Lecture 7

Sampling Distribution

Let be independent identically distributed (i.i.d.) random

variables. If we want to estimate the population mean , we take

sample , and use

∑

to estimate .

Definition: A statistic is a function of the observable random

variables in a sample and known constants.

Example:

∑

All statistics have probability distributions, which we call their

sampling distributions.

Example: Toss a die three times.

Theorem: Let . Then

.

Note:

√ .

Proof:

Example: (#7.11) A forester studying the effects of fertilization on

certain pine forests is interested in estimating the average basal

area of pine trees. In studying basal areas, he has discovered that

these measurements are normally distributed with standard

deviation appr. 4 sq. inches. If the forester samples n = 9 trees, find

the probability that the sample mean will be within 2 sq. in. of the

population mean.

Solution:

Theorem: Let . Then

√ , and ∑

.

Proof: done.

Example: . Find a number b such that

(∑

) .

Solution:

Suppose we want to make an inference about the population

variance based on from normal population. We

estimate by the sample variance:

∑

Theorem: Let . Then

.

Also and are independent random variables.

Proof:

Definition: Let and . Then, if Z and W are

independent,

√ is said to have a t distribution with

degrees of freedom.

Example: (#7.30) Let and . Z and W are

independent, then

√ .

(a) Find ;

(b) Given

.

Show that and

;

Solution:

Definition: Let and

be independent. Then

or

(F distribution)

Suppose we want to compare the variances of two normal

populations. We take a sample of size from one population, and

a sample of size from another.

We look at

.

Theorem:

.

Proof:

Example: If we take independent samples of size and

from two normal populations with equal population

variances, find b such that (

) .

Solution:

Example: (#7.34) Show

Solution:

Central Limit Theorem

We already showed that if is a random sample from any

distribution with mean and variance , then and

.

Now we want to develop an approximation for the sampling

distribution of (regardless of the distribution of the population).

Consider random sample of size n from exponential distribution

with mean 10:

{

Sample

size

Average of 1000

sample means Variance of

1000 sample

means

n = 5 9.86 10 19.63 20

n = 25 9.95 10 3.93 4

Theorem (CLT): Let be i.i.d. with and

.

Define ∑

√

√

Then ∫

√

for all u.

In other words, is asymptotically normally distributed with mean

and variance .

CLT can be applied to from any distribution provided that

and are both finite and n is large.

Example: (#7.42) The fracture strength of tempered glass averages

14 and has standard deviation 2.

(a) What is the probability that the average fracture strength of

100 randomly selected pieces of this glass exceeds 14.5?

(b) Find the interval that includes, with probability 0.95, the

average fracture strength of 100 randomly selected pieces of

this glass.

Solution:

Normal Approximation to Binomial Distribution

Example: (#7.80) The median age of residents of the US is 31

years, If a survey of 100 randomly selected US residents is to be

taken, what is the appr. probability that at least 60 will be under 31

years of age?

Solution: