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Chapter
9
Estimating the
Value of a
Parameter
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Section
9.1
Estimating a
Population
Proportion
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A point estimate is the value of a statistic
that estimates the value of a parameter.
For example, the point estimate for the
x
population proportion is p̂ = , where x is
n
the number of individuals in the sample
with a specified characteristic and n is the
sample size.
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Parallel Example 1: Calculating a Point Estimate for the
Population Proportion
In July of 2008, a Quinnipiac University Poll asked
1783 registered voters nationwide whether they
favored or opposed the death penalty for persons
convicted of murder. 1123 were in favor.
Obtain a point estimate for the proportion of registered
voters nationwide who are in favor of the death
penalty for persons convicted of murder.
9-4
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Solution
Obtain a point estimate for the proportion of registered
voters nationwide who are in favor of the death
penalty for persons convicted of murder.
1123
p̂ =
= 0.63
1783
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Parallel Example 1: Calculating a Point Estimate for the
Population Proportion
The Gallup Organization conducted a poll in which a
simple random sample of 1016 Americans 18 and
older were asked, “Do you consider the amount of
federal income tax you have to pay is fair?” Of the
1016 adult Americans surveyed, 558 said yes. Obtain
a point estimate for the proportion of Americans 18
and older who believe the amount of federal income
tax they pay if fair.
9-6
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A confidence interval for an unknown
parameter consists of an interval of
numbers based on a point estimate.
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 – α)·100%.
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For example, a 95% level of confidence
(α = 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 not to include
the parameter.
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Confidence interval estimates for the population
proportion 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:
Level of confidence: As the level of confidence
increases, the margin of error also increases.
Sample size: As the size of the random sample
increases, the margin of error decreases.
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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Sampling Distribution of p̂
For a simple random sample of size n, the
sampling distribution of p̂ is approximately
normal with mean m p̂ = p and standard deviation
s p̂ =
p(1- p) , provided that np(1 – p) ≥ 10.
n
NOTE: We also require that each trial be independent;
when sampling from finite populations (n ≤ 0.05N).
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Interpretation of a Confidence Interval
A (1 – α)·100% confidence interval
indicates that (1 – α)·100% of all simple
random samples of size n from the
population whose parameter is unknown
will result in an interval that contains the
parameter. (This does NOT mean we are
(1 – α)·100% percent sure of our estimate)
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Constructing a (1 – α)·100% Confidence
Interval for a Population Proportion
Suppose that a simple random sample of size n is taken
from a population. A (1 – α)·100% confidence interval
for p is given by the following quantities
Lower bound:
p̂(1- p̂)
p̂ - za 2 ×
n
p̂(1- p̂)
p̂ + za 2 ×
n
Note: It must be the case that np̂(1- p̂) ³ 10 and
Upper bound:
n ≤ 0.05N to construct this interval.
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Margin of Error
The margin of error, E, in a (1 – α) 100%
confidence interval for a population proportion is
given by
p̂(1- p̂)
E = za ×
n
2
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Parallel Example 2: Constructing a Confidence Interval for a
Population Proportion
In July of 2008, a Quinnipiac University Poll asked
1783 registered voters nationwide whether they
favored or opposed the death penalty for persons
convicted of murder. 1123 were in favor.
Obtain a 90% confidence interval for the proportion of
registered voters nationwide who are in favor of the
death penalty for persons convicted of murder.
9-15
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Solution
•
•
p̂ = 0.63
np̂(1- p̂) = 1783(0.63)(1- 0.63) = 415.6 ³ 10
and the sample size is definitely less than 5% of
the population size
•
α = 0.10 so zα/2 = z0.05 = 1.645
•
0.63(1 0.63)
Lower bound: 0.63  1.645
 0.61
1783
Upper bound: 0.63  1.645 0.63(1 0.63)  0.65
1783
•
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Solution
We are 90% confident that the proportion of
registered voters who are in favor of the death
penalty for those convicted of murder is between
0.61and 0.65.
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Example
In a survey conducted by Princeton Survey Research
Associates International, 800 randomly sampled 16- to
17-years-old living in the United States were asked
whether they have ever used their cell phone to text
while driving. Of the 800 teenagers, 272 indicated that
they text while driving. Obtain a 95% confidence
interval for the proportion of 16- to 17-year-olds who
text while driving.
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Example
For the previous problem, determine the effect on the
margin of error by increasing the level of confidence
from 95% to 99%.
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Example
A Retirement Confidence Survey of 1153 workers and
retirees in the United States 25 years or age and older
conducted by Employee Benefit Research Institute in
January 2010 found that 496 had less than $10,000 in
savings.
a) Obtain a point estimate for the population proportion
of workers and retirees in the United States 25 years
and older who have less than $10,000 in savings.
b) Construct a 95% confidence interval for the population
proportion of workers and retirees in the United States
25 years and older who have less than $10,000 in
savings. Interpret the interval.
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Sample size needed for a specified margin of
error, E, and level of confidence (1 – α):
2
æ za 2 ö
n = p̂(1- p̂) ç
÷
è E ø
Problem: The formula uses p̂ which depends
on n, the quantity we are trying to determine!
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Two possible solutions:
1. Use an estimate of p̂ based on a pilot study or an
earlier study.
2. Let p̂ = 0.5 which gives the largest possible value of
n for a given level of
confidence and a given
margin of error.
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Sample Size Needed for Estimating the
Population Proportion p
The sample size required to obtain a (1 – α)·100%
confidence interval for p with a margin of error E is
given by
2
æ za 2 ö
n = p̂(1- p̂) ç
÷
è E ø
(rounded up to the next integer), where p̂ is a prior
estimate of p.
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Sample Size Needed for Estimating the
Population Proportion p
If a prior estimate of p is unavailable, the sample size
required is
2
z 2 
n  0.25 
 E 
rounded up to the next integer.
The margin of error should always be expressed as a
decimal when using Formulas (4) and (5).

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Parallel Example 4: Determining Sample Size
A sociologist wanted to determine the percentage of
residents of America that only speak English at
home. What size sample should be obtained if she
wishes her estimate to be within 3 percentage points
with 90% confidence assuming she uses the estimate
obtained from the Census 2000 Supplementary
Survey of 82.4%?
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Solution
•
E = 0.03
•
z 2  z0.05  1.645
•
p̂ = 0.824
•
1.645 2
n  0.824(1 0.824)
  436.04
 0.03 
We round this value up to 437. The sociologist must
survey 437 randomly selected American residents.
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Example
An economist wants to know if the proportion of the
U.S. population who commutes to work via
carpooling is on the rise. What size sample should
be obtained if the economist wants an estimate
within 2 percentage points of the true proportion with
90% if:
a) the economist uses the 2009 estimate of 10%
obtained from the American Community Survey?
b) the economist does not use any prior estimate?
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Example
An urban economist wishes to estimate the
proportion of Americans who own their homes.
What size sample should be obtained if he wishes the
estimate to be within 0.02 with 90% confidence if:
a) he uses a 2010 estimate of 0.669 obtained from
the U.S. Census Bureau?
b) he does not use any prior estimate?
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Section
9.2
Estimating a
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 μ.
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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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Solution
The sample mean is
2.46  2.47 
x
17
 2.45
 2.464
The point estimate of μ is 2.464 grams.
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Student’s t-Distribution
Suppose that a simple random sample of size n is
taken from a population. If the population from
which the sample is drawn follows a normal
distribution, the distribution of
x
t
s
n
follows Student’s t-distribution with n – 1
degrees of freedom where x is the sample mean
and s is the sample standard deviation.
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Parallel Example 2: Comparing the Standard Normal
Distribution to the t-Distribution Using Simulation
a)
Obtain 1,000 simple random samples of size n = 5 from a
normal population with μ = 50 and σ = 10.
b)
Determine the sample mean and sample standard deviation
for each of the samples.
Compute z 
c)
d)
x 

n
and
x 
for each sample.
t
s
n
Draw a histogram for both z and t.

