This document discusses an analysis of variance (ANOVA) study conducted by Burke Marketing Services to evaluate potential new versions of a children's dry cereal. The experimental design and ANOVA were used to test differences between the cereal versions and make a product recommendation. The document provides an introduction to ANOVA, including how it can test for differences between three or more population means. It also outlines the assumptions of ANOVA, how to calculate test statistics like mean squares, and how to conduct an F-test to determine whether population means are equal or not.

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統計學 Spring 2010
授課教師：統計系余清祥
日期：2010年3月16日
第四週：變異數分析
與實驗設計

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Chapter 13
Chapter 13
STATISTICS
STATISTICS in
in PRACTICE
PRACTICE
 Burke Marketing Services,
Inc., is one of the most
experienced market research
firms in the industry.
 In one study, a firm retained
Burke to evaluate potential new versions of a
children’s dry cereal.
 Analysis of variance was the statistical method used to
study the data obtained from the taste tests.
 The experimental design employed by Burke and the
subsequent analysis of variance were helpful in
making a product design recommendation.

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Chapter 13, Part A
Chapter 13, Part A
Analysis of Variance and Experimental Design
Analysis of Variance and Experimental Design
 Introduction to Analysis of Variance
 Analysis of Variance: Testing for the Equality
of k Population Means
 Multiple Comparison Procedures

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Introduction
Introduction to Analysis of Variance
to Analysis of Variance
Analysis of Variance (ANOVA) can be used to test
for the equality of three or more population means.
Analysis of Variance
Analysis of Variance (ANOVA) can be used to test
(ANOVA) can be used to test
for the equality of three or more population means.
for the equality of three or more population means.
Data obtained from observational or experimental
studies can be used for the analysis.
Data obtained from observational or experimental
Data obtained from observational or experimental
studies can be used for the analysis.
studies can be used for the analysis.
We want to use the sample results to test the
following hypotheses:
We want to use the sample results to test the
We want to use the sample results to test the
following hypotheses:
following hypotheses:
H
H0
0:
: 
1
1
=
=

2
2
=
=

3
3
=
=
. . .
. . . =
= 
k
k
H
Ha
a:
: Not all population means are equal
Not all population means are equal

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Introduction to Analysis of Variance
Introduction to Analysis of Variance
H
H0
0:
: 
1
1
=
=

2
2
=
=

3
3
=
=
. . .
. . . =
= 
k
k
H
Ha
a:
: Not all population means are equal
Not all population means are equal
If H0 is rejected, we cannot conclude that all population
means are different.
If
If H
H0
0 is rejected, we cannot conclude that all population
is rejected, we cannot conclude that all population
means are different.
means are different.
Rejecting H0 means that at least two population means
have different values.
Rejecting
Rejecting H
H0
0 means that at least two population means
means that at least two population means
have different values.
have different values.

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 Sampling Distribution of Given H0 is True
x
x
Introduction to Analysis of Variance
Introduction to Analysis of Variance

 1
x1
x 3
x3
x
2
x2
x
Sample means are close together
Sample means are close together
because there is only
because there is only
one sampling distribution
one sampling distribution
when
when H
H0
0 is true.
is true.
2
2
x
n

 
2
2
x
n

 

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Introduction to Analysis of Variance
Introduction to Analysis of Variance
 Sampling Distribution of Given H0 is False
x
x
3
3 1
x1
x 2
x2
x
3
x3
x 1
1 2
2
Sample means come from
Sample means come from
different sampling distributions
different sampling distributions
and are not as close together
and are not as close together
when
when H
H0
0 is false.
is false.

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For each population, the response variable is
normally distributed.
For each population, the response variable is
For each population, the response variable is
normally distributed.
normally distributed.
Assumptions for Analysis of Variance
Assumptions for Analysis of Variance
The variance of the response variable, denoted  2,
is the same for all of the populations.
The variance of the response variable, denoted
The variance of the response variable, denoted 
 2
2
,
,
is the same for all of the populations
is the same for all of the populations.
.
The observations must be independent.
The observations must be independent.
The observations must be independent.

