Problems and Answers for Chapter 8
k-Sample Tests of Hypothesis: The
Analysis of Variance
1. In order to better understand the sums of squares of analysis of variance, do the following
exercise. Four samples of size 2 are given in the following table:
I
II
III
IV
2
4
9
5
3
9
5
3
(a) Express each of the eight observations as the sum of three components, i.e.,
Xij = X .. + (X i. − X .. ) + (Xij − X i. ).
(b) Compute the SSTotal , SSTreat , and SSError terms directly from the components and show
that SSTotal is equal to the sum of SSTreat and SSError .
(c) Compute the three SS values by the computational formulas and show that the values
obtained are the same as those obtained in part (a).
1. (a) X 1. = 3, X 2. = 7, X 3. = 6, X 4. = 4, and X .. = 5. So, for example,
X11 = 5 + (3 − 5) + (2 − 3) = 2
X12 = 5 + (3 − 5) + (4 − 3) = 4
X21 = 5 + (7 − 5) + (9 − 7) = 9
(b) SSTotal = 50, SSTreat = 20, and SSError = 30. Use the theoretical formulas to determine
the various sums of squares.
SSTotal =
(Xij − X .. )2
i
SSTreat =
i
SSError =
j
j
i
(X i. − X .. )2
j
(Xij − X i. )2
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
66
(c) Use the computational formulas to determine the various sums of squares, for example,
SSTotal =
i
2
Xij
−
j
T..2
.
N
2. Sickle cell anemia is a genetically transmitted disease. Normal human red blood cells are
“biconcave discs” meaning that they are shaped like microscopic breath mints. People who
are homozygous for the sickle cell gene have “sickled” red blood cells. These cells are curved
like a sickle or are irregularly shaped and are often pointed. In the smallest blood vessels
sickled cells tend to form clots and hamper normal blood ﬂow. Sickled blood cells are also
fragile and rupture easily. Those who are heterozygous for sickle cell anemia may exhibit mild
signs of the disease, but these are very rarely serious.
As part of a research study on sickle cell anemia and its relation to other diseases, researchers measured hemoglobin levels (g/100 ml) in 30 subjects divided into three groups of
10: normal (control), sickle cell disease (homozygotes), and sickle cell trait (heterozygotes)
subjects. Are there signiﬁcant diﬀerences in hemoglobin levels among the three groups?
Normal subjects
Sickle cell disease
Sickle cell trait
13.6
1.6
9.8
1.6
12.7
1.5
Mean
St. dev.
SSTreat = 78.87
SSError = 66.33
2. The hypotheses are H0 : µN = µD = µT versus Ha : At least one pair of µi ’s are diﬀerent.
From the table below, since 16.03 4.26, reject H0 . There are signiﬁcant diﬀerences in
hemoglobin levels among the groups.
Source
SS
df
MS
F
c.v.
78.87
66.33
2
27
39.44
2.46
16.03
4.26
145.20
29
Sickle cell status
Error
Total
Use Bonferroni t tests to separate means. The table below summarizes the results.
Bonferroni t
P value
Normal − Disease
Normal − Trait
Disease − Trait
5.42
< 0.001
1.28
> 0.20
−4.13
< 0.001
The normal and sickle cell trail subjects are not diﬀerent, but both of these groups diﬀer
from the sickle cell disease subjects in hemoglobin level.
9.8a
12.7b
13.6b .
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
67
3. In a project for a botany class, 15 sunﬂower seeds were randomly assigned to and planted
in pots whose soil had been subjected to one of three fertilizer treatments. Twelve of the
seeds germinated and the table below shows the height of each plant (in cm) two weeks after
germination. Are there signiﬁcant diﬀerences in heights among treatments? Analyze with
ANOVA and Bonferroni t tests, if necessary.
Treatment 1
Treatment 2
Treatment 3
23
27
32
34
26
28
33
35
38
25
26
33
3. The hypotheses are H0 : µ1 = µ2 = µ3 versus Ha : At least one pair of µi ’s are diﬀerent. The
summary data and ANOVA table are given below.
Treatment 1
Treatment 2
Treatment 3
29.0
4
32.0
5
28.0
3
X i.
ni
N = 12,
i
Xij = 360,
j
i
2
Xij
= 11046
j
Source
SS
df
MS
F
c.v.
Treatments
Error
36
210
2
9
18.0
23.3
0.77
4.26
Total
246
11
Since 0.77 < 4.26, accept H0 . The treatments are not signiﬁcantly diﬀerent in their ability
increase growth in sunﬂowers. Further analysis is not warranted.
4. In a study of lead concentration in breast milk of nursing mothers, subjects were divided into
ﬁve age classes. Determine if there was a signiﬁcant diﬀerence in the mean lead concentration
(in µg/dl) among the classes. Analyze the summary data below using ANOVA and appropriate
mean separation techniques if necessary. (Based on data reported in Bassam Younes et al.,
“Lead concentrations in breast milk of nursing mothers living in Riyadh,” Annals of Saudi
Medicine, 1995, 15(3): 249–251.)
