As I understand UK Schools teach that the Standard Deviation is found using:
whereas US Schools teach:
(at a basic level anyway).
This has caused a number of my students problems in the past as they have searched on the Internet, but found the wrong explanation.
Why the difference?
With simple datasets say 10 values, what degree of error will there be if the wrong method is applied (eg in an exam)?
The first formula is the population standard deviation and the second formula is the the sample standard deviation. The second formula is also related to the unbiased estimator of the variance - see wikipedia for further details.
I suppose (here) in the UK they don't make the distinction between sample and population at high school. They certainly don't touch concepts such as biased estimators.
Because nobody has yet answered the final question--namely, to quantify the differences between the two formulas--let's take care of that.
For many reasons, it is appropriate to compare standard deviations in terms of their ratios rather than their differences. The ratio is
$$s_n / s = \sqrt{\frac{N-1}{N}} = \sqrt{1 - \frac{1}{N}} \approx 1 - \frac{1}{2N}.$$
The approximation can be viewed as truncating the (alternating) Taylor series for the square root, indicating the error cannot exceed $|\binom{1/2}{2}N^{-2}|$ = $1 / (8 N^2)$. This establishes that the approximation is more than good enough (for our purposes) once $N$ is $2$ or larger.
It is immediate that the two SD estimates are within (about) 10% of each other once $N$ exceeds $5$, within 5% once $N$ exceeds $10$, and so on. Clearly, for many purposes these discrepancies are so small that it does not matter which formula is used, especially when the SD is intended for describing the spread of data or for making semi-quantitative assessments or predictions (such as in employing the 68-95-99.7 rule of thumb). The discrepancies are even less important when comparing SDs, such as when comparing the spreads of two datasets. (When the datasets are equinumerous, the discrepancies effectively vanish altogether and both formulas lead to identical conclusions.) Arguably, these are the forms of reasoning we are trying to teach beginning students, so if the students are becoming concerned about which formula to use, that could be taken as a sign that the text or the class is failing to emphasize what is really important.
We might want to pay some attention to the case of very small $N$. Here, people may be using $t$ tests instead of $z$ tests, for instance. In that case, it is essential to employ whichever formula for the standard deviation is used by one's table or software. (This is not a matter of one formula being wrong or right; it's just a consistency requirement.) Most tables use $s$, not $s_n$: this is the one place in the elementary syllabus where the text and teacher need to be clear about which formula to use.
I am not sure this is purely a US vs. British issue. The rest of this page is excerpted from a faq I wrote.(http://www.graphpad.com/faq/viewfaq.cfm?faq=1383).
How to compute the SD with n-1 in the denominator
Compute the square of the difference between each value and the sample mean.
Add those values up.
Divide the sum by n-1. The result is called the variance.
Take the square root to obtain the Standard Deviation.
Why n-1?
Why divide by n-1 rather than n when computing a standard deviation? In step 1, you compute the difference between each value and the mean of those values. You don't know the true mean of the population; all you know is the mean of your sample. Except for the rare cases where the sample mean happens to equal the population mean, the data will be closer to the sample mean than it will be to the true population mean. So the value you compute in step 2 will probably be a bit smaller (and can't be larger) than what it would be if you used the true population mean in step 1. To make up for this, divide by n-1 rather than n.v This is called Bessel's correction.
But why n-1? If you knew the sample mean, and all but one of the values, you could calculate what that last value must be. Statisticians say there are n-1 degrees of freedom.
When should the SD be computed with a denominator of n instead of n-1?
Statistics books often show two equations to compute the SD, one using n, and the other using n-1, in the denominator. Some calculators have two buttons.
The n-1 equation is used in the common situation where you are analyzing a sample of data and wish to make more general conclusions. The SD computed this way (with n-1 in the denominator) is your best guess for the value of the SD in the overall population.
If you simply want to quantify the variation in a particular set of data, and don't plan to extrapolate to make wider conclusions, then you can compute the SD using n in the denominator. The resulting SD is the SD of those particular values. It makes no sense to compute the SD this way if you want to estimate the SD of the population from which those points were drawn. It only makes sense to use n in the denominator when there is no sampling from a population, there is no desire to make general conclusions.
The goal of science is almost always to generalize, so the equation with n in the denominator should not be used. The only example I can think of where it might make sense is in quantifying the variation among exam scores. But much better would be to show a scatterplot of every score, or a frequency distribution histogram.
This is Bessel's Correction. The US version is showing the formula for the sample standard deviation, where the UK version above is the standard deviation of the sample.
Since N is the number of points in the data set, one could argue that by calculating the mean one has reduced the degree of freedom in the data set by one (since one introduced a dependency into the data set), so one should use N-1 when estimating the standard deviation from a data set for which one had to estimate the mean before.
