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Chapter 9: One- and Two-Sample Estimation
Problems:
9.1 Introduction:
·
Suppose we have a population with some unknown
parameter(s).
Example: Normal(,)
 and  are parameters.
·
We need to draw conclusions (make inferences) about the
unknown parameters.
·
We select samples, compute some statistics, and make
inferences about the unknown parameters based on the
sampling distributions of the statistics.
* Statistical Inference
(1) Estimation of the parameters (Chapter 9)
 Point Estimation
 Interval Estimation (Confidence Interval)
(2) Tests of hypotheses about the parameters (Chapter 10)
9.3 Classical Methods of Estimation:
Point Estimation:
A point estimate of some population parameter  is a single
value ˆ of a statistic ̂ . For example, the value X of the statistic
x computed from a sample of size n is a point estimate of the
population mean .
Interval Estimation (Confidence Interval = C.I.):
An interval estimate of some population parameter  is an
interval of the form (ˆL ,ˆU), i.e, ˆL << ˆU . This interval contains
the true value of  "with probability 1", that is P(ˆL<<ˆU)=1
ˆ
is called a (1)100% confidence interval
ˆ(L ˆU, )ˆL= <<
U
(C.I.)
for .
 ˆ1
is called the confidence coefficient
L
 ˆU = lower confidence limit

= upper confidence limit
  =0.1, 0.05, 0.025, 0.01 (0<<1)
9.4 Single Sample: Estimation of the Mean ():
Recall:
E( X )   X  
Var( X )   X2
  
X ~ N ,

n



2
n
X 
~ N(0,1) (2 is known)
/ n
X 
T
~ t(n  1) (2 is unknown)
S/ n
Z
We use the sampling distribution of X to
make inferences about .
Notation:
Za is the Z-value leaving an area of a to
the right; i.e., P(Z>Za)=a or equivalently,
P(Z<Za)=1a
Point Estimation of the Mean ():
n
· The sample mean X   X i / n
is a "good" point estimate for .
i 1
Interval Estimation (Confidence Interval) of the Mean ():
(i) First Case: 2 is known:
Result:
n
If
X   Xi / n
i 1
is the sample mean of a random sample of size n
from a population (distribution) with mean  and known variance
2, then a (1)100% confidence interval for  is :
( X  Z
2
 X  Z
2

n
, X  Z
2


n
)
n
 X  Z     X  Z 
n
n
2
2
where Z  is the Z-value leaving an area
2
of /2 to the right; i.e., P(Z> Z )=/2, or
2
equivalently, P(Z< Z  )=1/2.
2
Note:


We are (1)100% confident that   ( X  Z 
, X  Z
)
2
n
2
n
Example 9.2:
The average zinc concentration recorded from a sample of zinc
measurements in 36 different locations is found to be 2.6
gram/milliliter. Find a 95% and 99% confidence interval (C.I.) for
the mean zinc concentration in the river. Assume that the
population standard deviation is 0.3.
Solution:
= the mean zinc concentration in the river.
Population
Sample
=??
n=36
=0.3
=2.6
First, a point estimate for  is X =2.6.
(a) We want to find 95% C.I. for .
 = ??
95% = (1)100%
 0. 95 = (1)
 =0.05
 /2 = 0.025
Z = Z
0.025
= 1.96
A 95% C.I. for  is

2
X  Z
2

X  Z
2

n

n
   X  Z
2

n
 0.3 
 0.3 
2.6  (1.96)
    2.6  (1.96)

 36 
 36 
 2.6  0.098 <  < 2.6 + 0.098
 2.502 <  < 2.698
  ( 2.502 , 2.698)
We are 95% confident that  ( 2.502 , 2.698).
(b) Similarly, we can find that (Homework) A 99% C.I. for  is
2.471 <  < 2.729
  ( 2.471 , 2.729)
We are 99% confident that  ( 2.471 , 2.729)
Notice that a 99% C.I. is wider that a 95% C.I..
Note:
Error
|-----------|
______|____________________________|____
|----------------------|--------------------------|
Theorem 9.1:
If X is used as an estimate of
, we can then be (1)100% confident that the error (in
estimation) will not exceed Z  
2
n
Example:
In Example 9.2, we are 95% confident that the sample mean
X  2.6 differs from the true mean  by an amount less than
Z
2

 0.3 
 (1.96)
  0.098
n
 36 
Note:

