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Chapter 8
Confidence Intervals
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.
Confidence Interval
• A confidence interval for the population
mean is an interval constructed around the
sample mean so that the interval will contain
the population mean with certain probability.
• The probability that the interval contains the
population parameter is called a confidence
level.
• The probability for the interval not to contain
the population mean is 1- confidence level,
denoted by a.
• a=1- confidence level
• confidence level=1-a
8-2
Example 8.1: The Car Mileage Case
Suppose an automaker conducts mileage
tests on a sample of 50 of its new midsize cars and obtains the sample mean
with x =31.56. Assuming population
standard deviation σ=0.8. Please
compute the 95 percent confidence
interval of the population mean.
8-3
Confidence Interval vs a
• The probability that the confidence interval
will contain the population mean, the
confidence level , is denoted by 1 - a
• If a =0.05, what is the confidence level?
• 1-0.05=0.95
• If the confidence level is 99%, a =0.01?
8-4
z-Based Confidence Intervals for a Population
Mean: Population σ is Known
– If a population is normally distributed with
mean m and standard deviation σ, then the
sampling distribution of x is normal with
mean mx = m and standard deviation  x =  n
– x is approximately normally distributed if
population mean m standard deviation σ
both exist and n≥30.
8-5
z-Based Confidence Intervals for a Population
Mean: Population σ is Known
• If a population has standard deviation 
(known),
• and if the population is normal or if
sample size is large (n  30), then …
• … a (1-a) confidence interval for m is
x  za 2


 
=  x - za 2
, x  za 2

n 
n
n

8-6
t and Right Hand Tail Areas
The definition of the
critical value Zα
The area to
the right if 1-α
Zα
Zα is the percentile such that
P(Z< Zα) =1-α
8-7
Find out the critical values Zα/2
Confidence level
0.90
0.95
0.99
0.96
0.94
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
=
=
=
=
=
Z0.05=
Z0.025=
Z0.005 = Z0.02 =
α
α/2
P(Z<Zα/2 )
=1-α/2
Zα/2
Z0.03 =
8-8
Find out the critical values Zα/2
Confidence level
α
α/2
P(Z<Zα/2 )
=1-α/2
Zα/2
0.90
0.1
0.95
0.05
0.96
0.04
0.94
0.06
0.05
0.025 0.005 0.02
0.03
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
P(Z<Zα/2 )
=0.95
=0.975
=0.995
=0.98
=0.97
Z0.05=
Z0.025=
Z0.005 = Z0.02 =
Z0.03 =
2.575
1.88
1.645 1.96
0.99
0.01
2.05
8-9
1- a Confidence Interval
• If the population is normal or if n≥30, and
population standard deviation is know, the
1- a confidence interval for population mean
is
 

 x  za 2

n


– The normal point za/2 gives a right hand tail area
under the standard normal curve equal to a/2
8-10
Example 8.1: The Car Mileage Case
Suppose an automaker conducts mileage
tests on a sample of 50 of its new midsize cars and obtains the sample mean
with x =31.56. Assuming population
standard deviation σ=0.8. Please
compute the 95 percent confidence
interval of the population mean.
x = 31.56;σ = 0.8;n = 50
x  z
/
0.025
n

 

=  x  1.96

n



 

=  x - 1.96
, x  1.96

n
n


= 31.34,31.78
8-11
99% Confidence Interval
• A bank manager developed a new
system to reduce the service time.
Suppose the new service time has a
normal distribution with known standard
deviation 2.47 minutes. The mean of a
sample of 10 randomly selected
customers is 5.46. Please construct an
99% confidence interval of the
population mean.
8-12
99% Confidence Interval
• For a 99% confidence level, 1 – a =
0.99, so a = 0.01, and a/2 = 0.005
– Reading between table entries
z0.005 = 2.575
• The 99% confidence interval is
  

 

, x  2.575
 x  2.575
 =  x - 2.575

n
n
n

 

=
8-13
99% Confidence Interval
x = 5.46;σ = 2.47;n = 10
 

x  z0.005 x  =  x  2.575 
n


 

=  x - 2.575
, x  2.575
n
n 

= 3.45,7.47
8-14
Notes on the Example
• The confidence interval can be expression as
x - E, x  E

margin of error E = za 2 n
• The length of confidence interval is equal to
2E.
• The 99% confidence interval is slightly wider
than the 95% confidence interval
– The higher the confidence level, the bigger the
critical value Zα/2, the wider the interval.
8-15
The Effect of a on Confidence Interval
Width
za/2 = z0.025 = 1.96
za/2 = z0.005 = 2.575
8-16
t-Based Confidence Intervals for a
Mean:  Unknown
• If  is unknown (which is usually the
case), we can construct a confidence
interval for m based on the sampling
distribution of
t=
x -m
s
n
• If the population is normal, then for any
sample size n, this sampling distribution
is called the t distribution
8-17
t-Based Confidence Intervals for a
Mean:  Unknown
• If the sampled population is normally
distributed with mean m, then a (1a)100%
confidence interval for m is
x  ta
s
2
n
• The result applies if sample size is ≥30
• ta/2 is the t point giving a right-hand tail area
of a/2 under the t curve having n1 degrees of
freedom
8-18
The t Distribution
• The curve of the t distribution is similar to
that of the standard normal curve
– Symmetrical and bell-shaped
– The t distribution is more spread out than
the standard normal distribution
– The spread of the t is given by the number
of degrees of freedom
• Denoted by df
• For a sample of size n, there are one fewer
degrees of freedom, that is,
df = n – 1
8-19
Degrees of Freedom and the
t-Distribution
As the number of degrees of freedom increases, the spread
of the t distribution decreases and the t curve approaches
the standard normal curve
8-20
The t Distribution and Degrees of
Freedom
• As the sample size n increases, the
degrees of freedom also increases
• As the degrees of freedom increase, the
spread of the t curve decreases
• As the degrees of freedom increases
indefinitely, the t curve approaches the
standard normal curve
– If n ≥ 30, so df = n – 1 ≥ 29, the t curve is
very similar to the standard normal curve
8-21
t and Right Hand Tail Areas
• Use a t point denoted by ta
– ta is the point on the horizontal axis under
the t curve that gives a right hand tail equal
to a
– So the value of ta in a particular situation
depends on the right hand tail area a and
the number of degrees of freedom
• df = n – 1
• a = 1 – a , where 1 – a is the specified
confidence coefficient
8-22
Definition of the critical value tα
8-23
Using the t Distribution
• Example: Find ta for a sample of size
n=15 and right hand tail area of 0.025
– For n = 15, df = 14, t0.025=2.145
– a = 0.025
• Note that a = 0.025 corresponds to a
confidence level of 0.95
– In Table 8.3, along row labeled 14 and
under column labeled 0.025, read a table
entry of 2.145
– So ta = 2.145
8-24
Using the t Distribution
Continued
8-25
Find out the critical values tα/2
Confidence level
Df=10,
CL=0.9
Df=20,
CL=0.95
Df=25,
CL=0.99
Df=30,
CL=0.98
Df=40,
CL=0.99
α
α/2
0.1
0.05
0.01
0.02
0.01
0.05
0.025
0.005
0.01
0.005
P(t>tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
=α/2
=
=
=
=
=
tα/2
t
0.05
=1.812
and degree of
freedom
8-26
Find out the critical values Zα/2
Confidence level
Df=10,
CL=0.9
Df=20,
CL=0.95
Df=25,
CL=0.99
Df=30,
CL=0.98
Df=40,
CL=0.99
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
P(t<tα/2 )
=?
=
=
=
=
=
tα/2
t 0.05 =
t 0.025 =
t 0.005 =
t 0.01 =
t 0.005 =
1.812
2.086
2.787
2.457
2.704
and degree of
freedom
α
α/2
8-27
• Suppose that, in order to reduce risk, a large bank
has decided to initiate a policy limiting the mean
debt-to-equity ratio for its portfolio of commercial
loans to 1.5. In order to estimate the mean debt-toequity ratio of its load portfolio, the bank randomly
selects a sample of 15 of its commercial loan
accounts. The sample mean and standard deviation
of the sample is 1.3433 and 0.1921. The population
has normal distribution. Please construct a 95%
confidence interval for the mean of the debt-to-equity
ratio.
8-28
Example 8.4 Debt-to-Equity Ratios
• Estimate the mean debt-to-equity ratio of the
loan portfolio of a bank
• Select a random sample of 15 commercial
loan accounts
– x = 1.3433
– s = 0.1921
– n = 15
• Want a 95% confidence interval for the ratio
• Assume all ratios are normally distributed but
σ unknown
8-29
Example 8.4 Debt-to-Equity Ratios
Continued
• Have to use the t distribution
• At 95% confidence, 1 – a = 0.95 so a = 0.05
and a/2 = 0.025
• For n = 15, df = 15 – 1 = 14
• Use the t table to find ta/2 for df = 14, ta/2 =
t0.025 = 2.145
• The 95% confidence interval:
x  t0.025
s
0.1921
= 1.3433  2.145
n
15
= 1.3433  0.1064 = 1.2369,1.4497
8-30
Length of Confidence Interval
• The confidence interval can be expression as
x - E, x  E

margin of error equal to E = za 2 n
or E = ta 2 s
n
• The length of confidence interval is equal to
2E.
8-31
Sample Size Determination (z)
If σ is known, then a sample of size of
at least
2
 za 2
n = 
E




will result in a confidence interval such
so that x is within E units of m with
probability 100(1-a)% .
8-32
Sample Size Determination (t)
If σ is unknown and is estimated from s,
then a sample of size of at least
 ta 2 s 

n = 
 E 
2
will give an interval so that x is within E
units of m, with 100(1-a)% confidence.
The number of degrees of freedom for
the ta/2 point is the size of the
preliminary sample minus 1.
8-33
• A bank manager developed a new system to reduce
the service time. Suppose the new service time has a
normal distribution with known standard deviation
2.47 minutes. How large the sample should be if the
manager wants to be 99% confident that sample
mean is within 0.5 minute of mu (the population
mean) .
•
 za 2
n = 
 E
2
  2.575  2.47 
 = 
 = 161.8
0.5

 
2
round up to 162
8-34
• A new alert system is installed for air traffic controllers. It is
hoped that the mean “alert time” for the new equipment is less
than 8 seconds. In order to test the equipment 15 randomly
selected air traffic controllers are trained to use the machine
and their alert times for a simulated collision course are
recorded. The sample alert times has a mean of 7.4 seconds
and s=1.026. Supposed the alert times are normally distributed,
please construct 95% confidence interval for the mean alert
time of the machine. Can we be 95% confident that mu is less
than 8 seconds? Determine the sample size needed to make
us 95% confident that the sample mean is within a margin of
error of 0.3 second of mu.
• t0.025 =2.145, for degree of freedom 14; rockville
2
 ta 2 s   2.145 1.026 
 = 
n = 
 = 53.8  54
0.3

 E  
2
8-35
Selecting an Appropriate Confidence
Interval for a Population Mean
8-36