Download 8.2.1 Confidence Interval Estimates for Population

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8.1 Introduction
The field of statistical inference consist of those
methods used to make decisions or to draw
conclusions about a population. These methods
utilize the information contained in a sample from the
population in drawing conclusions.
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Estimator
• In statistics, the method used
Estimate
• The value that obtained from a sample
I have a sample of 5 numbers and I take the average.
The estimator is taking the average of the sample.
The estimator of the mean.
Let say, the average = 4
the estimate.

Point estimate – an estimate of the parameter
using a single number.
◦ E.g :
x is the point estimate for 
 In choosing the point estimators, we have to
depends on the properties of the estimators.
 Unbiased
 Consistent
 Efficient
 Sufficient
2

x
 x  x2  ...  xn 
E(x )  E  1

n


1
  E  x1   E  x2   ...  E  xn  
n
1
  1   2  ...   n 
n
1 n

   i 
n  i 1


1
 n    
n


Example 8.2
If a random sample size n is taken from a population with mean,  and
variance  2, hence x (mean sample) is a consistent estimator for  .
 x  x  ...  xn 
V x V  1 2

n


1
 2 V  x1  x2  ...  xn  
n
1
 2 V  x1   V  x2   ...  V  xn  
n
1
 2 2  2  ...   2 
n
1
2
2
 2  n  
n
n


 lim var ˆ  0
n 
2
lim
0
n  n
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.
Example 8.3
Sample mean is an efficient estimator compares to sample median in
estimating the population mean.
Proof:
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
Variance for sample mean, var x 

n
 2 if n is odd
Variance for sample median, var x  
 2
if n is even

2

 
 
This gives var x  var x
if
n2
Thus, significantly, mean is more efficient in estimating 
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Sufficient
•If it used all the sample’s information
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X
2
S
2
n
1

 X I  X 
n  1 i 1
P̂
2
Definition 8.1: An Interval Estimate
In interval estimation, an interval is constructed around
the point estimate and it is stated that this interval is
likely to contain the corresponding population
parameter.
Definition 8.2: Confidence Level and Confidence Interval
Each interval is constructed with regard to a given confidence
level and is called a confidence interval. The confidence level
associated with a confidence interval states how much confidence
we have that this interval contains the true population parameter. The
confidence level is denoted by 1   100% .
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The (1   )100% Confidence Interval of Population Mean, 
(i) x  z

2
n
if  is known and normally distributed population

 

or  x  z
   x  z

2
2
n
n


s
(ii) x  z
if  is unknown, n large (n  30)
2
n
s
s 

or  x  z
   x  z

2
2
n
n


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s
(iii) x  tn 1,
if  is unknown, normally distributed population
2
n
and small sample size  n  30 
s
s 

or  x  tn 1,
   x  tn 1,

2
2
n
n

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If a random sample of size n  20 from a normal population
with the variance  2  225 has the mean x  64.3, construct
a 95% confidence interval for the population mean, .
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It is known that, n  20,   x  64.3 and   15
For 95% CI,
95%  100(1 –  )%
1 –  0.95
  0.05

2
 0.025
z  z0.025  1.96
2
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  
Hence, 95% CI  x  z 

n

2 
 15 
 64.3  1.96 

 20 
 64.3  6.57
 [57.73, 70.87]
@
57.73    70.87
Thus, we are 95% confident that the mean of random variable
is between 57.73 and 70.87
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Example 8.5 :
A publishing company has just published a new textbook. Before the
company decides the price at which to sell this textbook, it wants to know the
average price of all such textbooks in the market. The research department at
the company took a sample of 36 comparable textbooks and collected the
information on their prices. This information produced a mean price RM 70.50
for this sample. It is known that the standard deviation of the prices of
all such textbooks is RM4.50. Construct a 90% confidence interval for the mean
price of all such college textbooks.
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solution
It is known that, n  36,   x  RM70.50 and   RM 4.50
For 90% CI,
90%  100(1 –  )%
1 –  0.90
  0.1