Age Class
≤ 20
21–25
26–30
31–35
≥ 36
Total
ni
Xi
si
Ti
4
0.515
0.14
2.060
19
0.777
0.41
14.763
13
0.697
0.31
9.061
3
0.832
0.33
2.496
8
1.315
0.65
10.520
47
0.828
38.900
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
68
4. The hypotheses are H0 : Mean concentrations are identical in all age classes versus Ha : at
least one pair of µi ’s are diﬀerent. Test with α = 0.05. Using Formula 8.5,
(ni − 1)s2i = 3 × 0.142 + 18 × 0.412 + 12 × 0.312 + 2 × 0.332 + 7 × 0.652 = 7.413.
SSError =
i
Using Formula 8.4,
SSAge =
T2
i.
ni
−
T..2
= 2.562.
N
Source
SS
df
MS
F
c.v.
Age
Error
2.562
7.413
4
42
0.6405
0.1765
3.63
2.61
Total
9.975
46
Since 3.63 > 2.61, reject H0 . At least some of the age classes have
5 diﬀerent lead concentrations. Use Bonferroni t tests to separate the means. There are 2 = 10 comparisons
to make. To compare the ﬁrst two age classes, the hypotheses would be H0 : µ≤20 = µ21–25
versus Ha : µ≤20 = µ21–25 . The others are similar. The test statistic is
X i. − X j.
Bonferroni t = .
1
1
MSE ni + nj
Test at the α = 0.05
10 = 0.005 level with ν = N − k = 47 − 5 = 42. From Table C.4, the
critical values of the test statistic are ±2.963. The table below summarizes the results. We
see that the mean lead concentration in the ≥ 36 age class is signiﬁcantly diﬀerent from the
concentration in the ≤ 20, 21–25, and 25–30 age classes.
Comparison
µ≤20 and µ21–25
µ≤20 and µ25–30
µ≤20 and µ30–35
µ≤20 and µ≥36
µ21–25 and µ25–30
µ21–25 and µ30–35
µ21–25 and µ≥36
µ26–30 and µ30-35
µ26–30 and µ≥36
µ30–35 and µ≥36
Bonferroni t
Result
−1.134
−0.758
−0.988
−3.110
0.529
−0.211
−3.038
−0.502
−3.274
−1.698
Accept H0
Accept H0
Accept H0
Reject H0
Accept H0
Accept H0
Reject H0
Accept H0
Reject H0
Accept H0
5. The laughing kookaburra, Dacelo novaguineae, is a large terrestrial kingﬁsher native to eastern
Australia. Laughing kookaburras live in cooperatively breeding groups of two to eight birds in
which a dominant pair is assisted by helpers that are its oﬀspring from previous seasons. These
helpers are involved in all aspects of clutch, nestling, and ﬂedgling care. Does increased group
size lead to an increase in mean ﬂedging success? Use the data below to determine whether
there was a signiﬁcant diﬀerence in the mean number of ﬂedgings among groups of diﬀerent
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
69
sizes. Use mean separation techniques if required. (Based on data interpolated from Sarah
Legge “The eﬀect of helpers on reproductive success in the laughing kookaburra,” Journal of
Animal Ecology, 2000, 69: 714–724.)
Group size
2
3
4
5
X
s
n
1.15
1.16
46
1.43
1.09
28
1.66
1.55
29
1.67
0.73
15
SSTreat = 5.81
SSTotal = 173.17
5. The hypotheses are H0 : µ1 = µ2 = µ3 = µ4 versus Ha : At least one pair of µi ’s are diﬀerent.
Source
SS
df
MS
F
c.v.
Group size
Error
5.81
167.36
3
114
1.94
1.47
1.32
2.70
Total
173.17
117
Since 1.32 < 2.70, accept H0 . There is no evidence for a signiﬁcant diﬀerence in ﬂedging
success for breeding groups of various sizes. The researcher suggests two possible reasons for
this: (1) Helpers may be unable to prevent early siblicide which often occurs shortly after
hatching and (2) if provisioning young is costly, then the beneﬁts of reducing workloads as
group size increases may outweigh the beneﬁts of ﬂedging more young.
6. An instructor in a second-term calculus course wishes to determine whether the year in college
has any eﬀect on the performance of his students on their ﬁnal exam. The table below lists
the exam grade (out of 150) for students categorized by year. Are there signiﬁcant diﬀerences
in performance among years? Analyze with ANOVA and Bonferroni t tests, if necessary.
First year
Sophomore
Junior
122
111
104
118
113
98
129
111
127
117
97
113
123
130
72
91
71
72
96
121
6. The hypotheses are H0 : µ1 = µ2 = µ3 versus Ha : At least one pair of µi ’s are diﬀerent. The
summary data and ANOVA table are given below:
X i.
ni
First year
Sophomore
Junior
114.8
9
116.0
5
87.2
6
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
N = 20,
i
Xij = 2136,
j
i
70
2
Xij
= 234, 852.
j
Source
SS
df
MS
F
c.v.
