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Mean in std. deviation?


Guest Grantys

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Guest Grantys

Why do you have to use the mean in a standard deviation formula? Cant you use any other average? Only asking as it is part of my coursework in the advantages and disadvantades of the averages. Any help on this topic too would be most welcome!

 

Thanks,

 

Grant.

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What do you mean, how is it derived?

 

One wishes to have a measure of how spread out the data is. There are several ways of doing this (eg interquartile range), one is the standard deviation. It is a measure of how far things are away from the mean.

 

Sadly one cannot just add up the (directed) distances from the mean of the sample points, as the answer will just be zero. One could add up the abs values of all the distances, indeed this is a recognized statistic. However, the preferred measure is the variance:

 

E(X^2) - E(X)^2,

 

It is the average of the sum of the squares of the distances from the origin.

 

E meaning expectation, and X being the r.v.

 

This has the benefit of giving more weight to those numbers which are further from the mean than those closer.

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the mean in the formula is the sample mean, or if u have the population, then use the population mean.

 

the actual form of standard deviation differs if ur calculating the population SD , or estimating the population from the sample SD

 

in general. the formula where you divide by (n-1) gives you the unbiased estimator for SD if you only have a sample of the population

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whats the difference between dividing by (n-1) and n ? i mean, obviously they'd give different answers, but why are there two different variations for one forumla? whats the difference between the two?

if you dont have a whole population data, and only a random sample, then u use the n-1 formula.

 

an estimator T(x) is called unbiased for theta

if

E(T(x))=theta... i.e the average of the estimators is the true value.

 

unbiased estimators are generally the best estimators.

 

if we find

 

E(S^2) where s^2 is the formula with just n in it.

 

then u get the value to be

 

(1-1/n)sigma^2 which is not equal to the true variance we are extimating. in this case its asymptotically unbiased. since as n gets bigger, the estimator becomes closer to the real value.

but a simple transformation gets us a proper unbiased one

 

take S'^2=n/(n-1) S^2 where s^2 is from above

 

that gives you E(s'^2)=sigma^2 (due to linearity of expectancy)

 

and that is called the sample variance.

 

dont worry too much about it.

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The E I meant is in fact

 

[math]\mathbb{E}[/math]

 

not a sigma

 

as bloodhound says, though apparently in response to a different remark, n and n-1 are used in different contexts depending on whether you want an unbaised estimator of the population variance or not.

 

ah, he's added a further explanation now....

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where that E thing means the sum of

x is the number

x-bar is the average

and n is the number of numbers.... like if you have 5 pieces of data then n = 5

Matt, thanx :)

 

so what did he mean then by "where that E thing means the sum of"

 

I thought that was represented by Sigma?

 

or was I mislead?

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