Why is the mean of sample variances not equal to the population variance?

#1
Several texts say that the mean of all values of an unbiased estimator is equal to the parameter it is estimating.

They also say that the sample variance using Bessel's correction is an unbiased estimator of the population variance.

Here's what I'm confused about:

Consider a population with N=3, let's say {3,5,13}.
The population mean is \(\frac{3+5+13}{3} = 7\)
The population variance is \(\frac{(3-7)^2+(5-7)^2+(13-7)^2}{3} = 18.6...\)

There are three possible samples of n=2: {3,5}, {5,13}, and {3,13}.

Sample mean of {3,5} is 4.
Sample variance using Bessel's correction is \(\frac{(3-4)^2+(5-4)^2}{2-1} = 2\)

Sample mean of {5,13} is 9.
Sample variance using Bessel's correction is \(\frac{(5-9)^2+(13-9)^2}{2-1} = 32\)

Sample mean of {3,13} is 8.
Sample variance using Bessel's correction is \(\frac{(3-8)^2+(13-8)^2}{2-1} = 50\)

But the mean of the sample variances is \(\frac{2+32+50}{3}=28\neq18.6...\)

Am I doing something wrong? If I'm calculating correctly, have I misunderstood what it means for an estimator to be unbiased? Any help would be appreciated.
 
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#2
For instance, McClave, Benson and Sincich's Statistics for Business and Economics says, "If the sampling distribution of a sample statistic has a mean equal to the population parameter the statistic is intended to estimate, the statistic is said to be an unbiased estimator of the parameter."

Can anybody tell me where I'm going wrong?
 

Dason

Ambassador to the humans
#3
The assumption is that the sampling is done with replacement. There is a finite population correction factor you can multiply by to correct for sampling done without replacement from a finite population.