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From: eeb@ukc.UUCP (E.E.Bassett)
Newsgroups: net.math.stat
Subject: Re: Normal distribution probability problem.
Message-ID: <587@eagle.ukc.UUCP>
Date: Wed, 15-Jan-86 07:11:45 EST
Article-I.D.: eagle.587
Posted: Wed Jan 15 07:11:45 1986
Date-Received: Thu, 23-Jan-86 20:52:17 EST
References: <2792@ut-ngp.UUCP>
Reply-To: eeb@ukc.UUCP (E.E.Bassett)
Distribution: net
Organization: U of Kent at Canterbury, Canterbury, UK
Lines: 55


	Two questions from the same set of facts.  Assume you have two
	populations, A and B, for which you have full information (i.e.,
	the value of every event in each population).
	
		1)  If you draw a sample of size n = 1 from each population,
	what is the probability that the sample from population A is larger
	than the sample from population B?
	
		2)  Now assume the only information you have about the two
	populations is based on samples of, say, size n = 10.  Thus you have
	the mean and standard deviation of a sample from each population, but
	know nothing about the individual events within the populations.  Now
	what is the probability that a single sample drawn from population A
	will be larger than one drawn from population B?
	
		I think the answer to (1) is simply a z-value and the
	probability is the area under the normal curve.  But I haven't a clue
	on how to work it out if you only have sample data.  I'm interested in
	theory as well as an algorithm.  Any help will be appreciated.
	
				Thanks. 
	
	
The answer to (1) is easy: yes, the probability is what you term
a z-value. To be specific, let A and B represent the random variables
from the two populations; let A be distributed N(meana, vara) and
B N(meanb,varb). (Note that the second parameter shown is the variance
rather than its square root, the s.d.)
Then, since linear combinations of normals are themselves normal,
   
          A - B  is N (meana - meanb , vara + varb) ,
and the probability you require is simply
               
          Pr (A - B > 0).

Substituting the (known) values for the means and variances of A 
and B is then easy.

The answer to (2) is more tricky.  Since you have sample information
about the distributions you can estimate the means and standard deviations
in the usual way, i.e. by the sample means and sample standard deviations.
Substituting these in the formula obtained for (1) will now give an
estimate of the probability you require.
Of course, many statisticians - particularly the Bayesian variety - will
argue that one should be able to express odds on A exceeding B in the
situation described in (2). So stick in the odd prior distribution or
two (actually four, if you can take the means and variances as
independent), turn a handle or four and out comes the posterior
probability. Haven't tried it, but it looks rather Fisher-Behrens'ish
to me.

   Eryl Bassett
   Univ. of Kent
   Canterbury, U.K.