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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 Between-Treatments Estimate of Population
Variance
 Within-Treatments Estimate of Population
Variance
 Comparing the Variance Estimates: The F Test
 ANOVA Table

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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 Analysis of variance can be used to test for the
equality of k population means.
 The hypotheses tested is
H0:
Ha: Not all population means are equal
where mean of the jth population.
k


 

 
2
1
j


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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 Sample data
= value of observation i for treatment j
= number of observations for treatment j
= sample mean for treatment j
= sample variance for treatment j
= sample standard deviation for treatment j
ij
x
j
n
j
x
2
j
s
j
s

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 Statisitcs
 The sample mean for treatment j
 The sample variance for treatment j
Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
j
n
i
ij
j
n
x
x
j


 1
1
)
(
1
2
2





j
n
i
j
ij
j
n
x
x
s
j

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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 The overall sample mean
where nT = n1 + n2 +. . . + nk
 If the size of each sample is n, nT = kn then
T
k
j
n
i
ij
n
x
x
j

 

1 1
kn
x
k
n
x
kn
x
x
k
j
ij
k
j
n
i
ij
k
j
n
i
ij
j
j


 
 
 



1
1 1
1 1
/

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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 Between-Treatments Estimate of Population
Variance
 The sum of squares due to treatments (SSTR)
 The mean square due to treatments (MSTR)
1
)
(
1
2





k
x
x
n
MSTR
k
j
j
j



k
j
j
j x
x
n
1
2
)
(

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Analysis of Variance:
Analysis of Variance:
Testing for the Equality of
Testing for the Equality of k
k Population Means
Population Means
 Within-Treatments Estimate of Population Variance
 The sum of squares due to error (SSE)
 The mean square due to error (MSE)



k
j
j
j s
n
1
2
)
1
(
k
n
s
n
MSE
T
k
j
j
j




1
2
)
1
(

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Between
Between-
-Treatments Estimate
Treatments Estimate
of Population Variance
of Population Variance
 A between-treatment estimate of  2 is called the
mean square treatment and is denoted MSTR.
2
1
( )
MSTR
1
k
j j
j
n x x
k




 2
1
( )
MSTR
1
k
j j
j
n x x
k





Denominator represents
Denominator represents
the
the degrees of freedom
degrees of freedom
associated with SSTR
associated with SSTR
Numerator is the
Numerator is the
sum of squares
sum of squares
due to treatments
due to treatments
and is denoted SSTR
and is denoted SSTR

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 The estimate of  2 based on the variation of the
sample observations within each sample is called
the mean square error and is denoted by MSE.
Within
Within-
-Samples Estimate
Samples Estimate
of Population Variance
of Population Variance
k
n
s
n
T
k
j
j
j




1
2
)
1
(
MSE
k
n
s
n
T
k
j
j
j




1
2
)
1
(
MSE
Denominator represents
Denominator represents
the
the degrees of freedom
degrees of freedom
associated with SSE
associated with SSE
Numerator is the
Numerator is the
sum of squares
sum of squares
due to error
due to error
and is denoted SSE
and is denoted SSE

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Comparing the Variance Estimates:
Comparing the Variance Estimates: The
The F
F Test
Test
 If the null hypothesis is true and the ANOVA
assumptions are valid, the sampling distribution of
MSTR/MSE is an F distribution with MSTR d.f.
equal to k - 1 and MSE d.f. equal to nT - k.
 If the means of the k populations are not equal, the
value of MSTR/MSE will be inflated because MSTR
overestimates  2.
 Hence, we will reject H0 if the resulting value of
MSTR/MSE appears to be too large to have been
selected at random from the appropriate F
distribution.

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
F
F = MSTR/MSE
= MSTR/MSE
H
H0
0:
: 
1
1
=
=

2
2
=
=

3
3
=
=
. . .
. . . =
= 
k
k

H
Ha
a: Not all population means are equal
: Not all population means are equal
 Hypotheses
 Test Statistic

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 Rejection Rule
where the value of F is based on an
F distribution with k  1 numerator d.f.
and nT  k denominator d.f.
Reject
Reject H
H0
0 if
if p
p-
-value
value <
< 

p-value Approach:
Critical Value Approach: Reject
Reject H
H0
0 if
if F
F >
> F
F


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Sampling Distribution of MSTR/MSE
Sampling Distribution of MSTR/MSE
 Rejection Region
Do Not Reject H0
Do Not Reject H0
Reject H0
Reject H0
MSTR/MSE
MSTR/MSE
Critical Value
Critical Value
F
F
Sampling Distribution
Sampling Distribution
of MSTR/MSE
of MSTR/MSE



