Chapter 12
ONE-WAY ANALYSIS OF VARIANCE
We wish to compare the mean responses for several populations where the
levels of a single explanatory variable define the populations.
Example: One-Way ANOVA Logic
The two figures provide frequency plots for data obtained by taking
independent random samples of size 10 from three populations. The three
populations each had a normal distribution (normality is one of the
assumptions of ANOVA) and the population means were 60, 65, and 70,
respectively. So the population means are indeed not all equal. In Scenario I
(Figure 11.2): the population standard deviations were all equal to 1.5. In
Scenario II (Figure 11.3) the population standard deviations were all equal to 3.
Another assumption for ANOVA is that the populations have equality standard
deviations.
a) Just looking at the
frequency plots, which of
the two scenarios do you
think would provide more
evidence that at least one
of the population means is
different from the others?
b) The samples summary statistics, produced using MINTAB, for all six
samples are provided next:
Scenario I
N
MEAN MEDIAN STDEV
Sample 1
Sample 2
Sample 3
10 60.6
10 64.5
10 70.2
60.7
64.7
70.3
1.6
1.1
2.2
Scenario II N MEAN MEDIAN STDEV
Sample 1
10 60.9
60.3 3.5
Sample 2
10 66.9
66.9 2.9
Sample 3
10 69.4
71.1 3.7
MIN
MAX
Q1
Q3
58.6
62.3
66.3
62.8 59.1 62.2
66.1 64.0 65.1
73.3 69.0 71.6
MIN
55.7
60.6
61.2
MAX
Q1
Q3
67.3 58.2 63.6
70.3 65.8 69.1
73.2 66.5 72.2
How did the sample means compare to the mean of the population from
which the sample was generated?
How did the sample standard deviations in Scenario I compare to those in
Scenario II?
Solution
(a) It is a lot more obvious that
the three different
samples in Scenario I are
from different
populations. In Scenario
II there is a lot more
overlap between the
populations.
(b)
Each sample mean is not exactly equal to the mean of the population
from which the sample was generated. The two Sample 1 means of
60.6 and 60.9 were not equal to each other nor equal to the
population mean of 60.
The samples in Scenario I were generated from populations whose
natural variation within each population was smaller compared to the
natural variation within each population in Scenario II.
For each scenario, even though the population standard deviations
were equal, the sample standard deviations were not exactly equal,
but they were comparable. So the sample means do vary, from about
60 to about 70, for each scenario. There is a good deal of variation
between the sample means.
c) What about Scenario III?
In which we have frequency plots for three independent samples of size 10
each taken from a normal population with mean of 65 and a standard
deviation of 1.5. So in this Scenario III, the population means are indeed all
equal.
Do the sample means vary?
Is there variation within each
sample?
Does the data in Scenario III
provide evidence that the
population means are not all
equal?
Although the population means were all equal, there is still a small amount of
variation between the sample means. The variation within each sample seems
to mask any slight variation there is in the sample means. The data in Scenario
III do not provide evidence that the population means are different.
What We’ve Learned?
As seen in Example, the decision about equality of the population means will
be based on examining the variability between the sample means. This
between variability will be contrasted to how much natural variation there is
within the groups.
The test-statistic actually used to make the decision is called an F-statistic and
can be loosely viewed as follows:
Variation BETWEEN the sample means
F
Natural Variation WITHIN the samples
One-Way ANOVA Logic Revisited
Recall the three scenarios that were presented in Example above.
Scenario I:
Independent random samples from three normal populations
with population means of 60, 65, and 70, respectively, and
population standard deviations all equal to 1.5.
Scenario II:
Independent random samples from three normal populations
with population means of 60, 65, and 70, respectively, and
population standard deviations all equal to 3.
Scenario III:
Independent random samples from three normal populations
with population means all equal to 65 and population
standard deviations all equal to 1.5.
The test in ANOVA is essentially the ratio of two measures of variation in the
sample data. The variation between the sample means is compared to the
natural variation of the observations within the samples via their ratio that is
called the F-statistic.
F
Variation BETWEEN the sample means MSB

Natural Variation WITHIN the samples MSW
Formally these two measures of variation are called mean squares, with the
numerator referred to as the mean square between the groups (MSB) and the
denominator referred to as the mean square within the groups (MSW).
The larger the variation between the sample means (MSB), as compared to the
natural variation within the samples (MSW), the more support there is for a
difference in the population means. F-tests are one-sided tests with the
direction of extreme being to the right
The variation between the sample means was greatest for Scenarios I and II as
compared to Scenario III. The natural variation within the samples was
greatest for Scenario II as compared to Scenarios I and III.
Q: So which of the three scenarios would you expect to result in the largest
value of the F-statistic?
The table below provides the values of the F-statistic for the one-way ANOVA
test of equality of the population means
Scenario
Scenario I: H 1 is true
Scenario II: H 1 is true
Scenario II: H 0 is true
Value of the F-statistic for
testing H 0 : 1   2  3
p-value
F  80.4
p-value  0
p-value  0.01
p-value  0.84
F  16.4
F  0.17
Note the value of the F-statistic is smallest and the p-value the largest when
the null hypothesis is true (Scenario III).
For Scenarios I and II, the population means are different, but the smaller
population standard deviation in Scenario I accentuates the differences by
producing a larger F-ratio and an extremely small p-value.
The larger value of the F-statistic corresponds to more evidence (based on the
data) that the population means are not all equal.
Let's Do It! 12.1 Would We Reject the Null Hypothesis?
Two sets of side-by-side boxplots are shown. Each set represents the boxplots
based on independent random samples selected from three normal
populations with possibly different population means but equal population
standard deviations.
