Chapter 8
8-1
Confidence Interval Estimation
Learning Objectives
Confidence Intervals
In this chapter, you learn:
To construct and interpret confidence interval estimates
for the mean and the proportion
How to determine the sample size necessary to
develop a confidence interval for the mean or
proportion
when Population Standard Deviation σ is Known
when Population Standard Deviation σ is Unknown
Confidence Intervals for the Population
Proportion, p
Determining the Required Sample Size
How to use confidence interval estimates in auditing
Chap 8-1
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Content of this chapter
Confidence Intervals for the Population
Mean, µ
Point Estimates
Point and Interval Estimates
A point estimate is a single number,
We can estimate a
Population Parameter …
a confidence interval provides additional
information about variability
Lower
Confidence
Limit
Point Estimate
Chap 8-2
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Upper
Confidence
Limit
with a Sample
Statistic
(a Point Estimate)
Mean
µ
X
Proportion
π
p
Width of
confidence interval
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Chap 8-4
Chapter 8
8-2
Confidence Intervals
Confidence Interval Estimate
An interval gives a range of values:
How much uncertainty is associated with a
point estimate of a population parameter?
Takes into consideration variation in sample
statistics from sample to sample
An interval estimate provides more
information about a population characteristic
than does a point estimate
Based on observations from 1 sample
Such interval estimates are called confidence
intervals
Stated in terms of level of confidence
Chap 8-5
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Estimation Process
Random Sample
Population
(mean, µ, is
unknown)
Mean
X = 50
Gives information about closeness to
unknown population parameters
Can never be 100% confident
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Chap 8-6
General Formula
I am 95%
confident that
µ is between
40 & 60.
The general formula for all
confidence intervals is:
Point Estimate ± (Critical Value)(Standard Error)
Sample
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Chap 8-8
Chapter 8
8-3
Confidence Level
Confidence Level, (1-α)
(continued)
Suppose confidence level = 95%
Also written (1 - α) = 0.95
A relative frequency interpretation:
Confidence Level
Confidence for which the interval
will contain the unknown
population parameter
In the long run, 95% of all the confidence
intervals that can be constructed will contain the
unknown true parameter
A percentage (less than 100%)
A specific interval either will contain or will
not contain the true parameter
No probability involved in a specific interval
Chap 8-9
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Confidence Interval for µ
(σ Known)
Confidence Intervals
Assumptions
Population standard deviation σ is known
Population is normally distributed
If population is not normal, use large sample
Confidence
Intervals
Population
Mean
Confidence interval estimate:
Population
Proportion
X±Z
σ Known
σ Unknown
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σ
n
where X is the point estimate
Z is the normal distribution critical value for a probability of α/2 in each tail
σ/ n is the standard error
Chap 8-11
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Chap 8-12
Chapter 8
8-4
Finding the Critical Value, Z
Common Levels of Confidence
Z = ± 1.96
Consider a 95% confidence interval:
1− α = 0.95
Commonly used confidence levels are 90%,
95%, and 99%
Confidence
Level
α
= 0.025
2
Z units:
α
= 0.025
2
Z= -1.96
X units:
Z= 1.96
0
Lower
Confidence
Limit
Point Estimate
Upper
Confidence
Limit
Chap 8-13
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80%
90%
95%
98%
99%
99.8%
99.9%
α/2
Intervals
extend from
σ
X+Z
n
1− α
x
x1
x2
Confidence Intervals
Basic Business Statistics, 10/e
1.28
1.645
1.96
2.33
2.58
3.08
3.27
Chap 8-14
Example
µx = µ
σ
n
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0.80
0.90
0.95
0.98
0.99
0.998
0.999
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
α/2
to
X−Z
Z value
1− α
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Intervals and Level of Confidence
Sampling Distribution of the Mean
Confidence
Coefficient,
(1-α)x100%
of intervals
constructed
contain µ;
Determine a 95% confidence interval for the
true mean resistance of the population.
(α)x100% do
not.
Chap 8-15
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Chap 8-16
Chapter 8
8-5
Example
Interpretation
(continued)
We are 95% confident that the true mean
resistance is between 1.9932 and 2.4068
ohms
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
Although the true mean may or may not be
in this interval, 95% of intervals formed in
this manner will contain the true mean
σ
X± Z
n
Solution:
= 2.20 ± 1.96 (0.35/ 11)
= 2.20 ± 0.2068
1.9932 ≤ µ ≤ 2.4068
Chap 8-17
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Confidence
Intervals
σ Known
If the population standard deviation σ is
unknown, we can substitute the sample
standard deviation, S
Population
Proportion
This introduces extra uncertainty, since
S is variable from sample to sample
So we use the t distribution instead of the
normal distribution
σ Unknown
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Chap 8-18
Confidence Interval for µ
(σ Unknown)
Confidence Intervals
Population
Mean
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Chap 8-19
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Chap 8-20
Chapter 8
8-6
Confidence Interval for µ
(σ Unknown)
Student’s t Distribution
(continued)
Assumptions
The t is a family of distributions
Population standard deviation is unknown
Population is normally distributed
If population is not normal, use large sample
The t value depends on degrees of
freedom (d.f.)
