Download One-Way Analysis of Variance: Comparing Several Means

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CHAPTER
22
(Alan Schein Photography/CORBIS)
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One-Way Analysis of
Variance: Comparing
Several Means
In this chapter we cover. . .
Comparing several means
The analysis of variance F
test
Using technology
The idea of analysis of
variance
The ANOVA model
Some details of ANOVA*
The two-sample t procedures of Chapter 17 compare the means of two populations or the mean responses to two treatments in an experiment. Of course,
studies don’t always compare just two groups. We need a method for comparing
any number of means.
EXAMPLE 22.1 Cars, SUVs, and pickup trucks
Sport-utility vehicles (SUVs) and pickup trucks have replaced cars in many American driveways. SUVs and trucks are often larger and heavier than cars and so are
under attack for reasons of safety and fuel economy. Do SUVs and trucks have lower
gas mileage than cars? Table 22.1 contains data on the highway gas mileage (in miles
per gallon) for 31 midsize cars, 31 SUVs, and 14 standard-size pickup trucks. The gas
mileages and vehicle classifications are compiled by the Environmental Protection
Agency.1 Because the cars are two-wheel-drive, the table describes only two-wheeldrive SUVs and trucks. The popular four-wheel-drive versions get poorer mileage.
Figure 22.1 shows side-by-side stemplots of the gas mileages for the three types of
vehicle. We used the same stems in all three for easier comparison. It does appear
that gas mileage decreases as we move from cars to SUVs to pickups.
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
Here are the means, standard deviations, and five-number summaries for the three
types of vehicle, from software:
N
Mean Median
Midsize 31 27.903 27.000
SUV
31 22.677 22.000
Pickup 14 21.286 20.000
StDev Minimum Maximum
Q1
Q3
2.561 23.000 33.000 26.000 30.000
3.673 17.000 29.000 19.000 26.000
2.758 17.000 26.000 19.750 23.500
We will use the mean to describe the center of the gas mileage distributions. Mean
gas mileage goes down as we move from midsize cars to SUVs to pickup trucks. The
differences among the means are not large. Are they statistically significant?
TABLE 22.1 Highway gas mileage for 2003 model vehicles
Midsize Cars
Model
Acura 3.5TL
Audi A6
Audi A8
Buick Century
Buick Regal
Cadillac CTS
Cadillac Seville
Chevrolet Monte Carlo
Chrysler Sebring
Honda Accord
Hyundai Sonata
Hyundai XG350
Infiniti I35
Jaguar S-Type
Jaguar Vanden Plas
Kia Optima
Lexus ES300
Lexus GS300
Lincoln LS
Mercedes-Benz E320
Mercedes-Benz E500
Mitsubishi Diamante
Mitsubishi Galant
Nissan Altima
Nissan Maxima
Saab 9-5
Saturn L200
Saturn L300
Toyota Camry
Volkswagen Passat
Volvo S80
Sport-Utility Vehicles
MPG
29
27
25
29
27
26
27
32
30
33
30
26
26
26
24
30
29
25
26
27
23
25
27
29
26
29
32
29
32
31
28
Model
Cadillac Escalade
Chevrolet Avalanche
Chevrolet Blazer
Chevrolet Suburban
Chevrolet Tahoe
Chevrolet Tracker
Chevrolet Trailblazer
Chrysler PT Cruiser
Dodge Durango
Ford Escape
Ford Expedition
Ford Explorer
Honda CR-V
Hyundai Santa Fe
Infiniti QX4
Isuzu Axiom
Isuzu Rodeo
Jeep Grand Cherokee
Jeep Liberty
Kia Sorento
Lincoln Navigator
Mazda Tribute
Mitsubishi Montero
Nissan Pathfinder
Nissan Xterra
Saturn Vue
Suzuki Vitara
Toyota 4Runner
Toyota Highlander
Toyota Rav4
Volvo XC 90
Standard Pickup Trucks
MPG
18
18
23
18
18
26
22
25
19
28
19
19
28
27
21
21
23
21
24
20
17
28
22
21
24
28
25
20
27
29
24
Model
Chevrolet Silverado
Dodge Dakota
Dodge Ram
Ford Explorer
Ford F150
Ford Ranger
GMC Sierra
Lincoln Blackwood
Mazda B2300
Mazda B3000
Mazda B4000
Nissan Frontier
Toyota Tacoma
Toyota Tundra
MPG
20
19
20
20
20
26
20
17
26
22
21
23
25
19
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Comparing several means
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0
0
000
000000
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17
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0
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0
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00000
0
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0
0
00
Pickup
Figure 22.1 Side-by-side stemplots comparing the highway gas mileages of midsize
cars, sport-utility vehicles, and standard-size pickup trucks, from Table 22.1.
Comparing several means
Call the mean highway gas mileages for the three populations of vehicles µ1 for
midsize cars, µ2 for SUVs, and µ3 for pickups. The subscript reminds us which
group a parameter or statistic describes. To compare these three population
means, we might use the two-sample t test several times:
r Test H0: µ1 = µ2 to see if the mean miles per gallon for midsize cars
differs from the mean for SUVs.
r Test H0: µ1 = µ3 to see if midsize cars differ from pickup trucks.
r Test H0: µ2 = µ3 to see if SUVs differ from pickups.
The weakness of doing three tests is that we get three P -values, one for
each test alone. That doesn’t tell us how likely it is that three sample means are
spread apart as far as these are. It may be that x 1 = 27.903 and x 3 = 21.286 are
significantly different if we look at just two groups but not significantly different
if we know that they are the smallest and the largest means in three groups. As
we look at more groups, we expect the gap between the smallest and the largest
sample mean to get larger. (Think of comparing the tallest and shortest person
in larger and larger groups of people.) We can’t safely compare many parameters
by doing tests or confidence intervals for two parameters at a time.
The problem of how to do many comparisons at once with some overall
measure of confidence in all our conclusions is common in statistics. This is the
problem of multiple comparisons. Statistical methods for dealing with multiple
comparisons usually have two steps:
1. An overall test to see if there is good evidence of any differences among
the parameters that we want to compare.
2. A detailed follow-up analysis to decide which of the parameters differ and
to estimate how large the differences are.
multiple comparisons
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
The overall test, though more complex than the tests we met earlier, is often
reasonably straightforward. The follow-up analysis can be quite elaborate. In
our basic introduction to statistical practice, we will concentrate on the overall
test, along with data analysis that points to the nature of the differences.
The analysis of variance F test
We want to test the null hypothesis that there are no differences among the
mean highway gas mileages for the three vehicle types:
H0: µ1 = µ2 = µ3
The alternative hypothesis is that there is some difference, that not all three
population means are equal:
Ha : not all of µ1 , µ2 , and µ3 are equal
analysis of variance F test
The alternative hypothesis is no longer one-sided or two-sided. It is “manysided,” because it allows any relationship other than “all three equal.” For example, Ha includes the case in which µ2 = µ3 but µ1 has a different value.
The test of H0 against Ha is called the analysis of variance F test. Analysis
of variance is usually abbreviated as ANOVA. The ANOVA F test is almost
always carried out by software that reports the test statistic and its P-value.
EXAMPLE 22.2 Cars, SUVs, and pickup trucks: ANOVA
As we will soon see, software tells us that for the gas mileage data in Table 22.1,
F = 31.61 with P-value P < 0.001. There is very strong evidence that the three
types of vehicle do not all have the same mean gas mileage.
The F test does not say which of the three means are significantly different. It
appears from our preliminary data analysis that SUVs and pickups have similar fuel
economy and that both get distinctly poorer mileage than midsize cars. The gap of
more than 5 miles per gallon between midsize cars and the other two types of vehicle
is large enough to be of practical interest.
Our conclusion: There is strong evidence (P < 0.001) that the means are not all
equal. The most important difference among the means is that midsize cars have
better gas mileage than SUVs and pickups.
Example 22.2 illustrates our approach to comparing means. The ANOVA
F test (done by software) assesses the evidence for some difference among the
population means. In most cases, we expect the F test to be significant. We
would not undertake a study if we did not expect to find some effect. The formal
test is nonetheless important to guard against being misled by chance variation.
We will not do the formal follow-up analysis that is often the most useful part
of an ANOVA study. Follow-up analysis would allow us to say which means
differ and by how much, with (say) 95% confidence that all our conclusions are
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The analysis of variance F test
correct. We rely instead on examination of the data to show what differences
are present and whether they are large enough to be interesting.
APPLY YOUR KNOWLEDGE
400
600
22.1 Do fruit flies sleep? Mammals and birds sleep. Insects such as fruit flies
rest, but is this rest sleep? Biologists now think that insects do sleep.
One experiment gave caffeine to fruit flies to see if it affected their rest.
We know that caffeine reduces sleep in mammals, so if it reduces rest in
fruit flies that’s another hint that the rest is really sleep. The paper
reporting the study contains a graph similar to Figure 22.2 and states,
“Flies given caffeine obtained less rest during the dark period in a
dose-dependent fashion (n = 36 per group, P < 0.0001).”2
(a) The explanatory variable is amount of caffeine, in milligrams per
milliliter of blood. The response variable is minutes of rest
(measured by an infrared motion sensor) during a 12-hour dark
period. Outline the design of this experiment.
(b) The P-value in the report comes from the ANOVA F test. What
means does this test compare? State in words the null and
alternative hypotheses for the test in this setting. What do the
graph and the statistical test together lead you to conclude?
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Rest (min)
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1 mg/ml
2.5 mg/ml
5 mg/ml
Caffeine
Figure 22.2 Bar graph comparing the mean rest of fruit flies given different amounts
of caffeine, for Exercise 22.1.
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
22.2 Road rage. A random-digit-dialing survey asked a random sample of
1382 drivers questions about their driving. Responses to a number of
questions were then combined to yield a measure of “angry/threatening
driving,” commonly called “road rage.” What driver characteristics go
with road rage? There were no significant differences among races or
levels of education. What about the effect of the driver’s age? Here are
the mean responses for three age-groups:3
<30 yr
30–55 yr
>55 yr
2.22
1.33
0.66
The report says that F = 34.96, with P < 0.01.
(a) What are the null and alternative hypotheses for the ANOVA F
test? Be sure to explain what means the test compares.
(b) Based on the sample means and the F test, what do you conclude?
Using technology
Any technology used for statistics should perform analysis of variance.
Figure 22.3 displays ANOVA output for the data of Table 22.1 from the Minitab
statistical software, the Excel spreadsheet program, and the TI-83 graphing calculator.
Minitab and Excel give the sizes of the three samples and their means. These
agree with those in Example 22.1. The most important part of all three outputs
reports the F test statistic, F = 31.61, and its P-value. Minitab sensibly reports
P as 0.000, that is, 0 to three decimal places. Excel and the TI-83 show that P
is about 0.0000000001. (The E-10 in these displays means move the decimal
point 10 digits to the left.) There is very strong evidence that the three types
of vehicle do not all have the same mean gas mileage.
All three outputs also report sums of squares (SS) and mean squares (MS).
We don’t need this information now. The optional section “Some Details of
ANOVA” at the end of this chapter explains these terms.
Minitab also gives confidence intervals for all three means that help us see
which means differ and by how much. The SUV and pickup intervals overlap,
and the midsize interval lies well above them on the mileage scale. These are
95% confidence intervals for each mean separately. We are not 95% confident
that all three intervals cover the three means. This is another example of the
peril of multiple comparisons.
APPLY YOUR KNOWLEDGE
22.3 Logging in the rainforest. How does logging in a tropical rainforest
affect the forest several years later? Researchers compared forest plots in
Borneo that had never been logged (Group 1) with similar plots nearby
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Using technology
Minitab
Excel
TI-83
Figure 22.3 Analysis of variance output for the gas mileage data from Minitab,
Excel, and the TI-83.
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
that had been logged 1 year earlier (Group 2) and 8 years earlier
(Group 3). Although the study was not an experiment, the authors
explain why we can consider the plots to be randomly selected. The
data appear in Table 22.2. The variable Trees is the count of trees in a
plot; Species is the count of tree species in a plot. The variable Richness
is the ratio of number of species to number of individual trees,
calculated as Species/Trees.4
TABLE 22.2 Data from a study of logging in Borneo
Observation
Group
Trees
Species
Richness
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
27
22
29
21
19
33
16
20
24
27
28
19
12
12
15
9
20
18
17
14
14
2
17
19
18
4
22
15
18
19
22
12
12
22
18
22
20
15
21
13
13
19
13
19
15
11
11
14
7
18
15
15
12
13
2
15
8
17
4
18
14
18
15
15
10
12
0.81481
0.81818
0.75862
0.95238
0.78947
0.63636
0.81250
0.65000
0.79167
0.48148
0.67857
0.78947
0.91667
0.91667
0.93333
0.77778
0.90000
0.83333
0.88235
0.85714
0.92857
1.00000
0.88235
0.42105
0.94444
1.00000
0.81818
0.93333
1.00000
0.78947
0.68182
0.83333
1.00000
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Using technology
Figure 22.4 Excel output for analysis of variance on the number of trees in forest
plots, for Exercise 22.3.
(a) Make side-by-side stemplots of Trees for the three groups. Use stems
0, 1, 2, and 3 and split the stems (see page 17). What effects of
logging are visible?
(b) Figure 22.4 shows Excel ANOVA output for Trees. What do the
group means show about the effects of logging?
(c) What are the values of the ANOVA F statistic and its P-value?
What hypotheses does F test? What conclusions about the effects
of logging on number of trees do the data lead to?
22.4 Dogs, friends, and stress. If you are a dog lover, perhaps having your
dog along reduces the effect of stress. To examine the effect of pets in
stressful situations, researchers recruited 45 women who said they were
dog lovers. The EESEE story “Stress among Pets and Friends” describes
the results. Fifteen of the subjects were randomly assigned to each of
three groups to do a stressful task alone (the control group), with a good
friend present, or with their dog present. The subject’s mean heart rate
during the task is one measure of the effect of stress. Table 22.3 contains
the data.
(a) Make stemplots of the heart rates for the three groups (round to the
nearest whole number of beats). Do any of the groups show outliers
or extreme skewness?
(b) Figure 22.5 gives the Minitab ANOVA output for these data. Do
the mean heart rates for the groups appear to show that the presence
of a pet or a friend reduces heart rate during a stressful task?
(c) What are the values of the ANOVA F statistic and its P-value?
What hypotheses does F test? Briefly describe the conclusions you
draw from these data. Did you find anything surprising?