Years
Error
3308.8
3418.4
2
17
1654.41
201.08
8.23
3.52
Total
6727.2
19
Since 8.23 > 3.52, reject H0 . There are signiﬁcant diﬀerences in the ﬁnal exam grades
by year. Use Bonferroni t tests to separate means. The table below summarizes the results.
Bonferroni t
P value
First year − Sophomore
First year − Junior
Sophomore − Junior
−0.15
> 0.50
3.69
< 0.005
3.35
< 0.005
The ﬁrst year and sophomore are not diﬀerent, but both of these classes diﬀer from the
juniors in ﬁnal exam grades.
87.2a
114.8b
116.0b .
7. To test the eﬀectiveness of various denture adhesives, an instrument called the TA.XT2i Texture Analyzer made by Texture Technologies Corp. was used to measure the amount of force
required to separate dentures from a mouth and gum cast. The force required for separation was recorded in decigrams. The adhesives were: A) karaya, B) karaya with sodium
borate, C) carboxymethylcellulose sodium (32%) and ethylene oxide homopolymer, and D)
carboxylmethylcellulose sodium (49%) and ethylene oxide homopolymer. Are there any signiﬁcant diﬀerences among these denture adhesives holding abilities? Analyze with ANOVA
and DMRT, if necessary.
Ti.
Xi
s2i
CSSi
A
B
C
D
71
79
80
72
88
66
76
70
90
80
76
82
75
81
60
66
74
58
80
82
91
95
84
72
456
76
62
310
474
79
46
230
414
69
83.2
416
504
84
66.8
334
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
71
7. The hypotheses are H0 : µA = µB = µC = µD versus Ha : At least one pair of µi ’s are diﬀerent.
Test with α = 0.05.
(1848)2
SSTotal = 144294 −
= 1998
24
CSSi = 1290
SSError =
i
SSTreat = 1998 − 1290 = 708
Source
SS
df
MS
F
c.v.
Adhesives
Error
708
1290
3
20
236.0
64.5
3.66
3.10
Total
1998
23
Since 3.66 > 3.10, reject H0 . At least some of the denture adhesives are signiﬁcantly
diﬀerent. Use a Duncan’s multiple range test to locate these diﬀerences.
MSE
64.5
=
= 3.28.
n
6
Number of means
rp
SSRp
2
3
4
2.950
9.676
3.097
10.158
3.190
10.463
Adhesives C, A, and B are not signiﬁcantly diﬀerent. A, B, and D are not signiﬁcantly
diﬀerent. Only C and D are signiﬁcantly diﬀerent.
C
A
B
D
69a
76ab
79ab
84b
8. Zonation can be observed in microhabitats such as the intertidal shoreline of a bay. A study
was conducted on the distribution of the common gastropod Bembecium at various levels
about three meters apart in the intertidal zone at Polka Point. Several 0.25 m2 quadrats were
placed along zones 1, 2, and 3 (zone 2 is mean water level) and the number of Bembecium in
each were counted.
Zone 1
Rank
Zone 2
Rank
Zone 3
0
2
12
16
16
21
23
29
33
38
43
47
54
67
4
6
16
27
41
Sum
Sum
Sum
Rank
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
72
(a) State the appropriate null and alternative hypotheses for these data.
(b) Rank all the observations (use midranks when necessary) and compute the sums of the
sample ranks Ri .
(c) Compute the Kruskal-Wallis test statistic H for these data and determine whether H0
should be rejected.
(d ) If necessary, carry out paired comparisons.
8. (a) H0 : All three zones have the same median numbers of Bembecium per quadrat. Ha : At
least one zone has a diﬀerent median number of Bembecium per quadrat.
(b) See the table below. Note that N = 19.
Zone 1
Rank
Zone 2
Rank
Zone 3
Rank
0
2
12
16
16
21
23
1
2
5
7
7
9
10
29
33
38
43
47
54
67
12
13
14
16
17
18
19
4
6
16
27
41
3
4
7
11
15
Sum
41
Sum
109
Sum
40
Ave. rank
5.86
Ave. rank
15.57
Ave. rank
8.00
(c) ν = df = k − 1 = 2. If α = 0.05, then the c.v. = 5.99. From the table above,
k
Ri N + 1 2
12
ni
−
N (N + 1)
ni
2
i=1
12 7(5.86 − 10)2 + 7(15.57 − 10)2 + 5(8 − 10)2 = 11.29.
=
19(20)
H=
Since H > 5.99, reject H0 . There is evidence that the median distance moved by at least
one group of snails is diﬀerent from the others.
(d ) Comparisons are required since H0 was rejected. n1 = n2 = 7, n3 = 5, and N = 19. The
= 0.0167.
average ranks are listed in the table above. k = 3. Let α = 0.05, so α = 0.05
(32)
α
So 1 − 2 = 0.9917. Thus, for all comparisons the c.v. = 2.40. For all 3 comparisons the
hypotheses are H0 : The median numbers of Bembecium in the ith and jth zones are the
same versus Ha : The median numbers of Bembecium in the ith and jth zones are diﬀerent.
