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Chapter 6
Review
6-1 Preparation
6-2 The Standard Normal Distribution
6-3 Applications of Normal Distributions
6-4 Sampling Distributions and Estimators
6-5 The Central Limit Theorem
6-6 Assessing Normality
6-7 Normal as Approximation to Binomial
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-1
Example
Bone mineral density test scores are normally distributed with
a mean of 0 and a standard deviation of 1. . These scores
can be helpful in identifying the presence of osteoporosis.
The result of the test is can be measured as z scores.
If a randomly selected adult undergoes a bone density test,
find the probability that the result is a reading less than 1.27.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-2
Look at Table A-2
The are from the left hand side of the curve to 1.27 is .8980
note that the hundredths are on top.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-3
Here is a picture that describes
this
P( z  1.27)  0.8980
The probability of random adult having a bone
density less than 1.27 is 0.8980.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-4
Example 2
Using the same bone density test, find the probability that a
randomly selected person has a result above –1.00 (which is
considered to be in the “normal” range of bone density
readings.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-5
Example 2 – continued
Since we are looking for the area to the right we have to
subtract.
The probability of a randomly selected adult having a bone
density above –1 is 0.8413.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-6
Example 3
Find this probability of getting a bone density reading
between –1.00 and –2.50. This indicates the subject has
osteopenia.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-7
Example 3
1. Find the area to the left of z = –2.50
0.0062.
2. Find the area to the left of z = –1.00
0.1587.
3. The area between z = –2.50 and z = –1.00 is the difference
between these areas.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-8
Example 4
Find the 95th Percentile. This is the z score which
separates the lower 95%.
5% or 0.05
This is an example of finding a z score from an area.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-9
Solution
5% or 0.05
1.645
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-10
Example
Find the value of z0.025. This means find the 97.5
percentile.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-11
Solution
The notation z0.025 is used to represent the z score
with an area of 0.025 to its right.
Referring back to the bone density example,
z0.025 = 1.96.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-12
Chapter 6
Normal Probability Distributions
6-1 Preparation
6-2 The Standard Normal Distribution
6-5 The Central Limit Theorem
6-6 Assessing Normality
6-7 Normal as Approximation to Binomial
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-13
6-3 Applications of Normal Distributions
Example
Tall Clubs International has a requirement that women
must be at least 70 inches tall.
Given that women have normally distributed heights
with a mean of 63.8 inches and a standard deviation of
2.6 inches, find the percentage of women who satisfy
that height requirement.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-14
Solution
Convert to a z score and use Table A-2 or technology to
find the shaded area.
z
x

70  63.8

 2.38
2.6
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-15
Solution
Draw the normal distribution and shade the region.
The area to the right of 2.38 is 0.008656, and so about
0.87% of all women meet the requirement.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-16
Example
Men’s heights are normally distributed with a mean of 69.5 inches
and a standard deviation of 2.4 inches.
When designing aircraft cabins, what ceiling height will allow 95% of
men to stand without bumping their heads?
This is an example of finding a z score from an area.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-17
Solution
With z = 1.645, μ = 69.5, and σ = 2.4. we can solve for x.
x     z    69.5  1.645 2.4  73.448 inches
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-18
6-4 Sampling Distributions and Estimators
Example
2
x
Roll a die 5 times. Find the mean , variance s,
and the proportion of odd numbers of the
results.
What will happen if you continue this process?
What will the mean of all of your sample means
approach?
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-19
Example – Elevators
Assume male weights follow a normal distribution with a mean of 182.9
lb and a standard deviation of 40.8 lb.
Suppose an elevator has a maximum capacity of 16 passengers with a
total weight of 2500 lb.
Assuming a worst case scenario in which the passengers are all male,
what are the chances the elevator is overloaded?
a.
Find the probability that 1 randomly selected male has a weight
greater than 156.25 lb.
b.
Find the probability that a sample of 16 males have a mean weight
greater than 156.25 lb (which puts the total weight at 2500 lb,
exceeding the maximum capacity).
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-20
Solution
a.
Find the probability that 1 randomly selected male has a weight
greater than 156.25 lb.
Use the methods presented in Section 6.3. We can convert to a z
score and use Table A-2.
z
x

156.25  182.9

 0.65
40.8
Using Table A-2, the area to the right is 0.7422.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-21
Solution
b.
Find the probability that a sample of 16
males have a mean weight greater than
156.25 lb.
Since the distribution of male weights is
assumed to be normal, the sample mean will
also be normal.
 x   x  182.9
 x 40.8
x 