EESEE
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
TABLE 22.3 Mean heart rates during stress with a pet (P),
with a friend (F), and for the control group (C)
Group
Rate
Group
Rate
Group
Rate
P
F
P
C
C
P
F
F
P
F
C
C
C
F
F
69.169
99.692
70.169
80.369
87.446
75.985
83.400
102.154
86.446
80.277
90.015
99.046
75.477
88.015
92.492
P
C
P
C
C
F
F
C
P
P
F
F
C
F
P
68.862
87.231
64.169
91.754
87.785
91.354
100.877
77.800
97.538
85.000
101.062
97.046
62.646
81.600
72.262
C
C
P
P
P
C
C
C
F
F
F
P
P
F
P
84.738
84.877
58.692
79.662
69.231
73.277
84.523
70.877
89.815
98.200
76.908
69.538
70.077
86.985
65.446
F
14.08
P
0.000
Analysis of Variance for Beats
Source
DF
SS
MS
Group
2
2387.7
1193.8
Error
42
3561.3
84.8
Total
44
5949.0
Individual 95% CIs For Mean
Based on Pooled StDev
Level
Control
Friend
Pet
N
15
15
15
Mean
82.524
91.325
73.483
Pooled StDev =
9.208
StDev
9.242
8.341
9.970
72.0
80.0
88.0
96.0
Figure 22.5 Minitab output for the data in Table 22.3 on heart rates (beats per
minute) during stress, for Exercise 22.4. The “Control” group worked alone, the
“Friend” group had a friend present, and the “Pet” group had a pet dog present.
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The idea of analysis of variance
(a )
(b)
Figure 22.6 Boxplots for two sets of three samples each. The sample means are the
same in (a) and (b). Analysis of variance will find a more significant difference among
the means in (b) because there is less variation among the individuals within those
samples.
The idea of analysis of variance
Here is the main idea for comparing means: what matters is not how far apart
the sample means are but how far apart they are relative to the variability of
individual observations. Look at the two sets of boxplots in Figure 22.6. For
simplicity, these distributions are all symmetric, so that the mean and median
are the same. The centerline in each boxplot is therefore the sample mean.
Both sets of boxplots compare three samples with the same three means. Like
the three vehicle types in Example 22.1, the means are different but not very
different. Could differences this large easily arise just due to chance, or are they
statistically significant?
r The boxplots in Figure 22.6(a) have tall boxes, which show lots of
variation among the individuals in each group. With this much variation
among individuals, we would not be surprised if another set of samples
gave quite different sample means. The observed differences among the
sample means could easily happen just by chance.
r The boxplots in Figure 22.6(b) have the same centers as those
in Figure 22.6(a), but the boxes are much shorter. That is, there is much
less variation among the individuals in each group. It is unlikely that any
sample from the first group would have a mean as small as the mean of the
second group. Because means as far apart as those observed would rarely
arise just by chance in repeated sampling, they are good evidence of real differences among the means of the three populations we are sampling from.
You can use the One-Way ANOVA applet to demonstrate the analysis of
variance idea for yourself. The applet allows you to change both the group
means and the spread within groups. You can watch the ANOVA F statistic
and its P-value change as you work.
This comparison of the two parts of Figure 22.6 is too simple in one way.
It ignores the effect of the sample sizes, an effect that boxplots do not show.
Small differences among sample means can be significant if the samples are
very large. Large differences among sample means may fail to be significant if
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CHAPTER 22 r One-Way Analysis of Variance: Comparing Several Means
the samples are very small. All we can be sure of is that for the same sample size,
Figure 22.6(b) will give a much smaller P-value than Figure 22.6(a). Despite
this qualification, the big idea remains: if sample means are far apart relative to
the variation among individuals in the same group, that’s evidence that something other than chance is at work.
THE ANALYSIS OF VARIANCE IDEA
Analysis of variance compares the variation due to specific sources with
the variation among individuals who should be similar. In particular,
ANOVA tests whether several populations have the same mean by
comparing how far apart the sample means are with how much variation
there is within the samples.
one-way ANOVA
It is one of the oddities of statistical language that methods for comparing means are named after the variance. The reason is that the test works by
comparing two kinds of variation. Analysis of variance is a general method
for studying sources of variation in responses. Comparing several means is the
simplest form of ANOVA, called one-way ANOVA. One-way ANOVA is the
only form of ANOVA that we will study.
THE ANOVA F STATISTIC
The analysis of variance F statistic for testing the equality of several
means has this form:
F =
F distribution
variation among the sample means
variation among individuals in the same sample
We give more detail later. Because ANOVA is in practice done by software,
the idea is more important than the detail. The F statistic can only take values that are zero or positive. It is zero only when all the sample means are
identical and gets larger as they move farther apart. Large values of F are evidence against the null hypothesis H0 that all population means are the same.
Although the alternative hypothesis Ha is many-sided, the ANOVA F test is
one-sided because any violation of H0 tends to produce a large value of F .
How large must F be to provide significant evidence against H0 ? To answer questions of statistical significance, compare the F statistic with critical
values from an F distribution. The F distributions are described on page 456
in Chapter 17. A specific F distribution is specified by two parameters: a numerator degrees of freedom and a denominator degrees of freedom. Table D in
the back of the book contains critical values for F distributions with various
degrees of freedom.
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The idea of analysis of variance
EXAMPLE 22.3 Using the F table
Look again at the software output for the gas mileage data in Figure 22.3. All three
outputs give the degrees of freedom for the F test, labeled “df ” or “DF.” There are
2 degrees of freedom in the numerator and 73 in the denominator.
In Table D, find the numerator degrees of freedom 2 at the top of the table. Then
look for the denominator degrees of freedom 73 at the left of the table. There is
no entry for 73, so we use the next smaller entry, 50 degrees of freedom. The upper
critical values for 2 and 50 degrees of freedom are
p
Critical
value
0.100
0.050
0.025
0.010
0.001
2.41
3.18
3.97
5.06
7.96
Figure 22.7 shows the F density curve with 2 and 50 degrees of freedom and the
upper 5% critical value 3.18. The observed F = 31.61 lies far to the right on this
curve. We see from the table that F = 31.61 is larger than the 0.001 critical value,
so P < 0.001.
Software, of course, uses the F distribution with 2 and 73 degrees of freedom to
calculate P-values. Excel also tells us how large F must be to be significant at the
α = 0.05 level. This is the 5% critical value for F with 2 and 73 degrees of freedom.
It is 3.122. Approximating the P-value from Table D is usually adequate in practice,
as it is in this example.
F density curve, df = 2 and 50
Area = 0.05
0
1
3.18
10.00
Figure 22.7 The density curve of the F distribution with 2 degrees of freedom in the
numerator and 50 degrees of freedom in the denominator, for Example 22.3. The
value F ∗ = 3.18 is the 0.05 critical value.
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The degrees of freedom of the F statistic depend on the number of means
we are comparing and the number of observations in each sample. That is, the
F test takes into account the number of observations. Here are the details.
DEGREES OF FREEDOM FOR THE F TEST
We want to compare the means of I populations. We have an SRS of size
n i from the ith population, so that the total number of observations in all
samples combined is
N = n1 + n2 + · · · + n I
If the null hypothesis that all population means are equal is true, the
ANOVA F statistic has the F distribution with I − 1 degrees of freedom
in the numerator and N − I degrees of freedom in the denominator.
EXAMPLE 22.4 Degrees of freedom for F
In Examples 22.1 and 22.2, we compared the mean highway gas mileage for three
types of vehicle, so I = 3. The three sample sizes are
n 1 = 31
n 2 = 31
n 3 = 14
The total number of observations is therefore
N = 31 + 31 + 14 = 76
The ANOVA F test has numerator degrees of freedom
I −1=3−1=2
and denominator degrees of freedom
N − I = 76 − 3 = 73
These degrees of freedom are given in the outputs in Figure 22.3.
APPLY YOUR KNOWLEDGE
22.5 Logging in the rainforest, continued. Exercise 22.3 compares the
number of tree species in rainforest plots that had never been logged
(Group 1) with similar plots nearby that had been logged 1 year earlier
(Group 2) and 8 years earlier (Group 3).
(a) What are I , the n i , and N for these data? Identify these quantities
in words and give their numerical values.
(b) Find the degrees of freedom for the ANOVA F statistic. Check
your work against the Excel output in Figure 22.4.
(c) For these data, F = 11.43. What does Table D tell you about the
P-value of this statistic?
22.6 Dogs, friends, and stress, continued. Exercise 22.4 compares the
mean heart rates for women performing a stressful task under three
conditions.
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611
(a) What are I , the n i , and N for these data? State in words what each
of these numbers means.
(b) Find both degrees of freedom for the ANOVA F statistic. Check
your results against the Minitab output in Figure 22.5.
(c) The output shows that F = 14.08. What does Table D tell you
about the significance of this result?
22.7 Smoking among French men. A study of smoking categorized a sample
of French men aged 20 to 60 years as nonsmokers (146 men), former
smokers (125 men), moderate smokers (104 men), or heavy smokers
(84 men). One set of response variables was the amount of several types
of food consumed, in grams.5
(a) What are the degrees of freedom for an ANOVA F statistic
that compares the mean weights of a food consumed in the
4 populations? Use Table D with 3 and 200 degrees of freedom to
give an approximate P-value for each of the following foods:
(b) For milk, F = 3.58.
(c) For eggs, F = 1.08.
(d) For vegetables, F = 6.02.
The ANOVA model
Like all inference procedures, ANOVA is valid only in some circumstances.
Here is the model that describes when we can use ANOVA to compare population means.
Where are the details?
THE ANOVA MODEL
r We have I independent SRSs, one from each of I populations.
r The i th population has a Normal distribution with unknown mean
µi . The means may be different in the different populations. The
ANOVA F statistic tests the null hypothesis that all of the
populations have the same mean:
H0: µ1 = µ2 = · · · = µ I
Ha : not all of the µi are equal
r All of the populations have the same standard deviation σ , whose
value is unknown.
The ANOVA model has I + 1 parameters, the I population means and
the standard deviation σ .
The first two requirements are familiar from our study of the two-sample
t procedures for comparing two means. As usual, the design of the data production is the most important foundation for inference. Biased sampling or
confounding can make any inference meaningless. If we do not actually draw
Papers reporting scientific
research must be short. Brevity
allows researchers to hide
details about their data. Did
they choose their subjects in a
biased way? Did they report
data on only some of their
subjects? Did they try several
statistical analyses and report
the one that looked best? The
statistician John Bailar
screened more than 4000
papers for the New England
Journal of Medicine. He says,
“When it came to the
statistical review, it was often
clear that critical information
was lacking, and the gaps
nearly always had the practical
effect of making the authors’
conclusions look stronger than
they should have.”
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robustness
separate SRSs from each population or carry out a randomized comparative experiment, it is often unclear to what population the conclusions of inference
apply. This is the case in Example 22.1, for example. ANOVA, like other inference procedures, is often used when random samples are not available. You
must judge each use on its merits, a judgment that usually requires some knowledge of the subject of the study in addition to some knowledge of statistics. We
might consider the 2003 model year vehicles in Example 22.1 to be samples
from vehicles of their types produced in recent years.
Because no real population has exactly a Normal distribution, the usefulness
of inference procedures that assume Normality depends on how sensitive they
are to departures from Normality. Fortunately, procedures for comparing means
are not very sensitive to lack of Normality. The ANOVA F test, like the t procedures, is robust. What matters is Normality of the sample means, so ANOVA
becomes safer as the sample sizes get larger, because of the central limit theorem
effect. Remember to check for outliers that change the value of sample means
and for extreme skewness. When there are no outliers and the distributions are
roughly symmetric, you can safely use ANOVA for sample sizes as small as 4
or 5. (Don’t confuse the ANOVA F , which compares several means, with the
F statistic discussed in Chapter 17, which compares two standard deviations
and is not robust against non-Normality.)
The third assumption is annoying: ANOVA assumes that the variability of
observations, measured by the standard deviation, is the same in all populations. You may recall from Chapter 17 (page 454) that there is a special version
of the two-sample t test that assumes equal standard deviations in both populations. The ANOVA F for comparing two means is exactly the square of this
special t statistic. We prefer the t test that does not assume equal standard deviations, but for comparing more than two means there is no general alternative
to the ANOVA F . It is not easy to check the assumption that the populations
have equal standard deviations. Statistical tests for equality of standard deviations are very sensitive to lack of Normality, so much so that they are of little
practical value. You must either seek expert advice or rely on the robustness of
ANOVA.
How serious are unequal standard deviations? ANOVA is not too sensitive
to violations of the assumption, especially when all samples have the same or
similar sizes and no sample is very small. When designing a study, try to take
samples of the same size from all the groups you want to compare. The sample standard deviations estimate the population standard deviations, so check
before doing ANOVA that the sample standard deviations are similar to each
other. We expect some variation among them due to chance. Here is a rule of
thumb that is safe in almost all situations.
CHECKING STANDARD DEVIATIONS IN ANOVA
The results of the ANOVA F test are approximately correct when the
largest sample standard deviation is no more than twice as large as the
smallest sample standard deviation.
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EXAMPLE 22.5 Do the standard deviations allow ANOVA?
In the gas mileage study, the sample standard deviations for midsize cars, SUVs, and
pickups are
s 1 = 2.561
s 2 = 3.673
s 3 = 2.758
These standard deviations easily satisfy our rule of thumb. We can safely use
ANOVA to compare the mean gas mileage for the three vehicle types.
The report from which Table 22.1 was taken also contained data on 22 subcompact cars. Can we use ANOVA to compare the means for all four types of vehicle?
The standard deviation for subcompact cars is s 4 = 5.262 miles per gallon. This is
slightly more than twice the smallest standard deviation:
largest s
5.262
=
= 2.05
smallest s
2.561
Our rule of thumb is conservative, so many statisticians would go ahead with
ANOVA in this circumstance.
A large standard deviation is often due to skewness or outliers. It turns out that
the government’s “subcompact” category includes three rather exotic cars, a Bentley,
a Ferrari, and a Maserati. All three get poor gas mileage. The standard deviation
for the remaining 19 subcompact cars is s = 3.304, well within the guidelines for
ANOVA.
EXAMPLE 22.6 Which color attracts beetles best?
To detect the presence of harmful insects in farm fields, we can put up boards covered
with a sticky material and examine the insects trapped on the boards. Which colors
attract insects best? Experimenters placed six boards of each of four colors at random
locations in a field of oats and measured the number of cereal leaf beetles trapped.
Here are the data:6
Board color
Blue
Green
White
Yellow
Insects trapped
16
37
21
45
11
32
12
59
20
20
14
48
21
29
17
46
14
37
13
38
7
32
20
47
We would like to use ANOVA to compare the mean numbers of beetles that would
be trapped by all boards of each color. Because the samples are small, we plot the
data in side-by-side stemplots in Figure 22.8. Minitab output for ANOVA appears
Blue
Green
White
0 7
0
0
0
1 146
1
1 2347
1
2 01
2 09
2 01
2
Yellow
3
3 2277
3
3 8
4
4
4
4 5678
5
5
5
5 9
Figure 22.8 Side-by-side
stemplots comparing the
counts of insects attracted by
six boards for each of four
board colors, for Example
22.6.
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in Figure 22.9. The yellow boards attract by far the most insects (x 4 = 47.167), with
green next (x 2 = 31.167) and blue and white far behind.