Let e be the maximum amount of the error, that is e  Z 
,
n
2
2
then:





e  Z
 n  Z
 n   Z 
2
n
2
e


2
e 
Theorem 9.2:
If X is used as an estimate of , we can then be (1)100%
confident that the error (in estimation) will not exceed a specified



amount e when the sample size is n   Z  


e
 2 
2
Note:
1. All fractional values ofn  (Z   / e) 2 are rounded up to the
2
next whole number.
2. If  is unknown, we could take a preliminary sample of
size n30 to provide an estimate of . Then using
S
9.2
n
2
 ( X i  X ) /( n  1)
i 1
as an approximation for  in Theorem
we could determine approximately how many observations
are needed
Example
9.3: to provide the desired degree of accuracy.
How large a sample is required in Example 9.2 if we want to be
95% confident that our estimate of  is off by less than 0.05?
Solution:
We have = 0.3 , Z   1.96 , e=0.05. Then by Theorem 9.2,
2
2
2

 

0
.
3

n   Z    1.96 
  138.3  139


0.05 
 2 e 
Therefore, we can be 95% confident that a random sample of
size n=139 will provide an estimate X differing from  by an
amount less than e=0.05.
Interval Estimation (Confidence Interval) of the Mean ():
(ii) Second Case: 2 is unknown:
Recall:
X 
T

~ t(n  1)

S/ n
Result:
n
n
If X   X i / n and S   ( X i  X ) 2 /( n  1) are the sample mean
i 1
i 1
and the sample standard deviation of a random sample of size
n from a normal population (distribution) with unknown variance
2, then a (1)100% confidence interval for  is :
( X  t
2
X  t
2
X  t
2
S
S
, X  t
)
n
n
2
S
n
S
S
   X  t
n
n
2
where t  is the t-value with =n1 degrees of freedom leaving
2
an area of /2 to the right; i.e., P(T> t  )=/2, or equivalently,
P(T< t  )=1/2.
2
2
Example 9.4:
The contents of 7 similar containers of sulfuric acid are 9.8, 10.2,
10.4, 9.8, 10.0, 10.2, and 9.6 liters. Find a 95% C.I. for the mean
of all such containers, assuming an approximate normal
distribution.
Solution:
n
.n=7
X   X i / n  10.0
i 1
First, a point estimate for  is
S
n
2
 ( X i  X ) /( n  1)  0.283
i 1
n
X   X i / n  10.0
i 1
Now, we need to find a confidence interval for .
 = ??
95%=(1)100%  0. 95=(1)  =0.05  /2=0.025
t  = t0.025 =2.447 (with =n1=6 degrees of freedom)
2
A 95% C.I. for  is
X  t
2
 X  t
2
S
n
0.025

6
S
S
   X  t
n
n
2

 0.283 
 0.283 
    10.0  (2.447 )

 10.0  (2.447 )
 7 
 7 
 10.0  0.262<  < 10.0 + 0.262
 9.74 <  < 10.26
  ( 9.74 , 10.26)
We are 95% confident that  ( 9.74 , 10.26).
t0.025=2.447
9.5 Standard Error of a Point Estimate:
·
The standard
error of an estimator is its standard
n
deviation. X   X i / n
· We use i 1
as a point estimator of , and we used
X
the sampling distribution of
to make a (1)100% C.I. for
 X  / n
.
X
(X)   / n
·
The standard deviation of
, which is
,is called
the standard error of . We
 write s.e.
X  Z  C.I.
 Xfor ,
Z when
s.e( X )2 is known, is
· Note: a (1)100%
n
2
2
· Note: a (1)100% C.I. for , when 2 is unknown and the
S
X  t  is  X  t  ŝ.e(X)
distribution is normal,
n
2
2
(=n1 df)
9.7 Two Samples: Estimating the Difference between Two
Means (12):
Recall: For two independent samples:

 X1 X 2  1  2  E( X1  X 2 )




2
X1  X 2
X

1X2
Z
 12
n1

 22
n2
 Var( X 1  X 2 )
  X2 1  X 2 
 12
n1
( X 1  X 2 )  ( 1   2 )
 12
n1

 22
n2

 22
n2
~N(0,1)
Point Estimation of 12:
 X1  X 2 is a "good" point estimate for 12.
Confidence Interval of 12:
(i) First Case: 12 and  22 are known:
 Z

( X1  X 2 )  (1   2 )
~ N(0,1)
 

n1 n 2
2
1
2
2
Result:
a (1)100% confidence interval for 12 is :
( X1  X 2 )  Z 
2
 12
n1
or ( X 1  X 2 )  Z 
2