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ANOVA Table
ANOVA Table
SST is partitioned
SST is partitioned
into SSTR and SSE.
into SSTR and SSE.
SST
SST’
’s degrees of freedom
s degrees of freedom
(d.f.) are partitioned into
(d.f.) are partitioned into
SSTR
SSTR’
’s
s d.f. and
d.f. and SSE
SSE’
’s
s d.f.
d.f.
Treatment
Treatment
Error
Error
Total
Total
SSTR
SSTR
SSE
SSE
SST
SST
k
k –
– 1
1
n
nT
T –
– k
k
n
nT
T -
- 1
1
MSTR
MSTR
MSE
MSE
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares
MSTR/MSE
MSTR/MSE
F
F

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ANOVA Table
ANOVA Table
SST divided by its degrees of freedom nT – 1 is the
overall sample variance that would be obtained if we
treated the entire set of observations as one data set.
SST divided by its degrees of freedom
SST divided by its degrees of freedom n
nT
T –
– 1
1 is the
is the
overall sample variance that would be obtained if we
overall sample variance that would be obtained if we
treated the entire set of observations as one data set.
treated the entire set of observations as one data set.
With the entire data set as one sample, the formula
for computing the total sum of squares, SST, is:
With the entire data set as one sample, the formula
With the entire data set as one sample, the formula
for computing the total sum of squares, SST, is:
for computing the total sum of squares, SST, is:
2
1 1
SST ( ) SSTR SSE
j
n
k
ij
j i
x x
 
   
 2
1 1
SST ( ) SSTR SSE
j
n
k
ij
j i
x x
 
   


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ANOVA Table
ANOVA Table
ANOVA can be viewed as the process of partitioning
the total sum of squares and the degrees of freedom
into their corresponding sources: treatments and error.
ANOVA can be viewed as the process of partitioning
ANOVA can be viewed as the process of partitioning
the total sum of squares and the degrees of freedom
the total sum of squares and the degrees of freedom
into their corresponding sources: treatments and error.
into their corresponding sources: treatments and error.
Dividing the sum of squares by the appropriate degrees
of freedom provides the variance estimates and the F
value used to test the hypothesis of equal population
means.
Dividing the sum of squares by the appropriate degrees
Dividing the sum of squares by the appropriate degrees
of freedom provides the variance estimates and the
of freedom provides the variance estimates and the F
F
value used to test the hypothesis of equal population
value used to test the hypothesis of equal population
means.
means.

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 Example:
 National Computer Products, Inc. (NCP),
manufactures printers and fax machines at plants
located in Atlanta, Dallas, and Seattle.
 Object: To measure how much employees at these
plants know about total quality management.
 A random sample of six employees was selected from
each plant and given a quality awareness examination.

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 Data
 Let
= mean examination score for population 1
= mean examination score for population 2
= mean examination score for population 3
1

2

3


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Test
Test for the Equality of
for the Equality of k
k Population Means
Population Means
 Hypotheses
H0: = =
Ha: Not all population means are equal
 In this example
1. dependent or response variable : examination score
2. independent variable or factor : plant location
3. levels of the factor or treatments : the values of a
factor selected for investigation, in the NCP
example the three treatments or three population
are Atlanta, Dallas, and Seattle.
1
 2
 3


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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
Three assumptions
1. For each population, the response variable is normally
distributed. The examination scores (response variable)
must be normally distributed at each plant.
2. The variance of the response variable, , is the same for
all of the populations. The variance of examination scores
must be the same for all three plants.
3. The observations must be independent. The examination
score for each employee must be independent of the
examination score for any other employee.
2


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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 ANOVA Table
 p-value = 0.003 < α = .05. We reject H0.

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 Example: Reed Manufacturing
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
Janet Reed would like to know if
there is any significant difference in
the mean number of hours worked per
week for the department managers
at her three manufacturing plants
(in Buffalo, Pittsburgh, and Detroit).

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 Example: Reed Manufacturing
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
A simple random sample of five
managers from each of the three
plants was taken and the number of
hours worked by each manager for
the previous week is shown on the
next slide.
Conduct an F test using α = .05.

