Changes to Standard Deviation?

1. The Bob

Changes to Standard Deviation??

How many of you know that Standard Deviation has changed.

It used to be:$$\sqrt{\frac{\Sigma(x_i - \overline{x})^2}{n}}$$

And now it is:$$\sqrt{\frac{\Sigma(x_i - \overline{x})^2}{n - 1}}$$

It is the Variance of data but square rooted:

$$s^2 = \frac{\Sigma(x_i - \overline{x})^2}{n - 1}$$ convertd to: $$s = \sqrt{\frac{\Sigma(x_i - \overline{x})^2}{n - 1}}$$

Not really anything important, just wanted people to know and comment (if necessary) on the fact that it has changed.

The Bob (2004 ©)

Last edited: Oct 9, 2004
2. Hurkyl

15,987
Staff Emeritus
Actually, both formulae are used... I forget the reasons for using n instead of n-1, though.

3. mathman

6,754
It depends on what you are using for the mean. If you know the mean, then you divide by n. If you estimate the mean from the sample, then you use n-1, because the estimated mean has a statistical error.

4. The Bob

I do understand that both are still used but I didn't realise why until:
- Mathman came along and said why.

Cheers guys.

The Bob (2004 ©)

5. Ethereal

0
Why does using n-1 instead correct the error? Is this negligible for large values of n?

646
7. bjr_jyd15

75
n-1

n-1 is used for samples in order to adjust for the variability of the data set which does not included all possible events.

using n tends to produce an undersestimate of the population variance. So we use n-1 in the denominator to provide the appropriate correction for this tency.

to sum up:
when using populations, use n as the denominator.
else
use n-1

hope that helps!

8. Ontoplankton

148
The factor (n-1) is used to make the sample variance an "unbiased estimator" of the population variance.

There's no particular reason you need an unbiased estimator, though. For example, if you want to minimize mean squared error, it turns out that it's much better to use (n+1) instead of (n-1) (in case of a normal distribution). See Jaynes, chapter 17.