 10.2
n
16
Converting to z:
156.25  182.9
z
 2.61
10.2
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-22
Solution
b. Find the probability that a sample of 16 males have a
mean weight greater than 156.25 lb.
While there is 0.7432 probability that any given male will weigh
more than 156.25 lb, there is a 0.9955 probability that the
sample of 16 males will have a mean weight of 156.25 lb or
greater.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-23
Review
Binomial Probability Distribution
1. The procedure must have a fixed number of trials.
2. The trials must be independent.
3. Each trial must have all outcomes classified into two categories
(commonly, success and failure).
4. The probability of success remains the same in all trials.
Solve by binomial probability formula, Table A-1, or technology.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-24
Approximation of a Binomial Distribution
with a Normal Distribution
np  5
nq  5
then   np ,   npq and the
random variable has
a
distribution.
(normal)
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-25
Example – NFL Coin Toss
In 431 NFL football games that went to over time, the teams
that won the coin toss went on to win 235 of those games.
What is the probability of such an event occurring assuming
there is a 0.5 probability of winning a game after winning the
coin toss?
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-26
Solution
If the coin-toss method is fair, teams winning the toss would
win about 50% of the games (we’d expect 215.5 wins in 431
overtime games).
The given problem involves a binomial distribution with n = 431
trials and an assumed probability of success of p = 0.5.
Use the normal approximation to the binomial distribution.
Step 1: The conditions check:
np  431 0.5   215.5  5
nq  431 0.5   215.5  5
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-27
Example – NFL Coin Toss
Step 2: Find the mean and standard deviation of the normal
distribution:
  np  431 0.5  215.5
  npq  431 0.5 0.5  10.38027
Step 3: We want the probability of at least 235 wins,
so x = 235.
Step 4: The vertical strip will go from 234.5 to 235.5.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-28
Example – NFL Coin Toss
Step 5: We will shade the area to the right of 234.5.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-29
Example – NFL Coin Toss
Step 6: Find the z score and use technology or Table A-2 to
determine the probability.
z
x

234.5  215.5

 1.83
10.380270
The probability is 0.0336 for the coin flip winning team to win at
least 235 games.
This probability is low enough to suggest the team winning coin
flip has an unfair advantage.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-30
Don’t Forget
When we use the normal distribution (which is a continuous
probability distribution) as an approximation to the binomial
distribution (which is discrete), a continuity correction is
made to a discrete whole number x in the binomial
distribution by representing the discrete whole number x by
the interval from
x – 0.5 to x + 0.5
(that is, adding and subtracting 0.5).
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-31
Example
In the Chapter Problem we noted that a Pew Research
Center poll of 1007 randomly selected adults showed that
85% of respondents know what Twitter is. The sample
results are n = 1007 and pˆ  0.70.
a. Find the margin of error E that corresponds to a 95% confidence
level.
b. Find the 95% confidence interval estimate of the population
proportion p.
c. Based on the results, can we safely conclude that more than 75%
of adults know what Twitter is?
d. Assuming that you are a newspaper reporter, write a brief
statement that accurately describes the results and includes all of
the relevant information.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-32
Example - Continued
Requirement check: simple random sample; fixed
number of trials, 1007; trials are independent; two
outcomes per trial; probability remains constant. Note:
number of successes and failures are both at least 5.
a) Use the formula to find the margin of error.
ˆˆ
pq
E  z 2
 1.96
n
E  0.0220545
 0.85 0.15
1007
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-33
Example - Continued
b) The 95% confidence interval:
pˆ  E  p  pˆ  E
0.85  0.0220545  p  0.85  0.0220545
0.828  p  0.872
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-34
Example - Continued
c) Based on the confidence interval obtained in part
(b), it does appear that more than 75% of adults
know what Twitter is.
Because the limits of 0.828 and 0.872 are likely to
contain the true population proportion, it appears
that the population proportion is a value greater than
0.75.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-35
Example - Continued
d) Here is one statement that summarizes the results:
85% of U.S. adults know what Twitter is. That
percentage is based on a Pew Research Center poll
of 1007 randomly selected adults.
In theory, in 95% of such polls, the percentage
should differ by no more than 2.2 percentage points
in either direction from the percentage that would be
found by interviewing all adults in the United States.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-36
Example
Many companies are interested in knowing the
percentage of adults who buy clothing online.
How many adults must be surveyed in order to be 95%
confident that the sample percentage is in error by no
more than three percentage points?
a. Use a recent result from the Census Bureau: 66%
of adults buy clothing online.
b. Assume that we have no prior information suggesting
a possible value of the proportion.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-37
Example - Continued
a) Use
pˆ  0.66 and qˆ  1  pˆ  0.34
  0.05 so z 2  1.96
E  0.03
z 