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1
1
2
2
3
3
4
4
5
5
48
48
54
54
57
57
54
54
62
62
73
73
63
63
66
66
64
64
74
74
51
51
63
63
61
61
54
54
56
56
Plant
Plant 1
1
Buffalo
Buffalo
Plant 2
Plant 2
Pittsburgh
Pittsburgh
Plant
Plant 3
3
Detroit
Detroit
Observation
Observation
Sample
Sample Mean
Mean
Sample Variance
Sample Variance
55
55 68
68 57
57
26.0
26.0 26.5
26.5 24.5
24.5
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
H0:  1= 2= 3
Ha: Not all the means are equal
where:
  1 = mean number of hours worked per
 week by the managers at Plant 1
 2 = mean number of hours worked per
week by the managers at Plant 2
  3 = mean number of hours worked per
week by the managers at Plant 3
1. Develop the hypotheses.
1. Develop the hypotheses.
 p -Value and Critical Value Approaches

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2. Specify the level of significance.
2. Specify the level of significance.  = .05
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 p -Value and Critical Value Approaches
3. Compute the value of the test statistic.
3. Compute the value of the test statistic.
MSTR = 490/(3 - 1) = 245
SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490
(Sample sizes are all equal.)
Mean Square Due to Treatments
= (55 + 68 + 57)/3 = 60
x

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3. Compute the value of the test statistic.
3. Compute the value of the test statistic.
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
MSE = 308/(15 - 3) = 25.667
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
Mean Square Due to Error
(continued)
F = MSTR/MSE = 245/25.667 = 9.55
 p -Value and Critical Value Approaches

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Treatment
Treatment
Error
Error
Total
Total
490
490
308
308
798
798
2
2
12
12
14
14
245
245
25.667
25.667
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares
9.55
9.55
F
F
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 ANOVA Table

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Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
5. Determine whether to reject
5. Determine whether to reject H
H0
0.
.
We have sufficient evidence to conclude that the
mean number of hours worked per week by
department managers is not the same at all 3
plant.
The p-value < .05, so we reject H0.
With 2 numerator d.f. and 12 denominator
d.f.,the p-value is .01 for F = 6.93. Therefore, the
p-value is less than .01 for F = 9.55.
 p - Value Approach
4. Compute the
4. Compute the p
p –
–value.
value.

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Slide
Slide
Slide
5. Determine whether to reject
5. Determine whether to reject H
H0
0.
.
Because F = 9.55 > 3.89, we reject H0.
 Critical Value Approach
4. Determine the critical value and rejection rule.
4. Determine the critical value and rejection rule.
Reject H0 if F > 3.89
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
We have sufficient evidence to conclude that the
mean number of hours worked per week by
department managers is not the same at all 3
plant.
Based on an F distribution with 2 numerator
d.f. and 12 denominator d.f., F.05 = 3.89.

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Slide
Slide
Slide
Test for the Equality of
Test for the Equality of k
k Population Means
Population Means
 Summary

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Slide
Slide
Multiple Comparison Procedures
Multiple Comparison Procedures
 Suppose that analysis of variance has provided
statistical evidence to reject the null hypothesis of
equal population means.
 Fisher’s least significant difference (LSD)
procedure can be used to determine where the
differences occur.

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Slide
Slide
Slide
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
 Test Statistic
1 1
MSE( )
i j
i j
x x
t
n n



1 1
MSE( )
i j
i j
x x
t
n n



 Hypotheses
j
i
a
j
i
H
H






:
:
0

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Slide
Slide
Slide
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
where the value of ta/2 is based on a
t distribution with nT  k degrees of freedom.
 Rejection Rule
Reject
Reject H
H0
0 if
if p
p-
-value
value <
< a
a
p-value Approach:
Critical Value Approach:
Reject
Reject H
H0
0 if
if t
t <
< 
t
ta
a/2
/2 or
or t
t >
> t
ta
a/2
/2

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Slide
Slide
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i -
- x
xj
j
 Test Statistic
/2
1 1
LSD MSE( )
i j
t n n

 
/2
1 1
LSD MSE( )
i j
t n n

 
where

i j
x x

i j
x x
Reject
Reject H
H0
0 if > LSD
if > LSD

i j
x x

i j
x x
 Hypotheses
 Rejection Rule
j
i
a
j
i
H
H






:
:
0

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Slide
Slide
Slide
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i 
 x
xj
j
 Example: Reed Manufacturing
Recall that Janet Reed wants to know
if there is any significant difference in
the mean number of hours worked per
week for the department managers
at her three manufacturing plants.
Analysis of variance has provided
statistical evidence to reject the null
hypothesis of equal population means.
Fisher’s least significant difference (LSD) procedure
can be used to determine where the differences
occur.

