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Copyright © 2004 Pearson Education, Inc.
Chapter 11
Analysis of Variance
11-1 Overview
11-2 One-Way ANOVA
11-3 Two-Way ANOVA
Copyright © 2004 Pearson Education, Inc.
Slide 2
Slide 3
Section 11-1 & 11-2
Overview and One-Way
ANOVA
Created by Erin Hodgess, Houston, Texas
Copyright © 2004 Pearson Education, Inc.
Overview
Slide 4
Analysis of variance(ANOVA) is a
method for testing the hypothesis
that three or more population means
are equal.
For example:
H0: µ 1 = µ2 = µ3 = . . . µk
H1: At least one mean is different
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ANOVA methods
require the F-distribution
Slide 5
1. The F-distribution is not symmetric; it
is skewed to the right.
2. The values of F can be 0 or positive,
they cannot be negative.
3. There is a different F-distribution for
each pair of degrees of freedom for the
numerator and denominator.
Critical values of F are given in Table A-5
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F - distribution
Figure 11-1
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Slide 6
One-Way ANOVA
Slide 7
An Approach to Understanding ANOVA
1. Understand that a small P-value (such as
0.05 or less) leads to the rejection of the null
hypothesis of equal means. With a large Pvalue (such as greater than 0.05), fail to reject
the null hypothesis of equal means.
2. Develop an understanding of the underlying
rationale by studying the example in this
section.
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Slide 8
3. Become acquainted with the nature
of the SS (sum of squares) and MS
(mean square) values and their role
in determining the F test statistic,
but use statistical software packages
or a calculator for finding those
values.
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Definition
Slide 9
Treatment (or factor)
A treatment(or factor) is a property or
characteristic that allows us to
distinguish the different populations
from another.
Use computer software or TI-83 Plus for
ANOVA calculations if possible
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One-Way ANOVA
Slide 10
Assumptions
1. The populations have approximately normal
distributions.
2. The populations have the same variance 
(or standard deviation  ).
2
3. The samples are simple random samples.
4. The samples are independent of each other.
5. The different samples are from populations
that are categorized in only one way.
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Procedure for testing:
Slide 11
H0: µ1 = µ2 = µ3 = . . .
1. Use STATDISK, Minitab, Excel, or a TI83
Calulator to obtain results.
2. Identify the P-value from the display.
3. Form a conclusion based on these
criteria:
 If P-value  , reject the null hypothesis of
equal means.
 If P-value >  , fail to reject the null
hypothesis of equal means.
Copyright © 2004 Pearson Education, Inc.
Example:
Readability Scores
Slide 12
Given the readability scores summarized in Table 11-1 and
a significance level of  = 0.05, use STATDISK, Minitab,
Excel, or a TI-83 PLUS calculator to test the claim that the
three samples come from populations with means that are
not all the same.
H0: 1 = 2 = 3
H1: At least one of the means is different from the others.
Please refer to the displays on the next slide.
The displays all show a P-value of 0.000562 or 0.001.
Because the P-value is less than the significance level of
 = 0.05, we reject the null hypothesis of equal means.
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Example:
Readability Scores
Copyright © 2004 Pearson Education, Inc.
Slide 13
Example:
Readability Scores (cont’d)
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Slide 14
Example:
Readability Scores
Slide 15
Given the readability scores summarized in Table 11-1 and
a significance level of  = 0.05, use STATDISK, Minitab,
Excel, or a TI-83 PLUS calculator to test the claim that the
three samples come from populations with means that are
not all the same.
There is sufficient evidence to support the claim that the
three population means are not all the same. We
conclude that those books have readability levels that are
not all the same.
Copyright © 2004 Pearson Education, Inc.
ANOVA
Fundamental Concept
Slide 16
Estimate the common value of  2 using
1. The variance between samples (also
called variation due to treatment) is an
estimate of the common population variance  2
that is based on the variability among the sample
means.
2. The variance within samples (also called
variation due to error) is an estimate of the
common population variance  2 based on the
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sample variances.
Relationships Among
Components of ANOVA
Figure 11-2
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Slide 17
ANOVA
Fundamental Concept
Slide 18
Test Statistic for One-Way ANOVA
F=
variance between samples
variance within samples
A excessively large F test statistic is
evidence against equal population means.
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Calculations with
Equal Sample Sizes
Slide 19
Variance between samples = nsx
2
2
where sx = variance of samples means
Variance within samples = sp
2
where sp2 = pooled variance (or the mean
of the sample variances)
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Example:
Sample Calculations
Use Table 11-2 to
calculate the
variance between
samples, variance
within samples,
and the F test
statistic.
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Slide 20
Example:
Sample Calculations
Slide 21
Use Table 11-2 to calculate the variance between samples,
variance within samples, and the F test statistic.
s2x = (5.5)2 + (6.0)2 + (6.0)2 – [(5.5+6.0+6.0)2/3]
3–1
s2x = 102.25 – 102.08333 = 0.0833
2
ns2x = 4(0.0833) = 0.3332
Copyright © 2004 Pearson Education, Inc.
Example:
Sample Calculations
Slide 22
Use Table 11-2 to calculate the variance between samples,
variance within samples, and the F test statistic.
s2p = 3.0 + 2.0 + 2.0 = 2.333
3
F= ns2x =
s 2p
0.3332 = 0.1428
2.3333
Copyright © 2004 Pearson Education, Inc.
