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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.
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Theorem: Let . Then
.
Note:
√ .
Proof:
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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:
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Theorem: Let . Then
√ , and ∑
.
Proof: done.
Example: . Find a number b such that
(∑
) .
Solution:
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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:
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Definition: Let and . Then, if Z and W are
independent,
√ is said to have a t distribution with
degrees of freedom.
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Example: (#7.30) Let and . Z and W are
independent, then
√ .
(a) Find ;
(b) Given
.
Show that and
;
Solution:
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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:
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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:
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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:
{
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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.
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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:
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Normal Approximation to Binomial Distribution
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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: