Chapter 12
Analysis of Variance
12-2 One-Way ANOVA
1. a. One-way analysis of variance is appropriate for these data because they represent three or
more populations categorized by a single characteristic that distinguishes the populations
from each other. The distinguishing characteristic in this case is epoch.
b. One-way analysis of variance tests the equality of two or more population means by
analyzing sample variances. It finds a difference in the population means if the variance
between the sample means is larger than can be expected considering the variance within the
samples.
3. We should reject the hypothesis that the three epochs have the same mean skull breadth. There
is sufficient evidence to conclude that at least one of the means is different from the others.
NOTE: When testing the hypothesis that three or more groups have the same mean, the test statistic
is F, where F is the ratio is the ratio of the variance between the groups to the variance within the
groups as defined in the text. This manual generally uses the generic notation F = s 2B /s 2p . As in
previous chapters, the superscripts and subscripts (the numerator df and denominator df) may be
used to identify which F distribution to look up in the tables.
5. Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 33
C.V. F = Fα = F0.05 = 3.3158
calculations:
F = s 2B /s 2p
0.05
= 669.0011/70.6481
F
1
3.3158
= 9.4695 [TI-83/84+]
P-value = P( F332 >9.4695) = 0.0006 [TI-83/84+]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that μ1 = μ2 = μ3. There is
sufficient evidence to reject the claim that the three books have the same mean Flesch
Reading Ease score.
7. Ho: μ1 = μ2 = μ3 = μ4
H1: at least one μi is different
α = 0.05 and dfnum = 3, dfden = 156
C.V. F = Fα = F0.05 = 2.6626 [Excel]
calculations:
F = s 2B /s 2p
= 11.99995/34.08497
= 0.35206 [Excel]
3
P-value = P( F156
>0.3521) = 0.7877 [Excel]
2
33
0.05
1
2.6626
3
F 156
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CHAPTER 12 Analysis of Variance
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that μ1 = μ2 = μ3 = μ4.
There is not sufficient evidence to reject the claim that the mean weight loss is the same for
all four diets. Given the four mean losses that range from 2.1 lbs to 3.3 lbs, it appears that
one year of following the diet does not result in a weight loss worth the effort. The effort
may be justified, however, when other potential benefits are considered – viz., maintaining
one’s weight (i.e., not gaining weight), enjoying a healthy lifestyle, etc.
9. Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 32
C.V. F = Fα = F0.05 = 3.3158
calculations:
F = s 2B /s 2p
0.05
= 0.42400.933/7981.129
= 5.313 [SPSS]
F
1
3.3158
2
P-value = P( F32 >5.313) = 0.010 [SPSS]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that μ1 = μ2 = μ3. There is
sufficient evidence to reject the claim that PG, PG-13 and R movies have the same gross
amount.
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32
NOTE FOR THE REMAINING EXERCISES IN THIS SECTION: This section is calculation-oriented.
Do not get so involved with the formulas that you miss the concepts. This manual arranges the
calculations to promote both computational efficiency and understanding of the underlying
principles. The following notation is used.
k = the number of groups
ni = the number of scores in group i (where i = 1,2,…,k)
x i = the mean of group i
s i2 = the variance of group i
x = the overall mean of all the scores in all the groups
= (Σni x i )/Σni = the (weighted) mean of the group means
= Σ x i /k = simplified form when each group has equal size n
2
B
s = the variance between the groups
= Σni( x i – x )2/(k-1)
= n Σ( x i – x )2/(k-1) = ns 2x = simplified form when each group has equal size n
s 2p = the variance within the groups
= (Σdfi s i2 )/Σdfi
= Σ s i2 /k = simplified form when each group has equal size n
numerator df = k-1
denominator df = Σdfi
= k(n-1) = simplified form when each group has equal size n
2 2
F = s B /s p = (variance between groups)/(variance within groups)
P-value = Fcdf(F, 99, numerator df, denominator df) from the TI-83/84+ calculator
One-Way ANOVA SECTION 12-2
265
As a crude check against errors, and to help get a feeling for the problem, always verify that the
following “overall” values are realistic in that
x is a value between the lowest and the highest of the x i values.
s 2p is a value between the lowest and the highest of the s i2 values.
11. Since each group has equal size n=10, use the simplified form of the calculations.
The following preliminary values are identified.
n
Σx
Σx2
x
s2
k=3
small
10
4315
1918833
431.5
6323.389
medium
10
3906
1620816
390.6
10570.267
x = Σxi /k
n = 10
s 2x = Σ(x i -x) 2 /(k-1)
= 630.723
large
10
3858
1764878
385.8
30717.956
= 402.633
s = Σsi /k
2
p
.
2
= 15870.537
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 27
C.V. F = Fα = F0.05 = 3.3541
0.05
calculations:
2 2
F = ns x /s p
F
1
3.3541
= 10(630.723)/15870.537
= 0.3974
P-value = Fcdf(0.3974,99,2,27) = 0.6759
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that μ1 = μ2 = μ3.
There is not sufficient evidence to reject the claim that the different car categories have the
same mean head injury values. No, these data do not suggest that larger cars are safer.
2
27
NOTE: Since subtracting the same value from each will not affect the differences between the
group means or any variances in the problem, that strategy may be used to reduce the
mathematical magnitude of each of the following problems. In Exercise 13, for example, one
could subtract 3 minutes from each time and deal only with the excess seconds over 3 minutes as
follows.
mile 1: 15 24 23 22 21
mile 2: 19 22 21 17 19
mile 3: 34 31 29 31 29
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CHAPTER 12 Analysis of Variance
13. Since each group has equal size n=5, use the simplified form of the calculations.
The following preliminary values are identified (all measurements in seconds).
mile 1
5
1005
202055
201.0
12.50
n
Σx
Σx2
x
s2
k=3
mile 2
5
998
199216
199.6
3.80
x = Σxi /k
n=5
= 203.8
s = Σsi /k
s = Σ(x i -x) /(k-1)
2
x
mile 3
5
1054
222200
210.8
4.20
2
p
2
= 37.24
.
2
= 6.8333
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 12
0.05
C.V. F = Fα = F0.05 = 3.8853
F
1
3.8853
calculations:
2 2
F = ns x /s p
= 5(37.24)/6.8333
= 27.2488
P-value = Fcdf(27.2488,99,2,12) = 3.45E-5 = 0.00003
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that μ1 = μ2 = μ3. There is
sufficient evidence to reject the claim that it takes the same time to ride each of the miles.
Yes; the data suggest that there may be a hill on mile 3.
2
12
15. The Minitab output is as follows
Level
king
menthol
filter
N
25
25
25
Mean
1.2560
0.8720
0.9160
StDev
0.2329
0.2424
0.2478
Source
Factor
Error
Total
DF
2
72
74
SS
2.2083
4.1856
6.3939
MS
1.1041
0.0581
F
18.99
P
0.000
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 72
C.V. F = Fα = F0.05 = 3.1504
calculations:
F = s 2B /s 2p
= 1.1041/0.0581
0.05
= 18.99 [Minitab]
F
1
3.1504
P-value = P( F722 >18.99) = 0.000 [Minitab]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that μ1 = μ2 = μ3. There is
sufficient evidence to reject the claim that the three types of cigarettes yield the mean
amount of nicotine. Yes; given that only the king size cigarettes are not filtered, it appears
(although not formally justifiable by the ANOVA) that the filters do make a difference.
2
72
17. The Tukey results indicate that the only significant difference (P-value = 0.021) is between μ1
the mean for the small cars) and μ3 (the mean for the large cars). The conclusion is the same as
that obtained with the Bonferroni results. The slightly larger Bonferroni P-value =0.024
indicates that, at least in this case, the Bonferroni method is more conservative (i.e., requires a
larger difference before declaring significance).
Two-Way ANOVA SECTION 12-3
267
12-3 Two-Way ANOVA
1. Two-way analysis of variance is appropriate for these data because they are categorized by two
characteristics that distinguish groups of data from each other. The distinguishing
characteristics in this case are gender and age bracket.
3. Because each cell contains the same number of observations (viz., 5), this is a balanced design.
5. Ho: there is no gender-age interaction
H1: there is gender-age interaction
α = 0.05 and dfnum = 1, dfden = 60
C.V. F = Fα = F0.05 = 4.0012
calculations:
F = MSInt/MSErr
0.05
= 1.995/8.275
= 0.24 [Minitab]
F
1
4.0012
P-value = P( F601 >0.24) = 0.625 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that there is no
interaction between gender and age bracket. There is not sufficient evidence to reject the
claim that heights are not affected by an interaction between a person’s gender and that
person’s age bracket.
1
60
7. Ho: μ1 = μ2 [there is no age effect]
H1: at least one age μi is different
α = 0.05 and dfnum = 1, dfden = 36
C.V. F = Fα = F0.05 = 4.0012
calculations:
F = MSAge/MSErr
0.05
= 24.379/8.275
= 2.95 [Minitab]
F
1
4.0012
P-value = P( F601 >2.95) = 0.625 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that age bracket
has an effect on height.
1
60
9. Ho: μ1 = μ2 [there is no type effect]
H1: at least one type μi is different
α = 0.05 and dfnum = 1, dfden = 12
C.V. F = Fα = F0.05 = 4.7472
calculations:
F = MSType/MSErr
0.05
= 34060.5/15133.6
= 2.25 [Minitab]
F
1
4.7472
P-value = P( F121 >2.25) = 0.159 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that type has an effect
on head injury measurements. There is not enough evidence to support the claim that the
type of car (foreign or domestic) has an effect on head injury measurements.
1
12
268
CHAPTER 12 Analysis of Variance
NOTE: In exercises 11-13 the n=12 values in each of the six cells corresponds to df=11 for each of
the six cells and df = 6(11) for the pooled estimate of the mean square error.
11. Ho: there is no subject-target interaction
H1: there is subject-target interaction
α = 0.05 and dfnum = 2, dfden = 66
C.V. F = Fα = F0.05 = 3.1359 [Statdisk]
calculations:
F = MSInt/MSErr
0.05
= 3.375/[not given]
= 3.73 [Statdisk]
F
1
3.1359
2
P-value = P( F66 >3.73) = 0.0291 [Statdisk]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that there is no subject-target
interaction. There is sufficient evidence to reject the claim that perceived self-esteem in
other target people is unaffected by an interaction between the subject self-esteem and the
target’s self-esteem.
2
66
13. Ho: μ1 = μ2 = μ3 [there is no subject effect]
H1: at least one size μi is different
α = 0.05 and dfnum = 2, dfden = 66
C.V. F = Fα = F0.05 = 3.1359 [Statdisk]
calculations:
F = MSSubject/MSErr
0.05
= 1.4306/[not given]
F
1
3.1359
= 1.5824 [Statdisk]
2
P-value = P( F66 >1.5824) = 0.2132 [Statdisk]]
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that there is a subject
effect. There is not sufficient evidence to support the claim that the perceived self-esteem
in other target people is affected by the subject’s own measure (low, medium or high) of
self-esteem.
2
66
NOTE FOR HAND CALCULATIONS: The formulas and principles in this section are logical
extensions of those in the previous section. In particular,
SSRow = Σni (x i - x) 2 for i = 1,2,3… [for each row]
SSCol = Σnj (x j - x)2 for j = 1,2,3… [for each column]
SSTot = Σ (x - x)2 for all the x’s
When there is only one observation per cell, the unexplained variation is
SSErr = SSTot – SSRow – SSCol, and there is not enough data to measure interaction.
When there is more than one observation per cell, the unexplained variation (i.e., the failure of
experimental units in the same cell to respond the same) is
SSErr = Σ (x - xij )2 = Σdfij sij2 [for each cell – i.e., for each i,j (row,col) combination]
and the interaction sum of squares is
SSInt = SSTot – SSRow – SSCol – SSErr.
