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Chapter
13
Comparing Three
or More Means
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Section
13.1
Comparing Three
or More Means
(One-Way
Analysis of
Variance)
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Analysis of Variance (ANOVA) is an
inferential method used to test the equality of
three or more population means.
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CAUTION!
Do not test H0: μ1 = μ2 = μ3 by conducting
three separate hypothesis tests, because the
probability of making a Type I error will be
much higher than α.
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Requirements of a One-Way ANOVA Test
1. There must be k simple random samples; one
from each of k populations or a randomized
experiment with k treatments.
2. The k samples are independent of each other;
that is, the subjects in one group cannot be
related in any way to subjects in a second group.
3. The populations are normally distributed.
4. The populations must have the same variance;
that is, each treatment group has the population
variance σ2.
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Testing a Hypothesis Regarding k = 3
H0: μ1 = μ2 = μ3
H1: At least one population mean is different from
the others
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Testing Using ANOVA
The methods of one-way ANOVA are robust, so
small departures from the normality requirement
will not significantly affect the results of the
procedure. In addition, the requirement of equal
population variances does not need to be strictly
adhered to, especially if the sample size for each
treatment group is the same. Therefore, it is
worthwhile to design an experiment in which the
samples from the populations are roughly equal
in size.
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Verifying the Requirement of Equal
Population Variance
The one-way ANOVA procedures may be used
if the largest sample standard deviation is no
more than twice the smallest sample standard
deviation.
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Parallel Example 1: Verifying the Requirements of ANOVA
The following data represent the weight (in grams)
of pennies minted at the Denver mint in 1990,1995,
and 2000. Verify that the requirements in order to
perform a one-way ANOVA are satisfied.
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1990
2.50
2.50
2.49
2.53
2.46
2.50
2.47
2.53
2.51
2.49
2.48
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1995
2.52
2.54
2.50
2.48
2.52
2.50
2.49
2.53
2.48
2.55
2.49
2000
2.50
2.48
2.49
2.50
2.48
2.52
2.51
2.49
2.51
2.50
2.52
Copyright © 2017, 2013, and 2010 Pearson Education, Inc.
Solution
1. The 3 samples are simple random samples.
2. The samples were obtained independently.
3. Normal probability plots for the 3 years
follow. All of the plots are roughly linear so
the normality assumption is satisfied.
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Solution
4. The sample standard deviations are
computed for each sample using Minitab and
shown on the following slide. The largest
standard deviation is not more than twice the
smallest standard deviation
(2•0.0141 = 0.0282 > 0.02430)
so the requirement of equal population
variances is considered satisfied.
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Descriptive Statistics
Variable
1990
1995
2000
N
11
11
11
Variable
1990
1995
2000
Minimum
2.4600
2.4800
2.4800
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Mean
2.4964
2.5091
2.5000
Median
2.5000
2.5000
2.5000
Maximum
2.5300
2.5500
2.5200
TrMean
2.4967
2.5078
2.5000
Q1
2.4800
2.4900
2.4900
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StDev
0.0220
0.0243
0.0141
Q3
2.5100
2.5300
2.5100
SE Mean
0.0066
0.0073
0.0043
The basic idea in one-way ANOVA is to
determine if the sample data could come from
populations with the same mean, μ, or suggests
that at least one sample comes from a population
whose mean is different from the others.
To make this decision, we compare the
variability among the sample means to the
variability within each sample.
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We call the variability among the sample means
the between-sample variability, and the
variability of each sample the within-sample
variability.
If the between-sample variability is large relative
to the within-sample variability, we have
evidence to suggest that the samples come from
populations with different means.
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ANOVA F-Test Statistic
The analysis of variance F-test statistic is given by
between - sample variability
F0 =
within - sample variability
mean square due to treatments
=
mean square due to error
MST
=
MSE
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Computing the F-Test Statistic
Step 1: Compute the sample mean of the combined data
set by adding up all the observations and
dividing by the number of observations. Call
this value x .
Step 2: Find the sample mean for each sample (or
treatment). Let x1 represent the sample mean of
sample 1, x2 represent the sample mean of
sample 2, and so on.
Step 3: Find the sample variance for each sample (or
treatment). Let s12 represent the sample
variance for sample 1, s22 represent the sample
variance for sample 2, and so on.
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Computing the F-Test Statistic
Step 4: Compute the sum of squares due to treatments,
SST, and the sum of squares due to error, SSE.
Step 5: Divide each sum of squares by its
corresponding degrees of freedom (k – 1 and
n – k, respectively) to obtain the mean squares
MST and MSE.
Step 6: Compute the F-test statistic:
mean square due to treatments MST
F0 

mean square due to error
MSE
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Solution
The results of the computations for the data that led to
the F-test statistic are presented in the following table.
ANOVA Table:
Source of Sum of Degrees of
Variation Squares Freedom
Treatment
Mean
Squares
0.0009
2
0.0005
Error
0.013
30
0.0004
Total
0.0139
32
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F-Test
Statistic
1.25
Decision Rule in the One-Way ANOVA Test
If the P-value is less than the level of
significance, α, reject the null hypothesis.
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Example
Compute the F-test statistic for the following data.
𝒙𝟏
𝒙𝟐
𝒙𝟑
4
7
10
5
8
10
6
9
11
6
7
11
4
9
13
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Example
Suppose the teeth researcher wans to determine if
there is a difference in the mean shear bond
strength among the four treatment groups at
the 𝛼 = 0.05 level of significance.
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Example
The data in the table represent the number of pods
on a random sample of soybean plants for
various plot types. An agricultural researcher
wants to determine if the mean numbers of
pods for each plot type are equal. Use the
𝛼 = 0.05 level of significance.
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Plot Type
Pods
Liberty
32
31
36
35
41
34
39
37
38
No till
34
30
31
27
40
33
37
42
39
Chisel
plowed
34
37
24
23
32
33
27
34
30
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