Introduction to Probability and Statistics for Engineers and Scientists

6.6Sampling from a Finite Population 217

The following corollary of Theorem 6.5.1 will be quite useful in the following chapters.

Corollary 6.5.2

LetXi,...,Xnbe a sample from a normal population with meanμ.IfX denotes the
sample mean andSthe sample standard deviation, then

√

n

(X−μ)

S

∼tn− 1

That is,

√

n(X−μ)/Shas at-distribution withn−1 degrees of freedom.

Proof

Recall that at-random variable withndegrees of freedom is defined as the distribution of

Z

√

χn^2 /n

whereZis a standard normal random variable that is independent ofχn^2 , a chi-square
random variable withndegrees of freedom. It then follows from Theorem 6.5.1 that
√
n(X−μ)/σ
√
S^2 /σ^2

=

√

n

(X−μ)

S

is at-random variable withn−1 degrees of freedom.

6.6Sampling from a Finite Population

Consider a population ofNelements, and suppose that pis the proportion of the
population that has a certain characteristic of interest; that is,Npelements have this
characteristic, andN(1−p) do not. A sample of sizenfrom this population is said to
be arandom sampleif it is chosen in such a manner that each of the

(N

n

)
population subsets
of sizenis equally likely to be the sample. For instance, if the population consists of the
three elementsa,b,c, then a random sample of size 2 is one that is chosen so that each
of the subsets{a,b},{a,c}, and{b,c}is equally likely to be the sample. A random subset
can be chosen sequentially by letting its first element be equally likely to be any of theN
elements of the population, then letting its second element be equally likely to be any of
the remainingN−1 elements of the population, and so on.
Suppose now that a random sample of sizenhas been chosen from a population of size
N. Fori=1,...,n, let

Xi=

{

1 if theith member of the sample has the characteristic

0 otherwise