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Central Limit Theorem
Sample Proportions
Requirements
• Random sample from population?
• np  10?
n(1  p )  10?
Conclusions
• Shape of sampling distribution is approximately normal
• Mean of the sampling distribution is equal to the population proportion
p  p
• Standard deviation of the sampling distribution is equal to the following
formula
p 
p (1  p )
n
“Watch Out” on page 266
Why is the CLT important??
• See problem 13-3 on page 267
Examples of CLT for Proportions
“AP Guidelines”
Suppose that 25% of all households
in Solon have a pet cat. What would
be the shape, mean, and standard
deviation of sample proportions of
sample size 100? Why?
• Check Requirements
•
•
•
Random sample from pop? Stated SRS (YES)
np  10?
n(1  p )  10?
100(.25)  25  10 (YES)
100(1  .25)  75  10
Requirements met
• Shape is approximately normal
•  p  .25
•   .25(1  .25)  0.0433
p
100
You take an SRS of 100 households
and find that 17 households have a
pet cat. Where does your sample
proportion fall in this sampling
distribution? Is your sample result
surprising? Do you think that 25% of
Solon households have a pet cat?
P( p  .17)
.17  .25
P( z 
)
0.0433
P( z  1.85)
.03234
Since the probability of getting a sample
proportion of 17% or less, if 25% of Solon
households have cats, is only 3.2%, I suspect that
less than 25% really have cats.
Example
• A study was conducted in 2000 which surveyed a random sample of men
between the ages of 45 and 54. The researchers reported that 71.3% of such
men are considered overweight. If a random sample of 90 men in this age
group is selected, what is the probability that more than 70% of them will be
overweight?
A study was conducted in 2000 which surveyed a random sample of men between the ages of
45 and 54. The researchers reported that 71.3% of such men are considered overweight. If a
random sample of 90 men in this age group is selected, what is the probability that more than
70% of them will be overweight?
•
Check requirements
•
•
•
Random sample from pop? Stated Random YES
np  10?
90(.713)  64.17  10
YES
n(1  p )  10? 90(1  .713)  25.83  10
Conclusions
• Shape of samp. dist. is approx. normal
•  p  .713
• p 
.713(1  .713)
 0.0477
90
P( p  .70)
.70  .713
P( z 
)
0.0477
P( z  0.27)
1  P( z  0.27)
1  .3936
.6064