Comparing zones 1 and 2:
R1 R2 n1 − n2 |5.86 − 15.57|
9.71
= 3.23.
z12 = =√
= 19(20) 9.0476
1
1
N (N +1)
1
1
+
12
7
7
12
n1 + n2
z12 > 2.40, so reject H0 : There is evidence that the median numbers of Bembecium in
zones 1 and 2 are diﬀerent.
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
73
Comparing zones 1 and 3:
|5.86 − 8.00|
2.14
= 0.65.
z13 = =√
19(20) 1
10.8571
1
12
7 + 5
z13 < 2.40, so accept H0 : There is no evidence that the median numbers of Bembecium in
zones 1 and 3 are diﬀerent.
Comparing zones 2 and 3:
|15.57 − 8.00|
7.57
= 2.30.
z23 = =√
19(20) 1
10.8571
1
+
12
7
5
z23 < 2.40, so accept H0 : There is no evidence that the median numbers of Bembecium in
zones 2 and 3 are diﬀerent.
9. The silver content (%Ag) of a number of Byzantium coins discovered in Cyprus was determined. Nine came from the ﬁrst coinage of the reign of King Manuel I, Comnenus (1143–1180);
seven came from a second coinage minted several years later; four from a third coinage; and
another seven from a fourth coinage. The question is whether there were signiﬁcant diﬀerences in the silver content of the coins minted at diﬀerent times during King Manuel’s reign.
(HSDS, #149)
I
5.9
6.2
6.4
6.6
6.8
6.9
7.0
7.2
7.7
Rank
II
6.6
6.9
8.1
8.6
9.0
9.2
9.3
Rank
III
4.5
4.6
4.9
5.5
Rank
IV
Rank
5.1
5.3
5.5
5.6
5.8
5.8
6.2
(a) State the appropriate null and alternative hypotheses for these data.
(b) Rank all the observations (use midranks when necessary) and compute the sums of the
sample ranks, Ri .
(c) Compute the Kruskal-Wallis test statistic H for these data and determine whether H0
should be rejected.
(d ) If necessary, carry out paired comparisons.
9. (a) H0 : All 4 coinages have the same median silver content. Ha : At least one of the coinages
has a diﬀerent median silver content.
(b) See the table. Note that N = 27.
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
I
Rank
II
Rank
III
Rank
IV
Rank
5.9
6.2
6.4
6.6
6.8
6.9
7.0
7.2
7.7
11
12.5
14
15.5
17
18.5
20
21
22
6.6
6.9
8.1
8.6
9.0
9.2
9.3
15.5
18.5
23
24
25
26
27
4.5
4.6
4.9
5.5
1
2
3
6.5
5.1
5.3
5.5
5.6
5.8
5.8
6.2
4
5
6.5
8
9.5
9.5
12.5
Sum
151.5
Sum
159
Sum
12.5
Sum
55
Ave. rank
16.83
Ave. rank
22.71
Ave. rank
3.13
Ave. rank
7.86
74
(c) df = k − 1 = 3. If α = 0.05, then the c.v. = 7.81. Using the table above,
k
Ri N + 1 2
12
H=
ni
−
N (N + 1)
ni
2
i=1
12 =
9(16.83 − 14)2 + 7(22.71 − 14)2 + 4(3.13 − 14)2 + 7(7.86 − 14)2
27(28)
= 21.28.
Since H = 21.28 > 7.81, reject H0 . There is evidence that at least one of the coinages has
a diﬀerent median silver content.
(d ) n1 = 9, n2 = 7, n3 = 4, n4 = 7, and N = 27. The average ranks are given in the table.
α
= 0.05
k = 4 and α = 0.05 so α = 0.05
6 = 0.0083. So 1 − 2 = 0.9958. Thus, for all
(42)
comparisons the c.v. = 2.64. For all comparisons we have H0 : The median silver content
in the ith and jth coinages are the same versus Ha : The median silver content in the ith
and jth coinages are diﬀerent.
Comparing coinages 1 and 2:
|16.83 − 22.71|
5.88
z12 = = √ = 1.47.
27(28) 1
16
1
12
9 + 7
z12 < 2.64, so accept H0 . There is no evidence that the median silver content in coinages
1 and 2 are diﬀerent.
Comparing coinages 1 and 3:
|16.83 − 3.13|
13.70
= 2.87.
z13 = =√
27(28) 1
22.75
1
12
9 + 4
z13 > 2.64, so reject H0 . There is evidence that the median silver content in coinages 1
and 3 are diﬀerent.
Comparing coinages 1 and 4:
|16.83 − 7.86|
8.97
z14 = = √ = 2.24
27(28) 1
16
1
12
9 + 7
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
75
z14 < 2.64, so accept H0 . There is no evidence that the median silver content in coinages
1 and 4 are diﬀerent.