Two-Way ANOVA SECTION 12-3
269
15. There are n=3 observations in each of the indicated whey-supplement cells.
WHEY
10%
20%
Σx = 13.9
Σx = 14.1
Σx2 = 64.45
Σx2 = 66.33
x = 4.6333
x = 4.7000
s2 = 0.0233
s2 = 0.0300
30%
Σx = 14.4
Σx2 = 69.26
x = 4.8000
s2 = 0.0700
E
Σx = 9.6
M
Σx2 = 30.74
E yes x = 3.2000
N
s2 = 0.0100
Σx = 11.1
Σx2 = 41.09
x = 3.7000
s2 = 0.0100
Σx = 15.1
Σx2 = 76.13
x = 5.0333
s2 = 0.0633
Σx = 16.3
Σx2 = 88.61
x = 5.4333
s2 = 0.0233
x = 3.8000
x = 4.1667
x = 4.8667
x = 5.1167
S
U
P
P
L
T
0%
Σx = 13.2
Σx2 = 58.10
no x = 4.4000
s2 = 0.0100
x = 4.6333
x = 4.3417
Σx = 107.7 x = 4.4875
Σx2 = 494.71 s2 = 0.4959
SSSupp = Σni (x i -x) 2 = 12(4.6333-4.4875)2 + 12(4.3417-4.4875)2 = 0.51041667
SSWhey = Σnj (x j -x)2 = 6(3.8-4.4875)2 + 6(4.1667-4.4875)2 + 6(4.8667-4.4875)2 + 6(6.1167-4.4875)2 =
SSTot = Σ (x -x)2 = df∙s2 = 23∙(0.4959) = 11.40625
2
SSErr = Σ (x -xij ) =
Σdfij ij2 =
s
6.69125
2(0.0100) + 2(0.0233) + 2(0.0300) + 2(0.0700)
2(0.0100) + 2(0.0100) + 2(0.0633) + 2(0.0233) = 0.4800
SSInt = SSTot – SSSupp – SSWhey – SSErr = 11.40625 – 0.51041 – 6.69125 – 0.48000 = 3.72459
This is the ANOVA table for the tests of hypotheses in this exercise.
Source
Supplement
Whey
Interaction
Error
Total
df
1
3
3
16
23
SS
0.51041
6.69125
3.72459
0.48000
11.40625
MS
0.51041
2.23042
1.24153
0.03000
F
17.0137
74.3472
41.3843
Ho: there is no whey-supplement interaction
H1: there is whey-supplement interaction
α = 0.05 [assumed] and dfnum = 3, dfden = 16
C.V. F = Fα = F0.05 = 3.2389
calculations:
F = MSInt/MSErr
0.05
= 1.24153/0.03000
= 41.3843
F
1
3.2389
P-value = Fcdf(41.3843,99,3,16)
= 9.11E-8 = 0.00000009
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that there is no interaction
between the whey and the supplement. There is sufficient evidence to conclude that the
ratings are affected by an interaction between the use of the supplement and the amount of
whey.
Because there is interaction, further testing for whey or supplement effects are not appropriate.
The presence of interaction means that the effect of the supplement, for example, depends
upon the amount of whey and cannot be determined in a single test that combines the different
whey amounts.
3
16
270
CHAPTER 12 Analysis of Variance
17. The table and preliminary calculations are as follows.
WHEY
S
0%
U no 4.4
P yes 3.3
P
x = 3.85
10%
20%
4.5
5.0
x = 4.75
4.6
3.8
x = 4.20
2
30%
4.6
5.4
x = 5.00
x = 4.525
x = 4.375
Σx = 35.6
Σx2 = 161.42
x = 4.45
s2 = 0.4286
SSSupp = Σni (x i -x) = 4(4.525-4.45) + 4(4.375-4.45) = 0.0450
2
2
SSWhey = Σnj (x j -x)2 = 2(3.85-4.45)2 + 2(4.20-4.45)2 + 2(4.75-4.45)2 + 2(5.00-4.45)2 = 1.6300
SSTot = Σ (x -x)2 = df∙s2 = 7∙(0.4286) = 3.000
SSErr = SSTot – SSSupp – SSWhey = 3.0000 – 0.0450 – 1.6300 = 1.3250
This is the ANOVA table for the tests of hypotheses in this exercise.
Source
Supplement
Whey
Error
Total
df
1
3
3
7
SS
0.0450
1.6300
1.3250
3.0000
MS
0.04500
0.54333
0.44167
F
0.1019
1.2302
Ho: μ1 = μ2 [there is no supplement effect]
H1: at least one supplement μi is different
α = 0.05 [assumed] and dfnum = 1, dfden = 3
C.V. F = Fα = F0.05 = 10.128
calculations:
F = MSSupp/MSErr
0.05
= 0.04500/0.44167
= 0.1019
F
1
10.128
P-value = Fcdf(0.1019,99,1,3) = 0.7684
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that the supplement
has an effect on the rating. There is not enough evidence to support the claim that use of
the supplement affects the ratings of the pancakes.
1
3
Ho: μ1 = μ2 = μ3 = μ4 [there is no whey effect]
H1: at least one age μi is different
α = 0.05 [assumed] and dfnum = 3, dfden = 3
C.V. F = Fα = F0.05 = 9.2766
calculations:
F = MSWhey/MSErr
0.05
= 0.54333/0.44167
= 1.2302
F
1
9.2766
P-value = Fcdf(1.2302,99,3,3) = 0.4327
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that whey has an
effect on the rating. There is not enough evidence to support the claim that the amount of
whey used affects the rating of the pancakes.
3
3
Statistical Literacy and Critical Thinking
271
Statistical Literacy and Critical Thinking
1. Since two of the three samples are from the same power source, the three sets of sample data
are not independent. Since one-way analysis of variance requires independent sets of sample
data, the one-way analysis of variance should not be used. Even if the three data sets came
from three independent power sources, if all the measurements were taken at precisely the
same time the data would consist of matched triples and the one-way analysis of variance
(since it ignores the time characteristic) would not be the optimal technique to use.
2. One-way analysis of variance is used when the data are categorized according to one factor,
while two-way analysis of variance is used when the data are categorized according to two
factors.
3. While the factors used to categorize the data may originally be measured at any level, the data
to be analyzed must be quantitative data for which meaningful means and standard deviations
can be calculated. MPAA rating is not a quantitative variable. Running time could be
analyzed using the factors of budget bracket and MPAA rating, but MPAA rating could not be
analyzed using the factors of budget bracket and running time bracket.
4. No. Samples must be representative of the population to which the inference is to be made. If
only less expensive cars are used in the crash tests, then the results may be used to make
inferences only about the population of less expensive cars and not the population of all cars.
Chapter Quick Quiz
1. The one-way analysis of variance is used to test the null hypothesis that three or more samples
are from populations with equal means.
2. One-way analysis of variance tests are right tailed. The test statistic is the ratio of the variance
between groups to the variance within groups, and we reject the hypothesis that the groups
have equal means in favor of them having means that are not all equal only if the variance
between the groups is significantly larger than the variance between groups.
3. In one-way analysis of variance tests, the P-value is the probability of chance alone producing
a test statistic as large as or larger than the calculated test statistic. The larger the calculated
test statistic, therefore, the smaller the P-value.
4. The value of test statistic is F = 8.98.
5. The null hypothesis is that the three books (by Rowling, Clancy and Tolstoy) have the same
mean grade level reading scores. Since P-value = 0.001 < 0.05, we reject the null hypothesis
and conclude that the three books do not have the same mean grade level reading score.
6. The final conclusion is that at least one of the three books has a mean grade level reading score
that is different from the others.
7. One-way analysis of variance is used when data are categorized according to one factor, while
two-way analysis of variance is used when data are categorized according to two factors.
272
CHAPTER 12 Analysis of Variance
8. There is not sufficient evidence to reject the claim that a student’s estimate is unaffected by
interaction between that student’s gender and that student’s major.
9. No. Since P-value = 0.395 > 0.05 for testing for a gender effect, conclude that there is not
sufficient evidence to support a claim that students’ estimates of length are affected by their
genders.
10. No. Since P-value = 0.876 > 0.05 for testing for a major effect, conclude that there is not
sufficient evidence to support a claim that students’ estimates of length are affected by their
majors.
Review Exercises
1. a. The age data from the sample was categorized according to one variable, diet. Specifically,
it was noted whether the subject with each age in the age participated in the Atkins, Ornish,
Weight Watchers or Zones diet.
b. The data was analyzed using a one-way analysis of variance.
c. Since P-value = 0.41 > 0.05, we fail to reject the claim that the subjects come from
populations with the same mean age. It appears that the mean ages of the four treatment
groups are about the same.
d. A small P-value like 0.001 < 0.05 would lead to rejection of the claim that the subjects come
from populations with the same mean age. The conclusion would be that the four treatment
groups did not have the same mean age. If such were the case, it would not be possible to
declare whether any observed differences between the groups were due to the different
treatments or to the difference in ages.
2. Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 29
C.V. F = Fα = F0.05 = 3.3277
calculations:
F = s 2B /s 2p
0.05
= 3083363/56373
F
1
3.3277
= 54.70 [Minitab]
2
P-value = P( F29 >54.70) = 0.000 [Minitab]
conclusion:
Reject Ho; there is sufficient evidence to reject the claim that μ1 = μ2 = μ3. There is
sufficient evidence to reject the claim that the different car categories (4 cylinder, 6
cylinder, 8 cylinder) have different weights. Yes, it does appear that cars with more
cylinders tend to weigh more – but a formal conclusion about such a relationship would
require additional statistical methodology.