Check that we can safely use ANOVA to test equality of the four means. The
largest of the four sample standard deviations is 6.795 and the smallest is 3.764. The
ratio
largest s
6.795
=
= 1.8
smallest s
3.764
is less than 2, so these data satisfy our rule of thumb. The shapes of the four distributions are irregular, as we expect with only 6 observations in each group, but there
are no outliers. The ANOVA results will be approximately correct.
There are I = 4 groups and N = 24 observations overall, so the degrees of freedom for F are
numerator:
denominator:
I −1= 4−1 = 3
N − I = 24 − 4 = 20
This agrees with Minitab’s results. The F statistic is F = 42.84, a large F
with P < 0.001. Despite the small samples, the experiment gives very strong evidence of differences among the colors. Yellow boards appear best at attracting leaf
beetles.
Analysis of Variance for Beetles
Source
DF
SS
MS
Color
3
4134.0
1378.0
Error
20
643.3
32.2
Total
23
4777.3
F
42.84
P
0.000
Individual 95% CIs For Mean
Based on Pooled StDev
Level
Blue
Green
White
Yellow
N
6
6
6
6
Pooled StDev =
Mean
14.833
31.167
16.167
47.167
5.672
StDev
5.345
6.306
3.764
6.795
12
24
36
48
Figure 22.9 Minitab ANOVA output for comparing the four board colors in
Example 22.6.
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The ANOVA model
APPLY YOUR KNOWLEDGE
22.8
Checking standard deviations. Verify that the sample standard
deviations for these sets of data do allow use of ANOVA to compare
the population means.
(a) The counts of trees in Exercise 22.3 and Figure 22.4.
(b) The heart rates of Exercise 22.4 and Figure 22.5.
22.9 Species richness after logging. Table 22.2 gives data on the species
richness in rainforest plots, defined as the number of tree species in a
plot divided by the number of trees in the plot. ANOVA may not be
trustworthy for the richness data. Do data analysis: make side-by-side
stemplots to examine the distributions of the response variable in the
three groups and also compare the standard deviations. What about
the data makes ANOVA risky?
22.10 Marital status and salary. Married men tend to earn more than single
men. An investigation of the relationship between marital status and
income collected data on all 8235 men employed as managers or
professionals by a large manufacturing firm in 1976. Suppose (this is
risky) we regard these men as a random sample from the population of
all men employed in managerial or professional positions in large
companies. Here are summary statistics for the salaries of these
men:7
ni
xi
si
Single
Married
Divorced
Widowed
337
$21,384
$5,731
7,730
$26,873
$7,159
126
$25,594
$6,347
42
$26,936
$8,119
(a) Briefly describe the relationship between marital status and salary.
(b) Do the sample standard deviations allow use of the ANOVA
F test? (The distributions are skewed to the right. We expect
right-skewness in income distributions. The investigators actually
applied ANOVA to the logarithms of the salaries, which are more
symmetric.)
(c) What are the degrees of freedom of the ANOVA F test?
(d) The F test is a formality for these data, because we are sure that
the P-value will be very small. Why are we sure?
(e) Single men earn less on the average than men who are or have
been married. Do the highly significant differences in mean salary
show that getting married raises men’s mean income? Explain your
answer.
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Some details of ANOVA*
Now we will give the actual formula for the ANOVA F statistic. We have
SRSs from each of I populations. Subscripts from 1 to I tell us which sample a
statistic refers to:
Population
Sample
size
Sample
mean
Sample
std. dev.
1
2
·
·
·
I
n1
n2
·
·
·
nI
x1
x2
·
·
·
xI
s1
s2
·
·
·
sI
You can find the F statistic from just the sample sizes n i , the sample means
x i , and the sample standard deviations s i . You don’t need to go back to the
individual observations.
The ANOVA F statistic has the form
F =
mean squares
variation among the sample means
variation among individuals within samples
The measures of variation in the numerator and denominator of F are called
mean squares. A mean square is a more general form of a sample variance. An
ordinary sample variance s 2 is an average (or mean) of the squared deviations
of observations from their mean, so it qualifies as a “mean square.”
The numerator of F is a mean square that measures variation among the I
sample means x 1 , x 2 , . . . , x I . Call the overall mean response (the mean of all
N observations together) x. You can find x from the I sample means by
x=
MSG
n1 x 1 + n2 x 2 + · · · + n I x I
N
The sum of each mean multiplied by the number of observations it represents
is the sum of all the individual observations. Dividing this sum by N, the total
number of observations, gives the overall mean x. The numerator mean square
in F is an average of the I squared deviations of the means of the samples
from x. We call it the mean square for groups, abbreviated as MSG.
MSG =
n 1 (x 1 − x)2 + n 2 (x 2 − x)2 + · · · + n I (x I − x)2
I −1
Each squared deviation is weighted by n i , the number of observations it represents.
* This
more advanced section is optional if you are using software to find the F statistic.
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Some details of ANOVA
The mean square in the denominator of F measures variation among individual observations in the same sample. For any one sample, the sample
2
variance s i does this job. For all I samples together, we use an average of the
2
individual sample variances. It is again a weighted average in which each s i is
weighted by one fewer than the number of observations it represents, n i − 1.
2
Another way to put this is that each s i is weighted by its degrees of freedom
n i − 1. The resulting mean square is called the mean square for error, MSE.
2
MSE =
2
MSE
2
(n 1 − 1)s 1 + (n 2 − 1)s 2 + · · · + (n I − 1)s I
N−I
“Error” doesn’t mean a mistake has been made. It’s a traditional term for chance
variation. Here is a summary of the ANOVA test.
THE ANOVA F TEST
Draw an independent SRS from each of I populations. The i th
population has the N(µi , σ ) distribution, where σ is the common
standard deviation in all the populations. The i th sample has size n i ,
sample mean x i , and sample standard deviation s i .
The ANOVA F statistic tests the null hypothesis that all I populations
have the same mean:
H0: µ1 = µ2 = · · · = µ I
Ha : not all of the µi are equal
The statistic is
F =
MSG
MSE
The numerator of F is the mean square for groups
MSG =
n 1 (x 1 − x)2 + n 2 (x 2 − x)2 + · · · + n I (x I − x)2
I −1
The denominator of F is the mean square for error
2
MSE =
2
2
(n 1 − 1)s 1 + (n 2 − 1)s 2 + · · · + (n I − 1)s I
N−I
When H0 is true, F has the F distribution with I − 1 and N − I
degrees of freedom.
The denominators in the recipes for MSG and MSE are the two degrees of
freedom I − 1 and N − I of the F test. The numerators are called sums of
squares, from their algebraic form. It is usual to present the results of ANOVA
in an ANOVA table. Output from software usually includes an ANOVA table.
sums of squares
ANOVA table
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EXAMPLE 22.7 ANOVA calculations: software
Look again at the three outputs in Figure 22.3 (page 603). Minitab and Excel give the
ANOVA table. The TI-83, with its small screen, gives the degrees of freedom, sums
of squares, and mean squares separately. Each output uses slightly different language.
The basic ANOVA table is:
Source of variation
df
SS
MS
F statistic
Variation among samples
Variation within samples
2
73
606.45
700.34
MSG = 303.22
MSE = 9.59
31.61
You can check that each mean square MS is the corresponding sum of squares SS
divided by its degrees of freedom df. The F statistic is MSG divided by MSE.
pooled standard deviation
Because MSE is an average of the individual sample variances, it is also called
2
the pooled sample variance, written as s p . When all I populations have the same
2
population variance σ 2 (ANOVA assumes that they do), s p estimates the common variance σ 2 . The square root of MSE is the pooled standard deviation s p .