2
 0.05
z  z0.05  1.65
2
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  
Hence, 90% CI  x  z 

n

2 
 4.50 
 70.50  1.65 

36


 70.50  1.24
 [ RM 69.26, RM 71.74]
Thus, we are 90% confident that the mean price of all such
college textbooks is between RM69.26 and RM71.74
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The (1   )100% Confidence Interval for p for Large Samples ( n  30)
pˆ  z
2
pˆ 1  pˆ 
n
or
pˆ  z
pˆ 1  pˆ 
2
n
 p  pˆ  z
pˆ 1  pˆ 
2
n
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Example 8.6
According to the analysis of Women Magazine in
June 2005, “Stress has become a common part of
everyday life among working women in Malaysia. The
demands of work, family and home place an
increasing burden on average Malaysian women”.
According to this poll, 40% of working women
included in the survey indicated that they had a little
amount of time to relax. The poll was based on a
randomly selected of 1502 working women aged 30
and above. Construct a 95% confidence interval for
the corresponding population proportion.
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Solution
Let p be the proportion of all working women age 30 and above,
who have a limited amount of time to relax, and let pˆ be the
corresponding sample proportion. From the given information,
n  1502 , pˆ  0.40, qˆ  1  pˆ  1 – 0.40  0.60

ˆ
Hence, 95% CI  p  z 
2 
ˆˆ
pq
n



 0.40(0.60)
 0.40  1.96 
1502

 0.40  0.02478



 [0.375, 0.425] or 37.5% to 42.5%
Thus, we can state with 95% confidence that the proportion of all
working women aged 30 and above who have a limited amount of
time to relax is between 37.5% and 42.5%.
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Definition 8.3:
If x is used as an estimate of  , we can be 100(1- )% confident
that the error | x   | will not exceed a specified amount E when the
sample size is
 z /2 
n

E


2
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Example 8.7:
A team of efficiency experts intends to use the mean of a random
sample of size n=150 to estimate the average mechanical
aptitude of assembly-line workers in a large industry (as
measured by a certain standardized test). If, based on
experience, the efficiency experts can assume that   6.2
for such data, what can they assert with probability 0.99
about the maximum error of their estimate?
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Solutions
Substituting n  150,   6.2, and z0.005  2.575 into the expression
for the maximum error, we get
E

z / 2
n
2.575(6.2)
 1.30
150
Thus, the efficiency experts can assert with probability 0.99 that their
error will be less than 1.30.
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x
If pˆ  is used as an estimate of p, we can assert with
n
(1-  )100% confidence that the error is less than z /2
If we set E  z /2
pˆ (1  pˆ )
.
n
pˆ (1  pˆ )
and solve for n, the appropriate
n
sample size is
2
 z /2 
n
 pˆ (1  pˆ )
 E 
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Example 8.8:
A study is made to determine the proportion of voters in a
sizable community who favor the construction of a nuclear
power plant. If 140 of 400 voters selected at random favor the
140
project and we use pˆ  400  0.35 as an estimate of the actual
proportion of all voters in the community who favor the project,
what can we say with 99% confidence about the maximum
error?
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Solution
Substituting n  400, pˆ  0.35, and z0.005  2.575 into the formula,
we get
E  z /2
pˆ (1  pˆ )
n
(0.35)(0.65)
 2.575
 0.061
400
Thus, if we use pˆ  0.35 as an estimate of the actual proportion of
voters in the community who favor the project, we can assert with
99% confidence that the error is less than 0.061.
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Example 8.9:
How large a sample required if we want to be 95% confident
that the error in using p̂ to estimate p is less than 0.05? If
pˆ  0.12, find the required sample size.
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Solution
2
 z0.025 
n
 pˆ (1  pˆ )
 E 
2
 1.96 

 0.12(0.88)  162
 0.05 
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