n
2
ˆˆ
pq
 2
E2
1.96   0.66  0.34 


2
 0.03 
2
To be 95% confident that
our sample percentage is
within three percentage
points of the true
percentage for all adults,
we should obtain a simple
random sample of 958
adults.
 957.839
 958
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-38
Example - Continued
b) Use
  0.05 so z 2  1.96
E  0.03
z 

n
 2
2
 0.25
E2
1.96   0.25


2
 0.03 
2
To be 95% confident that
our sample percentage is
within three percentage
points of the true
percentage for all adults,
we should obtain a simple
random sample of 1068
adults.
 1067.1111
 1068
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-39
Example
A common claim is that garlic lowers cholesterol levels. In
a test of the effectiveness of garlic, 49 subjects were
treated with doses of raw garlic, and their cholesterol
levels were measured before and after the treatment.
The changes in their levels of LDL cholesterol (in mg/dL)
have a mean of 0.4 and a standard deviation of 21.0.
Use the sample statistics of n = 49, x = 0.4, and s = 21.0 to
construct a 95% confidence interval estimate of the mean
net change in LDL cholesterol after the garlic treatment.
What does the confidence interval suggest about the
effectiveness of garlic in reducing LDL cholesterol?
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-40
Example - Continued
Requirements are satisfied: simple random
sample and n = 49 (i.e., n > 30).
95% implies α = 0.05.
With n = 49, the df = 49 – 1 = 48
Closest df is 50, two tails, so
= 2.009
Using
= 2.009, s = 21.0 and n = 49 the margin
of error is:
E  t 2

21.0
 2.009 
 6.027
n
49
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-41
Example - Continued
Construct the confidence interval: x  0.4, E  6.027
x E    x E
0.4  6.027    0.4  6.027
5.6    6.4
We are 95% confident that the limits of –5.6 and 6.4 actually do contain
the value of μ, the mean of the changes in LDL cholesterol for the
population.
Because the confidence interval limits contain the value of 0, it is very
possible that the mean of the changes in LDL cholesterol is equal to 0,
suggesting that the garlic treatment did not affect the LDL cholesterol
levels.
It does not appear that the garlic treatment is effective in lowering LDL
cholesterol.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-42
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample sizes.
(See the following slide for the cases n = 3 and n = 12.)
2. The Student t distribution has the same general symmetric bell
shape as the standard normal distribution but it reflects the greater
variability (with wider distributions) that is expected with small
samples.
3. The Student t distribution has a mean of t = 0 (just as the standard
normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with the
sample size and is greater than 1 (unlike the standard normal
distribution, which has σ = 1).
5. As the sample size n gets larger, the Student t distribution gets
closer to the normal distribution.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-43
Student t Distributions for
n = 3 and n = 12
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-44
Finding the Point Estimate
and E from a Confidence Interval
Point estimate of μ:
= (upper confidence limit) + (lower confidence limit)
2
Margin of Error:
= (upper confidence limit) – (lower confidence limit)
2
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-45
Finding a Sample Size for Estimating a
Population Mean
= population mean
= population standard deviation
x = sample mean
= desired margin of error
= z score separating an area of
the standard normal distribution
in the right tail of
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-46
Round-Off Rule for Sample Size n
If the computed sample size n is not a whole
number, round the value of n up to the next
larger whole number.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-47
Finding the Sample Size n
When σ is Unknown
1. Use the range rule of thumb (see Section 3-3) to
estimate the standard deviation as follows:
.
2. Start the sample collection process without knowing σ
and, using the first several values, calculate the sample
standard deviation s and use it in place of σ. The
estimated value of σ can then be improved as more
sample data are obtained, and the sample size can be
refined accordingly.
3. Estimate the value of σ by using the results of some
other earlier study.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-48
Example
Assume that we want to estimate the mean IQ score for the
population of statistics students. How many statistics students
must be randomly selected for IQ tests if we want 95%
confidence that the sample mean is within 3 IQ points of the
population mean?
α
α/2
zα/2
E
σ
= 0.05
= 0.025
= 1.96
= 3
= 15
With a simple random sample of only 97
statistics students, we will be 95%
confident that the sample mean is within
3 IQ points of the true population mean μ.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-49
Part 2: Key Concept
This section presents methods for
estimating a population mean. In addition
to knowing the values of the sample data
or statistics, we must also know the value
of the population standard deviation, σ.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-50
Confidence Interval for Estimating a
Population Mean (with σ Known)
μ
= population mean
x
= sample mean
σ
= population standard deviation
n
= number of sample values
E
= margin of error
z α/2
= critical z value separating an area of α/2 in the
right tail of the standard normal distribution
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-51
Confidence Interval for Estimating
a Population Mean (with σ Known)
1. The sample is a simple random sample. (All samples
of the same size have an equal chance of being
selected.)
2. The value of the population standard deviation σ is
known.
3. Either or both of these conditions is satisfied: The
population is normally distributed or n > 30.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-52
Confidence Interval for Estimating
a Population Mean (with σ Known)
x  E    x  E where E  z 2 
or
or

n
x E
x  E,x  E 
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-53
Example
People have died in boat and aircraft accidents because an
obsolete estimate of the mean weight of men was used.
In recent decades, the mean weight of men has increased
considerably, so we need to update our estimate of that
mean so that boats, aircraft, elevators, and other such
devices do not become dangerously overloaded.
Using the weights of men from a random sample, we obtain
these sample statistics for the simple random sample:
n = 40 and x = 172.55 lb.
Research from several other sources suggests that the
population of weights of men has a standard deviation given
by σ = 26 lb.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-54
Example - Continued
a. Find the best point estimate of the mean weight of the
population of all men.
b. Construct a 95% confidence interval estimate of the
mean weight of all men.
c. What do the results suggest about the mean weight of
166.3 lb that was used to determine the safe passenger
capacity of water vessels in 1960 (as given in the
National Transportation and Safety Board safety
recommendation M-04-04)?
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-55
Example - Continued
a. The sample mean of 172.55 lb is the best point estimate of
the mean weight of the population of all men.
b. A 95% confidence interval implies that α = 0.05, so zα/2= 1.96.
Calculate the margin of error.

26
E  z 2 
 1.96 
 8.0574835
n
40
Construct the confidence interval.
x E    x E
172.55  8.0574835    172.55  8.0574835
164.49    180.61
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-56
Example - Continued
c. Based on the confidence interval, it is possible that the
mean weight of 166.3 lb used in 1960 could be the
mean weight of men today.
However, the best point estimate of 172.55 lb suggests
that the mean weight of men is now considerably greater
than 166.3 lb.
Considering that an underestimate of the mean weight
of men could result in lives lost through overloaded
boats and aircraft, these results strongly suggest that
additional data should be collected.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-57
Choosing the Appropriate Distribution
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-58
Choosing the Appropriate Distribution
Use the normal (z)
distribution
σ known and
normally distributed
population or n > 30
Use t distribution
σ not known and
normally distributed
population or n > 30
Use a nonparametric
method or bootstrapping
Population is not normally
distributed and n ≤ 30
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-59
Chapter 7
Estimates and Sample Sizes
7-1 Review and Preview
7-2 Estimating a Population Proportion
7-3 Estimating a Population Mean
7-4 Estimating a Population Standard Deviation or
Variance
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-60
Key Concept
This section we introduce the chi-square
probability distribution so that we can construct
confidence interval estimates of a population
standard deviation or variance.
We also present a method for determining the
sample size required to estimate a population
standard deviation or variance.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-61
Chi-Square Distribution
In a normally distributed population with variance σ2 ,
assume that we randomly select independent samples of
size n and, for each sample, compute the sample variance
s2 (which is the square of the sample standard deviation s).
The sample statistic χ2 (pronounced chi-square) has a
sampling distribution called the chi-square distribution.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-62
Chi-Square Distribution
2
n

1
s


2 
2
where
s
s2
σ2
df
= sample size
= sample variance
= population variance
= n – 1 degrees of freedom
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-63
Properties of the Distribution
of the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike
the normal and Student t distributions. As the number of
degrees of freedom increases, the distribution becomes
more symmetric.
Chi-Square Distribution
Chi-Square Distribution for
df = 10 and df = 20
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-64
Properties of the Distribution
of the Chi-Square Statistic
2. The values of chi-square can be zero or positive, but it cannot
be negative.
3. The chi-square distribution is different for each number of
degrees of freedom, which is df = n – 1. As the number of
degrees of freedom increases, the chi-square distribution
approaches a normal distribution.
In Table A-4, each critical value of χ2 corresponds to an area
given in the top row of the table, and that area represents the
cumulative area located to the right of the critical value.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-65
Example
A simple random sample of 22 IQ scores is obtained.
Construction of a confidence interval for the population
standard deviation σ requires the left and right critical
values of χ2 corresponding to a confidence level of 95%
and a sample size of n = 22.
Find the critical χ2 values corresponding to a 95% level of
confidence.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-66
Example - Continued
Critical Values of the Chi-Square Distribution
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-67
Estimators of σ2
The sample variance s2 is the best point
estimate of the population variance σ2.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-68
Estimator of σ
The sample standard deviation s is a
commonly used point estimate of σ (even
though it is a biased estimate).
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-69
Confidence Interval for Estimating a
Population Standard Deviation or Variance
σ
σ2
s
s2
n
E
 L2
 R2
=
=
=
=
=
=
=
=
population standard deviation
population variance
sample standard deviation
sample variance
number of sample values
margin of error
2