Critical Value of F
Slide 23
Right-tailed test
Degree of freedom with k samples of the
same size n
numerator df = k – 1
denominator df = k(n – 1)
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Calculations with
Unequal Sample Sizes
Slide 24
ni(xi – x)2
F=
variance within samples
variance between samples
=
k –1
(ni – 1)s2i
(ni – 1)
where x = mean of all sample scores combined
k = number of population means being compared
ni = number of values in the ith sample
xi = mean values in the ith sample
2
si = variance of values in the ith sample
Copyright © 2004 Pearson Education, Inc.
Key Components of
ANOVA Method
Slide 25
SS(total), or total sum of squares, is a
measure of the total variation (around x) in
all the sample data combined.
SS(total) = (x – x)
Formula 11-1
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2
Key Components
of ANOVA Method
Slide 26
SS(treatment) is a measure of the variation between the
samples. In one-way ANOVA, SS(treatment) is sometimes
referred to as SS(factor). Because it is a measure of
variability between the sample means, it is also referred to
as SS (between groups) or SS (between samples).
SS(treatment) = n1(x1 – x)2 + n2(x2 – x)2 + . . . nk(xk – x)2
Formula 11-2
= ni(xi - x)2
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Key Components
of ANOVA Method
SS(error) is a sum of squares representing the
variability that is assumed to be common to all the
populations being considered.
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Slide 27
Key Components
of ANOVA Method
Slide 28
SS(error) is a sum of squares representing the
variability that is assumed to be common to all the
populations being considered.
2
3
SS(error) = (n1 –1)s + (n2 –1)s + (n3 –1)s . . . nk(xk –1)s2i
2
1
2
2
= (ni – 1)si2
Formula 11-3
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Key Components
of ANOVA Method
Slide 29
SS(total) = SS(treatment) + SS(error)
Formula 11-4
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Mean Squares (MS)
Slide 30
Sum of Squares SS(treatment) and SS(error)
divided by corresponding number of degrees
of freedom.
MS (treatment) is mean square for treatment,
obtained as follows:
MS (treatment) =
SS (treatment)
k–1
Formula 11-5
Copyright © 2004 Pearson Education, Inc.
Mean Squares (MS)
Slide 31
MS (error) is mean square for error, obtained
as follows:
MS (error) =
SS (error)
N–k
Formula 11-6
MS (total) =
SS (total)
N–1
Formula 11-7
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Test Statistic for ANOVA
with Unequal Sample Sizes
Slide 32
F=
MS (treatment)
MS (error)
Formula 11-8
df = k – 1
 Denominator df = N – k
 Numerator
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Example:
Readability Scores
Copyright © 2004 Pearson Education, Inc.
Slide 33
Slide 34
Section 11-3
Two-Way ANOVA
Created by Erin Hodgess, Houston, Texas
Copyright © 2004 Pearson Education, Inc.
Two-Way
Analysis of Variance
Slide 35
 Two-Way ANOVA involves two
factors.
 The data are partitioned into
subcategories called cells.
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Example: New York
Marathon Runners
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Slide 36
Definition
Slide 37
There is an interaction between two
factors if the effect of one of the
factors changes for different
categories of the other factor.
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Example: New York
Marathon Runners
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Slide 38
Assumptions
Slide 39
1. For each cell, the sample values come
from a population with a distribution that
is approximately normal.
2. The populations have the same variance 2.
3. The samples are simple random samples.
4. The samples are independent of each other.
5. The sample values are categorized
two ways.
6. All of the cells have the same number
of sample values.
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Slide 40
Two-Way ANOVA calculations are quite
involved, so we will assume that a
software package is being used.
Minitab and Excel can do two-way ANOVA, but
STATDISK and the TI-83 Plus calculator cannot.
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Slide 41
Figure 11-3
Procedure
for
Two-Way
ANOVA
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Example: New York
Marathon Runners
Slide 42
Given the Minitab printout from the New York Marathon
data, work through the procedure for two-way ANOVA.
Check for Interaction Effects:
H0: There are no interaction effects.
H1: There are interaction effects.
F = MS(interaction) = 10,521,304 = 1.17
MS(error)
9,028,477
The P-value is shown as 0.329, so we fail to reject the
null hypothesis of no interaction between the two
factors.
Copyright © 2004 Pearson Education, Inc.
Example: New York
Marathon Runners
Slide 43
Given the Minitab printout from the New York Marathon
data, work through the procedure for two-way ANOVA.
Check for Row Effects:
H0: There are no effects from the row factors.
H1: There are row effects.
F = MS(gender) = 15,225,413 = 1.69
MS(error) 9,028,477
The P-value is shown as 0.206, so we fail to reject the
null hypothesis of no effects from gender.
Copyright © 2004 Pearson Education, Inc.
Example: New York
Marathon Runners
Slide 44
Given the Minitab printout from the New York Marathon
data, work through the procedure for two-way ANOVA.
Check for Column Effects:
H0: There are no effects from the column factors.
H1: There are column effects.
F = MS(age) =
MS(error)
46,043,490 = 5.10
9,028,477
The P-value is shown as 0.014. We have significance at
the 0.05 level, but not the 0.01 level. We reject the null
hypothesis of no effects from age.
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Special Case: One Observation
Slide 45
Per Cell and No Interaction
There is no variation within individual cells and
those samples cannot be calculated.
If it seems reasonable to assume (based on
knowledge about the circumstances) that there
is no interaction between the two factors, make
the assumption and then proceed as before to
test the following two hypotheses separately:
H0 : There are no effects from the row factor.
H0 : There are no effects from the column factor
Copyright © 2004 Pearson Education, Inc.