It estimates the common standard deviation σ of observations in each group.
The Minitab and TI-83 outputs in Figure 22.3 give the value s p = 3.097.
The pooled standard deviation s p is a better estimator of the common σ
than any individual sample standard deviation s i because it combines (pools)
the information in all I samples. We can get a confidence interval for any of
the means µi from the usual form
estimate ± t ∗ SEestimate
using s p to estimate σ . The confidence interval for µi is
sp
xi ± t∗ √
ni
Use the critical value t ∗ from the t distribution with N − I degrees of freedom,
because s p has N − I degrees of freedom. These are the confidence intervals
that appear in Minitab ANOVA output.
EXAMPLE 22.8 ANOVA calculations: without software
We can do the ANOVA test comparing the gas mileages of midsize cars, SUVs, and
pickup trucks using only the sample sizes, sample means, and sample standard deviations. These appear in Example 22.1, but it is easy to find them with a calculator.
There are I = 3 groups with a total of N = 76 vehicles.
The overall mean of the 76 gas mileages in Table 22.1 (page 598) is
n1 x 1 + n2 x 2 + n3 x 3
N
(31)(27.903) + (31)(22.677) + (14)(21.286)
=
76
1865.984
=
= 24.552
76
x=
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Some details of ANOVA
The mean square for groups is
n 1 (x 1 − x)2 + n 2 (x 2 − x)2 + n 3 (x 3 − x)2
I −1
1
=
[(31)(27.903 − 24.552)2 + (31)(22.677 − 24.552)2
3−1
+ (14)(21.286 − 24.552)2 ]
606.424
=
= 303.212
2
MSG =
The mean square for error is
2
MSE =
2
2
(n 1 − 1)s 1 + (n 2 − 1)s 2 + (n 3 − 1)s 3
N−I
(30)(2.561 ) + (30)(3.6732 ) + (13)(2.7582 )
=
73
700.375
=
= 9.594
73
2
Finally, the ANOVA test statistic is
F =
MSG
303.212
=
= 31.60
MSE
9.594
Our work differs slightly from the output in Figure 22.3 because of roundoff error.
We don’t recommend doing these calculations, because tedium and roundoff errors
cause frequent mistakes.
APPLY YOUR KNOWLEDGE
The calculations of ANOVA use only the sample sizes n i , the sample
means x i , and the sample standard deviations s i . You can therefore
re-create the ANOVA calculations when a report gives these summaries
but does not give the actual data. These optional exercises ask you to do
the ANOVA calculations starting with the summary statistics.
22.11 Road rage. Exercise 22.2 describes a study of road rage. Here are the
means and standard deviations for a measure of “angry/threatening
driving” for random samples of drivers in three age-groups:
Age-group
Less than 30 years
30 to 55 years
Over 55 years
n
x
s
244
734
364
2.22
1.33
0.66
3.11
2.21
1.60
(a) The distributions of responses are somewhat right-skewed.
ANOVA is nonetheless safe for these data. Why?
(b) Check that the standard deviations satisfy the guideline for
ANOVA inference.
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(c) Calculate the overall mean response x, the mean squares MSG
and MSE, and the ANOVA F statistic.
(d) What are the degrees of freedom for F ? How significant are the
differences among the three mean responses?
22.12 Exercise and weight loss. What conditions help overweight people
exercise regularly? Subjects were randomly assigned to three
treatments: a single long exercise period 5 days per week; several
10-minute exercise periods 5 days per week; and several 10-minute
periods 5 days per week on a home treadmill that was provided to the
subjects. The study report contains the following information about
weight loss (in kilograms) after six months of treatment:8
Treatment
n
x
s
Long exercise periods
Short exercise periods
Short periods with equipment
37
36
42
10.2
9.3
10.2
4.2
4.5
5.2
(a) Do the standard deviations satisfy the rule of thumb for safe use of
ANOVA?
(b) Calculate the overall mean response x and the mean square for
groups MSG.
(c) Calculate the mean square for error MSE.
(d) Find the ANOVA F statistic and its approximate P-value. Is
there evidence that the mean weight losses of people who follow
the three exercise programs differ?
22.13 Weights of newly hatched pythons. A study of the effect of nest
temperature on the development of water pythons separated python
eggs at random into nests at three temperatures: cold, neutral, and
hot. Exercise 20.22 (page 555) shows that the proportions of eggs that
hatched at each temperature did not differ significantly. Now we will
examine the little pythons. In all, 16 eggs hatched at the cold
temperature, 38 at the neutral temperature, and 75 at the hot
temperature. The report of the study summarizes the data in the
common form “mean ± standard error” as follows:9
Temperature
n
Weight (grams) at hatching
Propensity to strike
Cold
Neutral
Hot
16
38
75
28.89 ± 8.08
32.93 ± 5.61
32.27 ± 4.10
6.40 ± 5.67
5.82 ± 4.24
4.30 ± 2.70
(a) We will compare the mean weights
√ at hatching. Recall that the
standard error of the mean is s / n. Find the standard deviations
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Statistics in Summary
of the weights in the three groups and verify that they satisfy our
rule of thumb for using ANOVA.
(b) Starting from the sample sizes n i , the means x i , and the standard
deviations s i , carry out an ANOVA. That is, find MSG, MSE, and
the F statistic, and use Table D to approximate the P-value. Is
there evidence that nest temperature affects the mean weight of
newly hatched pythons?
22.14 Python strikes. The data in the previous exercise also describe the
“propensity to strike” of the hatched pythons at 30 days of age. This is
the number of taps on the head with a small brush until the python
launches a strike. (Don’t try this with adult pythons.) The data are
again summarized in the form “sample mean ± standard error of the
mean.” Follow the outline in (a) and (b) of the previous exercise for
propensity to strike. Does nest temperature appear to influence
propensity to strike?
Chapter 22 SUMMARY
One-way analysis of variance (ANOVA) compares the means of several
populations. The ANOVA F test tests the overall H0 that all the
populations have the same mean. If the F test shows significant differences,
examine the data to see where the differences lie and whether they are large
enough to be important.
The ANOVA model states that we have an independent SRS from each
population; that each population has a Normal distribution; and that all
populations have the same standard deviation.
In practice, ANOVA inference is relatively robust when the populations are
non-Normal, especially when the samples are large. Before doing the F test,
check the observations in each sample for outliers or strong skewness. Also
verify that the largest sample standard deviation is no more than twice as
large as the smallest standard deviation.
When the null hypothesis is true, the ANOVA F statistic for comparing
I means from a total of N observations in all samples combined has the
F distribution with I − 1 and N − I degrees of freedom.
ANOVA calculations are reported in an ANOVA table that gives sums of
squares, mean squares, and degrees of freedom for variation among groups
and for variation within groups. In practice, we use software to do the
calculations.
STATISTICS IN SUMMARY
After studying this chapter, you should be able to do the following.
A. RECOGNITION
1. Recognize when testing the equality of several means is helpful in
understanding data.
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2.
3.
Recognize that the statistical significance of differences among
sample means depends on the sizes of the samples and on how much
variation there is within the samples.
Recognize when you can safely use ANOVA to compare means.
Check the data production, the presence of outliers, and the sample
standard deviations for the groups you want to compare.
B. INTERPRETING ANOVA
1. Explain what null hypothesis F tests in a specific setting.
2. Locate the F statistic and its P-value on the output of analysis of
variance software.
