1. science
2. mathematics
3. statistics

Comparing Three or More Means (One-Way Anaylsis of Variance)
are normally distributed with s = 1.2 minutes. In a study, the
chain reconfigured five restaurants to have a single line and
measured the wait times for 50 randomly selected customers.
The sample standard deviation was determined to be
s = 0.84 minute. Does the evidence indicate that the variability in wait time is less for a single line than for multiple
lines at the a = 0.05 level of significance? Reject H0
19. Heights of Baseball Players Data obtained from the National Center for Health Statistics show that men between the
ages of 20 and 29 have a mean height of 69.3 inches, with a
standard deviation of 2.9 inches. A baseball analyst believes
that the standard deviation of heights of major-league baseball players is less than 2.9 inches. The heights (in inches) of
20 randomly selected players are shown in the table.
72
74
71
72
76
70
77
75
72
72
77
72
75
70
73
73
75
73
74
74
Source: espn.com
C–19
(a) Verify that the data are normally distributed by
drawing a normal probability plot.
(19c) Do not
H0standard deviation. 2.059 inches
(b) Compute
thereject
sample
(c) Test the belief at the a = 0.01 level of significance.
20. NCAA Softball NCAA rules require the circumference
of a softball to be 12 ; 0.125 inches. A softball manufacturer bidding on an NCAA contract is shown to meet
the requirements for mean circumference. Suppose the
NCAA also requires that the standard deviation of the
softball circumferences not exceed 0.05 inch. A representative from the NCAA believes the manufacturer
does not meet this requirement. She collects a random
sample of 20 softballs from the production line and finds
that s = 0.09 inch. Do you believe the balls conform?
Use the a = 0.05 level of significance? Yes; reject H0
21. P-Values Determine the exact P-value of the hypothesis test in Problem 9. P-value ⴝ 0.0446
22. P-Values Determine the exact P-value of the hypothesis test in Problem 10. P-value ⴝ 0.3199
4. Comparing Three or More Means
(One-Way Analysis of Variance)
Preparing for This Section Before getting started, review the following:
• Completely randomized design (Section 1.5,
pp. 42–43)
• Comparing two population means (Section 11.2,
pp. 521–528)
• Nature of hypothesis testing (Section 10.1,
pp. 454–460)
• Normal probability plots (Section 7.4, pp. 354–358)
Objectives
Note to Instructor
It is recommended that hand computations be de-emphasized in this section.
Concentrate on the concepts, and let
technology do the number crunching.
Definition
In Other Words
In ANOVA, the null hypothesis is
always that the means of the different
populations are equal. The alternative
hypothesis is always that the mean of
at least one population is different from
the others.
• Boxplots (Section 3.5, pp. 161–164)
Verify the requirements to perform a one-way ANOVA
Test hypotheses regarding three or more means using
one-way ANOVA
In Section 11.2, we compared two population means. Just as we extended the
concept of comparing two population proportions (Section 11.3) to comparing
three or more population proportions (Tests for Homogeneity of Proportions,
Section 12.2), we now extend the concept of comparing two population means
to comparing three or more population means. The procedure for doing this is
called Analysis of Variance, or ANOVA for short.
Analysis of Variance (ANOVA) is an inferential method that is used to
test the equality of three or more population means.
For example, a family doctor might want to show the mean HDL (socalled good) levels of cholesterol of males in the age groups 20 to 29 years old,
40 to 49 years old, and 60 to 69 years old are different. To conduct a hypothesis test, we assume the mean HDL cholesterol of each age group is the same.
If we call the 20- to 29-year-olds population 1, 40- to 49-year-olds population
2, and 60 to 69-year-olds population 3, our null hypothesis would be
H0: m1 = m2 = m3
C–20
Topics to Discuss
versus the alternative hypothesis,
H1: At least one of the population means is different from the others
CAUTION!
Do not test H0: m1 = m2 = m3 by
conducting three separate hypothesis
tests, because the probability of
making a Type I error will be much
higher than a.
CAUTION
It is vital that individuals be
randomly assigned to treatments.
As another example, a medical researcher might want to compare the effect
different levels of an experimental drug have on hair growth. The researcher
might randomly divide a group of subjects into three different treatment groups.
Group 1 might receive a placebo once a day, group 2 might receive 50 mg of the
experimental drug once a day, and group 3 might receive 100 mg of the experimental drug once a day.The researcher then compares the mean numbers of new
hair follicles for each of the three treatment groups. The three different treatment groups correspond to three different populations.
It is tempting to test the null hypothesis H0: m1 = m2 = m3 by comparing the
population means two at a time using the techniques introduced in Section 11.2.
If we proceeded this way, we would need to test three different hypotheses:
H0: m1 = m2 and H0: m1 = m3 and H0: m2 = m3
Each hypothesis would have a probability of Type I error (rejecting the null
hypothesis when it is true) of a. If we used an a = 0.05 level of significance,
each hypothesis would have a 95% probability of rejecting the null hypothesis
when the alternative hypothesis is true (i.e. a 95% probability of making a correct decision). The probability that all three hypotheses correctly reject the null
hypothesis is 0.953 = 0.86 (assuming the tests are independent). There is a
1 - 0.953 = 1 - 0.86 = 0.14, or 14%, probability that at least one hypothesis
will lead to an incorrect rejection of H0 . A 14% probability of a Type I error is
much higher than the desired 5% probability. As the number of populations
that are to be compared increases, the probability of making a Type I error using
multiple t-tests for a given value of a also increases.
To address this problem, Sir Ronald A. Fisher (1890–1962) introduced the
method of analysis of variance. Although it seems strange to name a procedure
that is used to compare means analysis of variance, the justification for the name
will become clear as we see the procedure in action.
The procedure used in this section is called one-way analysis of variance because there is only one factor that distinguishes the various populations in the
study. For example, in comparing the mean HDL cholesterol levels of males, the
only factor that distinguishes the three groups is age. In comparing the effectiveness of the hair growth drug, the only factor that distinguishes the three groups
is the amount of the experimental drug received (0 mg, 50 mg, or 100 mg).
In performing one-way analysis of variance, care must be taken so that the
subjects are similar in all characteristics except for the level of the treatment.
Fisher stated that this is easiest to accomplish through randomization, which
neutralizes the effect of the uncontrolled variables. In the hair growth example,
the subjects should be similar in terms of eating habits, age, and so on, by randomly assigning the subjects to the three groups.
Verify the Requirements to Perform
a One-Way ANOVA
To perform a one-way ANOVA test, certain requirements must be satisfied.
Requirements of a One-Way ANOVA Test
1. There are k simple random samples from k populations.
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 have the same variance; that is, each treatment group
has population variance s2.
Comparing Three or More Means (One-Way Anaylsis of Variance)
C–21
Suppose we are testing a hypothesis regarding k = 3 population means so
that the null hypothesis is
H0: m1 = m2 = m3
and the alternative hypothesis is
H1: at least one of the population means is different from the others
Figure 1(a) shows the distribution of each population if the null hypothesis is
true, and Figure 1(b) shows what the distribution of each population might look
like if the alternative hypothesis is true.
Figure 1
m1 m2 m3
(a)
In Other Words
Try to design experiments that use
ANOVA so that each treatment group is
the same size.
Note to Instructor
It may seem contradictory that we require equal population variances in oneway ANOVA, but are unwilling to make this
assumption when comparing two means.
The reason lies in the fact that there is
no parametric alternative to one-way
ANOVA. The parametric alternative to
pooling is Welch’s t.
EXAMPLE 1
Table 1
Control Fenugreek
Garlic
m1
m3
(b)
m2
The methods of one-way ANOVA are robust, so small departures from the
requirement of normality 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.