 22
n2
 12
n1

 1   2  ( X 1  X 2 )  Z 
2
 12
n1

 22
n2
 22
n2
2
2
2
2 





or  ( X 1  X 2 )  Z  1  2 , ( X 1  X 2 )  Z  1  2 

n1 n2
n1 n2 
2
2


2
2


(ii) Second Case: 1 = 2 =2 is unknown:
 If 1 and 2 are unknown but 1 =  2 =2, then the pooled
estimate of 2 is
2
2
2
2
2
2
(n1  1) S1  (n2  1) S 2

n1  n2  2
S 2p
 where S12 is the variance of the 1-st sample and S 22 is the
2
variance of the 2-nd sample. The degrees of freedom of S p is
=n1+n22.

Result:
a (1)100% confidence interval for 12 is :
( X 1  X 2 )  t
2
S 2p
n1
or ( X 1  X 2 )  t S p
2

S 2p
n2
 1   2  ( X 1  X 2 )  t 
2
S 2p
n1

S 2p
n2
1
1
1
1

 1   2  ( X 1  X 2 )  t S p

n1 n2
n1 n2
2
or
( X 1  X 2 )  t S p
2
or
1
1

n1 n2


1
1
1
1
(X  X )  t S

 , ( X 1  X 2 )  t S p

1
2

p


n
n
n
n
1
2
1
2


2
2
where t  is the t-value with =n1+n22 degrees of freedom.
2
2
2
Example 9.6: (1st Case:  1 and  2 are known)
An experiment was conducted in which two types of engines, A
and B, were compared. Gas mileage in miles per gallon was
measured. 50 experiments were conducted using engine type A
and 75 experiments were done for engine type B. The gasoline
used and other conditions were held constant. The average gas
mileage for engine A was 36 miles per gallon and the average
for engine B was 42 miles per gallon. Find 96% confidence
interval for BA, where A and B are population mean gas
mileage for engines A and B, respectively. Assume that the
population standard deviations are 6 and 8 for engines A and B,
respectively.
Solution:
Engine A
Engine B
nA=50
nB=75
X A =36
X B=42
A=6
B=8
A point estimate for BA is X B  X A =4236=6.
 = ??
96% = (1)100%  0. 96 = (1)  =0.04  /2 = 0.02
Z  = Z0.02 = 2.05
2
A 96% C.I. for BA is
( X B  X A )  Z
2
( X B  X A )  Z
2
 B2
nB
 B2
nB

 A2
nA

 B   A  ( X B  X A )  Z
 A2
nA
2
 B2
nB

 A2
nA
(42  36)  Z 0.02
82
75

62
50
64 36
6  (2.05)

75 50
6  2.571
3.43 < BA < 8.57
We are 96% confident that BA (3.43, 8.57).
2
2


Example 9.7:
Case: 1 = 2 unknown) Reading Assignment
2
2
Example: (2nd Case:  1 =  2 unknown)
(2nd
To compare the resistance of wire A with that of wire B, an
experiment shows the following results based on two
independent samples (original data multiplied by 1000):
Wire A: 140, 138, 143, 142, 144, 137
Wire B: 135, 140, 136, 142, 138, 140
Assuming equal variances, find 95% confidence interval for
AB, where A (B) is the mean resistance of wire A (B).
Solution:
Wire A
nA=6
XA  140.67
Wire B
nB=6
XB  138.50
S2A=7.86690
S2B=7.10009
____________________________________
.
A point estimate for AB is X A  X B =140.67138.50=2.17.
95% = (1)100%  0. 95 = (1)  =0.05  /2 = 0.025
= df = nA+nB  2= 10
t  = t0.025 = 2.228
2
2
2
(
n

1
)
S

(
n

1
)
S
A
B
B
S 2p  A
n A  nB  2
(6  1)(7.86690)  (6  1)(7.10009)

 7.4835
662
S p  S 2p  7.4835  2.7356
A 95% C.I. for AB is
( X A  X B )  t S p
2
1
1
1
1

  A   B  ( X A  X B )  t S p

n A nB
n A nB
2
or ( X A  X B )  t S p
2
1
1

n A nB
1 1
(140.67  138.50)  (2.228) (2.7356 )

6 6
2.17  3.51890
1.35< AB < 5.69
We are 95% confident that AB (1.35, 5.69)