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Section
8.2
Distribution of the
Sample Proportion
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Point Estimate of a Population
Proportion
Suppose that a random sample of size n is obtained
from a population in which each individual either
does or does not have a certain characteristic. The
sample proportion, denoted (read “p-hat”) is
given by pˆ
x
pˆ 
n
where x is the number of individuals in the sample
with the specified characteristic. The sample

proportion is a statistic that estimates the

population proportion, p.
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8-2
Sampling Distribution of pˆ
For a simple random sample of size n with
population proportion p:

ˆ is
• The shape of the sampling distribution
of p
approximately normal provided np(1-p)≥10.
ˆ is
• The mean of the sampling distribution of p
 pˆ  p.

• The standard deviation of the sampling
distribution
ˆ is
of p
p(1 p)

 pˆ 
n


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8-3
Sampling Distribution of pˆ
• When sampling from finite populations, this
assumption is verified by checking that the sample
size n is no more than 5% of the population size N

(n ≤ 0.05N).
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8-4
Parallel Example 4: Compute Probabilities of a Sample
Proportion
According to the Centers for Disease Control and
Prevention, 18.8% of school-aged children, aged
6-11 years, were overweight in 2004.
(a) In a random sample of 90 school-aged children, aged
6-11 years, what is the probability that at least 19%
are overweight?
(b) Suppose a random sample of 90 school-aged
children, aged 6-11 years, results in 24 overweight
children. What might you conclude?
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8-5
Solution
•
•
•
n=90 is less than 5% of the population size
np(1-p)=90(.188)(1-.188)≈13.7≥10
pˆ is approximately normal with mean=0.188 and
standard deviation = (0.188)(1 0.188)  0.0412
90

(a) In a random sample of 90 school-aged children, aged
6-11 years, what
 is the probability that at least 19%
are overweight?
0.19  0.188
Z
 0.0485 , P(Z>0.05)=1-0.5199=0.4801
0.0412
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8-6


Solution
• pˆ is approximately normal with mean=0.188 and
standard deviation = 0.0412
(b) Suppose a random sample of 90 school-aged
children, aged 6-11 years, results in 24 overweight
children. What might you conclude?
24
0.2667  0.188
pˆ 
 0.2667,
Z
1.91
90
0.0412
P(Z>1.91)=1-0.9719=0.028.
We would only expect to see about 3 samples in 100

resulting in a sample
proportion of 0.2667 or more.
This is an unusual sample if the true population
proportion is 0.188.
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8-7