Sample Standard Deviation vs. Population Standard Deviation I have an HP 50g graphing calculator and I am using it to calculate the standard deviation of some data. In the statistics calculation there is a type which can have two values:
Sample
Population
I didn't change it, but I kept getting the wrong results for the standard deviation. When I changed it to "Population" type, I started getting correct results!
Why is that? As far as I know, there is only one type of standard deviation which is to calculate the root-mean-square of the values!
Did I miss something?
 A: There are, in fact, two different formulas for standard deviation here: The population standard deviation $\sigma$ and the sample standard deviation $s$.
If $x_1, x_2, \ldots, x_N$ denote all $N$ values from a population, then the (population) standard deviation is 
$$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2},$$
where $\mu$ is the mean of the population.  
If $x_1, x_2, \ldots, x_N$ denote $N$ values from a sample, however, then the (sample) standard deviation is 
$$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2},$$
where $\bar{x}$ is the mean of the sample.
The reason for the change in formula with the sample is this: When you're calculating $s$ you are normally using $s^2$ (the sample variance) to estimate $\sigma^2$ (the population variance).  The problem, though, is that if you don't know $\sigma$ you generally don't know the population mean $\mu$, either, and so you have to use $\bar{x}$ in the place in the formula where you normally would use $\mu$.  Doing so introduces a slight bias into the calculation: Since $\bar{x}$ is calculated from the sample, the values of $x_i$ are on average closer to $\bar{x}$ than they would be to $\mu$, and so the sum of squares $\sum_{i=1}^N (x_i - \bar{x})^2$ turns out to be smaller on average than $\sum_{i=1}^N (x_i - \mu)^2$.  It just so happens that that bias can be corrected by dividing by $N-1$ instead of $N$.  (Proving this is a standard exercise in an advanced undergraduate or beginning graduate course in statistical theory.)  The technical term here is that $s^2$ (because of the division by $N-1$) is an unbiased estimator of $\sigma^2$. 
Another way to think about it is that with a sample you have $N$ independent pieces of information.  However, since $\bar{x}$ is the average of those $N$ pieces, if you know $x_1 - \bar{x}, x_2 - \bar{x}, \ldots, x_{N-1} - \bar{x}$, you can figure out what $x_N - \bar{x}$ is.  So when you're squaring and adding up the residuals $x_i - \bar{x}$, there are really only $N-1$ independent pieces of information there.  So in that sense perhaps dividing by $N-1$ rather than $N$ makes sense.  The technical term here is that there are $N-1$ degrees of freedom in the residuals $x_i - \bar{x}$.
For more information, see Wikipedia's article on the sample standard deviation.
