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Chapter 7, part C
VI. The Central Limit Theorem
We will use the CLT to determine the form of the
probability distribution of x
There are two cases, when the population
distribution is unknown (the most likely) and
when it is known.
A. The Theorem
“In selecting simple random samples of size n from a
population, the sampling distribution of x can be
approximated by a normal probability distribution
as the sample sizes become large.”
As a rule of thumb, if n30, the sample satisfies the
CLT. Try this link for an example of how the CLT
works with simply tossing a coin.
Usefulness
The beauty of the CLT is that even if you have no
idea the form of the probability distribution of the
overall population, sampling in large sizes will
give you a sampling distribution that is normal.
Then you can calculate probabilities like we’ve
already done.
Remember the SPEND example?
This portrays the distribution of the population and it
hardly looks like a normal distribution, does it?
Eric R. Dodge:
thanks to John Ottensman at
IUPUI.
By taking small samples, 500 times, the distribution begins to
look normal. But the CLT requires large samples.
Eric R. Dodge:
thanks to John Ottensman at
IUPUI.
Taking more samples fills in some of the gaps, but we’re still
using a small sample.
Watch what happens
to the distribution
when we take larger
samples.
Eric R. Dodge:
thanks to John Ottensman at
IUPUI.
Mean = 41.48, standard error = 8.13 minutes.
Eric R. Dodge:
thanks to John Ottensman at
IUPUI.
Mean = 41.34, standard error = 4.58 minutes.
We’ll talk more
about how
increasing n changes
things in a few
slides.
Eric R. Dodge:
thanks to John Ottensman at
IUPUI.
B. The Value of the Sampling
Distribution of x
When we take a sample, we calculate x , which will
almost certainly not be equal to .
Sampling error =
x
The practical value is to use the sampling
distribution to provide probability information
about the size of the sampling error.
EAI Example
The personnel director believes a sample
is an acceptable estimate of  if:
x
x    $500
So we need to find the probability of a sampling
error being no more than $500.
Probabilities
If we find this area, we just multiply by 2.
 x  730.30
51,300
E( x)=51,800
52,300
x
Calculating Probabilities
Find the z-score: z=(52,300-51,800)/730.3 = .68
Area from the standard normal table, z=.68 is .2518
so
 P x    $500  50.36%
C. Sample Size and Sampling
Distribution
There’s little more than a 50% chance that the
sampling error will be less than $500 in our
previous example. One way to improve our odds
is to take a larger sample.
As you saw in the earlier “SPEND” slides, the only
thing that changes is the standard error of the
sampling distribution.
The standard error decreases
Previously,

4000
x  ( ) 
 730.30
n
30
but if we take a sample size of n=100,

4000
x  ( ) 
 400
n
100
and the sampling distribution gets more narrow, just
like the SPEND examples when we increased n
from 8 to 25 to n=75.
And the probability increases
Find the z-score: z=(52,300-51,800)/400 = 1.25
Area from the standard normal table, z=1.25 is .3944
so
 P x    $500  78.88%
A larger sample size will provide a higher probability that
the value of x is within a specified distance of .