Number of observations that are free to vary after
sample mean has been calculated
Use Student’s t Distribution
Confidence Interval Estimate:
X ± t n-1
d.f. = n - 1
S
n
(where t is the critical value of the t distribution with n -1 degrees of
freedom and an area of α/2 in each tail)
Chap 8-21
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Degrees of Freedom (df)
Student’s t Distribution
Note: t
Idea: Number of observations that are free to vary
after sample mean has been calculated
Z as n increases
Standard
Normal
(t with df = ∞)
Example: Suppose the mean of 3 numbers is 8.0
Let X1 = 7
Let X2 = 8
What is X3?
Chap 8-22
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If the mean of these three
values is 8.0,
then X3 must be 9
(i.e., X3 is not free to vary)
t (df = 13)
t-distributions are bellshaped and symmetric, but
have ‘fatter’ tails than the
normal
t (df = 5)
Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2
(2 values can be any numbers, but the third is not free to vary
for a given mean)
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0
Chap 8-23
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t
Chap 8-24
Chapter 8
8-7
Student’s t Table
t distribution values
With comparison to the Z value
Upper Tail Area
df
.25
.10
Let: n = 3
df = n - 1 = 2
α = 0.10
α/2 = 0.05
.05
1 1.000 3.078 6.314
Confidence
t
Level
(10 d.f.)
2 0.817 1.886 2.920
α/2 = 0.05
3 0.765 1.638 2.353
The body of the table
contains t values, not
probabilities
0
Chap 8-25
1.325
1.310
1.28
0.90
1.812
1.725
1.697
1.645
0.95
2.228
2.086
2.042
1.96
0.99
3.169
2.845
2.750
2.58
Note: t
Z as n increases
Chap 8-26
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Confidence Intervals
A random sample of n = 25 has X = 50 and
S = 8. Form a 95% confidence interval for µ
Confidence
Intervals
tα/2 , n−1 = t 0.025,24 = 2.0639
Population
Mean
The confidence interval is
X ± t α/2, n-1
Z
____
1.372
Example
d.f. = n – 1 = 24, so
t
(30 d.f.)
0.80
2.920 t
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t
(20 d.f.)
Population
Proportion
S
8
= 50 ± (2.0639)
n
25
σ Known
σ Unknown
46.698 ≤ µ ≤ 53.302
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Chap 8-28
Chapter 8
8-8
Confidence Intervals for the
Population Proportion, π
Confidence Intervals for the
Population Proportion, π
(continued)
Recall that the distribution of the sample
proportion is approximately normal if the
sample size is large, with standard deviation
An interval estimate for the population
proportion ( π ) can be calculated by
adding an allowance for uncertainty to
the sample proportion ( p )
σp =
π (1− π )
n
We will estimate this with sample data:
p(1− p)
n
Chap 8-29
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Example
Confidence Interval Endpoints
Upper and lower confidence limits for the
population proportion are calculated with the
formula
p±Z
Chap 8-30
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A random sample of 100 people
shows that 25 are left-handed.
p(1− p)
n
Form a 95% confidence interval for
the true proportion of left-handers
where
Z is the standard normal value for the level of confidence desired
p is the sample proportion
n is the sample size
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Chap 8-32
Chapter 8
8-9
Example
Interpretation
(continued)
A random sample of 100 people shows
that 25 are left-handed. Form a 95%
confidence interval for the true proportion
of left-handers.
We are 95% confident that the true
percentage of left-handers in the population
is between
16.51% and 33.49%.
p ± Z p(1 − p)/n
Although the interval from 0.1651 to 0.3349
may or may not contain the true proportion,
95% of intervals formed from samples of
size 100 in this manner will contain the true
proportion.
= 25/100 ± 1.96 0.25(0.75) /100
= 0.25 ± 1.96 (0.0433)
0.1651 ≤ π ≤ 0.3349
Chap 8-33
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Determining Sample Size
Chap 8-34
Sampling Error
The required sample size can be found to reach
a desired margin of error (e) with a specified
level of confidence (1 - α)
Determining
Sample Size
For the
Mean
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The margin of error is also called sampling error
For the
Proportion
the amount of imprecision in the estimate of the
population parameter
the amount added and subtracted to the point
estimate to form the confidence interval
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Chap 8-36
Chapter 8
8-10
Determining Sample Size
Determining Sample Size
(continued)
Determining
Sample Size
For the
Mean
X±Z
σ
n
Determining
Sample Size
For the
Mean
Sampling error
(margin of error)
e=Z
σ
n
e=Z
Chap 8-37
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Determining Sample Size
σ
n
Now solve
for n to get
Z2 σ 2
n=
e2
Chap 8-38
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Required Sample Size Example
(continued)
To determine the required sample size for the
mean, you must know:
The desired level of confidence (1 - α), which
determines the critical Z value
If σ = 45, what sample size is needed to
estimate the mean within ± 5 with 90%
confidence?
n=
The acceptable sampling error, e
The standard deviation, σ
Z 2 σ 2 (1.645)2 (45)2
=
= 219.19
e2
52
So the required sample size is n = 220
(Always round up)
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Chap 8-40
Chapter 8
8-11
If σ is unknown
Determining Sample Size
(continued)
Determining
Sample Size
If unknown, σ can be estimated when
using the required sample size formula
Use a value for σ that is expected to be
at least as large as the true σ
For the
Proportion
Select a pilot sample and estimate σ with
the sample standard deviation, S
Chap 8-41
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Determining Sample Size
π (1− π )
e=Z
n
Now solve
for n to get
Z 2 π (1− π )
n=
e2
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Chap 8-42
Required Sample Size Example
(continued)
To determine the required sample size for the
proportion, you must know:
The desired level of confidence (1 - α), which
determines the critical Z value
How large a sample would be necessary
to estimate the true proportion defective in
a large population within ±3%, with 95%
confidence?
(Assume a pilot sample yields p = 0.12)
The acceptable sampling error, e
The true proportion of “successes”, π
Online Confidence intervals
Mean and CI online
π can be estimated with a pilot sample, if
necessary (or conservatively use π = 0.5)
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Chap 8-44
Chapter 8
8-12
Required Sample Size Example
Applications in Auditing
(continued)
Solution:
Six advantages of statistical sampling in
auditing
For 95% confidence, use Z = 1.96
e = 0.03
Sample result is objective and defensible
p = 0.12, so use this to estimate π
n=
Based on demonstrable statistical principles
Z 2 π (1− π ) (1.96)2 (0.12)(1− 0.12)
=
= 450.74
e2
(0.03)2
Provides sample size estimation in advance on an
objective basis
Provides an estimate of the sampling error
So use n = 451
Chap 8-45
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Applications in Auditing
(continued)
Can provide more accurate conclusions on the
population
Confidence Interval for
Population Total Amount
Point estimate:
Examination of the population can be time consuming and
subject to more nonsampling error
Samples can be combined and evaluated by different
auditors
Samples are based on scientific approach
Samples can be treated as if they have been done by a
single auditor
Population total = NX
Confidence interval estimate:
NX ± N ( t n−1 )
Objective evaluation of the results is possible
Based on known sampling error
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Chap 8-46
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S
n
N−n
N −1
(This is sampling without replacement, so use the finite population
correction in the confidence interval formula)
Chap 8-47
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Chap 8-48
Chapter 8
8-13
Confidence Interval for
Population Total: Example
Example Solution
N = 1000, n = 80,
A firm has a population of 1000 accounts and wishes
to estimate the total population value.
NX ± N ( t n−1 )
A sample of 80 accounts is selected with average
balance of $87.6 and standard deviation of $22.3.
S
n
X = 87.6, S = 22.3
N−n
N −1
= (1000 )(87.6) ± (1000 )(1.9905 )
Find the 95% confidence interval estimate of the total
balance.
22.3
80
1000 − 80
1000 − 1
= 87,600 ± 4,762.48
The 95% confidence interval for the population total
balance is $82,837.52 to $92,362.48
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Chap 8-49
Confidence Interval for
Total Difference
Chap 8-50
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Confidence Interval for
Total Difference
Point estimate:
Confidence interval estimate:
Total Difference = ND
ND ± N ( t n−1 )
Where the average difference, D, is:
n
∑D
D=
(continued)
SD
n
N−n
N −1
i
where
i=1
n
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2
i
where Di = audited value - original value
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n
∑ (D − D)
SD =
Chap 8-51
i=1
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n −1
Chap 8-52
Chapter 8
8-14
One-Sided Confidence Intervals
Application: find the upper bound for the
proportion of items that do not conform with
internal controls
Upper bound = p + Z
p(1− p) N − n
n
N −1
where
Z is the standard normal value for the level of confidence desired
p is the sample proportion of items that do not conform
n is the sample size
N is the population size
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Chap 8-53
Chapter Summary
Ethical Issues
A confidence interval estimate (reflecting
sampling error) should always be included
when reporting a point estimate
The level of confidence should always be
reported
The sample size should be reported
An interpretation of the confidence interval
estimate should also be provided
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Chap 8-54
Chapter Summary
(continued)
Introduced the concept of confidence intervals
Discussed point estimates
Developed confidence interval estimates
Created confidence interval estimates for the mean
(σ known)
Determined confidence interval estimates for the
mean (σ unknown)
Created confidence interval estimates for the
proportion
Determined required sample size for mean and
proportion settings
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Chap 8-55
Developed applications of confidence interval
estimation in auditing
Confidence interval estimation for population total
Confidence interval estimation for total difference
in the population
One-sided confidence intervals
Addressed confidence interval estimation and ethical
issues
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Chap 8-56