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Slide
Slide
For  = .05 and nT  k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
98
.
6
)
5
1
5
1
(
667
.
25
179
.
2
LSD 


/2
1 1
LSD MSE( )
i j
t n n

 
/2
1 1
LSD MSE( )
i j
t n n

 
MSE value was
MSE value was
computed earlier
computed earlier
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i 
 x
xj
j

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Slide
Slide
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i -
- x
xj
j
 LSD for Plants 1 and 2
• Conclusion
• Test Statistic

1 2
x x

1 2
x x = |55 - 68| = 13
Reject H0 if > 6.98

1 2
x x

1 2
x x
• Rejection Rule
• Hypotheses (A)
The mean number of hours worked at Plant 1 is
not equal to the mean number worked at Plant 2.
2
1
2
1
0
:
:






a
H
H

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Slide
Slide
Slide
 LSD for Plants 1 and 3
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i -
- x
xj
j
• Conclusion
• Test Statistic

1 3
x x

1 3
x x = |55  57| = 2
Reject H0 if > 6.98

1 3
x x

1 3
x x
• Rejection Rule
• Hypotheses (B)
There is no significant difference between the mean
number of hours worked at Plant 1 and the mean
number of hours worked at Plant 3.
3
1
3
1
0
:
:






a
H
H

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Slide
Slide
Slide
 LSD for Plants 2 and 3
Fisher
Fisher’
’s LSD Procedure
s LSD Procedure
Based on the Test Statistic
Based on the Test Statistic x
xi
i -
- x
xj
j
• Conclusion
• Test Statistic

2 3
x x

2 3
x x = |68 - 57| = 11
Reject H0 if > 6.98

2 3
x x

2 3
x x
• Rejection Rule
• Hypotheses (C)
The mean number of hours worked at Plant 2 is
not equal to the mean number worked at Plant 3.
3
2
3
2
0
:
:






a
H
H

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Slide
 The experimentwise Type I error rate gets larger
for problems with more populations (larger k).
Type I Error Rates
Type I Error Rates

EW
EW = 1
= 1 –
– (1
(1 –
– 
)
)(
(k
k –
– 1)!
1)!
 The comparisonwise Type I error rate  indicates
the level of significance associated with a single
pairwise comparison.
 The experimentwise Type I error rate EW is the
probability of making a Type I error on at least
one of the (k – 1)! pairwise comparisons.

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Slide
An Introduction to Experimental Design
An Introduction to Experimental Design
 A factor is a variable that the experimenter has
selected for investigation.
 A treatment is a level of a factor.
 Experimental units are the objects of interest in the
experiment.
 A completely randomized design is an
experimental design in which the treatments are
randomly assigned to the experimental units.
 If the experimental units are heterogeneous,
blocking can be used to form homogeneous groups,
resulting in a randomized block design.

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Slide
Slide
The between-samples estimate of 2 is referred
to as the mean square due to treatments (MSTR).
2
1
( )
MSTR
1
k
j j
j
n x x
k




 2
1
( )
MSTR
1
k
j j
j
n x x
k





 Between-Treatments Estimate of Population
Variance
Completely Randomized Design
Completely Randomized Design
denominator is the
denominator is the
degrees of freedom
degrees of freedom
associated with SSTR
associated with SSTR
numerator is called
numerator is called
the
the sum of squares due
sum of squares due
to treatments
to treatments (SSTR)
(SSTR)

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Slide
Slide
The second estimate of 2, the within-samples
estimate, is referred to as the mean square due to
error (MSE).
 Within-Treatments Estimate of Population
Variance
Completely Randomized Design
Completely Randomized Design
denominator is the
denominator is the
degrees of freedom
degrees of freedom
associated with SSE
associated with SSE
numerator is called
numerator is called
the
the sum of squares
sum of squares
due to error
due to error (SSE)
(SSE)
MSE 