Comparing coinages 2 and 3:
|22.71 − 3.13|
19.58
z23 = = 3.94.
=√
27(28) 1
24.75
1
12
7 + 4
z23 > 2.64 so reject H0 , There is evidence that the median silver content in coinages 2 and
3 are diﬀerent.
Comparing coinages 2 and 4:
|22.71 − 7.86|
14.85
z24 = = 3.50.
= √
27(28) 1
18
1
12
7 + 7
z24 > 2.64, so reject H0 . There is evidence that the median silver content in coinages 2
and 4 are diﬀerent.
Comparing coinages 3 and 4:
|3.13 − 7.86|
4.73
z34 = = 0.95.
=√
27(28) 1
24.75
1
12
4 + 7
z34 < 2.64, so accept H0 . There is no evidence that the median silver content in coinages
3 and 4 are diﬀerent.
10. (a) The data below are steady-state hemoglobin levels (g/100 ml) for patients with diﬀerent
types of sickle cell disease, these being HB SS, S/-thalassemia, and HB SC. Is there a
signiﬁcant diﬀerence in median steady-state hemoglobin levels for patients with diﬀerent
types of sickle cell disease? (HSDS, #310)
(b) If appropriate, carry out a set of paired comparisons.
HB SS
7.2
7.7
8.0
8.1
8.3
8.4
8.4
8.5
8.6
8.7
9.1
9.1
9.1
9.8
10.1
10.3
Rank
HB S/-thalassemia
8.1
9.2
10.0
10.4
10.6
10.9
11.1
11.9
12.0
12.1
Rank
HB SC
10.7
11.3
11.5
11.6
11.7
11.8
12.0
12.1
12.3
12.6
13.3
13.3
13.8
13.9
Rank
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
76
10. (a) H0 : All three populations have the same median versus Ha : At least one of the population
medians is diﬀerent from the others. Use the Kruskal-Wallis test. The rank sums and rank
averages are given in the table below. df = k − 1 = 2. If α = 0.05, then the c.v. = 5.99.
Using the table below,
k
R2
12
i
− 3(N + 1)
H=
N (N + 1)
ni
i=1
(152.5)2 (220.5)2 (447)2
12
− 3(40 + 1)
=
+
+
40(40 + 1)
16
10
14
12
=
(20, 587.6) − 123 = 27.64.
1640
Since H = 27.64 > 5.99, reject H0 . There is evidence that at least one population has a
signiﬁcantly diﬀerent median.
HB SS
7.2
7.7
8.0
8.1
8.3
8.4
8.4
8.5
8.6
8.7
9.1
9.1
9.1
9.8
10.1
10.3
Rank
HB S/-thalassemia
Rank
HB SC
Rank
8.1
9.2
10.0
10.4
10.6
10.9
11.1
11.9
12.0
12.1
4.5
16
17
20
21
23
24
30
31.5
33.5
10.7
11.3
11.5
11.6
11.7
11.8
12.0
12.1
12.3
12.6
13.3
13.3
13.8
13.9
22
25
26
27
28
29
31.5
33.5
35
36
37.5
37.5
39
40
447
1
2
3
4.5
6
7.5
9
10
11
13
13
13
13
15
18
19
R1
152.5
R2
220.5
R3
Ave. rank
9.53
Ave. rank
22.05
Ave. rank
31.93
(b) Comparisons are required since H0 was rejected. n1 = 16, n2 = 10, n3 = 14, and N = 40.
= 0.0167.
The average ranks are listed in the table above. k = 3. Let α = 0.05, so α = 0.05
(32)
α
So 1 − 2 = 0.9917. Thus, for all comparisons the c.v. = 2.40. For all 3 comparisons the
hypotheses are H0 : The medians of the ith and jth populations are the same versus Ha :
The medians of the ith and jth populations are diﬀerent.
Comparing HB SS and HB S/-thalassemia:
R 1 R2 n1 − n2 |9.53 − 22.05|
12.52
z12 = = 2.65.
=
= 40(41) 4.71
1
1
N (N +1)
1
1
+
12
16 + 10
12
n1
n2
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
77
z12 > 2.40, so reject H0 . There is evidence that the medians for HB SS and HB S/thalassemia are diﬀerent.
Comparing HB SS and HB SC:
|9.53 − 31.93|
22.40
z13 = =
= 5.23.
4.28
40(41) 1
1
+
12
16
14
z13 > 2.40, so reject H0 . There is evidence that the medians for HB SS and HB SC are
diﬀerent.
Comparing HB S/-thalassemia and HB SC:
|22.05 − 31.93|
9.88
z23 = = 2.04.
=
4.84
40(41) 1
1
+
12
10
14
z23 < 2.40, so accept H0 . The medians for HB S/-thalassemia and HB SC are not diﬀerent.
11. In order for Banksia serrata to germinate, ﬁre is required to open its seed cones. The frequency
and severity of ﬁre at a hillside site at Brown Lake on North Stradbroke Island varies by
location on the hill. In order to determine the relative densities of B. serrata at various
positions, a small study was conducted. At the hilltop, along the east-facing slope, and
in a level control area, several 100-meter transects were laid out. Every ten meters along
these transects, the species of the nearest tree in each quarter was recorded and the number
of B. serrata along each transect was determined. These data are given in the ﬁrst three
columns in the table below. The median girths (in cm) of the B. serrata were also determined
for each transect. These data appear in the last three columns of the table
Number of trees
Median girth (cm)
Hilltop
Slope
Control
Hilltop
Slope
Control
10
11
12
12
14
14
13
14
15
16
19
21
4
6
7
7
11
12
15
39
45
51
54
57
59
33
37
41
46
47
58
44
56
65
67
77
81
102
(a) Determine whether there is a signiﬁcant diﬀerence in the median numbers of trees at the
three locations. If appropriate, carry out a set of paired comparisons.
(b) Determine whether there is a signiﬁcant diﬀerence in the median girths of trees at the
three locations. If appropriate, carry out a set of paired comparisons.
11. (a) H0 : All 3 locations have the same number of B. serrata versus Ha : At least one of the
locations has diﬀerent numbers of B. serrata. The rank sums and rank averages are given
in the table below. df = k − 1 = 2. If α = 0.05, then the c.v. = 5.99. Using the table
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
78
below,
k
Ri N + 1 2
12
H=
ni
−
N (N + 1)
ni
2
i=1
12 6(9.25 − 10)2 + 6(15.58 − 10)2 + 7(5.86 − 10)2 = 9.79.
=
19(20)
Since H = 9.79 > 5.99, reject H0 . There is evidence that at least one location has a
diﬀerent number of B. serrata.
Number of trees
Hilltop
Rank
Slope
Rank
Control
Rank
10
11
12
12
14
14
5
6.5
9
9
13
13
13
14
15
16
19
21
11
13
15.5
17
18
19
4
6
7
7
11
12
15
1
2
3.5
3.5
6.5
9
15.5
R1
55.5
R2
93.5
R3
41
Ave. rank
9.25
Ave. rank
15.58
Ave. rank
5.86
Paired comparisons are required since H0 was rejected. n1 = n2 = 6, n3 = 7, and
N = 19. The average ranks are listed in the table above. k = 3. Let α = 0.05, so
= 0.0167. So 1 − α2 = 0.9917. Thus, for all comparisons the c.v. = 2.40. For all
α = 0.05
(32)
3 comparisons the hypotheses are H0 : The median numbers of B. serrata along transects
in the ith and jth locations are the same versus Ha : The median numbers of B. serrata
along transects in the ith and jth locations are diﬀerent.
Comparing Hilltop and Slope:
R1 R 2 n1 − n2 |9.25 − 15.58|
6.33
z12 = =
= 1.95.
= 19(20) 3.25
1
1
N (N +1)
1
1
+
12
6 + 6
12
n1
n2
z12 < 2.40, so accept H0 : There is no evidence that the median numbers of B. serrata
along transects on the Hilltop and Slope are diﬀerent.
Comparing Hilltop and Control:
|9.25 − 5.86|
3.39
z13 = =
= 1.08.
3.13
19(20) 1
1
12
6 + 7
z13 < 2.40, so accept H0 : There is no evidence that the median numbers of B. serrata
along transects on the Hilltop and Control are diﬀerent.
Comparing locations Slope and Control:
|15.58 − 5.86|
9.72
z23 = =
= 3.11.
3.13
19(20) 1
1
12
6 + 7
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
79
z23 > 2.40, so reject H0 : There is evidence that the median numbers of B. serrata along
transects on the Slope and Control are diﬀerent.
(b) H0 : All 3 locations have the same median girth B. serrata versus Ha : At least one of the
locations has a diﬀerent median girth. The rank sums and rank averages are given in the
table below. df = k − 1 = 2. If α = 0.05, then the c.v. = 5.99. Using the table below,
H=
12 6(9 − 10)2 + 6(5.83 − 10)2 + 7(14.43 − 10)2 = 7.82.
19(20)
Since H = 7.82 > 5.99, reject H0 . There is evidence that at least one location has a
diﬀerent median girth.
Median girth (cm)
Hilltop
Rank
Slope
Rank
Control
Rank
39
45
51
54
57
59
3
6
9
10
12
14
33
37
41
46
47
58
1
2
4
7
8
13
44
56
65
67
77
81
102
5
11
15
16
17
18
19
R1
54
R2
35
R3
101
Ave. rank
9.00
Ave. rank
5.83
Ave. rank
14.43
Comparisons are required since H0 was rejected. As in (a), the c.v. = 2.40. For all 3
comparisons the hypotheses are H0 : The median girths of B. serrata in the ith and jth
locations are the same versus Ha : The median girths of B. serrata in the ith and jth
locations are diﬀerent.
Comparing Hilltop and Slope: Making use of calculations in (a),
z12 =
|9.00 − 5.83|
= 0.98.
3.25
z12 < 2.40, so accept H0 : There is no evidence that the median girths of B. serrata on
the Hilltop and Slope are diﬀerent.
Comparing Hilltop and Control:
z13 =
|9.00 − 14.43|
= 1.73.
3.13
z13 < 2.40, so accept H0 : There is no evidence that the median girths of B. serrata on
the Hilltop and Control are diﬀerent.
Comparing locations Slope and Control:
z23 =
|5.83 − 14.43|
= 2.75.
3.13
z23 > 2.40, so reject H0 : There is evidence that the median girths of B. serrata on the
Slope and Control are diﬀerent.
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
80
12. Workers at a tree farm decided to test the eﬃcacy of three fertilizer mixtures on the growth of
Norway maple seedlings, Acer platanoides. The table below contains the heights of seedlings
(in feet) for the three fertilizer treatments. Determine if there are signiﬁcant diﬀerences in
the heights among the three treatments.
Fertilizer mixture
Ti.
CSSi
si
i
2
Xij
= 203.49,
A
B
C
2.0
2.1
2.4
2.8
2.9
3.1
3.2
3.7
3.8
4.1
2.3
2.9
1.5
1.2
1.9
1.9
3.4
2.1
2.6
2.4
3.1
1.5
2.2
2.9
1.7
2.1
2.8
1.5
2.8
2.2
30.1
4.61
0.72
22.2
3.82
0.61
22.8
3.20
0.60
T.. = 75.1,
SSTotal = 15.49
j
12. The hypotheses are H0 : µA = µB = µC versus Ha : At least one pair of µi ’s are diﬀerent.
CSSi = 11.63
SSError =
i
SSTreat = 15.49 − 11.63 = 3.86
Source
SS
df
MS
F
c.v.
Fertilizer mixtures
Error
3.86
11.63
2
27
1.93
0.43
4.49
3.35
Total
15.49
29
Since 4.49 > 3.35, reject H0 . At least one treatment mean is signiﬁcantly diﬀerent from
the others. Use DMRT to locate the diﬀerences.
MSE
0.43
=
= 0.21.
n
10
Number of means
rp
SSRp
2
3
2.919
0.613
3.066
0.644
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
81
Treatment A is signiﬁcantly diﬀerent from treatments B and C.
B
C
A
2.22a
2.28a
3.01b
13. A topic of recent public health interest is whether there is a measurable eﬀect of passive
smoking on pulmonary health, that is, exposure to cigarette smoke in the atmosphere among
nonsmokers. White and Froeb studied this question by measuring pulmonary function in a
number of ways in the following six groups:
Nonsmokers (NS): People who themselves did not smoke and were not exposed to cigarette
smoke either at home or on the job.
Passive Smokers (PS): People who themselves did not smoke and were not exposed to
cigarette smoke in the home, but were employed for 20 or more years in an enclosed working
area that routinely contained tobacco smoke.
Noninhaling Smokers (NI): People who smoked pipes, cigars, or cigarettes, but who did not
inhale.
Light Smokers (LS): People who smoked and inhaled 1–10 cigarettes per day for 20 or more
years. (Note: There are 20 cigarettes in a pack.)
Moderate Smokers (MS): People who smoked and inhaled 11–39 cigarettes per day for 20
or more years.
Heavy Smokers (HS): People who smoked and inhaled 40 or more cigarettes per day for 20
or more years.
A principal measure used by the authors to assess pulmonary function was forced midexpiratory ﬂow (FEF) and it is of interest to compare FEF in the six groups. Analyze
appropriately. (Based on data reported in James White and Herman Froeb, “Small-airways
dysfunction in non-smokers chronically exposed to tobacco smoke,” New England Journal of
Medicine, 1980, 302: 720–723.)
FEF data for smoking and nonsmoking males
Group
Mean FEF
(liters/sec)
St. dev. FEF
(liters/sec)
ni
NS
PS
NI
LS
MS
HS
3.78
3.30
3.32
3.23
2.73
2.59
0.79
0.77
0.86
0.78
0.81
0.82
200
200
50
200
200
200
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
82
13. The hypotheses are H0 : µNS = µPS = µNI = µLS = µMS = µHS versus Ha : At least one pair
of µi ’s are diﬀerent.
X i.
n
SSError =
HS
MS
LS
PS
NI
NS
2.59
200
2.73
200
3.23
200
3.30
200
3.32
50
3.78
200
(ni − 1)s2i
i
= 199(0.79)2 + 199(0.77)2 + 49(0.86)2 + 199(0.78)2 + 199(0.81)2 + 199(0.82)2
SSTreat
= 663.87.
(Ti. )2 (T.. )2
=
−
ni
N
i
(200 × 3.78)2 (200 × 3.30)2
(200 × 2.59)2
+
+ ··· +
200
200
200
(200 × 3.78 + 200 × 3.30 + · · · + 200 × 2.59)2
−
1050
= 184.38.
=
Source
SS
df
MS
F
c.v.
Group
Error
184.38
663.87
5
1044
36.88
0.64
57.63
2.27
Total
848.25
1049
Since 57.63 2.27, reject H0 . At least one pair of µi ’s are not equal.
Number of means
rp
SSRp
2
3
4
5
6
2.800
2.240
2.947
2.358
3.045
2.436
3.116
2.493
3.172
2.538
6!
There are 2!4!
= 15 possible comparisons. We list a few below. Comparing NS to NI, there
is a signiﬁcant diﬀerence because
2 · 200 · 50
|3.78 − 3.32|
= 4.112 > 2.240.
200 + 50
Comparing NI to PS, there is no signiﬁcant diﬀerence because
2 · 50 · 200
= 0.179 < 2.240.
|3.32 − 3.30|
50 + 200
Comparing PS to LS, there is no signiﬁcant diﬀerence because
2 · 200 · 200
= 0.990 < 2.240.
|3.30 − 3.23|
200 + 200
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
83
Comparing LS to MS, there is a signiﬁcant diﬀerence because
2 · 200 · 200
|3.23 − 2.73|
= 7.071 > 2.240.
200 + 200
Comparing MS to HS, there is no signiﬁcant diﬀerence because
2 · 200 · 200
|2.73 − 2.59|
= 1.980 < 2.240.
200 + 200
And so on. NS is signiﬁcantly diﬀerent from all others. NI, PS, and LS are not signiﬁcantly diﬀerent from each other, but are diﬀerent from the other three. MS and HS are not
signiﬁcantly diﬀerent from each other, but are diﬀerent from the other four.
2.59a
2.73a
3.23b
3.30b
3.32b
3.78c
14. To obtain a preliminary measure of plant species richness in various habitats, 64 m2 quadrats
were used at three locations on North Stradbroke Island: Brown Lake, 18-Mile Swamp, and
Pt. Lookout. The number of diﬀerent species in each quadrat were recorded.
(a) Is there a signiﬁcant diﬀerence in median plant species richness among these locations?
Brown Lake
18-Mile Swamp
Pt. Lookout
14
15
18
19
20
23
11
12
12
13
14
17
16
20
22
24
29
(b) Carry out a set of paired comparisons at the α = 0.05 level or explain why it is inappropriate to do so.
14. (a) H0 : Median plant species richness is the same at all three locations versus Ha : At least
one location has a diﬀerent median from the others. Use the Kruskal-Wallis test. The
rank sums and rank averages are given in the table below. df = k − 1 = 2. If α = 0.05,
then the c.v. = 5.99. Using the table below,
k
R2
12
i
− 3(N + 1)
H=
N (N + 1)
ni
i=1
(61)2 (24.5)2 (67.5)2
12
− 3(17 + 1)
=
+
+
17(17 + 1)
6
6
5
12
=
(1631.46) − 54 = 9.98.
306
Since H = 9.98 > 5.99, reject H0 . There is evidence that at least one location has a
signiﬁcantly diﬀerent median.
Problems and Answers for Chapter 8. k-Sample Tests of Hypothesis: The Analysis of Variance
Brown Lake
Rank
18-Mile Swamp
14
15
18
19
20
23
5.5
7
10
11
12.5
15
11
12
12
13
14
17
R1
61
Ave. rank
10.17
Rank
Pt. Lookout
Rank
1
2.5
2.5
4
5.5
9
16
20
22
24
29
8
12.5
14
16
17
R2
24.5
R3
67.5
Ave. rank
4.08
Ave. rank
13.5
84
(b) Comparisons are required since H0 was rejected. n1 = n2 = 6, n3 = 5, and N = 17. The
= 0.0167.
average ranks are listed in the table above. k = 3. Let α = 0.05, so α = 0.05
(32)
α
So 1 − 2 = 0.9917. Thus, for all comparisons the c.v. = 2.40. For all 3 comparisons the
hypotheses are H0 : The median species richness of the ith and jth locations are the same
versus Ha : The median species richness of the ith and jth locations are diﬀerent.
Comparing Brown Lake and 18-Mile Swamp:
R1 R 2 n1 − n2 |10.17 − 4.08|
6.09
=
z12 = = 2.09.
= 17(18) 2.92
1
1
N (N +1)
1
+
+ 1
12
6
6
12
n1
n2
z12 < 2.40, so accept H0 . Medians at Brown Lake and 18-Mile Swamp are not signiﬁcantly
diﬀerent.
Comparing Brown Lake and Pt. Lookout:
|10.17 − 13.5|
3.33
z13 = =
= 1.09.
3.06
17(18) 1
1
+
12
6
5
z13 < 2.40, so accept H0 . Medians at Brown Lake and Pt. Lookout are not signiﬁcantly
diﬀerent.
Comparing 18-Mile Swamp and Pt. Lookout:
|4.08 − 13.5|
9.42
z23 = =
= 3.08.
3.06
17(18) 1
1
+
12
6
5
z23 > 2.40, so reject H0 . Plant species richness is signiﬁcantly diﬀerent at 18-Mile Swamp
and Pt. Lookout.