Conference rusure::math

    
    some more notes to look at in case it might be relevent:
    
       406   TOOLS::STAN         14-DEC-1985     0  Math Notes Statistics
       738     NAC::PICKETT      23-JUL-1987     1  Statistical Inference
       823  SPYDER::TURNER        3-FEB-1988     1  Statistical Packages Available?
      1065   BEING::POSTPISCHIL  21-APR-1989     9  Statistics Books
      1145  MILKWY::JANZEN       30-OCT-1989    17  Chi-Square X^2 calculation (sta
      1184  DWOVAX::YOUNG        22-JAN-1990    10  HELP! Compound Statistical Esti
      1394  SMEGIT::ARNOLD        5-MAR-1991    24  Elementary (?) Statistics
      1447    ATSE::GOODWIN      23-MAY-1991    23  Statistics can be (mis)used to 
      1481    CSSE::NEILSEN      15-AUG-1991    11  another statistical conundrum
      1483  SOLVIT::DESMARAIS    22-AUG-1991     4  Statistical Data Anal in the Co
      1489    MEIS::SCHAUBER      4-SEP-1991     3  Statistical Functions Lib.
      1610  MINDER::WIGLEYA      18-MAY-1992     2  Wanted: Sanity check onmy rust
      1625    EPIK::FINNERTY     10-JUN-1992     4  Computing the tstatistic for r
      1665  LARVAE::TREVENNOR_A  17-SEP-1992     4  Calculating a statistical likel
      1710  MARVA2::RAK          13-JAN-1993     8  Needed: Slick StatisticTrick
      1735  GOCELT::RAK          26-MAR-1993     0  Statistical Mean andVariance T
     End of requested listing
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    

    I'm going to answer your second question first about the proper
    "average" variance to use.

    Brief answer: use the sum of the variances divided by the n.  You can
    take the square root of that and get an "average" standard deviation,
    if you wish.

    Explanation:  If you have a bunch of ordinary variables x1, x2, ..., xn
    if you want to take the "average" you add them together and then you
    "rescale" them to the same units as the original by dividing by n.
    What you end up with is something that is a kind of rough stand in
    for the seperate values, which we will call x*.

    Now imagine that you do the same thing, over and over again.  Each time
    you take the average.  Each set of variables xi is a sample.  There
    will be a distribution of values that each of the xi will take (perhaps
    each of the xi will have the same distribution of values, perhaps not).
    In other words, we are now talking about a set of *random* variables
    which we will call X1, X2, ..., XN.  The x* (the average) produced with
    each sample, also will have a distribution associated with it (which
    can, in principle at least be derived from the distributions of the
    Xi), and is therefore "really" a random variable X*.  The mean (pretend
    that doesn't mean the same thing as "average") and the
    variance/standard deviation are descriptors of the distributions (one
    being an indicator of the "location" of the distribution and the others
    being an indicator of the "scale" of the distribution).  We want to
    know what the mean and the variance of X* are.

    Let's add one more random variable, X+, which is the sum of the Xi's.
    X* is then X+/n.  What is the mean of X*?  Well the mean is a "linear
    operator".  That has two important consequences:

	1) The mean of the sum of random variables is the sum of their
	   means.  ( E(A+B) = E(A) + E(B) ).

	2) The mean of a random variable divided by a constant is the
	   same as dividing the mean of the random variable by the
	   constant.  ( E(A/c) = E(A)/c )

   So the mean of X+ is the sum of the means of the Xi, and the mean of
   X* (= X+/n) is the mean-of-X+ divided by n.  This is what justifies
   the procedure of finding the "average" mean by taking the average of
   the means.

   What about the *variance* of our average value (X*)?  Well, it turns out
   that the variance is *also* linear (actually not surprising when you
   remember that the variance is a mean: its the mean of the squared
   deviations from the overall mean), so the same reasoning applies:
   the variance of the average is the average of the variances.

   The standard deviation is *not* linear, however.  Since it does
   represent a "scale" you can multiply it by some scalar and get something
   meaningful, but you should never add standard deviations to each other,
   it doesn't mean anything.

   Note that since the "average" is the same thing as the "mean", the above
   says that you will get the same result however you divide up the
   samples into subsamples and take the averages of the averages.  You
   will get the mean and the variance of the whole thing.

   I'll try to address your other problem as soon as I have a moment.

