Introduction to Probability and Statistics for Engineers and Scientists

214 Chapter 6: Distributions of Sampling Statistics

wherex=

∑n

i= 1 xi/n. It follows from this identity that

(n−1)S^2 =

∑n

i= 1

Xi^2 −nX^2

Taking expectations of both sides of the preceding yields, upon using the fact that for any
random variableW,E[W^2 ]=Var(W)+(E[W])^2 ,

(n−1)E[S^2 ]=E

[ n

∑

i= 1

Xi^2

]

−nE[X^2 ]

=nE[X 12 ]−nE[X^2 ]

=nVar(X 1 )+n(E[X 1 ])^2 −nVar(X)−n(E[X])^2

=nσ^2 +nμ^2 −n(σ^2 /n)−nμ^2

=(n−1)σ^2

or

E[S^2 ]=σ^2

That is, the expected value of the sample variance S^2 is equal to the population
varianceσ^2.

6.5Sampling Distributions from a Normal Population

LetX 1 ,X 2 ,...,Xnbe a sample from a normal population having meanμand varianceσ^2.
That is, they are independent andXi∼N(μ,σ^2 ),i=1,...,n. Also let

X=

∑n

i= 1

Xi/n

and

S^2 =

∑n

i= 1

(Xi−X)^2

n− 1

denote the sample mean and sample variance, respectively. We would like to compute
their distributions.