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Chapter 6
Confidence Intervals
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Chapter Outline
 6.1 Confidence Intervals for the Mean (Large
Samples)
 6.2 Confidence Intervals for the Mean (Small
Samples)
 6.3 Confidence Intervals for Population Proportions
 6.4 Confidence Intervals for Variance and Standard
Deviation
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Section 6.1
Confidence Intervals for the Mean (Large Samples)
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Section 6.1 Objectives
 Find a point estimate and a margin of error
 Construct and interpret confidence intervals for the
population mean
 Determine the minimum sample size required when
estimating μ
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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
Parameter…
Mean: μ
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with Sample
Statistic
x
Example: Point Estimate for Population
μ
Market researchers use the number of sentences per advertisement
as a measure of readability for magazine advertisements. The
following represents a random sample of the number of sentences
found in 50 advertisements. Find a point estimate of the
population mean, . (Source: Journal of Advertising Research)
9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 25
17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7
14 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20
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Solution: Point Estimate for Population
μ
The sample mean of the data is
x 620
x

 12.4
n
50
Your point estimate for the mean length of all magazine
advertisements is 12.4 sentences.
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Interval Estimate
Interval estimate
 An interval, or range of values, used to estimate a population
parameter.
Point estimate
(
12.4
•
)
Interval estimate
How confident do we want to be that the interval estimate
contains the population mean μ?
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Level of Confidence
Level of confidence c
 The probability that the interval estimate contains the population
parameter.
c is the area under the
standard normal curve
between the critical values.
c
½(1 – c)
½(1 – c)
-zc
z=0
Critical values
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z
zc
Use the Standard
Normal Table to find the
corresponding z-scores.
The remaining area in the tails is 1 – c .
Level of Confidence
 If the level of confidence is 90%, this means that we are 90%
confident that the interval contains the population mean μ.
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
-zc = -1.645
z=0
zc =zc1.645
The corresponding z-scores are +1.645.
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z
Sampling Error
Sampling error
 The difference between the point estimate and the actual
population parameter value.
 For μ:
 the sampling error is the difference
 μ is generally unknown

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x varies from sample to sample
x–μ
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
When n  30, the sample
standard deviation, s, can
be used for .
 Sometimes called the maximum error of estimate or error
tolerance.
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Example: Finding the Margin of Error
Use the magazine advertisement data and a 95% confidence
level to find the margin of error for the mean number of
sentences in all magazine advertisements. Assume the sample
standard deviation is about 5.0.
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Solution: Finding the Margin of Error
 First find the critical values
0.95
0.025
0.025
zc
-zc = -1.96
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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 ≥
30.)
Solution: Finding the Margin of Error
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.
5.0
50
 1.4
You are 95% confident that the margin of error for the
population mean is about 1.4 sentences.
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Confidence Intervals for the Population
Mean
A c-confidence interval for the population mean μ

x E   x E
where E  zc

n
 The probability that the confidence interval contains μ is c.
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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 , if known.
Otherwise, if n  30, find the
sample standard deviation s and
use it as an estimate for .
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In Symbols
x
x
n
(x  x )2
s
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.
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In Symbols
Use the Standard
Normal Table.
E  zc

n
Left endpoint: x  E
Right endpoint: x  E
Interval:
xE  xE
Example: Constructing a Confidence
Interval
Construct a 95% confidence interval for the mean number of
sentences in all magazine advertisements.
Solution: Recall x  12.4and E = 1.4
Left Endpoint:
Right Endpoint:
 11.0
 13.8
xE
 12.4  1.4
xE
 12.4  1.4
11.0 < μ < 13.8
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Solution: Constructing a Confidence
Interval
11.0 < μ < 13.8
11.0
(
12.4 13.8
•
)
With 95% confidence, you can say that the population mean
number of sentences is between 11.0 and 13.8.
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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.
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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 = 1.645
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zc =zc1.645
z
Solution: Constructing a Confidence
Interval σ Known
 Margin of error:
E  zc

 1.645 
n
 Confidence interval:
1.5
20
 0.6
Left Endpoint:
Right Endpoint:
 22.3
 23.5
xE
 22.9  0.6
xE
 22.9  0.6
22.3 < μ < 23.5
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Solution: Constructing a Confidence
Interval σ Known
22.3 < μ < 23.5
22.3
(
x E
Point estimate
22.9
•
x
23.5
)
xE
With 90% confidence, you can say that the mean age of all
the students is between 22.3 and 23.5 years.
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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.
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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.
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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  is
 zc 
n

E


2
 If  is unknown, you can estimate it using s provided you have a
preliminary sample with at least 30 members.
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Example: Sample Size
You want to estimate the mean number of sentences in a magazine
advertisement. How many magazine advertisements must be
included in the sample if you want to be 95% confident that the
sample mean is within one sentence of the population mean?
Assume the sample standard deviation is about 5.0.
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Solution: Sample Size
 First find the critical values
0.95
0.025
0.025
zc
-zc = -1.96
z=0
zc = 1.96
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zczc= 1.96
z
Solution: Sample Size
zc = 1.96
  s = 5.0
E=1
 zc   1.96  5.0 
n

  96.04

1

 E  
2
2
When necessary, round up to obtain a whole number.
You should include at least 97 magazine advertisements in
your sample.
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Section 6.1 Summary
 Found a point estimate and a margin of error
 Constructed and interpreted confidence intervals for the
population mean
 Determined the minimum sample size required when
estimating μ
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