We can verify the requirement of normality by constructing normal probability plots. The requirement of equal population variances is more difficult to
verify. However, a general rule of thumb is as follows:
Verifying the Requirement of Equal Population Variances
The one-way ANOVA procedures may be used provided that the largest
sample standard deviation is no more than two times larger than the smallest sample standard deviation.
Testing the Requirements of One-Way ANOVA
Problem: Researcher Jelodar Gholamali wanted to determine the effectiveness of various treatments on glucose levels of diabetic rats. He randomly assigned diabetic albino rats into four treatment groups. Group 1 rats served as a
control group and were fed a regular diet. Group 2 rats were served a regular diet supplemented with a herb, fenugreek. Group 3 rats were served a
regular diet supplemented with garlic. Group 4 rats were served a regular
Onion
diet supplemented with onion. The basis for the study is that Persian folk299.7
lore states that diets supplemented with fenugreek, garlic, or onion help to
258.3
treat diabetes. After 15 days of treatment, the blood glucose was measured
in milligrams per deciliter (mg/dL). The results presented in Table 1 are
286.8
based on the results published in the article. Verify that the requirements
244.0
to perform one-way ANOVA are satisfied. Source: Jelodar, G. A., et.al., Effect
288.1
229.1
177.4
296.8
240.7
202.2
267.8
239.4
163.1
256.7
207.7
184.7
292.1
225.7
197.9
282.9
230.8
164.6
297.1
260.3
206.6
193.9
249.9
283.8
213.3
158.1
265.1
267.1
of Fenugreek, Onion and Garlic on Blood Glucose and Histopathology of Pancreas of Alloxan-Induced Diabetic Rats, Indian Journal of Medical Science,
2005;59:64–69.
C–22
Topics to Discuss
Approach: We must verify the previously listed requirements.
Solution
1. The rats were randomly assigned to each treatment group.
2. None of the subjects selected is related in any way, so the samples are independent.
3. Figure 2 shows the normal probability plots for all four treatment
groups. All the normal probability plots are roughly linear, so we conclude that the sample data come from populations that are approximately normally distributed.
Figure 2
(b)
(a)
(d)
(c)
4. The sample standard deviations for each sample are computed using
MINITAB and presented in Figure 3. The largest standard deviation is
21.18 mg/dL, and the smallest standard deviation is 13.49 mg/dL. Because the largest standard deviation is not more than two times larger
than the smallest standard deviation 12 # 13.49 = 26.98 7 21.182, the requirement of equal population variances is satisfied.
Figure 3
Descriptive Statistics: Control, Fenugreek, Garlic, Onion
Variable
Control
Fenugreek
Garlic
Onion
N
8
8
8
8
N*
0
0
0
0
Mean
278.56
224.16
180.24
271.00
SE Mean
5.31
4.77
6.03
7.49
StDev
15.03
13.49
17.06
21.18
Minimum
256.70
206.60
158.10
244.00
Q1
262.18
209.10
163.48
252.06
Median
283.35
227.40
181.05
266.10
Q3
291.10
237.25
196.90
294.53
Maximum
296.80
240.70
202.20
299.70
Because all four requirements are satisfied, we can perform a one-way
ANOVA.
Test Hypotheses Regarding Three or More
Means Using One-Way ANOVA
The computations in performing any analysis of variance test are tedious, so virtually all researchers use statistical software to conduct the test.When using software, it
is easiest to use the P-value approach. The nice thing about P-values is that the decision rule is always the same, regardless of the type of hypothesis being tested.
Decision Rule in the One-Way ANOVA Test
If the P-value is less than the level of significance, a, reject the null hypothesis.
EXAMPLE 2
Performing One-Way ANOVA Using Technology
Problem: The researcher in Example 1 wishes to determine if there is a difference in the mean glucose among the four treatment groups. Conduct the test
at the a = 0.05 level of significance.
Comparing Three or More Means (One-Way Anaylsis of Variance)
C–23
Approach: We will use MINITAB, Excel, and a TI-84 Plus graphing calculator to test the hypothesis. If the P-value is less than the level of significance, we
reject the null hypothesis.
Result: The researcher wants to show there is a difference in the mean glucose
among the four treatment groups. The null hypothesis is always a statement of no
difference. In this case it is a statement that the mean glucose among the four treatment groups is the same. So the null hypothesis is
H0: mcontrol = mfenugreek = mgarlic = monion
versus the alternative hypothesis
H1: at least one of the population means is different from the others
Figure 4(a) shows the output from MINITAB, Figure 4(b) shows the
output from Excel, and Figure 4(c) shows the output from a TI-84 Plus
graphing calculator.
Figure 4
One-Way ANOVA: Control, Fenugreek, Garlic, Onion
Source
Factor
Error
Total
DF
3
28
31
S 16.94
SS
50091
8032
58122
MS
16697
287
R Sq 86.18%
F
58.21
P
0.000
R Sq (adj) 84.70%
Individual 95% CIs For Mean
Based on Pooled StDev
Level
Control
Fenugreek
Garlic
Onion
N
8
8
8
8
Mean
278.56
224.16
180.24
271.00
StDev --+---------+---------+---------+------15.03
(---*--)
13.49
(--*---)
17.06 (---*--)
21.18
(--*---)
--+---------+---------+---------+-------
175
210
Pooled St Dev 16.94
(a) MINITAB Output
(b) Excel Output
P-value
(c) TI–84 Plus Output
245
280
C–24
Topics to Discuss
Notice that MINITAB indicates that the P-value is 0.000. This does not
mean that the P-value is 0, but instead means that the P-value is less than
0.0001. The output of Excel and the TI-84 Plus confirm this by indicating the
P-value is 3.7 * 10-12. Because the P-value is less than the level of significance, we reject the null hypothesis. There is sufficient evidence to conclude
that at least one of the population means for glucose levels is different from
the others.
Whenever you perform analysis of variance, it is always a good idea to present visual evidence that supports the conclusions of the test. Side-by-side boxplots are a great way to help see the results of the ANOVA procedure. Figure 5
shows the side-by-side boxplots of the data presented in Table 1. The boxplots
support the ANOVA results from Example 2.
Figure 5
Blood Glucose Levels
Control
Fenugreek
Garlic
Onion
150
175
200
225
250
Glucose (mg/dl)
275
300
Now Work Problems 11(a)–(d).
A Conceptual Understanding of One-Way ANOVA
Look again at Figure 4. You may have noticed that each of the three outputs
included an F-value 1F = 58.212. We now illustrate the idea behind the F-test
statistic. Remember, in testing any hypothesis, the null hypothesis is assumed
to be true until the evidence indicates otherwise. In testing the hypothesis
regarding k population means, we assume that m1 = m2 = Á = mk = m. That
is, we assume that all k samples come from the same normal population whose
mean is m and variance is s2. Table 2 shows the statistics that result by sampling from each of the k populations.
Table 2
Sample Size
Sample Mean
Sample Standard
Deviation
1
n1
x1
s1
2
n2
x2
s2
3
n3
x3
s3
Population
o
o
o
o
k
nk
xk
sk
Comparing Three or More Means (One-Way Anaylsis of Variance)
C–25
The computation of the F test statistic requires that we understand mean
squares. A mean square is an average (mean) of squared values. For example, any
variance is a mean square. The F-test statistic is the ratio of two mean squares.
If the null hypothesis is true, then each treatment group comes from the same
population whose mean is m and whose variance is s2. The sample mean of the
entire set of data (all treatment groups combined) is a good estimate of m. We will
call this sample mean x. Similarly, the sample mean of Sample 1 (or treatment 1)
will be x1 , the sample mean of Sample 2 (or treatment 2) will be x2 , and so on.
