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Chapter 6
Section 6.1 Objectives
Chapter 6:
 Understanding point estimates and interval estimates and
Confidence
Intervals
a margin of error
 Construct and interpret confidence intervals for the
population mean
 Determine the minimum sample size required when
estimating μ
Chapter Outline:
6.1 Confidence Intervals for the Mean (Large Samples)
6.3 Confidence Intervals for Population Proportions
Point Estimate for Population μ
Point Estimate
 A single value estimate for a population parameter
 Most unbiased point estimate of the population mean μ
is the sample mean x
Estimate Population with Sample
Parameter…
Statistic
Mean: μ
x
Larson/Farber 5th ed
Example: Point Estimate for Population μ
A social networking website allows its users to add
friends, send messages, and update their personal profiles.
The following represents a random sample of the number
of friends for 40 users of the website. Find a point
estimate of the population mean, µ. (Source: Facebook)
140 105 130 97 80 165 232 110 214 201 122
98 65 88 154 133 121 82 130 211 153 114
58 77 51 247 236 109 126 132 125 149 122
74 59 218 192 90 117 105
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Chapter 6
Solution: Point Estimate for Population μ
Interval estimate
 An interval, or range of values, used to estimate a
population parameter.
The sample mean of the data is
x
Interval Estimate
x 5232

 130.8
n
40
Left endpoint
Right endpoint
Point estimate
146.5
x  130.8
115.1
(
Your point estimate for the mean number of friends for
all users of the website is 130.8 friends.
115
)
120
125
130
135
140
145
150
Interval estimate
How confident do we want to be that the interval estimate
contains the population mean μ?
Level of Confidence
Common Levels of Confidence
Level of confidence c
 The probability that the interval estimate contains the
population parameter.
c is the area under the
c
standard normal curve
between the critical values.
½(1 – c)
½(1 – c)
–zc
z=0
Critical values
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
z
zc
Use the Standard
Normal Table to find the
corresponding z-scores.
The remaining area in the tails is 1 – c .
Larson/Farber 5th ed
If the level of confidence is 90%, this means that we are
90% confident that the interval contains the population
mean μ.
zc
z = 0 zc =zc1.645
–zc = –1.645
z
The corresponding z-scores are ±1.645.
2
Chapter 6
Common Levels of Confidence
Common Levels of Confidence
If the level of confidence is 95%, this means that we are
95% confident that the interval contains the population
mean, μ.
The horizontal segments
represent 90% confidence
intervals for different
samples of the same size.
In the long run, 9 of every
10 such intervals will
contain μ.
0.95
0.025
0.025
μ
zc = z
 1.96
c
z=0
zc =z1.96
c
z
The corresponding z-scores are ± 1.96.
Margin of Error
Common Levels of Confidence
If the level of confidence is 99%, this means that we are
99% confident that the interval contains the population
mean, μ.
0.99
0.005
0.005
zc = z2.575
z = 0 zc = z2.575
c
c
The corresponding z-scores are ± 2.575.
Larson/Farber 5th ed
z
The difference between the point estimate and the actual population
parameter value is called the sampling error.
When μ is estimated, the sampling error is the difference μ –x.
Since μ is usually unknown, the maximum value for the error can
be calculated using the level of confidence.
Given a level of confidence, the margin of error (sometimes called
the maximum error of estimate or error tolerance) E is the greatest
possible distance between the point estimate and the value of the
parameter it is estimating.
E  z cσ x  z c σ
n
When n  30, the sample standard
deviation, s, can be used for .
3
Chapter 6
Margin of Error
Example: Finding the Margin of Error
Margin of error
 The greatest possible distance between the point
estimate and the value of the parameter it is estimating
for a given level of confidence, c.
 Denoted by E.
E  zcσ x  zc
σ
n
Use the social networking website data and a 95%
confidence level to find the margin of error for the
mean number of friends for all users of the website.
Assume the sample standard deviation is about 53.0.
When n ≥ 30, the sample
standard deviation, s, can
be used for σ.
 Sometimes called the maximum error of estimate or
error tolerance.
Solution: Finding the Margin of Error
Solution: Finding the Margin of Error
 First find the critical values
0.95
z=0
zczc= 1.96
z
95% of the area under the standard normal curve falls
within 1.96 standard deviations of the mean. (You
can approximate the distribution of the sample means
with a normal curve by the Central Limit Theorem,
because n = 40 ≥ 30.)
Larson/Farber 5th ed

0.025
0.025
zc
–zc = –1.96
E  zc
n
 1.96 
 zc
s
n
You don’t know σ, but
since n ≥ 30, you can
use s in place of σ.
53.0
40
 16.4
You are 95% confident that the margin of error for the
population mean is about 16.4 friends.
4
Chapter 6
Constructing a Confidence Interval
• A c-confidence interval for the population mean μ is
x  E < μ < x + E.
• The probability that the confidence interval contains μ is c.
To construct a 95% confidence interval for the mean number of
friends for all users of the website.
Constructing a Confidence Interval
114.4 < μ < 147.2
•
Now Recall x  130.8 and E ≈ 16.4
Left Endpoint:
Right Endpoint:
x E
xE
 130.8  16.4
 130.8  16.4
 114.4
 147.2
With 95% confidence, you can say that the population
mean number of friends is between 114.4 and 147.2.
114.4 < μ < 147.2
Constructing Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean
(n ≥ 30 or σ known with a normally distributed population)
In Words
1. Find the sample statistics n and
x.
2. Specify σ, if known.
Otherwise, if n ≥ 30, find the
sample standard deviation s and
use it as an estimate for σ.
Larson/Farber 5th ed
In Symbols
x
s
x
n
(x  x )2
n 1
Constructing Confidence Intervals for μ
In Words
3. Find the critical value zc that
corresponds to the given
level of confidence.
4. Find the margin of error E.
5. Find the left and right
endpoints and form the
confidence interval.
In Symbols
Use the Standard
Normal Table or
technology.

Ez
c
n
Left endpoint: x  E
Right endpoint: x  E
Interval:
x E   x E
5
Chapter 6
Example: Constructing a Confidence Interval
σ Known
A college admissions director wishes to estimate the mean
age of all students currently enrolled. In a random sample
of 20 students, the mean age is found to be 22.9 years.
From past studies, the standard deviation is known to be
1.5 years, and the population is normally distributed.
Construct a 90% confidence interval of the population
mean age.
Solution: Constructing a Confidence Interval
σ Known
 First find the critical values
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
–zc = –1.645
z = 0 zc =zc1.645
z
zc = 1.645
Solution: Constructing a Confidence Interval
σ Known
Solution: Constructing a Confidence Interval
σ Known
 Margin of error:
22.3 < μ < 23.5
E  zc

n
 1.645 
1.5
20
 0.6
22.3
(
 Confidence interval:
Left Endpoint:
x E
Right Endpoint:
xE
 22.9  0.6
 22.9  0.6
 22.3
 23.5
x E
Point
estimate
22.9
23.5
x
xE
•
)
With 90% confidence, you can say that the mean
age of all the students is between 22.3 and 23.5
years.
22.3 < μ < 23.5
Larson/Farber 5th ed
6
Chapter 6
Interpreting the Results
 μ is a fixed number. It is either in the confidence interval
or not.
 Incorrect: “There is a 90% probability that the actual
mean is in the interval (22.3, 23.5).”
 Correct: “If a large number of samples is collected and a
confidence interval is created for each sample,
approximately 90% of these intervals will contain μ.
Sample Size
 Given a c-confidence level and a margin of error E, the
minimum sample size n needed to estimate the
population mean µ is
z
n c 
 E 
2
Interpreting the Results
The horizontal segments
represent 90% confidence
intervals for different
samples of the same size.
In the long run, 9 of every
10 such intervals will
contain μ.
μ
Example: Sample Size
You want to estimate the mean number of friends for all
users of the website. How many users must be included in
the sample if you want to be 95% confident that the
sample mean is within seven friends of the population
mean? Assume the sample standard deviation is about
53.0.
 If σ is unknown, you can estimate it using s, provided
you have a preliminary sample with at least 30
members.
Larson/Farber 5th ed
7
Chapter 6
Solution: Sample Size
Solution: Sample Size
 First find the critical values
zc = 1.96
0.95
zc
–zc = –1.96
z=0
2
zczc= 1.96
E=7
 z    1.96  53.0 
n c  
  220.23
7

 E  
0.025
0.025
σ ≈ s ≈ 53.0
z
zc = 1.96
Section 6.1 Summary
2
When necessary, round up to obtain a whole number.
You should include at least 221 users in your sample.
Section 6.3 Objectives
 Found a point estimate and a margin of error
 Find a point estimate for the population proportion
 Constructed and interpreted confidence intervals for the
 Construct a confidence interval for a population
population mean
 Determined the minimum sample size required when
estimating μ
proportion
 Determine the minimum sample size required when
estimating a population proportion
Larson/Farber 5th ed
8
Chapter 6
Point Estimate for Population p
Population Proportion
 The probability of success in a single trial of a binomial
experiment.
 Denoted by p
Point Estimate for p
 The proportion of successes in a sample.
 Denoted by
x number of successes in sample

n
sample size

pˆ 

read as “p hat”
Point Estimate for Population p
Estimate Population with Sample
Parameter…
Statistic
Proportion: p
p̂
Point Estimate for q, the population proportion of failures
 Denoted by qˆ  1  p
ˆ
 Read as “q hat”
Example: Point Estimate for p
In a survey of 1000 U.S. adults, 662 said that it is
acceptable to check personal e-mail while at work. Find a
point estimate for the population proportion of U.S. adults
who say it is acceptable to check personal e-mail while at
work. (Adapted from Liberty Mutual)
Solution: n = 1000
pˆ 
Confidence Intervals for p
A c-confidence interval for a population proportion p
•
pˆ  E  p  pˆ  E
where E  zc
pq
ˆˆ
n
•The probability that the confidence interval contains p is c.
and x = 662
x 662

 0.662  66.2%
n 1000
Larson/Farber 5th ed
9
Chapter 6
Constructing Confidence Intervals for p
In Words
In Symbols
1. Identify the sample statistics n
and x.
2. Find the point estimate p̂.
3. Verify that the sampling
distribution of p̂ can be
approximated by a normal
distribution.
4. Find the critical value zc that
corresponds to the given level of
confidence c.
Constructing Confidence Intervals for p
In Words
In Symbols
E  zc
5. Find the margin of error E.
pˆ 
x
n
npˆ  5, nqˆ  5
6. Find the left and right
endpoints and form the
confidence interval.
Left endpoint: p̂  E
Right endpoint: p̂  E
Interval:
Solution: Confidence Interval for p
In a survey of 1000 U.S. adults, 662 said that it is
acceptable to check personal e-mail while at work.
Construct a 95% confidence interval for the population
proportion of U.S. adults who say that it is acceptable to
check personal e-mail while at work.
 Verify the sampling distribution of
Solution: Recall pˆ  0.662
• Margin of error:
Larson/Farber 5th ed
pˆ  E  p  pˆ  E
Use the Standard
Normal Table or
technology.
Example: Confidence Interval for p
qˆ  1  pˆ  1  0.662  0.338
pq
ˆˆ
n
p̂can be
approximated by the normal distribution
npˆ  1000  0.662  662  5
nqˆ  1000  0.338  338  5
E  zc
pq
(0.662)  (0.338)
ˆˆ
 1.96
 0.029
n
1000
10
Chapter 6
Solution: Confidence Interval for p
Solution: Confidence Interval for p
 Confidence interval:
Left Endpoint:
pˆ  E
 0.633 < p < 0.691
Right Endpoint:
Point
estimate
pˆ  E
 0.662  0.029
 0.662  0.029
 0.633
 0.691
p̂  E
Sample Size
Example: Sample Size
 Given a c-confidence level and a margin of error E, the
minimum sample size n needed to estimate p is
2
 This formula assumes you have an estimate for
and qˆ.
 If not, use p
ˆ  0.5and
qˆ  0.5.
p̂  E
With 95% confidence, you can say that the population
proportion of U.S. adults who say that it is acceptable
to check personal e-mail while at work is between
63.3% and 69.1%.
0.633 < p < 0.691
z 
ˆ ˆ c 
n  pq
E
p̂
p̂
You are running a political campaign and wish to
estimate, with 95% confidence, the population
proportion of registered voters who will vote for your
candidate. Your estimate must be accurate within 3%
of the true population proportion. Find the minimum
sample size needed if
1. no preliminary estimate is available.
Solution:
Because you do not have a preliminary estimate
for pˆ , use pˆ  0.5 and qˆ  0.5.
Larson/Farber 5th ed
11
Chapter 6
Solution: Sample Size
 c = 0.95
zc = 1.96
E = 0.03
2
2
z 
 1.96 
ˆ ˆ  c   (0.5)(0.5) 
n  pq
  1067.11
 0.03 
E
Round up to the nearest whole number.
With no preliminary estimate, the minimum sample
size should be at least 1068 voters.
Example: Sample Size
You are running a political campaign and wish to
estimate, with 95% confidence, the population
proportion of registered voters who will vote for your
candidate. Your estimate must be accurate within 3% of
the true population proportion. Find the minimum
sample size needed if
ˆ  0.31.
2. a preliminary estimate gives p
Solution:
Use the preliminary estimate
pˆ  0.31
qˆ  1  pˆ  1  0.31  0.69
Solution: Sample Size
 c = 0.95
zc = 1.96
2
 Found a point estimate for the population proportion
E = 0.03
2
z 
 1.96 
ˆ ˆ  c   (0.31)(0.69) 
n  pq
  913.02
 0.03 
E
Round up to the nearest whole number.
With a preliminary estimate of pˆ  0.31, the
minimum sample size should be at least 914 voters.
Need a larger sample size if no preliminary estimate
is available.
Larson/Farber 5th ed
Section 6.3 Summary
 Constructed a confidence interval for a population
proportion
 Determined the minimum sample size required when
estimating a population proportion
12