Re: Standard Deviation Question....

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Re: Standard Deviation Question....

You might remember that on most  non graphing calculators that would handle
statistics you have  two different standard deviations to choose from--SDn and
SDn-1.  My TI30 gives the two answers as described below by Bogdan D Doytchinov;
however, the corresponding VARIANCES are actually  61.333333 for SDn   and  69 for
SDn-1.  The standard deviation in each case is squared.  I believe there was some
confusion between standard deviation and variance.  I would suggest that this be
confirmed by referring to any good statistics book as suggested below.  I am
neither a statistician nor a mathematician, but I do use statistics in my field.
It's always best to double check answers you get here with non web sources--books.
I will quote from Mendenhall,  __Introduction to Probability and Statistics, 1987.
It just happened to be sitting close by on the floor of my closet.  I quote:

"The variance of a population of N measurements y1, y2, ..., yn is defined to be
the average of the square of the deviations of the measurements about their mean mu
(Greek letter).  The population variance is denoted by sigma^2 (p. 37)" (and then a
formula is given, which I do not know how to recreate clearly via email).  The
Greek letter sigma is the normal way to denote "standard deviation."

"The variance of a sample of n measurements y1, y2, ... , yn is defined to be the
sum of the squared deviations of the measurements about their mean bar y [line over
the y] divided by (n-1).  The sample variance is denoted by S^2 and is given by the
formula ...(p. 37)"  That's s squared as opposed to the above sigma squared.

And "The standard deviation of a set of n measurements y1, y2, y3, ... yn is equal
to the positive square root of the variance (p. 38)."

Now, I know nothing about how a TI92 works, but on my TI86 when I enter the numbers
into a list and calculate the stats using the OneVa function, the sample standard
deviation is listed as Sx and the population standard deviation is sigma x.
Doesn't the 92 do something similar?  Good luck--D

PS  As I said above, it's best to check this for yourself in some books of your
choice.  That way you will have your best chance of actually understanding this
stuff.

Dear Mike:

> In a sense, the TI-92 is correct. You see, there are two different
> definitions of standard deviation:
>
> 1. Suppose you have 9 cards in a bag, with the numbers 4,7,10,11,12,14,15,21,32
> written on them. Then you start pulling out cards and random, and
> replacing them back.
> This experiment is modeling a random variable with mean 14 and variance
> 7.83. To compute the variance, you subtract the mean from each value,
> square the results, add them and divide by n=9 (the number of
> equiprobable outcomes), finally you take a square root.
> Thus, I understand why you expect the answer to your question to be 7.83.
>
> 2. Now look at the situation you have. The numbers 4,7,10,11,12,14,15,21,32
> are (most probably) obtained as observations of an unknown random
> variable, whose mean and variance you do not know. (It need NOT be the random
> variable
> described in the paragraph above). You want to estimate the mean and the
> variance. The best estimate for the mean is the average of the empirical data.
> (This is called the "empirical mean", and it is different from
> the real mean which you don't know).
> The best estimate for the variance is the so-called "empirical standard
> deviation", which you obtain in the following way. You subtract the
> empirical mean from each value, square the results, add them and divide
> by (n-1)=8 (the number of equiprobable outcomes
> minus one), and finally you take a square root.Thus you get the answer
> 8.31, which the TI gives you. You can check that 8.31=7.83*sqrt(9/8).
>
> Why divide by (n-1) instead of n? The answer to this question is rather
> tricky. It has
> something to do with the so-called "unbiased estimates" in statistics.
> The rationale is that, since you dom't know the real mean, and are
> replacing it by the empirical mean, you
> "lose one degree of freedom". This can all be found in books on
> statistics, and has to do with some deep results like Cochran's theorem.
>
> I realize that this might not be very helpful, but at least it lets you
> know that there is a reason for the answer you are getting, other than a
> bug in the calculator.
>
> Yours,
>
> Bogdan Doytchinov
> ------------------------------------------
>
> Mike Herald <oasis9@HOTMAIL.COM> wrote:
>
> >Can someone explain to me why TI-92 gives me the wrong answer for
> >standard deviation.
> >
> >I put in:
> >stdDev({4,7,10,11,12,14,15,21,32})
> >
> >and I get 8.30662
> >
> >but the answer is 7.83
> >Try to keep in mind I kinda new at using a TI-92.  So, why do I get 8.31
> >instead of 7.83
> >
> >Mike
> >
> >
> >______________________________________________________
> >Get Your Private, Free Email at http://www.hotmail.com
> >



--

Douglas S. Oliver
Department of Anthropology
University of California
Riverside, CA 92521
e-mail: douglaso@citrus.ucr.edu
    or: dsoliver@earthlink.net
References: