79
Chapter 7

Estimation and Confidence
Inferential Statistics
We use sample data and sample statistics in order to obtain a
reasonable estimate of the corresponding population parameters
under study.

Estimation, error bounds, confidence interval and confidence
level.
Section 7.1

The Problem of Parameter Estimation
Seek to make inferences about the population based on sample
information.
Sample statistics  Population parameters,
Unbiased estimations need random samples,
Sample statistics represent the most likely values the
unknown population parameters may take.

Given a single random sample of size
n
and a single statistic
obtained from the sample, how do we make meaningful inference
about the corresponding population parameter?
80
Section 7.2 Estimating a Population Mean

Use a random sample of size
to compute x .
n
  x.
Population mean
 is in the interval ( x  margin of error, x  margin of error)
Definition:
EB  Margin of error . (EB=Error Bound)
Fact:
 is in the interval ( x  EB, x  EB)  x is in (   EB,   EB)
Proof (graphically):
EB
(
)
x  EB
x

x  EB
EB
(
  EB
x

)
  EB
Algebraically:
If x  EB    x  EB , take case x    x  EB
(It is the same to take the case x  EB    x )
  x  EB

  EB  x
x    x    EB
Thus,   EB  x    EB .
81
 P( x  EB    x  EB)  P(   EB  x    EB)
 x2
2
)  N ( ,
)
By the CLT , X ~ N (  x ,
n
n
1

2
  EB
 Z ( / 2)

  EB


2
X
Z ( / 2)
0
Z
Definition:
Level of confidence = 1   = P( x  EB    x  EB)
Z ( 2) 
(   EB)   EB n



n
EB  Z  2


n
Interpretation of confidence level 1   :
1   is the fraction of samples in which the interval
( x  EB, x  EB) will contain  .
In another wards, “We are 1   confident that  is in
( x  EB, x  EB) for the sample with mean x .
82
Example 1. Find the 99% confidence interval for the population
mean  if the sample mean is x  84.2 , the sample size n  40 , and
population deviation is known from a previous study to be   7.3 .
Solution:
1    0.99 , x  84.2 , n  40 ,   7.3 , ( x  EB, x  EB )  ?
1    0.99 ,   0.01,

 0.005
2
7.3
7.3
  
EB  Z  
 Z (0.005) 
 2.575 
 2.97
40
40
2 n
( x  EB, x  EB)  (84.2  2.97, 84.2  2.97)  (81.23, 87.17)
That is, we are 99% confident that the population mean  is in the
interval (81.23, 87.17).
Example 2.
We want to estimate the mean weight of plastic
discarded by household in one week. How many household must
we randomly selected if we want to be 99% sure that the sample
mean is within 0.250 lb of the true population mean  ? Assume
population standard deviation   1.100 lb.
Solution:
1    0.99 , EB  0.250 ,   1.100 , n  ?
  
EB  Z  
2 n
83
2

2
2
    
n   Z      Z (0.005)(1.100)  =  2.575(1.100)  =128.3689
0.250
0.250
 

  2  EB  
n  129
Always round up.
To be 99% sure that the population mean  is with 0.25lb of the
sample mean, the sample size should at least 129 households.
Example 3. A sample of size 35 is taken from X ~ N (, 900) . Find
the confidence level if  is to be within 7.66 units of the sample
mean.
Solution:
n  35 , X ~ N (, 900)
hence   30 , EB  7.66 , 1    ? .
  
EB  Z  
2 n
   EB n 7.66 35

 1.51
 Z  

30
2
 
Z    1.51  P( Z  1.51)  1  
2
2
P(Z  1.51)  0.9345,

2
i.e. 1   0.9345
1    2(0.9345)  1
 0.869
The confidence level is about 87%.
1

2

2
1.51
84
Section 7.3
Estimating a Population Proportion
“Proportion of voters favoring such and such candidates”
Let
X
= the number of successes in a sample size n .
X ~ B(n, p) ,
where p is the probability of success or the population
proportion.
If np  10 , and n(1  p)  10 , then approximately
X ~ N (np, np(1  p)) .
Sample Proportion:
P 
X 1
p(1  p)
 X ~ N ( p,
)
n n
n
(Recall the properties of linear transformation:
If
X
1
X
n
, then
1
1
p(1  p)
1
1
 1   X  np  p and  1   X  np(1  P) 
X
X
n
n
n
n
n
n
n
sample success
p(1  p)
p (1  p )
)  N ( p ,
) , p 
n
n
n
EB n
   p   EB  p 
Z  

p (1  p )
p (1  p )
2
n

   p (1  p )
1
EB  Z  
2
n
2
P  ~ N ( p,
p   EB
Z ( / 2)

2
p
p   EB
0
Z ( / 2)
Z
85
Example 1. A sample size of 192 is taken from
X ~ B(1, p) .
Find
the 98% confidence interval for p if 100 success are observed.
Solution:
n  192 , 1    0.98 , P  
100
 0.521, ( p  EB, p  EB )  ?
192
0.521(1  0.521)
   p (1  p )
EB  Z  
 Z (0.01)
 2.33(0.036)  0.084
n
192
2
( p  EB, p  EB)  (0.521  0.084, 0.521  0.084)  (0.437, 0.605)
Example 2. If we want to be 95% confident that our sample survey
produces a proportion with an error no larger than 0.04, how large
must the sample be?
Solution:
1    0.95 , EB  0.04 , n  ?
   p(1  p)
EB  Z  
n
2
 z ( 2)  
n
p (1  p) .
 EB 
2

To have the least n value to guarantee the 95% confident level
we need to maximize
p(1  p) .
The maximum is at
p  0.5 .
Thus,
 Z 0.025
 1.95996 
n
(0.5)(0.5)  
  600.23  601(round

 0.04 
 0.04 
2
2
up).