Finding an estimate of s2 is somewhat more complicated. One approach is
to estimate s2 by computing a measure of variation in sample means from one
treatment group to the next, weighted by the sample size of the corresponding
treatment group. We call this value the mean square due to treatment, denoted
MST. It is computed as follows:
MST =
n11x1 - x22 + n21x2 - x22 + Á + nk1xk - x22
k - 1
(1)
If the null hypothesis is true, then MST is an unbiased estimator of s2, the variance of the population.
A second approach to estimating s2 is to compute the sample variance for
each sample (or treatment), and then to find a weighted average of the sample
variances. We call this the mean square due to error, denoted MSE. The mean
square due to error is an unbiased estimator of s2 whether or not the null hypothesis is true. It is computed as follows:
MSE =
1n1 - 12s21 + 1n2 - 12s22 + Á + 1nk - 12s2k
n - k
(2)
where n is the total size of the sample. In other words, n = n1 + n2 + Á + nk .
The F-test statistic is the ratio of the two estimates of s2.
F =
mean square due to treatment
MST
=
mean square due to error
MSE
If the null hypothesis is true, both MST and MSE provide unbiased estimates
for s2, the population variance. So, if the null hypothesis is true, we would expect
the F-test statistic to be close to 1. However, if the null hypothesis is not true, at least
one of the sample means from a treatment group will be “far away” from x, the
sample mean of the entire data set.This will cause MST to be large relative to MSE,
which ultimately leads to an F-test statistic substantially larger than 1.
We now present the steps to be used in the computation of the F-test statistic.
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 s21
represent the sample variance for sample 1, s22 represent the sample variance for sample 2, and so on.
Step 4: Compute the mean square due to treatment, MST.
Step 5: Compute the mean square due to error, MSE.
Step 6: Compute the F-test statistic:
F =
mean square due to treatment
MST
=
mean square due to error
MSE
C–26
Topics to Discuss
EXAMPLE 3
Computing the F-Test Statistic
Problem: Compute the F-test statistic for the data presented in Example 1.
Approach: We follow Steps 1–6 just presented.
Solution
Step 1: Compute the mean of the entire data set.
x =
7633.30
288.1 + 296.8 + Á + 249.9 + 265.1
=
= 238.54
32
32
Step 2: Find the sample mean of each treatment. Call the control group population 1, the fenugreek group population 2, the garlic group population 3, and
the onion group population 4. Then
x1 =
288.1 + 296.8 + Á + 283.8
= 278.56
8
x2 =
229.1 + 240.7 + Á + 213.3
= 224.16
8
x3 =
177.4 + 202.2 + Á + 158.1
= 180.24
8
x4 =
299.7 + 258.3 + Á + 265.1
= 271.00
8
Step 3: Find the sample variance for each treatment group.
1288.1 - 278.5622 + 1296.8 - 278.5622 + Á + 1283.8 - 278.5622
= 225.77
8 - 1
1229.1 - 224.1622 + 1240.7 - 224.1622 + Á + 1213.3 - 224.1622
s22 =
= 181.99
8 - 1
1177.4 - 180.2422 + 1202.2 - 180.2422 + Á + 1158.1 - 180.2422
s23 =
= 291.03
8 - 1
1299.7 - 27122 + 1258.3 - 27122 + Á + 1265.1 - 27122
s24 =
= 448.58
8 - 1
s21 =
Step 4: Compute MST.
MST =
=
81278.56 - 238.5422 + 81224.16 - 238.5422 + 81180.24 - 238.5422 + 81271 - 238.5422
4 - 1
50,087.4112
3
= 16,695.80
Step 5: Compute MSE.
MSE =
=
18 - 12225.77 + 18 - 12181.99 + 18 - 12291.03 + 18 - 12448.48
32 - 4
8030.89
28
= 286.82
Step 6: Compute the F-test statistic.
F =
mean square due to treatment
16,695.80
MST
=
=
= 58.21
mean square due to error
MSE
286.84
This is the same result provided by the statistical software and graphing calculator in Example 2.
Comparing Three or More Means (One-Way Anaylsis of Variance)
C–27
Looking back at the formula for the mean square due to treatment, we notice
that if the null hypothesis is true the overall mean, x, should be close to the means
computed from each sample (treatment), x1 , x2 , and so on. If one or more of the
means computed from each sample is substantially different from the overall
mean, MST will be large, which in turn makes the F-statistic large. In Example 3,
we can see that the control group has a sample mean much larger than the overall
mean 1x1 = 278.56 versus x = 238.542, and the garlic group has a sample mean
much smaller than the overall mean 1x3 = 180.24 versus x = 238.542.
The results of the computations that lead to the F-test statistic are presented in an ANOVA table, the form of which is shown in Table 3.
Table 3
Figure 6
Area 0.05
F0.05,3,28 2.99
In Other Words
If we reject the null hypothesis when
doing ANOVA, we are rejecting the
assumption that the population means
are all equal. However, the test doesn’t
tell us which means differ.
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Treatment
50,087.41
k - 1 = 4 - 1 = 3
Error
8030.89
n - k = 32 - 4 = 28
Total
58,118.3
n - 1 = 32 - 1 = 31
Mean
Squares
F-Test
Statistic
16,695.80
58.21
286.82
Notice that the “sum of squares treatment” is the numerator of the computation for the mean square due to treatment. The “sum of squares error” is the
numerator of the computation for the mean square due to error. Each entry in
the mean square column is the sum of squares divided by the corresponding degrees of freedom. In addition,
Sum of squares total = sum of squares treatment + sum of squares error
This result is a consequence of the requirement of independence of the observations within the groups.
If we were conducting this ANOVA test by hand, we would compare the
F-test statistic with a critical F-value. The critical F-value is the F-value whose
area in the right tail is a with k - 1 degrees of freedom in the numerator and
n - k degrees of freedom in the denominator. The critical F-value for the claim
made in Example 1 at the a = 0.05 level of significance is F0.05, 3.28 L 2.99. Because the F-test statistic, 58.21, is greater than the critical F, reject the null hypothesis. See Figure 6.
Suppose the null hypothesis of equal population means is rejected. This
conclusion tells us that at least one of the population means is different from the
others, but we don’t know which one. We can determine which population
means differ using Tukey’s tests, which is not discussed in this text.
ASSESS YOUR UNDERSTANDING
Concepts and Vocabulary
1. The acronym ANOVA stands for _____ _____ _____.
2. What are the requirements to perform a one-way
ANOVA? Is the test robust?
3. What is the mean square due to treatment estimate of s2?
What is the mean square due to error estimate of s2?
4. Why does a large value of the F statistic provide evidence
against the null hypothesis H0: m1 = m2 = Á = mk?
Skill Building
In Problems 5 and 6, fill in the ANOVA table.
5.
Source of
Variation
Treatment
Error
Total
6.
Sum of
Squares
Degrees of
Freedom
387
2
8042
27
Mean
Squares
F-Test
Statistic
Source of
Variation
Sum of
Squares
Treatment
2814
3
Error
4915
36
Total
Degrees of
Freedom
Mean
Squares
F-Test
Statistic
C–28
Topics to Discuss
In Problems 7 and 8, determine the F-test statistic based on the given summary statistics. cHint: x =
7.
Population
Sample
Size
Sample
Mean
1
10
40
2
10
42
3
10
Population
Sample
Size
Sample
Mean
Sample
Variance
48
1
15
105
34
31
2
15
110
40
25
3
15
108
30
4
15
90
38
Sample
Variance
44
8.