In answering the questions below, remember that the F-test in ANOVA is
based on comparing the variation between the sample means to the natural
variation within the samples.
(a)
Based on the boxplots in Set I, do you think the null hypothesis
H 0 : 1   2   3 will be rejected using a one-way ANOVA F-test? Explain
your answer.
(b)
Based on the boxplots in Set II, do you think the null hypothesis
H 0 : 1   2   3 will be rejected using a one-way ANOVA F-test? Explain
your answer.
THE F-DISTRIBUTION
Recall that the test statistic in ANOVA is essentially the ratio of two measures
of variation in the sample data. The variation between the sample means is
compared to the natural variation of the observations within the samples via
their ratio that is called the F-statistic.
The probability distribution of the F-statistic is called an F-distribution.
The family of F-distributions is a family of skewed to the right distributions,
each with a minimum value of 0. F-distributions are indexed by a pair of
degrees of freedom, referred to as the numerator and denominator degrees of
freedom.
Example Working with the F-Distribution
A study was conducted to assess the effectiveness of the I = 3
different ads, Ad A, Ad B and Ad C. n = 43 subjects: 14 shown Ad A, 14 Ad B,
15 Ad C Score of effectiveness (larger score indicates more favorably). Is there
evidence at the 5% level to conclude that one ad is more effective than the
other two ads?
Suppose the ANOVA assumptions hold and the MSB =303.7 and the MSW =
120.4. Then,
The distribution of the F-statistic under HO is n Fdistribution with I – 1 = 2 and
n – I = 43 – 3 = 40 degrees of freedom.
Using Oneway ANOVA
command in
MINITAB:
Decision and Conclusion
At a 5% significance level we would accept H0. We conclude that it appears
the three ads are equally effective.
Let's Do It !Working with the F-Distribution
ANOVA performed to test the equality of I = 4 population means. Independent
random samples of size seven obtained from each of the four populations. The
F-test resulted in an observed test statistic value of 3.28.
A) Numerator degrees of freedom = I - 1 = ______ .
B) Denominator degrees of freedom = n - I = ______
C) Find the p-value for this test.
D) Complete the picture with the proper labeling for the distribution and
shade the area corresponding to the p-value.
E) Are the results statistically significant at the 5% level?
YES
NO
Explain:
Performing an ANOVA F-Test Using the TI
Example Three Treatments
Suppose we enter the data for the three headache drugs from Example
12.7 under the lists L1, L2, and L3. Then the following steps can be
used to perform the one-way ANOVA on these data:
Note: The term Factor represents the between source of variation. The
term Error represents the within source of variation.
Complete the ANOVA table
The p-value is 0.034. At 5% level, conclude that there appears to be a
difference in the mean time to relief for the three drugs.
MULTIPLE COMPARISONS*
In one-way ANOVA we are trying to compare many population means—that is,
we are trying to do multiple comparisons. However, we would like to do the
many comparisons at one time and still be able to attach to our conclusions
some overall measure of confidence.
The statistical approach to this problem of multiple comparisons is to first do
an overall test to assess if there are any differences between the population
means, and if we accept that there is a difference, then to do a follow-up
analysis that helps determine which of the means differ and estimates by how
much they differ.
The F-test described in the preceding sections is this overall test in ANOVA. In
this last section on one-way ANOVA we discuss the need for this overall F-test
in ANOVA and present some multiple comparison methods that are part of the
follow-up analysis should our decision be to reject H0.
When we perform an ANOVA to test if the population means are equal, we
must remember that if we reject the null hypothesis we can only conclude that
at least one of the population means appears to be different from the other
population means. Thus the ANOVA F-test is only the first step in our analysis.
If significant differences exist among the treatment means, we would like to
investigate which means are significantly different and perhaps measure how
different they are. We can determine where the differences among the means
seem to occur by conducting multiple comparisons.
Let's Do It! 1Three Treatments Revisited
(a) Report the confidence interval for each pairwise comparison:
P-vlaue for μ1 – μ2 : ______________ , Is Drug 1 and 2 Different?
P-value for μ1 – μ3: ______________ , Is Drug 1 and 3 Different?
P-value for
μ2 – μ3 :
______________, Is Drug 2 and 3 Different?
(b) State your conclusions regarding the differences between the
mean response for the three drug groups based on the Bonferroni multiple
comparison method.
Let’s Do It
A researcher wishes to try three different techniques to lower the blood
pressure of individuals diagnosed with high blood pressure. The subjects are
randomly assigned to three groups; the first group takes medication, the
second group exercises, and the third group follows a special diet (1:
Medication, 2: Exercise, 3: Diet). After four weeks, the reduction in each
person’s blood pressure is recorded.
A one-way ANOVA was calculated on reduction in blood pressure Using SPSS.
ANOVA
Sum of Squares
df
Mean Square
F
Between Groups
160.133
2
80.067
Within Groups
104.800
12
8.733
Total
264.933
14
Sig.
9.168
.004
Multiple Comparisons
(I) Technique
(J) Technique
1
2
8.000*
1.869
.003
3
4.200
1.869
.103
1
-8.000*
1.869
.003
3
-3.800
1.869
.147
1
-4.200
1.869
.103
2
3.800
1.869
.147
2
3
Mean Difference (I-J)
Std. Error
Sig.
a. F = _____
b. p-value= __________
c. Using   0.05, is there a significant difference between the three
techniques? Yes NO
d. If your answer in part b is yes, determine which techniques are
significantly different. (Check all that applies)
___Medication and Exercise
___Medication and Diet
___Exercise and Diet
End of chapter12