Sorry if this answer is simple but I was wondering why is there a difference between a population variance and a sample variance?

I understand The variance is calculated as:

$$\text{Var} = \frac{1}{N}(x_i-\mu)^2$$

and the sample variance is computed as

$$\text{Var}_s = \frac{1}{N-1}(x_i-\mu)^2$$

In real world data sets would you use the sample variance most of the time? What if the population is also a sample? How would you know? Also does the mean change when you are looking at a population or a sample?

  • $\begingroup$ For a population mean does not change. If the sample is also the population the mean is same for both. mean change for sample in general. for example if a population consists of 100 elements and three different people take samples of say, 10 elements each .in general no two people will select same elements. hence their means will be different. $\endgroup$ – SA-255525 Jun 6 '14 at 17:55
  • $\begingroup$ The sample variance is computed as $$\text{Var}_s = \frac{1}{N-1}(x_i-\bar{x})^2$$ and it is the possible difference between $\mu$ and $\bar{x}$ which leads to using the changed denominator to make the sample variance an unbiased estimator of the population variance $\endgroup$ – Henry Oct 26 '15 at 15:43

It is probably more understandable to refer to the sample variance as the unbiased estimator of the overall variance. Usually, the set of numbers from which we are estimating these values do not reflect the entire universe of possibilities; they are a sample from which we want to make some inferences. We want to use this sample to estimate the mean and variance not of the sample itself, but of the underlying distribution. Understanding this, and running through the algebra (which can be found here) you see that using the statistic known as the "sample variance" is the statistic, if calculated on a hundred gazillion separate samples from the underlying distribution, whose expected value is the variance of the underlying distribution. When an estimator's expected value is the actual value in which we are interested, we call it "unbiased". Using the statistic known as the"population" variance, if we were to apply it to a huge set of samples and take ITS expectation (mean) it would be slightly lower than the "true" variance.

If, however, you have the entire (finite) population, and not a sample from it, such as the distribution of a six-sided die, then the population variance is the statistic to use, as you are not estimating it from a sample, but calculating it from the complete probability space.


Population mean and variance deal with fixed length data (size of population). You just want to know average height and deviation of height in the fixed group of people, for instance. Sample mean and variance are similar characteristic but the data interpretation is different. You want to know average height and its variance in your city and took sample of people from your group as a "sample" to estimate it.

What you wrote $Var_s$ is an unbiased estimator of the variance. Sample variance usually is defined in the same way as population variance. It is a best estimator of real variance in certain sense, however it is biased. $Var_s$ is not the best but unbiased. I would suggest you to read specila literature to understand the notions of "best estimator" and "biased, unbiased estimator".


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