©nixi
.d
©ni
9. The following data represent a simple random sample
of n = 4 from three populations that are known to
be normally distributed. Verify that the F-test statistic is 2.04.
10. The following data represent a simple random sample of
n = 5 from three populations that are known to be normally distributed. Verify that the F-test statistic is 2.599.
Sample 1
Sample 2
Sample 3
73
67
72
82
77
80
82
66
87
19
81
67
77
30
97
83
96
Sample 1
Sample 2
Sample 3
28
22
25
23
25
24
30
17
27
23
Applying the Concepts
11. Corn Production The data in the table represent the numNW ber of corn plants in randomly sampled rows (a 17-foot by
5-inch strip) for various types of plot. An agricultural researcher wants to know whether the mean numbers of
plants for each plot type are different.
Sludge plot
Plot Type
Plot Type
Number of Plants
25
27
33
30
No Till
28
Spring Disk
27
Spring disk
32
30
33
35
34
34
No till
30
26
29
32
25
29
Sludge Plot
Source: Andrew Dieter and Brad Schmidgall. Joliet Junior College
(a) Write the null and alternative hypotheses.
(b) State the requirements that must be satisfied to use
the one-way ANOVA procedure.
(c) Use the following partial MINITAB output to test
the hypothesis at the a = 0.05 level of significance.
Reject H0
One-way ANOVA: Sludge Plot, Spring Disk, No Till
Source
Factor
Error
Total
DF
2
15
17
SS
84.11
88.83
172.94
MS
42.06
5.92
F
7.10
35
30
25
Number of Plants
(e) Verify that the F-test statistic is 7.10.
12. Soybean Yield 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 know
whether the mean numbers of pods for each plot type are
different.
P
0.007
(d) Shown are side-by-side boxplots of each type of plot.
Do these boxplots support the results obtained in
part (c)?
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
Source: Andrew Dieter and Brad Schmidgall, Joliet Junior College
Comparing Three or More Means (One-Way Anaylsis of Variance)
(a) Write the null and alternative hypotheses.
(b) State the requirements that must be satisfied to use
the one-way ANOVA procedure.
(c) Use the following MINITAB output to determine if
the number of births differs by day of the week
using the a = 0.01 level of significance:
(a) Write the null and alternative hypotheses.
(b) State the requirements that must be satisfied to use
the one-way ANOVA procedure.
(c) Use the following MINITAB output to test the hypothesis at the a = 0.05 level of significance.
Reject H0
One-way ANOVA: Liberty, No Till Chisel Plowed
Source
Factor
Error
Total
DF
2
24
26
MS
74.5
19.8
SS
149.0
474.7
623.6
F
3.77
P
0.038
C–29
One-way ANOVA: Mon, Tues, Wed, Thurs, Fri
Source
Factor
Error
Total
(d) Shown are side-by-side boxplots of each type of plot.
Do these boxplots support the results obtained in
part (c)?
DF
SS
MS
4 11507633 2876908
35 10270781 293451
39 21778414
F
9.80
P
0.000
(d) Shown are side-by-side boxplots of each type of plot.
Do these boxplots support the results obtained in
part (c)?
Chisel Plowed
Plot Type
Friday
No Till
Thursday
Day
Liberty
20
30
Number of Plants
40
(e) Verify that the F-test statistic is 3.77.
(f) Based on the boxplots, which type of plot appears to
have a significantly different mean number of plants?
Liberty or No Till
13. Births by Day of Week An obstetrician knew that there
were more live births during the week than on weekends.
She wanted to determine whether the mean number of
births was different for each of the five days of the week.
She randomly selected eight dates for each of the five days
of the week and obtained the following data:
Monday
Tuesday
Wednesday Thursday Friday
10,456
11,621
11,084
11,171
11,545
10,023
11,944
11,570
11,745
12,321
10,691
11,045
11,346
12,023
11,749
10,283
12,927
11,875
12,433
12,192
10,265
12,577
12,193
12,132
12,422
11,189
11,753
11,593
11,903
11,627
11,198
12,509
11,216
11,233
11,624
11,465
13,521
11,818
12,543
12,543
Source: National Center for Health Statistics
Wednesday
Tuesday
Monday
10,000
11,000
12,000
13,000
Number of Births
(e) Verify that the F-test statistic is 9.80.
(f) Based on the boxplots, which day of the week appears to have a significantly different number of
births? Monday
14. Punkin Chunkin The World Championship Punkin
Chunkin contest is held every fall in Millsboro, Delaware.
Contestants build devices meant to hurl 8 to 10-pound
pumpkins across a field. One class of entry is the air cannon, which must use compressed air to fire a pumpkin.
The following data represent a simple random sample of
distances that pumpkins have traveled (in feet) for the
years 2001 to 2004. Is there evidence to conclude that the
C–30
Topics to Discuss
mean distance that a pumpkin is fired is different for the
various years?
Financial
Energy
Utilities
10.76
12.72
11.88
15.05
13.91
5.86
17.01
6.43
13.46
3967.95
5.07
11.19
9.90
3596.09
19.50
18.79
3.95
3539.66
3970.00
8.16
20.73
3.44
3591.16
3877.68
10.38
9.60
7.11
6.75
17.40
15.70
2001
2002
2003
2004
3494.70
3881.54
3895.85
4224.00
3360.20
3232.74
4434.28
4065.60
3911.02
3696.19
3448.56
3124.40
3414.54
3665.89
3718.77
3816.64
3453.12
3631.78
Source: www.punkinchunkin.com/main.htm
(a) Write the null and alternative hypotheses.
(b) State the requirements that must be satisfied to use
the one-way ANOVA procedure.
(c) Use the following MINITAB output to test the claim
at the a = 0.05 level of significance.
One-way ANOVA: 2001, 2002, 2003, 2004
Source
Factor
Error
Total
DF
3
20
23
SS
659242
1559456
2218698
sectors and obtained the 5-year rates of return shown in
the following table (in percent):
F
2.82
MS
219747
77973
P
0.065
(d) Shown are side-by-side boxplots of each type of plot.
Do these boxplots support the results obtained in
part (c)?
Boxplot of 2001, 2002, 2003, 2004
2001
2002
2003
Source: Morningstar.com
(a) State the null and alternative hypothesis.
(b) Verify that the requirements to use the one-way
ANOVA procedure are satisfied. Normal probability
plots indicate that the sample data come from normal
populations.
(c) Test whether the mean rates of return are different at
the a = 0.05 level of significance. Do not reject H0
(d) Draw boxplots of the three sectors to support the results obtained in part (c).
16. Reaction Time In an online psychology experiment sponsored by the University of Mississippi, researchers asked
study participants to respond to various stimuli. Participants were randomly assigned to one of three groups. Subjects in group 1 were in the simple group. They were
required to respond as quickly as possible after a stimulus
was presented. Subjects in group 2 were in the go/no-go
group. These subjects were required to respond to a particular stimulus while disregarding other stimuli. Finally,
subjects in group 3 were in the choice group. They needed
to respond differently, depending on the stimuli presented. Depending on the type of whistle sound, the subject
must press a certain button. The reaction time (in seconds)
for each stimulus is presented in the table.
Simple
Go/No Go
Choice
0.430
0.588
0.561
0.498
0.375
0.498
0.480
0.409
0.519
0.376
0.613
0.538
0.402
0.481
0.464
0.329
0.355
0.725
Source: PsychExperiments; The University of Mississippi;
www.olemiss.edu/psychexps/
2004
3000 3200 3400 3600 3800 4000 4200 4400 4600
Distance (feet)
(e) Verify that the F-test statistic is 2.82.
15. Rates of Return A stock analyst wondered whether the
mean rate of return of financial, energy, and utility stocks
differed over the past 5 years. He obtained a simple random sample of eight companies from each of the three
The researcher wants to determine if the mean reaction
times for each stimulus are different.
(a) State the null and alternative hypotheses.
(b) Verify that the requirements to use the one-way
ANOVA procedure are satisfied. Normal probability
plots indicate that the sample data come from a normal population.
(c) Test whether the mean reaction times to the three stimuli differ at the a = 0.05 level of significance.
(d) Draw boxplots of the three stimuli to support the analytic results obtained in part (c).
(16a) H0: mS ⴝ mG ⴝ mC vs. H1: at least one mean is not equal
(16b) Do not reject H0
C–31
Comparing Three or More Means (One-Way Anaylsis of Variance)
17. Crash Data The Insurance Institute for Highway Safety conducts experiments in
which cars are crashed into a fixed barrier at 40 mph. In the Institute’s 40-mph offset
test, 40% of the total width of each vehicle strikes a barrier on the driver’s side. The
barrier’s deformable face is made of aluminum honeycomb, which makes the forces in
the test similar to those involved in a frontal offset crash between two vehicles of the
same weight, each going just less than 40 mph. Suppose you are in the market to buy a
new family car.You want to know whether the mean chest compression resulting from
this offset crash is the same for large family cars, passenger vans, and midsize utility vehicles. The following data were collected from the institute’s study.
Large
Family Cars
Hyundai XG350
Chest
Compression
(mm)
Chest
Compression
(mm)
Passenger Vans
33
Toyota Sienna
29
Midsize
Utility Vehicles
Chest
Compression
(mm)
Honda Pilot
31
Ford Taurus
28
Honda Odyssey
28
Toyota 4Runner
36
Buick LeSabre
28
Ford Freestar
27
Mitsubishi Endeavor
35
Chevrolet Impala
26
Mazda MPV
30
Nissan Murano
29
Chrysler 300
34
Chevrolet Uplander
26
Ford Explorer
29
Pontiac Grand Prix
34
Nissan Quest
33
Jeep Liberty
36
Toyota Avalon
31
Kia Sedona
21
Buick Randezvous
29
Source: Insurance Institute for Highway Safety
(17a) H0: mL ⴝ mP ⴝ mM vs. H1: at least one mean is not equal
(a) The researcher wants to know if the means for chest compression for each class
of vehicle differ. State the null and alternative hypotheses.
(b) Verify that the requirements to use the one-way ANOVA procedure are satisfied. Normal probability plots indicate that the sample data come from normal
populations.
(c) Test whether the mean chest compression for each vehicle type is different at
the a = 0.01 level of significance. Do not reject H0
(d) Draw boxplots of the three types of vehicle to support the analytic results obtained in part (c).
18. Crash Data The Insurance Institute for Highway Safety conducts experiments in
which cars are crashed into a fixed barrier at 40 mph. In the Institute’s 40-mph offset test, 40% of the total width of each vehicle strikes a barrier on the driver’s side.
The barrier’s deformable face is made of aluminum honeycomb, which makes the
forces in the test similar to those involved in a frontal offset crash between two vehicles of the same weight, each going just less than 40 mph. Suppose you are in the
market to buy a new family car. You want to know if the mean head injury resulting
from this offset crash is the same for large family cars, passenger vans, and midsize
utility vehicles. The following data were collected from the institute’s study.
Head
Injury
(hic)
Large
Family Cars
Passenger Vans
Head
Injury
(hic)
Midsize
Utility Vehicles
Head
Injury
(hic)
Hyundai XG350
264
Toyota Sienna
148
Honda Pilot
202
Ford Taurus
170
Honda Odyssey
238
Toyota 4Runner
216
Buick LeSabre
409
Ford Freestar
340
Mitsubishi Endeavor
186
Chevrolet Impala
204
Mazda MPV
693
Nissan Murano
517
Chrysler 300
149
Chevrolet Uplander
550
Ford Explorer
202
Pontiac Grand Prix
627
Nissan Quest
470
Kia Sorento
552
Toyota Avalon
166
Kia Sedona
332
Chevy Trailblazer
386
Source: Insurance Institute for Highway Safety
(18a) H0: mL ⴝ mP ⴝ mM vs. H1: at least one of the means is not equal
(a) The researcher wants to know whether the means for head injury for each class
of vehicle differ. State the null and alternative hypotheses.
C–32
Topics to Discuss
(b) Verify that the requirements to use the one-way
ANOVA procedure are satisfied. Normal probability
plots indicate that the sample data come from normal
populations.
(c) Test whether the mean head injury for each vehicle
type differs at the a = 0.01 level of significance. Do
not reject H0
(d) Draw boxplots of the three vehicle types to support
the analytic results obtained in part (c).
viduals in group 3 received the NCEP-1 Diet ( 610%
saturated fat, 12% monounsaturated fat, and 6%
polyunsaturated fat). After 28 days, their LDL cholesterol levels were recorded. The data in the following
table are based on this study.
Saturated Fat
Mediterranean
245
56
125
123
78
100
166
101
140
104
158
151
196
145
138
300
118
268
140
145
75
240
211
71
218
131
184
Texas
173
125
116
5.46
223
160
144
130
101
19. pH in Rain An environmentalist wanted to determine
if the mean acidity of rain differed among Alaska,
Florida, and Texas. He randomly selected six rain dates
at each of the three locations and obtained the data in
the following table.
Alaska
5.41
Florida
4.87
NCEP-1
5.39
5.18
6.29
177
4.90
4.40
5.57
193
83
135
5.14
5.12
5.15
224
263
144
149
150
130
4.80
4.89
5.45
5.24
5.06
5.30
Source: National Atmospheric Deposition Program
(19a) H0: mA ⴝ mF ⴝ mT vs. H1: at least one of the
means is not equal
(a) State the null and alternative hypothesis.
(b) Verify that the requirements to use the one-way
ANOVA procedure are satisfied. Normal probability
plots indicate that the sample data come from a normal population.
(c) Test if the mean pHs in the rainwater are different at
the a = 0.05 level of significance. Reject H0
(d) Draw boxplots of the pH in rain for the three states to
support the results obtained in part (c).
20. Lower Your Cholesterol Researchers Francisco
Fuentes and his colleagues wanted to determine the
most effective diet for reducing LDL cholesterol, the socalled “bad” cholesterol, among three diets: (1) a
saturated-fat diet, (2) the Mediterranean diet, and
(3) the U.S. National Cholesterol Education Program or
NCEP-1 Diet. The participants in the study were shown
to have the same levels of LDL cholesterol before the
study. Participants were randomly assigned to one of the
three treatment groups. Individuals in group 1 received
the saturated fat diet, which is 15% protein, 47% carbohydrates, 38% fat (20% saturated fat, 12% monounsaturated fat, and 6% polyunsaturated fat). Individuals in
group 2 received the Mediterranean diet, which is 47%
carbohydrates, 38% fat ( 610% saturated fat, 22% monounsaturated fat, and 6% polyunsaturated fat). Indi-
(20a) H0: mSF ⴝ mM ⴝ mNCE vs. H1: at least one of the means is
not equal
(a) Does the evidence suggest the cholesterol levels differ? State the null and alternative hypothesis.
(b) Verify that the requirements to use the one-way
ANOVA procedure are satisfied. Normal probability
plots indicate that the sample data come from normal
populations.
(c) Test if the mean LDL cholesterol levels are different at
the a = 0.05 level of significance. Reject H0
(d) Draw boxplots of the three LDL cholesterol levels for
the three groups to support the analytic results obtained in part (c).
21. Concrete Strength An engineer wants to know if the mean
strengths of three different concrete mix designs differ significantly. He randomly selects 9 cylinders that measure 6
inches in diameter and 12 inches in height in which mixture
67-0-301 is poured, 9 cylinders of mixture 67-0-400, and 9
cylinders of mixture 67-0-353. After 28 days, he measures
the strength (in pounds per square inch) of the cylinders.
The results are presented in the following table:
Mixture
67-0-301
Mixture
67-0-400
Mixture
67-0-353
3960
4090
4070
4120
4150
3820
4040
3830
4330
4640
3820
3750
3780
3940
4620
4190
4010
3990
3890
4080
3730
3850
4150
4320
3990
4890
4190
Comparing Three or More Means (One-Way Anaylsis of Variance)
(a) State the null and alternative hypotheses.
(b) Explain why we cannot use one-way ANOVA to test
these hypotheses.
22. Analyzing Journal Article Results Researchers (Brian G.
Feagan et al., Erythropoietin with Iron Supplementation
to Prevent Allogeneic Blood Transfusion in Total Hip
Joint Arthroplasty, Annals of Internal Medicine, Vol. 133,
No. 11) wanted to determine whether epoetin alfa was effective in increasing the hemoglobin concentration in patients undergoing hip arthroplasty. The researchers
screened patients for eligibility by performing a complete
medical history and physical of the patients. Once eligible
patients were identified, the researchers used a computergenerated schedule to assign the patients to the high-dose
epoetin group, low-dose epoetin group, or placebo group.
C–33
The study was double-blind. Based on an analysis of variance, it was determined that there were significant differences in the increase in hemoglobin concentration in the
three groups with a P-value less than 0.001. The mean increase in hemoglobin in the high-dose epoetin group was
19.5 g/L, the mean increase in hemoglobin in the low-dose
epoetin group was 17.2 g/L, and mean increase in hemoglobin in the placebo group was 1.2 g/L.
(a) Why do you think it was necessary to screen patients
for eligibility?
(b) Why was a computer-generated schedule used to assign patients to the various treatment groups?
(c) What does it mean for a study to be double-blind?
Why do you think the researchers desired a doubleblind study?
(d) Interpret the reported P-value.
(21a) H0: m67ⴚ0ⴚ301 ⴝ m67ⴚ0ⴚ400 ⴝ m67ⴚ0ⴚ353 vs. H1: at
least one of the means is not equal
Technology Step by Step
TI-83/84 Plus
ANOVA
Step 1: Enter the raw data into L1, L2, L3, and so on, for each population or
treatment.
Step 2: Press STAT, highlight TESTS, and select F:ANOVA(.
Step 3: Enter the list names for each sample or treatment after ANOVA(.
For example, if there are three treatments in L1, L2, and L3, enter
ANOVA(L1,L2,L3)
MINITAB
Excel
Press ENTER.
Step 1: Enter the raw data into C1, C2, C3, and so on, for each sample or
treatment.
Step 2: Select Stat, then highlight ANOVA, and select One-way (Unstacked).
Step 3: Enter the column names in the cell marked “Responses.” Click OK.
Step 1: Enter the raw data in columns A, B, C, and so on, for each sample or
treatment.
Step 2: Be sure the Data Analysis Tool Pak is activated. This is done by selecting the Tools menu and highlighting Add-Ins Á . Check the box for the
Analysis ToolPak and select OK. Select Tools, then highlight Data Analysis.
Select ANOVA: Single Factor and click OK.
Step 3: With the cursor in the “Input Range:” cell, highlight the data. Click
OK.
C–34
Topics to Discuss
Area
F
Table VII
F-Distribution Critical Values
Degrees of Freedom in the Numerator
Degrees of Freedom in the Denominator
Area in
Right Tail
1
2
3
4
5
6
7
8
1
0.100
0.050
0.025
0.010
0.001
39.86
161.45
647.79
4052.20
405284.00
49.59
199.50
799.50
4999.50
500000.00
53.59
215.71
864.16
5403.40
540379.00
55.83
224.58
899.58
5624.60
562500.00
57.24
230.16
921.85
5763.60
576405.00
58.20
233.99
937.11
5859.00
585937.00
58.91
236.77
948.22
5928.40
592873.00
59.44
238.88
956.66
5981.10
598144.00
2
0.100
0.050
0.025
0.010
0.001
8.53
18.51
38.51
98.50
998.50
9.00
19.00
39.00
99.00
999.00
9.16
19.16
39.17
99.17
999.17
9.24
19.25
39.25
99.25
999.25
9.29
19.30
39.30
99.30
999.30
9.33
19.33
39.33
99.33
999.33
9.35
19.35
39.36
99.36
999.36
9.37
19.37
39.37
99.37
999.37
3
0.100
0.050
0.025
0.010
0.001
5.54
10.13
17.44
34.12
167.03
5.46
9.55
16.04
30.82
148.50
5.39
9.28
15.44
29.46
141.11
5.34
9.12
15.10
28.71
137.10
5.31
9.01
14.88
28.24
134.58
5.28
8.94
14.73
27.91
132.85
5.27
8.89
14.62
27.67
131.58
5.25
8.85
14.54
27.49
130.62
4
0.100
0.050
0.025
0.010
0.001
4.54
7.71
12.22
21.20
74.14
4.32
6.94
10.65
18.00
61.25
4.19
6.59
9.98
16.69
56.18
4.11
6.39
9.60
15.98
53.44
4.05
6.26
9.36
15.52
51.71
4.01
6.16
9.20
15.21
50.53
3.98
6.09
9.07
14.98
49.66
3.95
6.04
8.98
14.80
