Download 12.1 Estimation

12.1 Estimation of the Difference between the Means
Motivating example:
Objective:
we want to determine the difference between the mean ages of the
customers shopping at the inner-city and the suburban.
1 : the mean age of all customers who shop at the inner-city store
 2 : the mean age of all customers who shop at the suburban store
 12 : the variance of the ages of all customers who shop at the inner-city store
 22 : the variance of the ages of all customers who shop at the suburban store
We want to estimate the difference
1  2 .
Let
x1,1 , x1, 2 ,, x1,36 :
the ages of 36 customers shopping at the inner-city store.
x2,1 , x2, 2 ,, x2, 49 :
the ages of 49 customers shopping at the suburban
store.
Then,
36
x1 
49
x
i 1
1,i
36
 40, x2 
x
i 1
2 ,i
49
 35
and
 x
36
s1 
i 1
The point estimate of
1,i
 x1 
36  1
 x
49
2
 9, s2 
i 1
 x2 
2
2 ,i
49  1
 10 .
1  2  x1  x2  40  35  5 .
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The point estimate of
 12  s12  92  81 .
The point estimate of
 22  s22  102  100
.
General Case:
1 : the mean of population 1
 2 : the mean of population 2
 12 : the variance of population 1
 22 : the variance of population 2
Let
x1,1 , x1, 2 ,, x1, n1 : the random sample from population 1
x2,1 , x2, 2 ,, x2, n2 : the random sample from population 2.
Then,
n1
x1 
x
1,i
i 1
: the sample mean of population 1.
n1
n2
x2 
x
i 1
n2
2 ,i
: the sample mean of population 2
and
 x
n1
s1 
i 1
1,i
 x1 
n1  1
2
: the standard deviation of population 1.
2
 x
n2
s2 
i 1
 x2 
2
2 ,i
n2  1
: the standard deviation of population 2.
The point estimate of u1  u 2
 x1  x2
Important Properties of X 1  X 2 :
X 1 : the sample statistic with possible value x1
X 2 : the sample statistic with possible value x2
Then,
 X  X  E X1  X 2   1  2
1
2
and

I.
2
X1  X 2
 Var X 1  X 2   E X 1  X 2   1   2  
2
 12
n1

 22
n2
Large Sample Case ( n1  30, n2  30 ):
Sampling Distribution of X 1  X 2 ( n1  30, n2  30 ):

X 1  X 2  N 1   2 ,

X
1
 X 2    X1  X 2
 X X
1
2
X1  X 2

X
1


 12  22 

 N  1   2 ,

n
n
1
2 

 X 2   1   2 



n1 n2
2
1
2
2
2
 N 0,1
Derivation of 1    100% confidence interval:
3
.
1    P Z  z 
2

  X  X      

2
1
2
 P 1
 z 

2
 X1  X 2


       X  X 

2
1
2
 P 1
 z 

2
 X1  X 2





1   2    X 1  X 2 

 P  z 
 z 

2
2
 X1  X 2


 P  z  X1  X 2  1   2    X 1  X 2   z  X1  X 2 
2
2


 P  X 1  X 2   z  X1  X 2  1   2   X 1  X 2   z  X1  X 2 
2
2


Thus, 1     100% confidence interval is
x1  x2   z  X  X  x1  x2   z
2
1
2
2
 12  22

n1 n2

 12  22
 12  22 
  x1  x2  z
 , x1  x2  z
 
2
2
n
n
n
n2 

1
2
1
1    100% confidence interval ( n1  30, n2  30 ):
 As
1,  2
are known,
x1  x2   z  X  X  x1  x2   z
2
1
2
2
 12  22

n1 n2

 12  22
 12  22 
  x1  x2  z
 , x1  x2  z
 
2
2
n
n
n
n2 

1
2
1
is a 1    100% confidence interval estimate of the population
4
1  2 .
difference
 As
1,  2
x1  x2   z
are unknown,
2
s X1  X 2  x1  x2   z
2
s12 s 22

n1 n2

s12 s 22
s12 s 22 
  x1  x2  z
 , x1  x2  z
 
2 n
2 n
n
n2 

1
2
1
is a 1    100% confidence interval estimate of the population
difference
1  2 .
Example (continue):
s X1  X 2
s12 s 22
9 2 10 2




 2.07
n1 n2
36 49
A 95% confidence interval for
x1  x2   z
2
1  2
is
s X1  X 2  40  35  z 0.025  2.07  5  1.96  2.07  5  4.06  0.94,9.06
II. Small Sample Case ( n1  30, n2  30 ):
Two assumptions are made:
1. Both populations have normal distribution.
2. The variance of the populations are equal (  1
2
Pooled estimate of
2
and
 X2
5
1X2
:
  22   2 )
2
 2 , denoted by s p ,
The pooled estimate of
 x
n1
s 2p 
n1  1s
 n2  1s

n1  n2  2
2
1
2
2
i 1
1,i
 x1    x 2,i  x 2 
2
n2
2
i 1
n1  n2  2
is a weighted average of the two sample variance
s12
and
s22 .
The estimate of
 X X
1
2
1 1
 12  22


  2   
n1 n2
 n1 n2 
is
s X1  X 2 
 1
1 
 .
s 2p 

 n1 n2 
1    100% confidence interval ( n1  30, n2  30 ):
x1  x2   t n n 2,
1
2
2
s X1  X 2  x1  x2   t n1  n2 2,
2
1 1
s 2p   
 n1 n2 

1
1 
2 1
2 1




 x1  x2  t n1  n2 2, s p    , x1  x2  t n1  n2 2, s p    
2
2

 n1 n2 
 n1 n2  
is a 1    100% confidence interval estimate of the population
difference
1  2 .
Note:
X 1  X 2  u1  u2  X 1  X 2  1   2 

 T n1  n2  2
s X1  X 2
1 1
s 2p   
 n1 n2 
6
Then, 1    100% confidence interval in small sample case can be
derived analogous to the one in large sample case.
Example:
Let
1 : the mean balance of checking account in Chekry Grove bank
 2 : the mean balance of checking account in Beechmont bank.
x1  1000, n1  12, s1  150
x 2  920, n2  10, s 2  120
Please find a 90% confidence interval for
1  2 .
[solution:]
s
2
p

n1  1s12  n2  1s22

n1  n2  2
11  150 2  9  120 2

 18855 .
12  10  2
Thus,
1 1
1
1
s X1 X 2  s 2p     18855    58.79 .
 12 10 
 n1 n2 
Then, a 90% confidence interval for
x1  x2   t n  n 2,
1
2
2
1  2
is
s X1  X 2  1000  920  t 20,0.05  58.79  80  1.725  58.79
 80  101.41
  21.41, 181.41
Online Exercise:
Exercise 12.1.1
Exercise 12.1.2
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