( )
n s
n k
j j
j
k
T
1 2
1
MSE 




( )
n s
n k
j j
j
k
T
1 2
1

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Slide
Slide
MSTR
SSTR
-

k 1
MSTR
SSTR
-

k 1
MSE
SSE
-

n k
T
MSE
SSE
-

n k
T
MSTR
MSE
MSTR
MSE
 ANOVA Table
Completely Randomized Design
Completely Randomized Design
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Treatments
Treatments
Error
Error
Total
Total
k
k 
 1
1
n
nT
T 
 1
1
SSTR
SSTR
SSE
SSE
SST
SST
n
nT
T 
 k
k

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Slide
Slide
AutoShine, Inc. is considering marketing a long-
lasting car wax. Three different waxes (Type 1, Type 2,
and Type 3) have been developed.
Completely Randomized Design
Completely Randomized Design
 Example: AutoShine, Inc.
In order to test the durability
of these waxes, 5 new cars were
waxed with Type 1, 5 with Type
2, and 5 with Type 3. Each car was then
repeatedly run through an automatic carwash until the
wax coating showed signs of deterioration.

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Slide
Slide
Completely Randomized Design
Completely Randomized Design
The number of times each car went through the
carwash is shown on the next slide. AutoShine,
Inc. must decide which wax to market. Are the
three waxes equally effective?
 Example: AutoShine, Inc.

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Slide
Slide
Slide
1
1
2
2
3
3
4
4
5
5
27
27
30
30
29
29
28
28
31
31
33
33
28
28
31
31
30
30
30
30
29
29
28
28
30
30
32
32
31
31
Sample Mean
Sample Mean
Sample Variance
Sample Variance
Observation
Observation
Wax
Wax
Type 1
Type 1
Wax
Wax
Type 2
Type 2
Wax
Wax
Type 3
Type 3
2.5
2.5 3.3
3.3 2.5
2.5
29.0
29.0 30.4
30.4 30.0
30.0
Completely Randomized Design
Completely Randomized Design

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Slide
Slide
 Hypotheses
Completely Randomized Design
Completely Randomized Design
where:
1 = mean number of washes for Type 1 wax
2 = mean number of washes for Type 2 wax
3 = mean number of washes for Type 3 wax
H0: 1=2=3
Ha: Not all the means are equal

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Slide
Slide
Because the sample sizes are all equal:
Completely Randomized Design
Completely Randomized Design
MSE = 33.2/(15  3) = 2.77
MSTR = 5.2/(3  1) = 2.6
SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2
SSTR = 5(29–29.8)2 + 5(30.4–29.8)2 + 5(30–29.8)2 = 5.2
 Mean Square Error
 Mean Square Between Treatments
  
1 2 3
( )/3
x x x x
  
1 2 3
( )/3
x x x x = (29 + 30.4 + 30)/3 = 29.8

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Slide
Slide
Slide
 Rejection Rule
Completely Randomized Design
Completely Randomized Design
where F.05 = 3.89 is based on an F distribution
with 2 numerator degrees of freedom and 12
denominator degrees of freedom
p-Value Approach: Reject H0 if p-value < .05
Critical Value Approach: Reject H0 if F > 3.89

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Slide
Slide
 Test Statistic
Completely Randomized Design
Completely Randomized Design
There is insufficient evidence to conclude that
the mean number of washes for the three wax
types are not all the same.
 Conclusion
F = MSTR/MSE = 2.6/2.77 = .939
The p-value is greater than .10, where F = 2.81.
(Excel provides a p-value of .42.)
Therefore, we cannot reject H0.

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Slide
Slide
Slide
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Treatments
Treatments
Error
Error
Total
Total
2
2
14
14
5.2
5.2
33.2
33.2
38.4
38.4
12
12
Completely Randomized Design
Completely Randomized Design
2.60
2.60
2.77
2.77
.939
.939
 ANOVA Table

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Slide
Slide
Slide
• For a randomized block design the sum of
squares total (SST) is partitioned into three
groups: sum of squares due to treatments, sum
of squares due to blocks, and sum of squares due
to error.
 ANOVA Procedure
Randomized Block Design
Randomized Block Design
SST = SSTR + SSBL + SSE
SST = SSTR + SSBL + SSE
• The total degrees of freedom, nT - 1, are partitioned
such that k - 1 degrees of freedom go to treatments,
b - 1 go to blocks, and (k - 1)(b - 1) go to the error
term.