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
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Histogram for z
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Histogram for t
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CONCLUSION:
• The histogram for z is symmetric and bell-shaped
with the center of the distribution at 0 and virtually all
the rectangles between –3 and 3. In other words, z
follows a standard normal distribution.
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CONCLUSION (continued):
• The histogram for t is also symmetric and bell-shaped
with the center of the distribution at 0, but the
distribution of t has longer tails (i.e., t is more
dispersed), so it is unlikely that t follows a standard
normal distribution. The additional spread in the
distribution of t can be attributed to the fact that we
use s to find t instead of σ. Because the sample
standard deviation is itself a random variable (rather
than a constant such as σ), we have more dispersion
in the distribution of t.
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Properties of the t-Distribution
1. The t-distribution is different for different degrees of
freedom.
2. The t-distribution is centered at 0 and is symmetric
about 0.
3. The area under the curve is 1. The area under the curve
to the right of 0 equals the area under the curve to the
left of 0, which equals 1/2.
4. As t increases or decreases without bound, the graph
approaches, but never equals, zero.
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Properties of the t-Distribution
5. The area in the tails of the t-distribution is a little
greater than the area in the tails of the standard normal
distribution, because we are using s as an estimate of σ,
thereby introducing further variability into the tstatistic.
6. As the sample size n increases, the density curve of t
gets closer to the standard normal density curve. This
result occurs because, as the sample size n increases,
the values of s get closer to the values of σ, by the Law
of Large Numbers.
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Parallel Example 3: Finding t-values
Find the t-value such that the area under the t-distribution
to the right of the t-value is 0.2 assuming 10 degrees of
freedom. That is, find t0.20 with 10 degrees of freedom.
9-43
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Solution
The figure to the left
shows the graph of the
t-distribution with 10
degrees of freedom.
The unknown value of t is labeled, and the area under
the curve to the right of t is shaded. The value of t0.20
with 10 degrees of freedom is 0.879.
9-44
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Example
Find the t-value such that the are under the t-distribution
to the right of the t-value is 0.10, assuming 15 degrees of
freedom (df). That is, find 𝑡0.10 with 15 degrees of
freedom.
9-45
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Example
Find the t-value such that the are under the t-distribution
to the left of the t-value is 0.05, assuming 6 degrees of
freedom (df).
t= -1.943
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Constructing a (1 – α)100% Confidence
Interval for μ
Provided
• sample data come from a simple random sample or
randomized experiment,
• sample size is small relative to the population size
(n ≤ 0.05N), and
• the data come from a population that is normally
distributed, or the sample size is large.
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Constructing a (1 – α)100% Confidence
Interval for μ
A (1 – α)·100% confidence interval for μ is given by
s
x  t 
n
2
Lower
bound:
Upper
bound:
s
x  t 
n
2
where t  is the critical value with n – 1 degrees of freedom.
2
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Example
Find the critical t-value that corresponds to 95%
confidence. Assume 16 degrees of freedom.
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Parallel Example: 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 μ = 50 and σ = 10. Construct
a 95% confidence interval for each sample. How many
of the samples result in intervals that contain μ = 50 ?
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Sample
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
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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)
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
SAMPLE
C16
C17
C18
C19
C20
C21
C22
C23
C24
C25
C26
C27
C28
C29
C30
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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,
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
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)
Note that 28 out of 30, or 93%, of the confidence
intervals contain the population mean μ = 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 μ.
Whether a confidence interval contains μ
depends solely on the sample mean, x .
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Parallel Example 4: Constructing a Confidence Interval about
a Population Mean
Construct a 99% confidence interval about the
population mean weight (in grams) of 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
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Weight (in grams) of Pennies
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•
• Lower bound:
s = 2.464 – 2.921
x  t 2 
n
0.02




17 
= 2.464 – 0.014 = 2.450
• Upper bound:

s = 2.464 + 2.921
x  t 2 
n
0.02




17 
= 2.464 + 0.014 = 2.478
We are 99% confident that the mean weight of pennies

minted after 1982 is between
2.450 and 2.478 grams.
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Example
Construct a 95% confidence interval for the mean miles
per gallon of a 2011 Ford Focus. Interpret the interval.
35.7 37.2 34.1 38.9 32.0 41.3 32.5 37.1 37.3
38.8 38.2 39.6 32.2 40.9 37.0 36.0
9-58
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Determining the Sample Size n
The sample size required to estimate the
population mean, µ, with a level of confidence
(1– α)·100% with a specified margin of error, E,
is given by
æ za × s ö
ç
÷
2
n=
çç E ÷÷
è
ø
2
where n is rounded up to the nearest whole
number.
9-59
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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 s = 0.02.
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• z 2  z0.005  2.575
• s = 0.02
• E = 0.005
æ za × s ö
2
æ 2.575(0.02) ö
ç
÷
2
n=
=ç
= 106.09
÷
çç E ÷÷ è 0.005 ø
è
ø
2
Rounding up, we find n = 107.
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Example
How large a sample size is required to estimate
the mean miles per gallon within 0.5 mile per
gallon with 95% confidence?
(remember s = 2.92)
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Example
Dr. Paul wants to estimate the mean serum HDL
cholesterol of all 20- to 29-year-old males. How
many subjects are needed to estimate the mean
serum HDL cholesterol of all 20- to 29-year-old
males within 1.5 points with 90% confidence,
assuming that s=12.5 based on earlier studies?
Suppose that Dr. Paul would prefer a 95%
confidence. How does the increase in
confidence affect the sample size required?
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Section
9.3
Estimating a
Population
Standard
Deviation
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Chi-Square Distribution
If a simple random sample of size n is obtained
from a normally distributed population with
mean μ and standard deviation σ, then
 
2
(n  1)s

2
2
has a chi-square distribution with n – 1 degrees
of freedom.
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Characteristics of the
Chi-Square Distribution
1. It is not symmetric.
2. The shape of the chi-square distribution
depends on the degrees of freedom, just like
the Student’s t-distribution.
3. As the number of degrees of freedom
increases, the chi-square distribution becomes
more nearly symmetric.
4. The values of χ2 are nonnegative (greater than
or equal to 0).
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Parallel Example 1: Finding Critical Values for the
Chi-Square Distribution
Find the chi-square values that separate the middle
95% of the distribution from the 2.5% in each tail.
Assume 18 degrees of freedom.
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Solution
Find the chi-square values that separate the middle
95% of the distribution from the 2.5% in each tail.
Assume 18 degrees of freedom.
χ20.975 = 8.231
χ20.025 = 31.526
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Example
Find the critical values χ2/2 and χ21- /2 for given
confidence interval and sample size:
a) 95% confidence, n = 25
b) 99% confidence, n = 14
c) 90% confidence, n = 17
9-70
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Example
Find the chi-square values that separate the middle
90% of the distribution from the 5% in each tail.
Assume 15 degrees of freedom.
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A (1 – )·100% Confidence Interval for σ2
If a simple random sample of size n is taken from a
normal population with mean μ and standard deviation
σ, then a (1 – α)·100% confidence interval for χ2 is
given by
Lower bound:
Upper bound:
(n 1)s2
2 2
(n 1)s2
12  2

where χ2/2 and χ21- /2 are found using n – 1 degrees of
freedom.

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A (1 – )·100% Confidence Interval for σ
To find a (1 – )·100% confidence interval about
σ, take the square root of the lower bound and
upper bound.
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Parallel Example 2: Constructing a Confidence Interval for a
Population Variance and Standard Deviation
One way to measure the risk of a stock is through the
standard deviation rate of return of the stock. The
following data represent the weekly rate of return (in
percent) of Microsoft for 15 randomly selected weeks.
Compute the 90% confidence interval for the risk of
Microsoft stock.
5.34 9.63 –2.38 3.54 –8.76 2.12 –1.95 0.27
0.15 5.84 –3.90 –3.80
2.85 –1.61 –3.31
Source: Yahoo!Finance
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Solution
A normal probability plot and boxplot indicate the data
is approximately normal with no outliers.
•s = 4.6974; s2 = 22.0659
•χ20.95 = 6.571 and χ20.05 = 23.685 for 15 – 1 = 14
•degrees of freedom
14(22.0659)
13.04
•
Lower bound:
23.685
14(22.0659)
•
Upper bound:
 47.01
6.571
We are 90%confident that the population standard
deviation rate of return of the stock is between 13.04
and 47.01.

Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
9-75
Example
The following shows the sale price of 12 randomly
selected 6-year-old Chevy Corvettes. Construct a 90%
confidence interval for the population variance and
standard deviation of the price of a 6-year-old Chevy
Corvette.
41,844
37,995
43,995
9-76
41,500
41,995
35,950
39,995
38,900
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
36,995
42,995
40,990
36,995
Example
The data represent the age (in weeks) at which babies
first crawl based on a survey of 12 mothers conducted
by some company. The data is normally distributed
and s = 10.00 weeks. Construct and interpret a 95%
confidence interval for the population standard
deviation of the age (in weeks) at which babies first
crawl.
52
26
9-77
30
39
44
28
35
39
26
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
47
37
56
Section
9.4
Putting It
Together:
Which Procedure
Do I Use?
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
9-79
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Example
Robert wishes to estimate the mean number of miles
that his Buick Lacrosse can be driven on a full tank of
gas. He fills up his car with regular unleaded gasoline
from the same gas station 10 times and records the
number of miles that he drives until his low-tank
indicator light comes on. He obtains the following.
Construct a 95% confidence interval for the mean
number of miles driven on a full tank of gas.
9-80
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Example
323.9
326.8
370.6
450.7
368.8
423.8
398.8
417.5
382.7
343.1
9-81
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Example
A simple random sample of size n=785 adults was
asked if they follow college football. Of the 785
surveyed, 275 responded that they did follow college
football. Construct a 95% confidence interval for the
population proportion of adults who follow college
football.
9-82
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Example
Suppose a sample of 30 students are given an IQ
test. If the sample has a standard deviation of 12.23
points, find a 99% confidence interval for the
population standard deviation.
9-83
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