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Posted: Mon Jul 14, 2008 7:43 am |
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Can someone give me a rule that tells me when to use n trials in the
denominator when determining variance and when to use n-1?
Openoffice calc (spreadsheet program) appears to use n-1, and yet some
of the references I've been studying use n. Obviously, the smaller
denominator will give a larger variance and standard deviation, so
clarification on this would be greatly appreciated.
Thanks in advance.
Henry Sun |
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| Ray Koopman... |
Posted: Mon Jul 14, 2008 8:17 am |
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On Jul 14, 10:43 am, henrysun... at (no spam) yahoo.com wrote:
Quote: Can someone give me a rule that tells me when to use n trials in the
denominator when determining variance and when to use n-1?
Openoffice calc (spreadsheet program) appears to use n-1, and yet some
of the references I've been studying use n. Obviously, the smaller
denominator will give a larger variance and standard deviation, so
clarification on this would be greatly appreciated.
Thanks in advance.
Henry Sun
It depends on what you want to talk about. If you divide by n you get
the variance of the data. If you divide by n-1 you get an estimate of
the variance of the population from which the data were drawn. So n
gets you a description of the data in hand, whereas n-1 gets you an
inference about the population that the data came from. |
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| Jack Tomsky... |
Posted: Mon Jul 14, 2008 8:40 am |
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Quote: On Jul 14, 10:43 am, henrysun... at (no spam) yahoo.com wrote:
Can someone give me a rule that tells me when to
use n trials in the
denominator when determining variance and when to
use n-1?
Openoffice calc (spreadsheet program) appears to
use n-1, and yet some
of the references I've been studying use n.
Obviously, the smaller
denominator will give a larger variance and
standard deviation, so
clarification on this would be greatly appreciated.
Thanks in advance.
Henry Sun
It depends on what you want to talk about. If you
divide by n you get
the variance of the data. If you divide by n-1 you
get an estimate of
the variance of the population from which the data
were drawn. So n
gets you a description of the data in hand, whereas
n-1 gets you an
inference about the population that the data came
from.
And in another twist, you divide by n+1 if you want the minimum mean-squared error estimate of the population variance.
Jack |
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Posted: Mon Jul 14, 2008 8:42 am |
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On Jul 14, 11:17 am, Ray Koopman <koop... at (no spam) sfu.ca> wrote:
Quote: On Jul 14, 10:43 am, henrysun... at (no spam) yahoo.com wrote:
Can someone give me a rule that tells me when to use n trials in the
denominator when determining variance and when to use n-1?
Openoffice calc (spreadsheet program) appears to use n-1, and yet some
of the references I've been studying use n. Obviously, the smaller
denominator will give a larger variance and standard deviation, so
clarification on this would be greatly appreciated.
Thanks in advance.
Henry Sun
It depends on what you want to talk about. If you divide by n you get
the variance of the data. If you divide by n-1 you get an estimate of
the variance of the population from which the data were drawn. So n
gets you a description of the data in hand, whereas n-1 gets you an
inference about the population that the data came from.
Thanks for the clarification. |
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| Ray Koopman... |
Posted: Mon Jul 14, 2008 2:22 pm |
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On Jul 14, 11:40 am, Jack Tomsky <jtom... at (no spam) ix.netcom.com> wrote:
Quote: [...] And in another twist, you divide by n+1 if you want the
minimum mean-squared error estimate of the population variance.
That's for a mesokurtic distribution, with mu4/m2^2 = 3.
In general, the divisor that minimizes the expected squared error
is n + k-2 - (k-3)/n, where k = mu4/m2^2. |
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| Jack Tomsky... |
Posted: Mon Jul 14, 2008 2:37 pm |
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Quote: On Jul 14, 11:40 am, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
[...] And in another twist, you divide by n+1 if
you want the
minimum mean-squared error estimate of the
population variance.
That's for a mesokurtic distribution, with mu4/m2^2 =
3.
In general, the divisor that minimizes the expected
squared error
is n + k-2 - (k-3)/n, where k = mu4/m2^2.
Ray, that's an interesting result, generalizing it beyond the normal distribution where the kurtosis is 3.
Jack |
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| Ray Koopman... |
Posted: Mon Jul 14, 2008 6:10 pm |
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Guest
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On Jul 14, 5:37 pm, Jack Tomsky <jtom... at (no spam) ix.netcom.com> wrote:
Quote: On Jul 14, 11:40 am, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
[...] And in another twist, you divide by n+1 if you
want the minimum mean-squared error estimate of the
population variance.
That's for a mesokurtic distribution, with mu4/m2^2 = 3.
In general, the divisor that minimizes the expected
squared error is n + k-2 - (k-3)/n, where k = mu4/m2^2.
Ray, that's an interesting result, generalizing it
beyond the normal distribution where the kurtosis is 3.
Jack
Thanks, Jack. I'm glad you could see through the typos and
incompleteness. For the record, "m2" should have been "mu2",
with mu2 & mu4 being the 2nd & 4th central moments. |
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