				    Topher

    Another way to look at your questions is via the methodology of 
    Statistical Process Control (SPC). As long as you're collecting
    independant samples from a stable process, you can check the mean and
    variance of each new sample against your past historical data. SPC
    provides a nice simple graphical technique to apply the tests. I'll
    attempt to describe the tests as best I can.
    
    The Mean test would be accomplished by testing the sample mean against
    a confidence interval (usually .9973) described by � �
    3*sigma/sq.root(n). Where n=50 in this case, and sigma is the estimate
    of the population sigma. The process mean is the mean of all the values
    in your database, or the mean of your sample means (where each sample
    represents 50 values). Sigma can be estimated from all your historical
    data using the standard formula,i.e., why deal with sample variances or
    standard deviations if you have all the historical data which you can
    calculate the total sigma on.
    
    The variance test can be accomplished by testing the variance against
    limits set by:    
    		      ______
    	Upper Limit = VAR(x)
                      ------ * Chi�(alpha/2,n-1)
                       n-1
    
                      ______
        Center Line = VAR(x)
    
    		      ______
    	Lower Limit = VAR(x)
                      ------ * Chi�(1-alpha/2,n-1)
                       n-1
    
    where alpha is the risk of a Type I error, 
    	Chi�(alpha,n) is the alpha percentage point of the chi-square
    		distribution with n degrees of freedom,
    	______	
     	VAR(x) is the mean variance
    
    Actually, it is more common to test the standard deviation rather than
    the variance in SPC.
    
    The SPC approach was started in the 20's by Shewhart. There are
    multiple `tests' one can do to the mean and sigma to look for anything
    unusual in the sample, all with the intent of keeping the alpha and
    beta risks of the tests as small as possible.
    
    Hope this is of some small help.
    
    Peter

Thanks for the information on SPC.  It will help us.  From your
explanation of the mean test, I wonder why the rejection region
is so small (.0037) compared to the conventional .05 in
textbook hypothesis testing.

Lam

RE: .4

    Different purposes.  The number concerned represents the number of
    "false positives" you are willing to tolerate.

    In conventional hypothesis testing the false positive is false
    rejection of the null hypothesis.  That is, you will tentatively
    conclude -- (tentatively because if it is important enough there will
    presumably be at least one replication, and if it is less important
    it will still probably lead to later contradictions which will cause
    it to be challenged -- only if it is completely unimportant will it
    not be re-examined) -- that something "interesting" is going on.

    The cost of a false positive is relatively low -- when it is not,
    stricter criteria are used (e.g., the second standard of .01).  A high
    alpha means fewer false negatives -- you are less likely to miss
    something interesting.

    In SPC if you get a "positive" then you throw away the batch, shut
    down the line for inspection, send everyone to the bomb shelters, order
    additional, costly tests or whatever.  A false positive is expensive
    and you cannot tolerate as high a level as you can in research.
    Therefore you set it low -- and 

    I'm not sure just where that particular standard value came from.  Its
    not clear that a standard value is appropriate at all.  In research the
    costs and the rewards are quite intangible.  It is, therefore hard to
    do detailed cost/benefit analysis to determine the proper alpha.  A
    rule of thumb is therefore appropriate (though whether the one we have
    ended up with is the one that we *should* be using is an interesting
    question).  In SPC, generally, the costs and benefits of various
    choices are much easier to quantify and so cost/benefit analysis would
    seem to be the way to go.

    (Probably argued in the SPC literature which I am almost totally
    ignorant of).


				Topher

    I think the previous reply (.5) summed it up quite nicely. I would only
    add the following:
    
    The .0037 is with only 1 `rule' applied. Very commonly, the Western
    Electric rules and/or Nelson rules are applied which increase alpha to
    a level ~.01 or greater. Examples are: 2 of 3 points beyond the 2-sigma
    limit, 6 consecutive points increasing or decreasing in the same
    direction. So, the alpha risk isn't really that low. The advantage of
    applying more rules (weighed against the increased alpha) is the
    greater `statistical power' of the test: the ability to reject Ho when
    H1 is true.
    
    Topher is right in that the economics of the situation should drive the
    choices. There is a chapter in a book called ``Introduction to
    STATISTICAL QUALITY CONTROL'' by Montgomery (publ.=Wiley) that is
    entitled "Economic Design of Control Charts". You can guess what this
    chapter is about!
    
    Peter