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Slide
Slide
Slide
MSTR
SSTR
-

k 1
MSTR
SSTR
-

k 1
MSTR
MSE
MSTR
MSE
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Treatments
Treatments
Error
Error
Total
Total
k
k 
 1
1
n
nT
T 
1
1
SSTR
SSTR
SSE
SSE
SST
SST
Randomized Block Design
Randomized Block Design
 ANOVA Table
Blocks
Blocks SSBL
SSBL b
b 
 1
1
(
(k
k –
– 1)(
1)(b
b –
– 1)
1)

SSBL
MSBL
-1
b

SSBL
MSBL
-1
b
MSE
SSE

 
( )( )
k b
1 1
MSE
SSE

 
( )( )
k b
1 1

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Slide
Randomized Block Design
Randomized Block Design
 Example: Crescent Oil Co.
Crescent Oil has developed three
new blends of gasoline and must
decide which blend or blends to
produce and distribute. A study
of the miles per gallon ratings of the
three blends is being conducted to determine if the
mean ratings are the same for the three blends.

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Slide
Slide
Randomized Block Design
Randomized Block Design
 Example: Crescent Oil Co.
Five automobiles have been
tested using each of the three
gasoline blends and the miles
per gallon ratings are shown on
the next slide.

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Slide
Slide
Slide
Randomized Block Design
Randomized Block Design
29.8
29.8 28.8
28.8 28.4
28.4
Treatment
Treatment
Means
Means
1
1
2
2
3
3
4
4
5
5
31
31
30
30
29
29
33
33
26
26
30
30
29
29
29
29
31
31
25
25
30
30
29
29
28
28
29
29
26
26
30.333
30.333
29.333
29.333
28.667
28.667
31.000
31.000
25.667
25.667
Type of Gasoline (Treatment)
Type of Gasoline (Treatment)
Block
Block
Means
Means
Blend X
Blend X Blend Y
Blend Y Blend Z
Blend Z
Automobile
Automobile
(Block)
(Block)

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Slide
Slide
 Mean Square Due to Error
Randomized Block Design
Randomized Block Design
MSE = 5.47/[(3  1)(5  1)] = .68
SSE = 62  5.2  51.33 = 5.47
MSBL = 51.33/(5  1) = 12.8
SSBL = 3[(30.333  29)2 + . . . + (25.667  29)2] = 51.33
MSTR = 5.2/(3  1) = 2.6
SSTR = 5[(29.8  29)2 + (28.8  29)2 + (28.4  29)2] = 5.2
The overall sample mean is 29. Thus,
 Mean Square Due to Treatments
 Mean Square Due to Blocks

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Slide
Slide
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Treatments
Treatments
Error
Error
Total
Total
2
2
14
14
5.20
5.20
5.47
5.47
62.00
62.00
8
8
2.60
2.60
.68
.68
3.82
3.82
 ANOVA Table
Randomized Block Design
Randomized Block Design
Blocks
Blocks 51.33
51.33 12.80
12.80
4
4

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Slide
Slide
 Rejection Rule
Randomized Block Design
Randomized Block Design
For  = .05, F.05 = 4.46
(2 d.f. numerator and 8 d.f. denominator)
p-Value Approach: Reject H0 if p-value < .05
Critical Value Approach: Reject H0 if F > 4.46

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Slide
Slide
 Conclusion
Randomized Block Design
Randomized Block Design
There is insufficient evidence to conclude that
the miles per gallon ratings differ for the three
gasoline blends.
The p-value is greater than .05 (where F =
4.46) and less than .10 (where F = 3.11). (Excel
provides a p-value of .07). Therefore, we cannot
reject H0.
F = MSTR/MSE = 2.6/.68 = 3.82
 Test Statistic

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Slide
Slide
Factorial Experiments
Factorial Experiments
 In some experiments we want to draw conclusions
about more than one variable or factor.
 Factorial experiments and their corresponding
ANOVA computations are valuable designs when
simultaneous conclusions about two or more
factors are required.
 For example, for a levels of factor A and b levels of
factor B, the experiment will involve collecting data
on ab treatment combinations.
 The term factorial is used because the
experimental conditions include all possible
combinations of the factors.