left-tailed critical value of
2

right-tailed critical value of
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-70
Confidence Interval for Estimating a
Population Standard Deviation or Variance
Requirements:
1. The sample is a simple random sample.
2. The population must have normally distributed
values (even if the sample is large).
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-71
Confidence Interval for Estimating a
Population Standard Deviation or Variance
Confidence Interval for the Population Variance σ2
n  1s

2
R
2

2
n  1s



Copyright © 2014, 2012, 2010 Pearson Education, Inc.
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Section 6.3-72
Confidence Interval for Estimating a
Population Standard Deviation or Variance
Confidence Interval for the Population
Standard Deviation σ
n  1s

2
R
2
 
n  1s
Copyright © 2014, 2012, 2010 Pearson Education, Inc.

2
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Section 6.3-73
Procedure for Constructing a
Confidence Interval for σ or σ2
1. Verify that the required assumptions are satisfied.
2. Using n – 1 degrees of freedom, refer to Table A-4
or use technology to find the critical values
 R2 and  L2 that correspond to the desired
confidence level.
3. Evaluate the upper and lower confidence interval limits
using this format of the confidence interval:
n  1s
 R2
2
 2
n  1s


2
 L2
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-74
Procedure for Constructing a
Confidence Interval for σ or σ2
4. If a confidence interval estimate of σ is desired, take the
square root of the upper and lower confidence interval
limits and change σ2 to σ.
5. Round the resulting confidence level limits. If using the
original set of data to construct a confidence interval,
round the confidence interval limits to one more decimal
place than is used for the original set of data. If using the
sample standard deviation or variance, round the
confidence interval limits to the same number of decimal
places.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-75
Caution
Confidence intervals can be used informally to compare
the variation in different data sets, but the overlapping of
confidence intervals should not be used for making
formal and final conclusions about equality of variances
or standard deviations.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-76
Example
A group of 22 subjects took an IQ test during part of a study.
The subjects had a standard deviation IQ score of 14.3.
Construct a 95% confidence interval estimate of σ, the
standard deviation of the population from which the sample
was obtained.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-77
Example - Continued
We must first verify the requirements are met.
1. We can treat the sample as a simple random sample.
2. The following display shows a histogram of the data and the
normality assumption can be reasonably met.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-78
Example - Continued
n = 22 so df = 22 – 1 = 21
Use table A-4 to find:
  10.283 and   35.479
2
L
2
R
 n  1 s 2
 R2
 22  114.3
35.479
2
n

1
s


2 
 L2
2

2
22  114.3


2
10.283
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Section 6.3-79
Example - Continued
Evaluation of the preceding expression yields:
121.0    417.6
2
Finding the square root of each part (before rounding), then
rounding to two decimal places, yields this 95% confidence
interval estimate of the population standard deviation:
11.0    20.4
Based on this result, we have 95% confidence that the limits
of 11.0 and 20.4 contain the true value of σ.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-80
Determining Sample Sizes
The procedures for finding the sample size necessary to
estimate σ2 are much more complex than the procedures
given earlier for means and proportions. Instead of using
very complicated procedures, we will use Table 7-2.
STATDISK also provides sample sizes. With STATDISK,
select Analysis, Sample Size Determination, and then
Estimate St Dev.
Minitab, Excel, and the TI-83/84 Plus calculator do not
provide such sample sizes.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-81
Determining Sample Sizes
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-82
Example
We want to estimate the standard deviation σ of all voltage
levels in a home. We want to be 95% confident that our
estimate is within 20% of the true value of σ.
How large should the sample be? Assume that the
population is normally distributed.
From Table 7-2, we can see that 95% confidence and an
error of 20% for σ correspond to a sample of size 48.
We should obtain a simple random sample of 48 voltage
levels form the population of voltage levels.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 6.3-83