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Assume we have a population with the values
2,3,4,5,6. This
population
has
μ = 4
and
σ
2
=
2 =
=
We now take all possible non ordered samples of size two from the
population and calculate the mean of each pair.
pair
2,3
2,4
2,5
2,6
3,4
3,5
3,6
4,5
4,6
5,6
x
2.5
3
3.5
4
3.5
4
4.5
4.5
5
5.5
The statistic
x
also has a mean, μ, and variance
σ
2
.
For the above pairs
μ=
[2.5 + 3 + 2(3.5) + 2(4) + 2(4.5) + 5 + 5.5]/10= 4
σ
2
= .75
Note that
μ = μ= 4.
In contrast, the variance of is less than the
variance of the population (2). The variances can be equated through
the following formulas:
for infinte populations
σ
2
=
n
2
σ
σ=
for finite populations
σ
2
=
1
N
n
N
n
2


σ
=
Since our population has N = 5 and n = 2 we obtain:
σ
2
=
o.75 =
.
For practical purposes we assume a population is infinite if n<.05N
The following is a Histogram of the above distribution of sample
means.
2
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 Fall '10
 RaySievers
 Statistics, Standard Error

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