Hi, my name is Rosalie. I have a question about sample variance.
To give an unbiased estimate of the population variance, the denominator of the sample variance should be (n-1) instead of n. I tried to convince myself by comparing the population variance and sample variance (with denominator n and n-1):
Consider the population {1, 2, 3, 4, 5}
Population mean mu = (1+2+3+4+5)/5=3
Population variance sigma =[(1-3)2 + (2-3)2+ ··· +(5-3)2]/5 = 2
Now, all the possible samples of size n = 4 drawn from the above population are considered.
Sample |
Sample Mean |
sum(x-xbar)2 /n |
sum(x-xbar)2 /(n-1) |
{1, 2, 3, 4} |
2.5 |
1.25 |
1.67 |
{1, 2, 3, 5} |
2.75 |
2.1875 |
2.92 |
{1, 2, 4, 5} |
3 |
2.5 |
3.3 |
{1, 3, 4, 5} |
3.25 |
2.1875 |
2.92 |
{2, 3, 4, 5} |
3.5 |
1.25 |
1.67 |
Mean |
3 |
1.875 |
2.502 |
It looks like the mean of the variance calculated by dividing by n (1.875) gives a better estimate of the population variance (2). Where did I go wrong? Thanks for your help.
Hi Rosalie,
The fact that the sample variance, calculated by dividing by n-1, is an unbiased estimator for the population variance is true if the population is infinite or you sample with replacement. In your example if you sample with replacement then you get 54 = 625 possible samples rather than the 5 you examined. Rather than doing an example that large I am going to repeat your calculation but using the population {1,2,3}and a sample of size 2.
Population mean mu = (1+2+3)/3 = 2
Population variance sigma =[(1-2)2 + (2-2)2+(3-2)2]/3 = 2/3
Now, all the possible samples of size n = 2 drawn from the above population with replacement are considered.
Sample |
Sample Mean |
sum(x-xbar)2 /n |
sum(x-xbar)2 /(n-1) |
{1, 1} |
1 |
0 |
0 |
{1, 2} |
1.5 |
0.25 |
0.5 |
{1, 3} |
2 |
1 |
2 |
{2, 1} |
1.5 |
0.25 |
0.5 |
{2, 2} |
2 |
0 |
0 |
{2, 3} |
2.5 |
0.25 |
0.5 |
| {3, 1} | 2 | 1 | 2 |
| {3, 2} | 2.5 | 0.25 | 0.5 |
| {3, 3} | 3 | 0 | 0 |
| Mean | 2 | 3/9 = 1/3 | 6/9 = 2/3 |
Penny