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Chapter
9
Estimating the
Value of a
Parameter Using
Confidence
Intervals
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Section
9.1
The Logic in
Constructing
Confidence Intervals
for a Population
Mean When the
Population Standard
Deviation Is Known
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Objectives
1. Compute a point estimate of the population mean
2. Construct and interpret a confidence interval for the
population mean assuming that the population
standard deviation is known
3. Understand the role of margin of error in
constructing the confidence interval
4. Determine the sample size necessary for estimating
the population mean within a specified margin of
error
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9-3
Objective 1
• Compute a Point Estimate of the Population
Mean
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A point estimate is the value of a statistic
that estimates the value of a parameter.
For example, the sample mean, x , is a
point estimate of the population mean .

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9-5
Parallel Example 1: Computing a Point Estimate
Pennies minted after 1982 are made from 97.5% zinc and 2.5%
copper. The following data represent the weights (in grams)
of 17 randomly selected pennies minted after 1982.
2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49
2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45
Treat the data as a simple random sample. Estimate the
population mean weight of pennies minted after 1982.
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9-6
Solution
The sample mean is
2.46  2.47 
x
17
 2.45
 2.464
The point estimate of  is 2.464 grams.
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9-7
Objective 2
• Construct and Interpret a Confidence Interval
for the Population Mean
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A confidence interval for an unknown
parameter consists of an interval of
numbers.
The level of confidence represents the
expected proportion of intervals that will
contain the parameter if a large number
of different samples is obtained. The
level of confidence is denoted
(1-)·100%.
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For example, a 95% level of confidence
(=0.05) implies that if 100 different
confidence intervals are constructed, each
based on a different sample from the same
population, we will expect 95 of the intervals
to contain the parameter and 5 to not include
the parameter.
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9-10
• Confidence interval estimates for the
population mean are of the form
Point estimate ± margin of error.
• The margin of error of a confidence
interval estimate of a parameter is a
measure of how accurate the point
estimate is.
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The margin of error depends on three factors:
1. Level of confidence: As the level of confidence
increases, the margin of error also increases.
2. Sample size: As the size of the random sample
increases, the margin of error decreases.
3. Standard deviation of the population: The more
spread there is in the population, the wider our
interval will be for a given level of confidence.
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9-12
The shape of the distribution of all possible
sample means will be normal, provided the
population is normal or approximately
normal, if the sample size is large (n≥30),
with
• mean
x  
• and standard deviation  x 


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
n
.
9-13
Because x is normally distributed, we
know 95% of all sample means lie
within 1.96 standard deviations of the
population mean, , and 2.5% of the
 sample means lie in each tail.

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95% of all sample means are in the interval
 1.96

n
 x    1.96

n
With a little algebraic manipulation, we can
rewrite this inequality and obtain:
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x 1.96 x    x 1.96 x .
It is common to write the 95% confidence
interval as
x 1.96 x
so that it is of the form
Point estimate ± margin of error.

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9-17
Parallel Example 2: Using Simulation to Demonstrate the
Idea of a Confidence Interval
We will use Minitab to simulate obtaining 30 simple
random samples of size n=8 from a population that is
normally distributed with =50 and =10. Construct
a 95% confidence interval for each sample. How
many of the samples result in intervals that contain
=50 ?
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Sample
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
Mean
47.07
49.33
50.62
47.91
44.31
51.50
52.47
59.62
43.49
55.45
50.08
56.37
49.05
47.34
50.33
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95.0% CI
40.14,
54.00)
42.40,
56.26)
43.69,
57.54)
40.98,
54.84)
37.38,
51.24)
44.57,
58.43)
45.54,
59.40)
52.69,
66.54)
36.56,
50.42)
48.52,
62.38)
43.15,
57.01)
49.44,
63.30)
42.12,
55.98)
40.41,
54.27)
43.40,
57.25)
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SAMPLE
C16
C17
C18
C19
C20
C21
C22
C23
C24
C25
C26
C27
C28
C29
C30
MEAN
44.81
51.05
43.91
46.50
49.79
48.75
51.27
47.80
56.60
47.70
51.58
47.37
61.42
46.89
51.92
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95%
37.88,
44.12,
36.98,
39.57,
42.86,
41.82,
44.34,
40.87,
49.67,
40.77,
44.65,
40.44,
54.49,
39.96,
44.99,
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CI
51.74)
57.98)
50.84)
53.43)
56.72)
55.68)
58.20)
54.73)
63.52)
54.63)
58.51)
54.30)
68.35)
53.82)
58.85)
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Note that 28 out of 30, or 93%, of the confidence
intervals contain the population mean =50.
In general, for a 95% confidence interval, any
sample mean that lies within 1.96 standard
errors of the population mean will result in a
confidence interval that contains .
Whether a confidence interval contains 
depends solely on the sample mean, x .
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Interpretation of a Confidence Interval
A (1-)·100% confidence interval indicates that, if we
obtained many simple random samples of size n from
the population whose mean, , is unknown, then
approximately (1-)·100% of the intervals will
contain .
For example, if we constructed a 99% confidence
interval with a lower bound of 52 and an upper bound
of 71, we would interpret the interval as follows:
“We are 99% confident that the population mean, ,
is between 52 and 71.”
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Constructing a (1- )·100% Confidence Interval
for ,  Known
Suppose that a simple random sample of size n is taken
from a population with unknown mean, , and known
standard deviation . A (1-)·100% confidence
interval for  is given by
Lower
Bound:
where z

