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CASE STUDY Illustration that the sample variance is a biased estimate of the population variance
Consider the population which consists of four counters marked zero, one, two, and three.
The mean score of the counters is equal to 0+1+2+3 divided by four because 6/4 = 1.5. The population mean therefore equals 1.5.
The variance of the population equals the sum total of all the deviations of population members from the population mean, squared and then divided by n. The population variance is the 1.25.
Consider all the samples of size 2 can be taken from this population the variances of all these samples of size two are shown in the table below.
Table 1: Variances of all the samples of size two from the population of 4 items numbered 0–3, using n as denominator of the computation
First counter
0
1
2
3
0
0
0.25
1
2.25
1
0.25
0
0.25
1
2
1
0.25
0
0.25
3
2.25
1
0.25
0
The mean variance of all the samples equals ∑(variances)/number of samples = 10/16
0.625.
This is not the same as the population variance therefore any sample variance will give us a biased estimate of the population variance. However this bias can be rectified if, instead of dividing by n when calculating the sample variances, we divide by n–1 (see Table 2).
Table 2: Variances of all the samples of size two from the population of 4 items numbered 0–3, using n-1 as denominator of the computation
First counter
0
1
2
3
0
0
0.5
2
4.5
1
0.5
0
0.5
2
2
2
0.5
0
0.5
3
4.5
2
0.5
0
The mean variance of all the samples equals ∑(variances)/number of samples = 20/16 = 1.25