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Slide
Slide
• The ANOVA procedure for the two-factor
factorial experiment is similar to the completely
randomized experiment and the randomized block
experiment.
 ANOVA Procedure
SST = SSA + SSB + SSAB + SSE
SST = SSA + SSB + SSAB + SSE
• The total degrees of freedom, nT  1, are partitioned
such that (a  1) d.f go to Factor A, (b  1) d.f go to
Factor B, (a  1)(b  1) d.f. go to Interaction, and
ab(r  1) go to Error.
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
• We again partition the sum of squares total (SST)
into its sources.

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Slide
Slide

SSA
MSA
-1
a

SSA
MSA
-1
a
MSA
MSE
MSA
MSE
Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Factor A
Factor A
Error
Error
Total
Total
a
a -
- 1
1
n
nT
T -
- 1
1
SSA
SSA
SSE
SSE
SST
SST
Factor B
Factor B SSB
SSB b
b -
- 1
1
ab
ab(
(r
r –
– 1)
1) 

SSE
MSE
( 1)
ab r


SSE
MSE
( 1)
ab r
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment

SSB
MSB
-1
b

SSB
MSB
-1
b
Interaction
Interaction SSAB
SSAB (
(a
a –
– 1)(
1)(b
b –
– 1)
1) 
 
SSAB
MSAB
( 1)( 1)
a b

 
SSAB
MSAB
( 1)( 1)
a b
MSB
MSE
MSB
MSE
MSAB
MSE
MSAB
MSE

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 Step 3 Compute the sum of squares for factor B
2
1 1 1
SST = ( )
a b r
ijk
i j k
x x
  

 2
1 1 1
SST = ( )
a b r
ijk
i j k
x x
  


2
1
SSA = ( . )
a
i
i
br x x


 2
1
SSA = ( . )
a
i
i
br x x



2
1
SSB = ( . )
b
j
j
ar x x


 2
1
SSB = ( . )
b
j
j
ar x x



Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
 Step 1 Compute the total sum of squares
 Step 2 Compute the sum of squares for factor A

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 Step 4 Compute the sum of squares for interaction
2
1 1
SSAB = ( . . )
a b
ij i j
i j
r x x x x
 
  
 2
1 1
SSAB = ( . . )
a b
ij i j
i j
r x x x x
 
  

Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
SSE = SST
SSE = SST –
– SSA
SSA –
– SSB
SSB -
- SSAB
SSAB
 Step 5 Compute the sum of squares due to error

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A survey was conducted of hourly wages
for a sample of workers in two industries
at three locations in Ohio. Part of the
purpose of the survey was to
determine if differences exist
in both industry type and
location. The sample data are shown
on the next slide.
 Example: State of Ohio Wage Survey
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment

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 Example: State of Ohio Wage Survey
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
Industry Cincinnati Cleveland Columbus
I 12.10 11.80 12.90
I 11.80 11.20 12.70
I 12.10 12.00 12.20
II 12.40 12.60 13.00
II 12.50 12.00 12.10
II 12.00 12.50 12.70

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 Factors
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
• Each experimental condition is repeated 3 times
• Factor B: Location (3 levels)
• Factor A: Industry Type (2 levels)
 Replications

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Source of
Source of
Variation
Variation
Sum of
Sum of
Squares
Squares
Degrees of
Degrees of
Freedom
Freedom
Mean
Mean
Squares
Squares F
F
Factor A
Factor A
Error
Error
Total
Total
1
1
17
17
.50
.50
1.43
1.43
3.42
3.42
12
12
.50
.50
.12
.12
4.19
4.19
 ANOVA Table
Factor B
Factor B 1.12
1.12 .56
.56
2
2
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
Interaction
Interaction .37
.37 .19
.19
2
2
4.69
4.69
1.55
1.55

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 Conclusions Using the p-Value Approach
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
(p-values were found using Excel)
Interaction is not significant.
•Interaction: p-value = .25 >  = .05
Mean wages differ by location.
•Locations: p-value = .03 <  = .05
Mean wages do not differ by industry type.
•Industries: p-value = .06 >  = .05

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 Conclusions Using the Critical Value
Approach
Two
Two-
-Factor Factorial Experiment
Factor Factorial Experiment
Interaction is not significant.
•Interaction: F = 1.55 < F = 3.89
Mean wages differ by location.
•Locations: F = 4.69 > F = 3.89
Mean wages do not differ by industry type.
•Industries: F = 4.19 < F = 4.75

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End of Chapter 13
End of Chapter 13