49.00
5
0.100
0.050
0.025
0.010
0.001
4.06
6.61
10.01
16.26
47.18
3.78
5.79
8.43
13.27
37.12
3.62
5.41
7.76
12.06
33.20
3.52
5.19
7.39
11.39
31.09
3.45
5.05
7.15
10.97
29.75
3.40
4.95
6.98
10.67
28.83
3.37
4.88
6.85
10.46
28.16
3.34
4.82
6.76
10.29
27.65
6
0.100
0.050
0.025
0.010
0.001
3.78
5.99
8.81
13.75
35.51
3.46
5.14
7.26
10.92
27.00
3.29
4.76
6.60
9.78
23.70
3.18
4.53
6.23
9.15
21.92
3.11
4.39
5.99
8.75
20.80
3.05
4.28
5.82
8.47
20.03
3.01
4.21
5.70
8.26
19.46
2.98
4.15
5.60
8.10
19.03
7
0.100
0.050
0.025
0.010
0.001
3.59
5.59
8.07
12.25
29.25
3.26
4.74
6.54
9.55
21.69
3.07
4.35
5.89
8.45
18.77
2.96
4.12
5.52
7.85
17.20
2.88
3.97
5.29
7.46
16.21
2.83
3.87
5.12
7.19
15.52
2.78
3.79
4.99
6.99
15.02
2.75
3.73
4.90
6.84
14.63
8
0.100
0.050
0.025
0.010
0.001
3.46
5.32
7.57
11.26
25.41
3.11
4.46
6.06
8.65
18.49
2.92
4.07
5.42
7.59
15.83
2.81
3.84
5.05
7.01
14.39
2.73
3.69
4.82
6.63
13.48
2.67
3.58
4.65
6.37
12.86
2.62
3.50
4.53
6.18
12.40
2.59
3.44
4.43
6.03
12.05
C–35
Comparing Three or More Means (One-Way Anaylsis of Variance)
Area
F
Table VII (continued)
F-Distribution Critical Values
Degrees of Freedom in the Numerator
Degrees of Freedom in the Denominator
Area in
Right Tail
9
10
15
20
30
60
120
1000
1
0.100
0.050
0.025
0.010
0.001
59.86
240.54
963.28
6022.5
602284.0
60.19
241.88
968.63
6055.8
605621.0
61.22
245.95
984.87
6157.3
615764.0
61.74
248.01
993.10
6208.7
620908.0
62.26
250.10
1001.4
6260.6
626099.0
62.79
252.20
1009.8
6313.0
631337.0
63.06
253.25
1014.0
6339.4
633972.0
63.30
254.19
1017.7
6362.7
636301.0
2
0.100
0.050
0.025
0.010
0.001
9.38
19.38
39.39
99.39
999.39
9.39
19.40
39.40
99.40
999.40
9.42
19.43
39.43
99.43
999.43
9.44
19.45
39.45
99.45
999.45
9.16
19.46
39.46
99.47
999.47
9.47
19.48
39.48
99.48
999.48
9.48
19.49
39.49
99.49
999.49
9.49
19.49
39.50
99.50
999.50
3
0.100
0.050
0.025
0.010
0.001
5.24
8.81
14.47
27.35
129.86
5.23
8.79
14.42
27.23
129.25
5.20
8.70
14.25
26.87
127.37
5.18
8.66
14.17
26.69
126.42
5.17
8.62
14.08
26.50
125.45
5.15
8.57
13.99
26.32
124.47
5.14
8.55
13.95
26.22
123.97
5.13
8.53
13.91
26.14
123.53
4
0.100
0.050
0.025
0.010
0.001
3.94
6.00
8.90
14.66
48.47
3.92
5.96
8.84
14.55
48.05
3.87
5.86
8.66
14.20
46.76
3.84
5.80
8.56
14.02
46.10
3.82
5.75
8.46
13.84
45.43
3.79
5.69
8.36
13.65
44.75
3.78
5.66
8.31
13.56
44.40
3.76
5.63
8.26
13.47
44.09
5
0.100
0.050
0.025
0.010
0.001
3.32
4.77
6.68
10.16
27.24
3.30
4.74
6.62
10.05
26.92
3.24
4.62
6.43
9.72
25.91
3.21
4.56
6.33
9.55
25.39
3.17
4.50
6.23
9.38
24.87
3.14
4.43
6.12
9.20
24.33
3.12
4.40
6.07
9.11
24.06
3.11
4.37
6.02
9.03
23.82
6
0.100
0.050
0.025
0.010
0.001
2.96
4.10
5.52
7.98
18.69
2.94
4.06
5.46
7.87
18.41
2.87
3.94
5.27
7.56
17.56
2.84
3.87
5.17
7.40
17.12
2.80
3.81
5.07
7.23
16.67
2.76
3.74
4.96
7.06
16.21
2.74
3.70
4.90
6.97
15.98
2.72
3.67
4.86
6.89
15.77
7
0.100
0.050
0.025
0.010
0.001
2.72
3.68
4.82
6.72
14.33
2.70
3.64
4.76
6.62
14.08
2.63
3.51
4.57
6.31
13.32
2.59
3.44
4.47
6.16
12.93
2.56
3.38
4.36
5.99
12.53
2.51
3.30
4.25
5.82
12.12
2.49
3.27
4.20
5.74
11.91
2.47
3.23
4.15
5.66
11.72
8
0.100
0.050
0.025
0.010
0.001
2.56
3.39
4.36
5.91
11.77
2.54
3.35
4.30
5.81
11.54
2.46
3.22
4.10
5.52
10.84
2.42
3.15
4.00
5.36
10.48
2.38
3.08
3.89
5.20
10.11
2.34
3.01
3.78
5.03
9.73
2.32
2.97
3.73
4.95
9.53
2.30
2.93
3.68
4.87
9.36
C–36
Topics to Discuss
Table VII (continued)
F-Distribution Critical Values
Degrees of Freedom in the Numerator
Degrees of Freedom in the Denominator
Area in
Right Tail
1
2
3
4
5
6
7
8
9
10
9
0.100
0.050
0.025
0.010
0.001
3.36
5.12
7.21
10.56
22.86
3.01
4.26
5.71
8.02
16.39
2.81
3.86
5.08
6.99
13.90
2.69
3.63
4.72
6.42
12.56
2.61
3.48
4.48
6.06
11.71
2.55
3.37
4.32
5.80
11.13
2.51
3.29
4.20
5.61
10.70
2.47
3.23
4.10
5.47
10.37
2.44
3.18
4.03
5.35
10.11
2.42
3.14
3.96
5.26
9.89
10
0.100
0.050
0.025
0.010
0.001
3.29
4.96
6.94
10.04
21.04
2.92
4.10
5.46
7.56
14.91
2.73
3.71
4.83
6.55
12.55
2.61
3.48
4.47
5.99
11.28
2.52
3.33
4.24
5.64
10.48
2.46
3.22
4.07
5.39
9.93
2.41
3.14
3.95
5.20
9.52
2.38
3.07
3.85
5.06
9.20
2.35
3.02
3.78
4.94
8.96
2.32
2.98
3.72
4.85
8.75
12
0.100
0.050
0.025
0.010
0.001
3.18
4.75
6.55
9.33
18.64
2.81
3.89
5.10
6.93
12.97
2.61
3.49
4.47
5.95
10.80
2.48
3.26
4.12
5.41
9.63
2.39
3.11
3.89
5.06
8.89
2.33
3.00
3.73
4.82
8.38
2.28
2.91
3.61
4.64
8.00
2.24
2.85
3.51
4.50
7.71
2.21
2.80
3.44
4.39
7.48
2.19
2.75
3.37
4.30
7.29
15
0.100
0.050
0.025
0.010
0.001
3.07
4.54
6.20
8.68
16.59
2.70
3.68
4.77
6.36
11.34
2.49
3.29
4.15
5.42
9.34
2.36
3.06
3.80
4.89
8.25
2.27
2.90
3.58
4.56
7.57
2.21
2.79
3.41
4.32
7.09
2.16
2.71
3.29
4.14
6.74
2.12
2.64
3.20
4.00
6.47
2.09
2.59
3.12
3.89
6.26
2.06
2.54
3.06
3.80
6.08
20
0.100
0.050
0.025
0.010
0.001
2.97
4.35
5.87
8.10
14.82
2.59
3.49
4.46
5.85
9.95
2.38
3.10
3.86
4.94
8.10
2.25
2.87
3.51
4.43
7.10
2.16
2.71
3.29
4.10
6.46
2.09
2.60
3.13
3.87
6.02
2.04
2.51
3.01
3.70
5.69
2.00
2.45
2.91
3.56
5.44
1.96
2.39
2.84
3.46
5.24
1.94
2.35
2.77
3.37
5.08
25
0.100
0.050
0.025
0.010
0.001
2.92
4.24
5.69
7.77
13.88
2.53
3.39
4.29
5.57
9.22
2.32
2.99
3.69
4.68
7.45
2.18
2.76
3.35
4.18
6.49
2.09
2.60
3.13
3.85
5.89
2.02
2.49
2.97
3.63
5.46
1.97
2.40
2.85
3.46
5.15
1.93
2.34
2.75
3.32
4.91
1.89
2.28
2.68
3.22
4.71
1.87
2.24
2.61
3.13
4.56
50
0.100
0.050
0.025
0.010
0.001
2.81
4.03
5.34
7.17
12.22
2.41
3.18
3.97
5.06
7.96
2.20
2.79
3.39
4.20
6.34
2.06
2.56
3.05
3.72
5.46
1.97
2.40
2.83
3.41
4.90
1.90
2.29
2.67
3.19
4.51
1.84
2.20
2.55
3.02
4.22
1.80
2.13
2.46
2.89
4.00
1.76
2.07
2.38
2.78
3.82
1.73
2.03
2.32
2.70
3.67
100
0.100
0.050
0.025
0.010
0.001
2.76
3.94
5.18
6.90
11.50
2.36
3.09
3.83
4.82
7.41
2.14
2.70
3.25
3.98
5.86
2.00
2.46
2.92
3.51
5.02
1.91
2.31
2.70
3.21
4.48
1.83
2.19
2.54
2.99
4.11
1.78
2.10
2.42
2.82
3.83
1.73
2.03
2.32
2.69
3.61
1.69
1.97
2.24
2.59
3.44
1.66
1.93
2.18
2.50
3.30
200
0.100
0.050
0.025
0.010
0.001
2.73
3.89
5.10
6.76
11.15
2.33
3.04
3.76
4.71
7.15
2.11
2.65
3.18
3.88
5.63
1.97
2.42
2.85
3.41
4.81
1.88
2.26
2.63
3.11
4.29
1.80
2.14
2.47
2.89
3.92
1.75
2.06
2.35
2.73
3.65
1.70
1.98
2.26
2.60
3.43
1.66
1.93
2.18
2.50
3.26
1.63
1.88
2.11
2.41
3.12
1000
0.100
0.050
0.025
0.010
0.001
2.71
3.85
5.04
6.66
10.89
2.31
3.00
3.70
4.63
6.96
2.09
2.61
3.13
3.80
5.46
1.95
2.38
2.80
3.34
4.65
1.85
2.22
2.58
3.04
4.14
1.78
2.11
2.42
2.82
3.78
1.72
2.02
2.30
2.66
3.51
1.68
1.95
2.20
2.53
3.30
1.64
1.89
2.13
2.43
3.13
1.61
1.84
2.06
2.34
2.99
Comparing Three or More Means (One-Way Anaylsis of Variance)
C–37
Table VII (continued)
F-Distribution Critical Values
Degrees of Freedom in the Numerator
Degrees of Freedom in the Denominator
Area in
Right Tail
12
15
20
25
30
40
50
60
120
1000
9
0.100
0.050
0.025
0.010
0.001
2.38
3.07
3.87
5.11
9.57
2.34
3.01
3.77
4.96
9.24
2.30
2.94
3.67
4.81
8.90
2.27
2.89
3.60
4.71
8.69
2.25
2.86
3.56
4.65
8.55
2.23
2.83
3.51
4.57
8.37
2.22
2.80
3.47
4.52
8.26
2.21
2.79
3.45
4.48
8.19
2.18
2.75
3.39
4.40
8.00
2.16
2.71
3.34
4.32
7.84
10
0.100
0.050
0.025
0.010
0.001
2.28
2.91
3.62
4.71
8.45
2.24
2.85
3.52
4.56
8.13
2.20
2.77
3.42
4.41
7.80
2.17
2.73
3.35
4.31
7.60
2.16
2.70
3.31
4.25
7.47
2.13
2.66
3.26
4.17
7.30
2.12
2.64
3.22
4.12
7.19
2.11
2.62
3.20
4.08
7.12
2.08
2.58
3.14
4.00
6.94
2.06
2.54
3.09
3.92
6.78
12
0.100
0.050
0.025
0.010
0.001
2.15
2.69
3.28
4.16
7.00
2.10
2.62
3.18
4.01
6.71
2.06
2.54
3.07
3.86
6.40
2.03
2.50
3.01
3.76
6.22
2.01
2.47
2.96
3.70
6.09
1.99
2.43
2.91
3.62
5.93
1.97
2.40
2.87
3.57
5.83
1.96
2.38
2.85
3.54
5.76
1.93
2.34
2.79
3.45
5.59
1.91
2.30
2.73
3.37
5.44
15
0.100
0.050
0.025
0.010
0.001
2.02
2.48
2.96
3.67
5.81
1.97
2.40
2.86
3.52
5.54
1.92
2.33
2.76
3.37
5.25
1.89
2.28
2.69
3.28
5.07
1.87
2.25
2.64
3.21
4.95
1.85
2.20
2.59
3.13
4.80
1.83
2.18
2.55
3.08
4.70
1.82
2.16
2.52
3.05
4.64
1.79
2.11
2.46
2.96
4.47
1.76
2.07
2.40
2.88
4.33
20
0.100
0.050
0.025
0.010
0.001
1.89
2.28
2.68
3.23
4.82
1.84
2.20
2.57
3.09
4.56
1.79
2.12
2.46
2.94
4.29
1.76
2.07
2.40
2.84
4.12
1.74
2.04
2.35
2.78
4.00
1.71
1.99
2.29
2.69
3.86
1.69
1.97
2.25
2.64
3.77
1.68
1.95
2.22
2.61
3.70
1.64
1.90
2.16
2.52
3.54
1.61
1.85
2.09
2.43
3.40
25
0.100
0.050
0.025
0.010
0.001
1.82
2.16
2.51
2.99
4.31
1.77
2.09
2.41
2.85
4.06
1.72
2.01
2.30
2.70
3.79
1.68
1.96
2.23
2.60
3.63
1.66
1.92
2.18
2.54
3.52
1.63
1.87
2.12
2.45
3.37
1.61
1.84
2.08
2.40
3.28
1.59
1.82
2.05
2.36
3.22
1.56
1.77
1.98
2.27
3.06
1.52
1.72
1.91
2.18
2.91
50
0.100
0.050
0.025
0.010
0.001
1.68
1.95
2.22
2.56
3.44
1.63
1.87
2.11
2.42
3.20
1.57
1.78
1.99
2.27
2.95
1.53
1.73
1.92
2.17
2.79
1.50
1.69
1.87
2.10
2.68
1.46
1.63
1.80
2.01
2.53
1.44
1.60
1.75
1.95
2.44
1.42
1.58
1.72
1.91
2.38
1.38
1.51
1.64
1.80
2.21
1.33
1.45
1.56
1.70
2.05
100
0.100
0.050
0.025
0.010
0.001
1.61
1.85
2.08
2.37
3.07
1.56
1.77
1.97
2.22
2.84
1.49
1.68
1.85
2.07
2.59
1.45
1.62
1.77
1.97
2.43
1.42
1.57
1.71
1.89
2.32
1.38
1.52
1.64
1.80
2.17
1.35
1.48
1.59
1.74
2.08
1.34
1.45
1.56
1.69
2.01
1.28
1.38
1.46
1.57
1.83
1.22
1.30
1.36
1.45
1.64
200
0.100
0.050
0.025
0.010
0.001
1.58
1.80
2.01
2.27
2.90
1.52
1.72
1.90
2.13
2.67
1.46
1.62
1.78
1.97
2.42
1.41
1.56
1.70
1.87
2.26
1.38
1.52
1.64
1.79
2.15
1.34
1.46
1.56
1.69
2.00
1.31
1.41
1.51
1.63
1.90
1.29
1.39
1.47
1.58
1.83
1.23
1.30
1.37
1.45
1.64
1.16
1.21
1.25
1.30
1.43
1000
0.100
0.050
0.025
0.010
0.001
1.55
1.76
1.96
2.20
2.77
1.49
1.68
1.85
2.06
2.54
1.43
1.58
1.72
1.90
2.30
1.38
1.52
1.64
1.79
2.14
1.35
1.47
1.58
1.72
2.02
1.30
1.41
1.50
1.61
1.87
1.27
1.36
1.45
1.54
1.77
1.25
1.33
1.41
1.50
1.69
1.18
1.24
1.29
1.35
1.49
1.08
1.11
1.13
1.16
1.22