3. Find the degrees of freedom for the F statistic from the number and
sizes of the samples. Use Table D of the F distributions to
approximate the P-value when software does not give it.
4. If the test is significant, use graphs and descriptive statistics to see
what differences among the means are most important.
Chapter 22 EXERCISES
22.15
22.16
22.17
(Shmuel Thaler/Index Stock Imager/
PictureQuest)
22.18
22.19
(Harry Sieplinga/HMS Images/The Image
Bank/Getty Images)
Exercises 22.15 to 22.19 describe situations in which we want to compare
the mean responses in several populations. For each setting, identify the
populations and the response variable. Then give I , the n i , and N.
Finally, give the degrees of freedom of the ANOVA F test.
Smoking and sleep. A study of the effects of smoking classifies
subjects as nonsmokers, moderate smokers, or heavy smokers. The
investigators interview a sample of 200 people in each group. Among
the questions is “How many hours do you sleep on a typical night?”
Which package design is best? A maker of detergents wants to
compare the attractiveness to consumers of six package designs. Each
package is shown to 120 different consumers who rate the
attractiveness of the design on a 1 to 10 scale.
Growing tomatoes. Do four tomato varieties differ in mean yield?
Grow 10 plants of each variety and record the yield of each plant in
pounds of tomatoes.
Strong concrete. The strength of concrete depends on the mixture of
sand, gravel, and cement used to prepare it. A study compares five
different mixtures. Workers prepare six batches of each mixture and
measure the strength of the concrete made from each batch.
Teaching sign language. Which of four methods of teaching
American Sign Language is most effective? Assign 10 of the 42
students in a class at random to each of three methods. Teach the
remaining 12 students by the fourth method. Record the students’
scores on a standard test of sign language after a semester’s study.
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Chapter 22 Exercises
22.20 Plant diversity. Does the presence of more species of plants increase
the productivity of a natural area, as measured by the total mass of
plant material? An experiment to investigate this question started as
follows:
In a 7-year experiment, we controlled one component of diversity, the
number of plant species, in 168 plots, each 9 m by 9 m. We seeded the
plots, in May 1994, to have 1, 2, 4, 8, or 16 species, with 39, 35, 29,
30, and 35 replicates, respectively.10
Total plant mass in each plot was measured in 2000.
(a) What null and alternative hypotheses does ANOVA test here?
(b) What are the values of N, I , and the n i ?
(c) What are the degrees of freedom of the ANOVA F statistic?
22.21 Plants defend themselves. When some plants are attacked by
leaf-eating insects, they release chemical compounds that attract other
insects that prey on the leaf-eaters. A study carried out on plants
growing naturally in the Utah desert demonstrated both the release of
the compounds and that they not only repel the leaf-eaters but attract
predators that act as the plants’ bodyguards.11 The investigators chose
8 plants attacked by each of three leaf-eaters and 8 more that were
undamaged, 32 plants of the same species in all. They then measured
emissions of several compounds during seven hours. Here are data
(mean ± standard error of the mean for eight plants) for one
compound. The emission rate is measured in nanograms (ng) per hour.
Group
Emission rate
(ng/hr)
Control
Hornworm
Leaf bug
Flea beetle
9.22 ± 5.93
31.03 ± 8.75
18.97 ± 6.64
27.12 ± 8.62
(a) Make a graph that compares the mean emission rates for the four
groups. Does it appear that emissions increase when the plant is
attacked?
(b) What hypotheses does ANOVA test in this setting?
(c) We do not have the full data. What would you look for in deciding
whether you can safely use ANOVA?
(d) What is the relationship between the standard error of the mean
(SEM) and the standard deviation for a sample? Do the standard
deviations satisfy our rule of thumb for safe use of ANOVA?
22.22 Does polyester decay? How quickly do synthetic fabrics such as
polyester decay in landfills? A researcher buried polyester strips in the
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soil for different lengths of time, then dug up the strips and measured
the force required to break them. Breaking strength is easy to measure
and is a good indicator of decay; lower strength means the fabric has
decayed.
Part of the study buried 20 polyester strips in well-drained soil in
the summer. Five of the strips, chosen at random, were dug up after
2 weeks; another 5 were dug up after 4 weeks, 8 weeks, and 16 weeks.
Here are the breaking strengths in pounds:12
2 weeks
4 weeks
8 weeks
16 weeks
(Alix Phanie, Rex Interstock/Stock
Connection/PictureQuest)
118
130
122
124
126
120
136
98
126
114
128
110
120
126
146
140
129
128
140
110
(a) Find the mean strength for each group and plot the means against
time. Does it appear that polyester loses strength consistently over
time after it is buried?
(b) Find the standard deviations for each group. Do they meet our
criterion for applying ANOVA?
(c) In Examples 17.2 and 17.3 (pages 441 and 445), we used the
two-sample t test to compare the mean breaking strengths for
strips buried for 2 weeks and for 16 weeks. The ANOVA F test
extends the two-sample t to more than two groups. Explain
carefully why use of the two-sample t for two of the groups was
acceptable but using the F test on all four groups is not acceptable.
22.23 Can you hear these words? To test whether a hearing aid is right for
a patient, audiologists play a tape on which words are pronounced at
low volume. The patient tries to repeat the words. There are several
different lists of words that are supposed to be equally difficult. Are the
lists equally difficult when there is background noise? To find out, an
experimenter had subjects with normal hearing listen to four lists with
a noisy background. The response variable was the percent of the
50 words in a list that the subject repeated correctly. The data set
contains 96 responses.13 Here are two study designs that could produce
these data:
Design A. The experimenter assigns 96 subjects to 4 groups at
random. Each group of 24 subjects listens to one of the lists. All
individuals listen and respond separately.
Design B. The experimenter has 24 subjects. Each subject listens
to all four lists in random order. All individuals listen and respond
separately.
Does Design A allow use of one-way ANOVA to compare the lists?
Does Design B allow use of one-way ANOVA to compare the lists?
Briefly explain your answers.
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Chapter 22 Exercises
Analysis of Variance for Percent
Source
DF
SS
MS
List
3
920.5
306.8
Error
92
5738.2
62.4
Total
95
6658.6
F
4.92
P
0.003
Individual 95% CIs For Mean
Based on Pooled StDev
Level
1
2
3
4
N
24
24
24
24
Pooled StDev =
Mean
32.750
29.667
25.250
25.583
StDev
7.409
8.058
8.316
7.779
7.898
24.0
28.0
32.0
36.0
Figure 22.10 Minitab ANOVA output for comparing the percents heard correctly
in four lists of words, for Exercise 22.24.
22.24 Can you hear these words? Figure 22.10 displays the Minitab output
for one-way ANOVA applied to the hearing data described in the
previous exercise. The response variable is “Percent,” and “List”
identifies the four lists of words. Based on this analysis, is there good
reason to think that the four lists are not all equally difficult? Write a
brief summary of the study findings.
22.25 How much corn should I plant? How much corn per acre should a
farmer plant to obtain the highest yield? Too few plants will give a low
yield. On the other hand, if there are too many plants, they will
compete with each other for moisture and nutrients, and yields will
fall. To find out, plant at different rates on several plots of ground and
measure the harvest. (Treat all the plots the same except for the
planting rate.) Use software to analyze these data from such an
experiment:14
Plants per acre
12,000
16,000
20,000
24,000
28,000
Yield (bushels per acre)
150.1
166.9
165.3
134.7
119.0
113.0
120.7
130.1
138.4
150.5
118.4
135.2
139.6
156.1
142.6
149.8
149.9
(Craig Aurness/CORBIS)
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(a) Do data analysis to see what the data appear to show about the
influence of plants per acre on yield and also to check the
conditions for ANOVA.
