To find mean wedivide the sum of no.s with 'n'for variance why 'n-1'

[dexterdev]

Hi all,
For finding average we take the sum of sequence numbers and divide by the number of elements. Why for variance this changes to number of elements minus 1.

-Devanand T

 

[Stephen Tashi]

The definition of the formula for the "sample variance" varies from book to book. Many books prefer the formula that uses division by n-1 because this is also the formula for the "unbiased estimator" of the population variance. Do you know about the definition of an "unbiased" estimator? - or the definition of an "estimator" for that matter?

 

[chiro]

To add to what Stephen Tashi said, there are general formulas for an arbitrary number of degrees of freedom.

If you do enough statistics you'll see n-1, n-2, n-4 instead of n-1 and the reason is because you are estimating a quantity using p known parameters which is why you divide by n-p to get un-biased estimator.

You'll learn this when you look at degrees of freedom in depth.

 

[dexterdev]

Thank you guys , I will see what an estimator and degrees of freedom etc...No idea on these.

 

[haruspex]

Let me try to show why n-1 is right in certain situations.
Suppose there is a total population N with unknown mean μ and variance σ². From this, you draw a sample S = {x_i} of size n. You can calculate the mean μ_S = Ʃx_i/n and variance σ²_S = Ʃ(x_i-μ_S)²/n of the sample (just treated as a set of numbers).
Consider the function V_S(y) = Ʃ(x_i-y)²/n.
If somehow you knew the real value of μ you could write down an unbiased estimate for σ² as V_S(μ).
The mean of the sample, μ_S, is that number y which minimises V_S(y). Since the true mean of the population is likely to be a little different, V_S(μ) tends to be greater than V_S(μ_S).
One way to think of this is that taking all the samples relative to μ_S removes one degree of freedom, leaving only n-1 degrees for how the samples are scattered around it.
Anyway, it's not hard to show that σ²_S*n/(n-1) is the least biased estimator for σ².
Sadly, the terminology is very misleading. "Sample variance" generally refers to the estimated variance of the population given the sample (i.e. the n-1 form), when it sounds like it should be the variance of the sample taken as an abstract set of numbers (the n form).