrochemicals is a successful international business, with several 
thousand employees in Research and Development world-wide.  We have two 
main Research sites, one at Jealott's Hill in the UK, and a second at 
Richmond, California, with a field trials network stretched across dozens 
of sites world-wide.
We currently have a vacancy for an environmental modeller . To ensure the 
safety of pesticides to man and the environment extensive data are 
collected, including the amounts of pesticide residue present on crops.  
In fact, one of the major activities in our Environmental Sciences 
function is performing crop residue trials.  These involve the 
application of pesticides to crops in field trials, and analysing the 
pesticide residue content of the edible crop parts.  Many of hundreds of 
such trials are performed every year covering variables such as 
countries, different crops, different application rates and timings, 
various crop parts and processed products, and of course different 
pesticides.  These data are ultimately used in conjunction with toxicity 
data to determine the safety of the use of the pesticide products to man, 
in order that government authorities can be satisfied that the proposed 
uses of the products can be permitted.
Since human safety is involved, the activity of crop residue work has 
naturally become very conservative in nature, based entirely on real 
measurements, many of which are essentially repetitive.  And since there 
has often been little scope for debate about what trials have to be done, 
there has been little reason to sit back and look at the data and seek to 
quantitatively understand it.  This contrasts with some other areas of 
scientific activity in the company, where a quantitative understanding 
(and if possible a theoretical understanding) of the data has become of 
paramount importance.  However it is clear that the complex datasets 
generated are amenable to understanding, since experienced pesticide 
residue chemists have a pretty good general idea of what the results will 
be before they do a trial, and a fair amount is known from other work on 
the fundamental processes involved.
You will run a project to develop a better quantitative understanding of 
the parameters which govern crop residue data in order to improve risk 
assessment and data generation processes.  There will be support from our 
mathematical modelling group in Environmental Sciences, who already work 
on other aspects of pesticide fate, transport and risk assessment in the 
environment, and our Statistics Group who provide a variety of services 
to R&D.;
Once some quantitative understanding has been gained, the practical 
applications will start to open up, and will need assessing.  These may 
include:
 - extrapolating results from major crops to related minor crops, from 
country to country and between chemically related products
 - predicting residue trial results for validation purposes, and building 
a scientific basis for reducing the repetitive element of the trialling 
work
 - predicting crop residues for early foresight of human risk assessment 
during novel chemical development programs
 - designing product application programs which will minimise or 
eliminate crop residues
 - refining ecological risk assessments for non-target organisms living 
in or feeding on treated crops
You will be a numerate self starter with good communication skills, keen 
to make the most of this exciting opportunity.
Jealott's Hill Research Station is located in pleasant surroundings 
between Bracknell and Maidenhead.  We offer a range of benefits including 
profit sharing, pension scheme, subsidised restaurant and lively sports 
and recreation club.
If you wish to apply for this vacancy please send your CV, with full 
career details including qualifications and experience to date, quoting 
reference Ecol 93 to Penny Hodge, Human Resources Department, Zeneca 
Agrochemicals, Jealott's Hill Research Station, Bracknell, Berkshire RG42 
6EY.

~~~ VACANCY FOR AN ENVIRONMENTAL MODELLER (TWO YEAR RESEARCH POSITION) 
~~~
ZENECA Agrochemicals is a successful international business, with several 
thousand employees in Research and Development world-wide.  We have two 
main Research sites, one at Jealott's Hill in the UK, and a second at 
Richmond, California, with a field trials network stretched across dozens 
of sites world-wide.
We currently have a vacancy for an environmental modeller . To ensure the 
safety of pesticides to man and the environment extensive data are 
collected, including the amounts of pesticide residue present on crops.  
In fact, one of the major activities in our Environmental Sciences 
function is performing crop residue trials.  These involve the 
application of pesticides to crops in field trials, and analysing the 
pesticide residue content of the edible crop parts.  Many of hundreds of 
such trials are performed every year covering variables such as 
countries, different crops, different application rates and timings, 
various crop parts and processed products, and of course different 
pesticides.  These data are ultimately used in conjunction with toxicity 
data to determine the safety of the use of the pesticide products to man, 
in order that government authorities can be satisfied that the proposed 
uses of the products can be permitted.
Since human safety is involved, the activity of crop residue work has 
naturally become very conservative in nature, based entirely on real 
measurements, many of which are essentially repetitive.  And since there 
has often been little scope for debate about what trials have to be done, 
there has been little reason to sit back and look at the data and seek to 
quantitatively understand it.  This contrasts with some other areas of 
scientific activity in the company, where a quantitative understanding 
(and if possible a theoretical understanding) of the data has become of 
paramount importance.  However it is clear that the complex datasets 
generated are amenable to understanding, since experienced pesticide 
residue chemists have a pretty good general idea of what the results will 
be before they do a trial, and a fair amount is known from other work on 
the fundamental processes involved.
You will run a project to develop a better quantitative understanding of 
the parameters which govern crop residue data in order to improve risk 
assessment and data generation processes.  There will be support from our 
mathematical modelling group in Environmental Sciences, who already work 
on other aspects of pesticide fate, transport and risk assessment in the 
environment, and our Statistics Group who provide a variety of services 
to R&D.;
Once some quantitative understanding has been gained, the practical 
applications will start to open up, and will need assessing.  These may 
include:
 - extrapolating results from major crops to related minor crops, from 
country to country and between chemically related products
 - predicting residue trial results for validation purposes, and building 
a scientific basis for reducing the repetitive element of the trialling 
work
 - predicting crop residues for early foresight of human risk assessment 
during novel chemical development programs
 - designing product application programs which will minimise or 
eliminate crop residues
 - refining ecological risk assessments for non-target organisms living 
in or feeding on treated crops
You will be a numerate self starter with good communication skills, keen 
to make the most of this exciting opportunity.
Jealott's Hill Research Station is located in pleasant surroundings 
between Bracknell and Maidenhead.  We offer a range of benefits including 
profit sharing, pension scheme, subsidised restaurant and lively sports 
and recreation club.
If you wish to apply for this vacancy please send your CV, with full 
career details including qualifications and experience to date, quoting 
reference Ecol 93 to Penny Hodge, Human Resources Department, Zeneca 
Agrochemicals, Jealott's Hill Research Station, Bracknell, Berkshire RG42 
6EY.

