Download The Practice of Statistics Third Edition Chapter 9

Daniel S. Yates
The Practice of Statistics
Third Edition
Chapter 9:
9.2 Sample Proportions
Copyright © 2008 by W. H. Freeman & Company
Essential Questions
• How do you describe the sampling distribution of a
sample proportion? (Shape, center and spread)
• How do you compute the mean and standard deviation
for the sampling distribution of ̂ ?
• What is the rule of thumb for the use of the standard
deviation of ̂ ?
• What are the conditions that are necessary to use the
Normal approximation to the sampling distribution of ̂ ?
• How do you use the Normal approximation to the
sampling distribution of ̂ to solve a probability problem
involving ̂ ?
Review from Chapter 8
• X is a random variable. It is the number of
successes.
• If X has a distribution of B(n, p), then
 X  np
and
 X  np(1  p)
• If
we multiply each X by a constant has the effect of
multiplying the means and the standard deviation by the
constant.
a bX  a  b X
and  a  bX  b X
The Sampling Distribution of a
Sample Proportion ̂
We want to estimate the proportion of success in the population.
We take a SRS form the population of interest.
The estimator in the sample proportion of successes:
̂ 
count of successes in sample X

size of sample
n
Values of X and ˆ will vary in repeated samples.
The count X is a random variable that follows the Binomial Distribution.
̂ is also a random variable but does not follow the Binomial Distributi on.
The Mean and Standard Deviation
of a Sample Proportion
We start with the following facts:
The definition of ˆ is
X 1
 X.
n n
The means and standard deviation of X is
 X  n and  X  n (1   )
So,
1
n
 ˆ  n  
 ˆ 
1
n 1   
n 1    

2
n
n
 1   
n
When to use the formula for the
standard deviation of ̂
When to use the Normal
Approximation for ̂
The Normal approximation is most accurate when ρ is close to ½ and is
least accurate when ρ is near 0 or 1.
The following rule of thumb will insure that Normal calculations are accurate
for most statistical purposes;
Example
If the true proportion of defectives
produced by a certain manufacturing
process is 0.08 and a sample of 400 is
chosen, what is the probability that the
proportion of defectives in the sample is
greater than 0.10?
Since nρ  400(0.08)  32 > 10 and
n(1-ρ) = 400(0.92) = 368 > 10,
it’s reasonable to use the normal
approximation.
Example
(continued)
 ˆ    0.08
 ˆ 
 (1   )
n
0.08(1  0.08)

 0.013565
400
ˆ   ˆ 0.10  0.08
Z

 1.47
 ˆ
0.013565
P(p > 0.1)  P(z > 1.47)
 1  0.9292  0.0708
Example
Suppose 3% of the people contacted by phone
are receptive to a certain sales pitch and buy
your product. If your sales staff contacts 2000
people, what is the probability that more than
100 of the people contacted will purchase your
product?
Clearly ρ = 0.03 and ̂ = 100/2000 = 0.05 so



0.05  0.03 
^
P(p > 0.05)  P  z >

(0.03)(0.97) 


2000


0.05  0.03 

 P z >
  P(z > 5.24)  0
0.0038145 

Example - continued
If your sales staff contacts 2000 people, what
is the probability that less than 50 of the
people contacted will purchase your product?
Now ρ = 0.03 and ̂ = 50/2000 = 0.025 so



0.025  0.03 
^
P(p  0.025)  P  z 

(0.03)(0.97) 


2000


0.025  0.03 
 P  z 
  P(z  1.31)  0.0951
0.0038145 

Classwork
• Textbook p.589 problem 9.25
• P.590 problem 9.26.