Re: Why, for Sample Standard Deviation, Divide by N-1, Instead of N?
- From: "Kevin E. Thorpe" <kevin.thorpe@xxxxxxxxxxx>
- Date: 15 Aug 2006 18:36:59 -0700
DarkProtoman wrote:
Kevin E. Thorpe wrote:
DarkProtoman wrote:
Why, when you calculate the sample standard deviation, you divide by
n-1, instead of n? I've heard someone said it has "nice mathematical
properties that make the math work out smoothly". What are those "nice
mathematical properties"? Thanks!!!!!
Supposing you have a random sample of observations from a population.
The sample variance \sum[(x_i - \bar{x})^2]/(n-1) is the unbiased
estimate
of the population variance. The standard deviation is the square root
of
this. It is actually the variance that has the "nice" properties
(unbiasedness,
its relation to the chi-squared distribution).
OK, then, what are the nice mathematical properties of the variance,
then?
I mentioned two.
It's unbiased.
Let s^2 be defined as above and xbar be the sample mean
computed on a sample from a normally distributed population
with mean mu and variance sigma^2. A theorem from
mathematical statistics states that xbar and s^2 are independent
random variables. Furthermore, (n-1)*s^2 / sigma^2 has a
chi-square distribution on n-1 degrees of freedom.
It is also this sample variance that ultimately gives the t-statistic
its distribution (again under normality).
--
Kevin E. Thorpe
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
.
- References:
- Why, for Sample Standard Deviation, Divide by N-1, Instead of N?
- From: DarkProtoman
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- From: Kevin E. Thorpe
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- From: DarkProtoman
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