Optech Solutions would like to annouce its new super efficient optimization
software.
One benchmark showed a speed up to about 12 seconds on a PC for a least
squares curve fit which hitherto had taken up to 30 hours on a SPARC system
using a state-of-the-art nonlinear optimization algorithm developed by
Floudas and Pardalos (1992)
You can find out more detail by visiting our web site at:
http://www.wbm.ca/users/optimize/
Thanks for your attention in this regard.
We look forward to being of service to you.
Jim Pulfer
-- 
Dr. Jim Pulfer
President
Optech Solutions
Box 123
Delisle,  SK
S0L  0P0    Canada
E-mail: optimize@eagle.wbm.ca
http://www.wbm.ca/users/optimize/

lepore@fido.econlab.arizona.edu wrote:
> 
> I have an economics problem that is a variation of the St. Petersberg
> Paradox.
> 
> A regular, six-sided, fair die will be rolled repeatedly until a 2 comes
> up.  When a 2 comes up on the nth roll, you will be paid 6^n dollars.
> The paradox comes in because it was assumed for a long time that people's
> willingness to pay to be in such a gamble should be based on its
> expected monetary value.  However, a game such as this has an infinite
> expected value, leading people to change their views and assume that
> people base their attitudes toward gambles on the expected utility (or
> satisfaction) a gamble will bring them.(I'm sorry for the bad
> explanation, but I'd rather  get to the meat of the problem)
> 
> Anyway, here's the problem.  If a person has a utility function
> u(x) = ln(x), what is the expected utility of this gamble?
> 
> Basically you are going to replace the payoff x that you used when
> calculating the expected value of the gamble with ln(x) for calculating
> the expected utility of the gamble.  I've reduced it to this:
> 
> Expected utility = (1/6)*ln(6)*[n*sum(5/6)^n-1].  I'm having trouble even
> verifying that this series even converges, much less what to.  Any help
> is appreciated.
> 
> Mike LePore
The sum from n=1 to infinity of n*x^(n-1) = (1-x)^-2 if abs(1-x) < 1.
To see this differentiate both sides of  
The sum from n=0 to infinity of x^n = (1-x)^-1.
So your series converges to (ln(6)/6)*(1-5/6)^-2 = 6*ln(6). 
Ellen Hertz

lepore@fido.econlab.arizona.edu wrote:
> 
> I have an economics problem that is a variation of the St. Petersberg
> Paradox.
> 
> A regular, six-sided, fair die will be rolled repeatedly until a 2 comes
> up.  When a 2 comes up on the nth roll, you will be paid 6^n dollars.
> The paradox comes in because it was assumed for a long time that people's
> willingness to pay to be in such a gamble should be based on its
> expected monetary value.  However, a game such as this has an infinite
> expected value, leading people to change their views and assume that
> people base their attitudes toward gambles on the expected utility (or
> satisfaction) a gamble will bring them.(I'm sorry for the bad
> explanation, but I'd rather  get to the meat of the problem)
> 
> Anyway, here's the problem.  If a person has a utility function
> u(x) = ln(x), what is the expected utility of this gamble?
> 
> Basically you are going to replace the payoff x that you used when
> calculating the expected value of the gamble with ln(x) for calculating
> the expected utility of the gamble.  I've reduced it to this:
> 
> Expected utility = (1/6)*ln(6)*[n*sum(5/6)^n-1].  I'm having trouble even
> verifying that this series even converges, much less what to.  Any help
> is appreciated.
> 
> Mike LePore
The sum from n=1 to infinity of n*x^(n-1) = (1-x)^-2 if abs(1-x) < 1.
To see this differentiate both sides of  
The sum from n=0 to infinity of x^n = (1-x)^-1.
So your series converges to (ln(6)/6)*(1-5/6)^-2 = 6*ln(6). 
Ellen Hertz

Hi My name is Abdulhamid Mukhtar.  I am in 12th grade taking Calculus I&II.;
 We bet points against our teach. Now our class have 120 pts.  If any one
of you knows every tough question in any field, Calculus would prefer,
would like to give us please email me at this address
amukhtar@mail.bcpl.lib.md.us.   
Thank you for taking your time to read this message.    

I want to say thanks to all the people who replied on 
my question on the F-distribution. They were all extreemly
helpful, and I've got what I wanted now.
These newsgroups are great!
Machiel