2
x  z 2 

n
Upper x  z  
 2
Bound:
n
is the critical Z-value.

Note: The sample size must be large (n≥30) or the
population must be normally distributed.
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Parallel Example 3: Constructing a Confidence Interval
Construct a 99% confidence interval about the
population mean weight (in grams) of pennies minted
after 1982. Assume =0.02 grams.
2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49
2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45
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Weight (in grams) of Pennies
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• z 2  2.575
• Lower bound:
0.02



x  z 2 
= 2.464-2.575 

17 
n
= 2.464-0.012 = 2.452
• Upper bound:


0.02



x  z 2 
= 2.464+2.575 

17 
n
= 2.464+0.012 = 2.476
We are 99% confident that the mean weight of pennies

 minted after 1982 is between
2.452 and 2.476 grams.
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Objective 3
• Understand the Role of the Margin of Error in
Constructing a Confidence Interval
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The margin of error, E, in a (1-)·100%
confidence interval in which  is known
is given by
E  z 2 

n
where n is the sample size.
Note: 
We require that the population from
which the sample was drawn be normally
distributed or the samples size n be greater
than or equal to 30.
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Parallel Example 5: Role of the Level of Confidence in the
Margin of Error
Construct a 90% confidence interval for the mean
weight of pennies minted after 1982. Comment on
the effect that decreasing the level of confidence has
on the margin of error.
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• z 2  1.645
• Lower bound:
0.02



x  z 2 
= 2.464-1.645 

17 
n
= 2.464-0.008 = 2.456
• Upper bound:


0.02



x  z 2 
= 2.464+1.645 

17 
n
= 2.464+0.008 = 2.472
We are 90% confident that the mean weight of pennies

 minted after 1982 is between
2.456 and 2.472 grams.
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Notice that the margin of error decreased from
0.012 to 0.008 when the level of confidence
decreased from 99% to 90%. The interval is
therefore wider for the higher level of
confidence.
Confidence
Level
Margin of
Error
Confidence
Interval
90%
0.008
(2.456, 2.472)
99%
0.012
(2.452, 2.476)
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Parallel Example 6: Role of Sample Size in the Margin
of Error
Suppose that we obtained a simple random sample of
pennies minted after 1982. Construct a 99%
confidence interval with n=35. Assume the larger
sample size results in the same sample mean, 2.464.
The standard deviation is still =0.02. Comment on
the effect increasing sample size has on the width of
the interval.
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• z 2  2.575
• Lower bound:
0.02



x  z 2 
= 2.464-2.575 

35 
n
= 2.464-0.009 = 2.455
• Upper bound:


0.02



x  z 2 
= 2.464+2.575 

35 
n
= 2.464+0.009 = 2.473
We are 99% confident that the mean weight of pennies

 minted after 1982 is between
2.455 and 2.473 grams.
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Notice that the margin of error decreased from
0.012 to 0.009 when the sample size increased
from 17 to 35. The interval is therefore
narrower for the larger sample size.
Sample
Size
17
Margin of
Error
Confidence
Interval
0.012
(2.452, 2.476)
35
0.009
(2.455, 2.473)
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Objective 4
• Determine the Sample Size Necessary for
Estimating the Population Mean within a
Specified Margin of Error
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Determining the Sample Size n
The sample size required to estimate the
population mean, , with a level of confidence
(1-)·100% with a specified margin of error,
E, is given by
2
z    
 2 
n  

E




where n is rounded up to the nearest whole
number.

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Parallel Example 7:
Determining the Sample Size
Back to the pennies. How large a sample would be
required to estimate the mean weight of a penny
manufactured after 1982 within 0.005 grams with
99% confidence? Assume =0.02.
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•
z 2  z0.005  2.575
• =0.02
• E=0.005
2
z    
2
•
 2  2.575(0.02) 
n  
 
  106.09

 E   0.005 


Rounding up, we find n=107.
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9-39