How to compute sample variance? Suppose we have a sample of size n= 9, where the sum equals 27, and the sum of the squaredvalues equals 113. Sample mean is 27/9=3 and mean of squared values is 113/9 = 12.56 Can we then say that the sample variance $S^2 = E(\overline X^2) - E(\overline X)^2 = 12.55 - 9 = 3.56$ ?
 A: Technically this is not the sample variance, not in the sense which most disciplines intend when they say "sample variance."  
You are calculating the empirical variance in your formula above for those $N=9$ data points, which would be the population variance if it consisted of all data points from the given population.  But since you say sample I'm assuming that's not the case.  So in this situation you would calculate the sample variance by dividing by $N-1$ (e.g. $8$, not $9$).  The formulas traditionally used for the sample variance are
$$\frac{1}{n-1}\sum_{i=1}^n \left(x_i - \bar x\right)^2 = \frac{1}{n-1}\left(\sum_{i=1}^n x_i^2 - \frac{1}{n}\left(\sum_{i=1}^n x_i\right)^2\right).$$
As a worked example, in your case $n=9$, $\sum_{i=1}^n x_i = 27$, and $\sum_{i=1}^n x_i^2 = 113$.  Therefore, by the second formula, the sample variance is
$$\frac{1}{9-1}\left(113 - \frac{1}{9}(27)^2\right) = \frac{1}{8}(32) = 4.$$
You can google "sample variance vs population variance" and look for a result in your discipline for more detail.     