(b) Carry out the ANOVA F test. State hypotheses; give F and its
P-value. What do you conclude?
(c) The observed differences among the mean yields in the samples
are quite large. Why are they not statistically significant?
22.26 Logging in the rainforest: species counts. Table 22.2 (page 604)
gives data on the number of trees per forest plot, the number of species
per plot, and species richness. Exercise 22.3 analyzed the effect of
logging on number of trees. Exercise 22.9 concludes that it would be
risky to use ANOVA to analyze richness. Use software to analyze the
effect of logging on the number of species.
(a) Make a table of the group means and standard deviations. Do the
standard deviations satisfy our rule of thumb for safe use of
ANOVA? What do the means suggest about the effect of logging
on the number of species?
(b) Carry out the ANOVA. Report the F statistic and its P-value and
state your conclusion.
22.27 Nematodes and tomato plants. How do nematodes (microscopic
worms) affect plant growth? A botanist prepares 16 identical planting
pots and then introduces different numbers of nematodes into the
pots. He transplants a tomato seedling into each pot. Here are data on
the increase in height of the seedlings (in centimeters) 16 days after
planting:15
Nematodes
(Eye of Science/Photo Researchers)
0
1,000
5,000
10,000
Seedling growth
10.8
11.1
5.4
5.8
9.1
11.1
4.6
5.3
13.5
8.2
7.4
3.2
9.2
11.3
5.0
7.5
(a) Make a table of means and standard deviations for the four
treatments. Make side-by-side stemplots to compare the
treatments. What do the data appear to show about the effect of
nematodes on growth?
(b) State H0 and Ha for the ANOVA test for these data, and explain
in words what ANOVA tests in this setting.
(c) Use software to carry out the ANOVA. Report your overall
conclusions about the effect of nematodes on plant growth.
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Chapter 22 Exercises
22.28 Earnings of athletic trainers (Optional). How much do newly hired
athletic trainers earn? Earnings depend on the educational and other
qualifications of the trainer and on the type of job. Here are summaries
from a sample survey that contacted institutions advertising for
trainers and asked about the people they hired:16
Type of
institution
High school
Clinic
College
n
Mean salary
x
Std. dev.
s
57
108
94
$26,470
$30,610
$30,019
$9507
$4504
$7158
Calculate the ANOVA table and the F statistic. Are there significant
differences among the mean salaries for the three types of job in the
population of all newly hired trainers? (Comments: The study authors
did an ANOVA. As is often the case, the published work does not
answer questions about the suitability of the analysis. We wonder if
responses from 60% of employers who advertised positions generate a
random sample of new hires. The actual data do not appear in the
report. The salary distributions are no doubt right-skewed; we hope
that strong skewness and outliers are absent because the subjects in
each group hold similar jobs. The large samples will then justify
ANOVA. Also, the sample standard deviations do not quite satisfy our
rule of thumb for safe use of ANOVA.)
22.29 Plant defenses (Optional). The calculations of ANOVA use only the
sample sizes n i , the sample means x i , and the sample standard
deviations s i . You can therefore re-create the ANOVA calculations
when a report gives these summaries but does not give the actual data.
Use the information in Exercise 22.21 to calculate the ANOVA table
(sums of squares, degrees of freedom, mean squares, and the F
statistic). Note that the report gives the standard error of the mean
(SEM) rather than the standard deviation. Are there significant
differences among the mean emission rates for the four populations of
plants?
22.30 F versus t (Optional). We have two methods to compare the means
of two groups: the two-sample t test of Chapter 17 and the ANOVA
F test with I = 2. We prefer the t test because it allows one-sided
alternatives and does not assume that both populations have the same
standard deviation. Let us apply both tests to the same data.
There are two types of life insurance companies. “Stock” companies
have shareholders, and “mutual” companies are owned by their
policyholders. Take an SRS of each type of company from those listed
in a directory of the industry. Then ask the annual cost per $1000 of
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insurance for a $50,000 policy insuring the life of a 35-year-old man
who does not smoke. Here are the data summaries:17
ni
xi
si
Stock
companies
Mutual
companies
13
$2.31
$0.38
17
$2.37
$0.58
(a) Calculate the two-sample t statistic for testing H0: µ1 = µ2
against the two-sided alternative. Use the conservative method to
find the P-value.
(b) Calculate MSG, MSE, and the ANOVA F statistic for the same
hypotheses. What is the P-value of F ?
(c) How close are the two P-values? (The square root of the F
statistic is a t statistic with N − I = n 1 + n 2 − 2 degrees of
freedom. This is the “pooled two-sample t” mentioned on page
454. So F for I = 2 is exactly equivalent to a t statistic, but it is a
slightly different t from the one we use.)
22.31 Exercising to lose weight (Optional). Exercise 22.12 describes a
randomized, comparative experiment that assigned subjects to three
types of exercise program intended to help them lose weight. Some of
the results of this study were analyzed using the chi-square test for
two-way tables, and some others were analyzed using one-way
ANOVA. For each of the following excerpts from the study report, say
which analysis is appropriate and explain how you made your choice.
(a) “Overall, 115 subjects (78% of 148 subjects randomized)
completed 18 months of treatment, with no significant difference
in attrition rates between the groups (P = .12).”
(b) “In analyses using only the 115 subjects who completed 18 months
of treatment, there were no significant differences in weight loss at
6 months among the groups.”
(c) “The duration of exercise for weeks 1 through 4 was significantly
greater in the SB compared with both LB and SBEQ groups
(P < .05). . . . However, exercise duration was greater in SBEQ
compared with both LB and SB groups for months 13 through 18
(P < .05).”
Chapter 22 MEDIA EXERCISES
Analysis of variance works by comparing the differences among the sample
means in several groups to the spread of observations within the groups. A
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Chapter 22 Media Exercises
set of means is significantly different if they are farther apart than random
variation would allow to happen just by chance. The extent of random
variation is assessed by the random spread among observations in the same
group. These exercises use an applet to demonstrate this ANOVA idea.
22.32 ANOVA compares several means. The One-Way ANOVA applet
displays the observations in three groups, with the group means
highlighted by black dots. When you open or reset the applet, the
scale at the bottom of the display shows that for these groups the
ANOVA F statistic is F = 31.74, with P < 0.001.
(a) The middle group has larger mean than the other two. Grab its
mean point with the mouse. How small can you make F ? What
did you do to the mean to make F -small? Roughly how significant
is your small F ?
(b) Starting with the three means aligned from your configuration at
the end of (a), drag any one of the group means either up or down.
What happens to F ? What happens to P ? Convince yourself that
the same thing happens if you move any one of the means, or if
you move one slightly and then another slightly in the opposite
direction.
22.33 ANOVA uses within-group variation. Reset the One-Way ANOVA
applet to its original state. As in Figure 22.6(b) (page 607), the
differences among the three means are highly significant (large F ,
small P ) because the observations in each group cluster tightly about
the group mean.
(a) Use the mouse to slide the Pooled Standard Error at the top of the
display to the right. You see that the group means do not change,
but the spread of the observations in each group increases. What
happens to F and P as the spread among the observations in each
group increases? What are the values of F and P when the slider is
all the way to the right? This is similar to Figure 22.6(a): variation
within groups hides the differences among the group means.
(b) Leave the Standard Error slider at the extreme right of its scale, so
that spread within groups stays fixed. Use the mouse to move the
group means apart. What happens to F and P as you do this?
APPLET
APPLET
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