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.
. Then ̅
Theorem: Let
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:
(∑
Solution:
. Find a number b such that
)
.
Suppose we want to make an inference about the population
variance
based on
from normal population. We
estimate
by the sample variance:
∑
Theorem: Let
Also ̅ and
Proof:
̅
. Then
are independent random variables.
.
Definition: Let
independent,
and
√
degrees of freedom.
. Then, if Z and W are
is said to have a t distribution with
Example: (#7.30) Let
independent, then
(a) Find
√
and
.
;
(b) Given
Show that
Solution:
. Z and W are
.
and
;
Definition: Let
and
or
be independent. Then
(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 Average of 1000
size
sample means
n=5
n = 25
Variance of
1000 sample
means
19.63
3.93
̅
9.86
9.95
10
10
Theorem (CLT): Let
.
Define
̅
20
4
be i.i.d. with
and
̅
∑
√
Then
√
∫
√
for all u.
In other words, ̅ is asymptotically normally distributed with mean
and variance
.
CLT can be applied to
and
from any distribution provided that
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: