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9-1
Estimation and Confidence Intervals
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9-2
When you have completed this chapter, you will be able to:
Define a point estimator, a point estimate, and desirable
properties of a point estimator such as
unbiasedness, efficiency, and consistency.
Define an interval estimator and an interval estimate
Define a confidence interval, confidence level, margin of
error, and a confidence interval estimate
Construct a confidence interval for the population mean
when the population standard deviation is known
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9-3
Construct a confidence interval for the population variance
when the population is normally distributed
Construct a confidence interval for the population mean
when the population is normally distributed and
the population standard deviation is unknown
Construct a confidence interval for a population proportion
Determine the sample size for attribute and
variable sampling
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Terminology
9-4
Point Estimate …is a single value (statistic) used to
estimate a population value (parameter)
Interval Estimate …states the range within which a
population parameter probably lies
Confidence Interval …is a range of values within which
the population parameter
is expected to occur
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9-5
Desirable properties of a point estimator
• efficient
… possible values are concentrated
close to the value of the parameter
• consistent
…values are distributed evenly on
both sides of the value of the parameter
• unbiased
…unbiased when the expected value equals the value
of the population parameter being estimated.
Otherwise, it is biased!
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Terminology
Standard error of the sample mean
…is the standard deviation
of the sampling distribution of the sample means
s
x
s
x
It is computed by
s = s
n
…is the symbol for the standard error of the
sample mean
…is the standard deviation of the population
n …is the size of the sample
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9-6
Standard Error
of the Means
If s is not known and n > 30,
the standard deviation of the sample(s)
is used to approximate the population
standard deviation
s = s
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x
Computed by…
n
9-7
9-8
…that determine the width of a confidence interval
are:
1. The sample size, n
2. The variability in the population,
usually estimated by s
3. The desired level of confidence
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Constructing
Confidence Intervals
9-9
IN GENERAL,
A confidence interval for a mean is computed
by:
 zα/2
s
n
x
Interpreting…
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Interpreting
Confidence Intervals
9 - 10
The Globe
Suppose that you read that
“…the average selling price
of a family home in
York Region is
$200
000 +/- $15000
at
95% confidence!”
This means…what?
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Interpreting
Confidence Intervals
The Globe
“…the average
selling price of a
family home in
York Region is
$200 000 +/$15 000 at 95%
confidence!”
9 - 11
In statistical terms, this means:
…that we are 95% sure that the
interval estimate obtained
contains the value of the
population mean.
Lower confidence limit is
$185 000 ($200 000 - $15 000)
Upper confidence limit is
$215 000 ($200 000 + $15 000)
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Also…
Interpreting
Confidence Intervals
The Globe
“…the mean
time to sell a
family home
in York
Region
is 40 days.
9 - 12
Your newspaper also reports
that…
You select a random
sample of 36 homes sold
during the past year,
and determine a
90% confidence interval
estimate
for the population mean
to be (31-39) days.
Do your sample results support the paper’s claim?
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Interpreting
Confidence Intervals
You select a
random sample
of 36 homes
sold during
the past year,
and determine a
90% confidence
interval estimate
for the population
mean to be
(31-39) days.
9 - 13
Lower confidence limit
is 31 days
Upper confidence limit
is 39 days
Our evidence does not support the
statement made by the newspaper,
i.e., the population mean is not 40 days,
when using a 90% interval estimate
There is a 10% chance (100%-90%)
that the interval estimate
does not contain the value
of the population mean!
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Interpreting
Confidence Intervals
9 - 14
90% Confidence Interval
… 10% chance of falling outside this interval
…or, focus on
tail areas …
i.e.  = 0.10
.05
.05
90%
31
39
 is the probability of a value falling
outside the confidence interval
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 15
Find the appropriate value of z:
P( X 
zs
X
zs
n
1
2
n
= .92
Locate Area on
the normal curve
Look up a= 0.46 in Table
to get the corresponding
z-score
This is a 92%
confidence interval
0.92
-1.75 0 1.75
Search in the centre of the
table for the area of 0.46
Z = +/- 1.75
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Constructing
Confidence Intervals
x
Common
Confidence
Intervals
 zα/2
s
n
s
95% C.I. for
the mean:
X  1.96
99% C.I. for
the mean:
s
X  2 .58
n
n
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 16
Also,
About95%
95%ofofthe
sample
means for
the constructed
a intervals
specifiedwill
sample
size
will lie
contain
thewithin
1.96 standard
parameter
being
deviations
estimated. of the
hypothesized
population mean.
Interval Estimates
9 - 17
If the population
standard deviation is
known or n > 30
If the population
standard deviation is
unknown and n<30
Use the z table…
Use the t-table…
n
x
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
x
 zα/2
s
 t α/2
s
n
More
on this
later…
9 - 18
The Dean of the Business School wants to
estimate the mean number of hours
worked per week by students.
A sample of 49 students
showed a mean of 24 hours
with a standard deviation of 4 hours.
What is the population mean?
Our best estimate is 24 hours.
This is a point estimate.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Find the 95 percent confidence
interval for the population mean.
9 - 19
Commonly denoted as 1-
 zα/2
s
Mean = 24 SD = 4 N = 49
n
Z = +/- 1.96
x
95% Confidence
Substitute
values:
24
+ 1.96
4
49
= 24 +/- 1.12
The Confidence Limits range from 22.88
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
to 25.12
Interval Estimates
90% confidence level
1- = 0.9
or  = 0.10
99% confidence level
1- = 0.99
or  = 0.010
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 20
Student’s t-distribution
….used for small sample
9 - 21
sizes
Characteristics
…like z, the t-distribution is continuous
…takes values between –4 and +4
…it is bell-shaped and symmetric about zero
…it is more spread out and flatter at the centre
than the z-distribution
…for larger and larger values of degrees of
freedom, the t-distribution becomes closer
and closer to the standard normal distribution
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Student’s t-distribution
Chart 9-1
Comparison of The Standard Normal Distribution
and the Student’s t Distribution
The t distribution should be flatter and more spread out
than the z distribution
t distribution
Z distribution
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9 - 22
Student’s t-distribution
9 - 23
…with df = 9 and 0.10 area in the upper tail…
t = 1.383
0.10
t
T -table
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Student’s t-distribution
df
1
2
3
4
5
6
7
8
9
10
11
9
Confidence Intervals
80%
90%
95%
98%
99%
Level of Significance for One-Tailed Test
0.100
0.050 0.025
0.010
0.005
0.10
Level of Significance for Two-Tailed Test
0.20
0.10
0.05
0.02
0.01
3.078
1.886
1.638
1.533
1.476
1.44
1.415
1.397
1.383
1.372
1.363
1.383
6.314
2.92
2.353
2.132
2.015
1.943
1.895
1.86
1.833
1.812
1.796
12.706
4.303
3.182
2.766
2.571
2.447
2.365
2.306
2.262
2.228
2.201
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.25
3.169
3.106
9 - 24
When?
9 - 25
…to use the z Distribution or the t Distribution
Population Normal?
NO
YES
n 30 or more?
NO
Use a
nonparametric
test
(see Ch16)
YES
Population standard
deviation known?
NO
Use the t
Use the z
distribution distribution
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
YES
Use the z
distribution
Student’s t-distribution
The Dean of the Business School wants to
estimate the mean number of hours worked
per week by students.
A sample of only 12 students showed a mean of
24 hours with a standard deviation of 4 hours.
Find the 95 percent confidence interval
for the population mean.
n is small
so use the t - Distribution
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 26
…sample of only 12 students
…a mean of 24 hours
Data
…a standard deviation of 4 hours
= 24 n = 12 s = 4
x
Formula
 tα/2
df = 12-1 = 11
X
Looking up 5% level of significance for
a two-tailed test with 11df, we find…
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 27
s
n
 = 1 – 95% = .05
Student’s t-distribution
df
1
2
3
4
5
6
7
8
9
10
11
11
Confidence Intervals
80%
90%
95%
98%
99%
Level of Significance for One-Tailed Test
0.100
0.050 0.025
0.010
0.005
Level of Significance for Two-Tailed Test
0.20
0.10
0.05
0.02
0.01
0.05
3.078
1.886
1.638
1.533
1.476
1.44
1.415
1.397
1.383
1.372
1.363
6.314
2.92
2.353
2.132
2.015
1.943
1.895
1.86
1.833
1.812
1.796
12.706
4.303
3.182
2.766
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.201
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.25
3.169
3.106
9 - 28
…sample of only 12 students
…a mean of 24 hours
Data
…a standard deviation of 4 hours
= 24 n = 12 s = 4
x
Formula
 tα/2
df = 12-1 = 11
9 - 29
s
n
 = 1 – 95% = .05
X
Looking up 5% level of significance for
t
= 2.201
0
025
.
a two-tailed test with 11df, we find…
4
24  2.201 12 = 24 +/- 2.54
The confidence limits range from 21.46 to 26.54
Compare these with earlier limits of 22.88 to 25.12
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Student’s t-distribution
The manager of the college cafeteria wants to
estimate the mean amount spent per customer
per purchase. A sample of 10 customers
revealed the following amounts spent:
$4.45 $4.05
$4.95 $3.25 $4.68
$5.75 $6.01
$3.99 $5.25 $2.95
Determine the 99% confidence interval
for the mean amount spent.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 30
Student’s t-distribution
$4.45 $4.05
$5.75 $6.01
Step 1
$4.95 $3.25 $4.68
$3.99 $5.25 $2.95
Determine the sample mean and standard deviation.
X
Step 2
9 - 31
= $4.53
s = $1.00
Enter the key data into the appropriate formula.
n = 10
x
Formula
 = 1-99% = .01
1.00
= 4.53  3.25 10
= $4.53 +/- $1.03
df = 10 – 1 = 9
 tα/2
s
n
We are 99% confident that the mean amount spent
per customer is between $3.50 and $5.56
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Constructing Confidence Intervals
for Population Proportions
A confidence interval for a population
proportion is estimated by:
Formula
pz
p (1  p )
n
p …is the symbol for the sample proportion
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 32
Constructing Confidence Intervals
for Population Proportions
A sample of 500 executives who own their
own home revealed 175 planned to sell their
homes and retire to Victoria.
Develop a 98% confidence interval for the
proportion of executives that plan to sell
and move to Victoria.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 33
Constructing Confidence Intervals
for Population Proportions
A sample of 500
executives who own
their own home
revealed 175
planned to sell their
homes and retire to
Victoria.
Develop a 98%
confidence interval
for the proportion
of executives…
Formula
p  z /2
ˆ
9 - 34
p(1  p)
n
n = 500 p = 175/500 = .35 z = 2.33
. 35  2 . 33
98% CL =
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
. 35 ( 1  . 35 )
500
. 35  . 0497
9 - 35
Finite-Population
Correction Factor
Used when n/N is 0.05 or more
Formula
sx =
s
n
N -n
N - 1
Correction
Factor
The attendance at the college hockey game last night
was 2700. A random sample of 250 of those in
attendance revealed that the average number of
drinks consumed per person was 1.8
with a standard deviation of 0.40.
Develop a 90% confidence interval estimate for the mean
number of drinks consumed per person.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 36
Finite-Population
Correction Factor
X
Formula
 Zα/2
s
N -n
N - 1
n
The attendance at the
college hockey game
last night was 2700.
N = 2700
n = 250 x = 1.8
A sample of 250 of
s = 0.40 /2 = 0.05
those in attendance
revealed that the
average number of
drinks consumed per Since 250/2700 >.05, use the correction factor
person was 1.8 with a
standard deviation of
.4
2700  250
0.40.
1.8  1.645 (
)(
)
Develop a 90%
2700 1
250
confidence interval
estimate.…
90% CL = 1 . 8  0 . 04
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9 - 37
Selecting the
Sample Size
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Factors
…that determine the sample size
are:
1. The degree of confidence selected
2. The maximum allowable error
3. The variation in the population
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9 - 38
Selecting the
Sample Size
Formula
9 - 39
 zα/2 s  2

n =
 E 
E … is the allowable error
Z …is the z-score for the chosen level of confidence
S …is the sample deviation of the pilot survey
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Selecting the
Sample Size
A consumer group would like to estimate the
mean monthly electricity charge for a single
family house in July (within $5) using a
99 percent level of confidence.
Based on similar studies the
standard deviation is estimated to be $20.00.
How large a sample is required?
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
9 - 40
Selecting the
Sample Size
Formula
 zα/2 s  2
 E 



A consumer group would
like to estimate the mean
monthly electricity charge
for a single family house
in July (within $5) using
a 99 percent level of
confidence.
Based on similar studies
the standard deviation is
estimated to be $20.00.
9 - 41
 2.58  20 2
= 
 5.00
2
=
(10.32)
=
106.5
90% CL = A minimum of
107 homes must be sampled.
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Selecting the
Sample Size
9 - 42
The Kennel Club wants to estimate the proportion of
children that have a dog as a pet.
Assume a 95% level of confidence and that the club
estimates that 30% of the children have a dog as a pet.
If the club wants the estimate to be
within 3% of the population proportion,
how many children would they
need to contact?
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Selecting the
Sample Size
New
The Kennel Club
wants to estimate the
proportion of children
that have a dog as a
pet.
Assume a 95% level
of confidence and
that the club estimates
that 30% of the
children have
a dog as a pet.
Formula
9 - 43
Z 2
n = p ( 1  p )  
E
 1 . 96  2
= . 3 (1  . 3 ) 

 . 03 
= (. 21 )( 65 . 33 ) 2
n = 896.4
A minimum of 897 children
must be sampled.
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9 - 44
9 - 45
This completes Chapter 9
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