2
29
Review Exercises
273
3. Ho: there is no type-size interaction
H1: there is type-size interaction
α = 0.05 and dfnum = 2, dfden = 12
C.V. F = Fα = F0.05 = 3.8853
calculations:
F = MSInt/MSErr
0.05
= 57385/155086
= 0.37 [Minitab]
F
1
3.8853
P-value = P( F122 >0.37) = 0.698 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that there is no
interaction between type and size. There is not sufficient evidence to reject the claim that
left femur crash loads are unaffected by an interaction between a car’s type (foreign or
domestic) and its size (small, medium or large).
2
12
4. Ho: μ1 = μ2 [there is no type effect]
H1: at least one type μi is different
α = 0.05 and dfnum = 1, dfden = 12
C.V. F = Fα = F0.05 = 4.7472
calculations:
F = MSType/MSErr
0.05
= 282752/155086
= 1.82 [Minitab]
F
1
4.7472
1
P-value = P( F12 >1.82) = 0.202 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that type has an effect
on left femur crash loads. There is not enough evidence to support the claim that the
type of car (foreign or domestic) has an effect on left femur crash loads..
1
12
5. Ho: μ1 = μ2 = μ3 [there is no size effect]
H1: at least one size μi is different
α = 0.05 and dfnum = 2, dfden = 12
C.V. F = Fα = F0.05 = 3.8853
calculations:
F = MSSize/MSErr
= 74067/155086
0.05
= 0.48 [Minitab]
F
1
3.8853
P-value = P( F122 >0.48) = 0.632 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that size has an effect
on left femur crash loads. There is not sufficient evidence to support the claim that the
size of a car (small, medium or large) has an effect on left femur crash loads.
2
12
274
CHAPTER 12 Analysis of Variance
6. These are precisely the values from Data Set 4. The Minitab output is as follows.
Level
N
king
25
menthol 25
non-menth25
Mean
15.720
14.960
14.800
StDev
0.936
4.168
4.233
Source
Factor
Error
Total
DF
2
72
74
SS
12.1
868.0
880.1
MS
6.0
12.1
F
0.50
P
0.608
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 72
C.V. F = Fα = F0.05 = 3.1504
calculations:
F = s 2B /s 2p
0.05
= 6.0/12.1
= 0.50 [Minitab]
F
1
3.1504
P-value = P( F722 >0.50) = 0.608 [Minitab]
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that μ1 = μ2 = μ3. There
is not sufficient evidence to reject the claim that the three types of cigarettes yield the same
mean amount of carbon monoxide. No; it appears that filters do not make a difference in
the amount of carbon monoxide – but a formal test using only two groups (filtered and
non-filtered) would provide a better test for answering that question.
2
72
7. There are n=4 observations in each of the indicated gender-smoking cells.
GENDER
male
female
SMOKE?
yes
no
Σx = 394.8
Σx = 391.8
Σx2 =38967.44 Σx2 =38378.44
x = 98.700
x = 97.950
s2 = 0.22667
s2 = 0.54333
Σx = 393.9
Σx2 =38789.69
x = 98.475
s2 = 0.12917
Σx = 393.0
Σx2 =38613.34
x = 98.250
s2 = 0.36333
x = 98.5875
x = 98.1000
x = 98.3250
x = 98.3625
Σx = 1573.5
x = 98.34375
Σx2 = 154748.91 s2 = 0.334625
SSGen = Σni (x i -x) 2 = 8(98.3250-98.34375)2 + 8(98.3625-98.34375)2 = 0.005625
SSSmo = Σnj (x j -x)2 = 8(98.5875-98.34375)2 + 8(98.1000-98.34375)2 = 0.950625
SSTot = Σ (x -x)2 = df∙s2 = 15∙(0.334625) = 5.019375
SSErr = Σ (x -xij )2 = Σdfij sij2 = 3(0.22667) + 3(0.54333)
3(0.12917) + 3(0.36333) = 3.78750
SSInt = SSTot – SSSupp – SSWhey – SSErr = 5.019375 – 0.005625 – 0.950625 – 3.78750 = 0.275625
This is the ANOVA table for the tests of hypotheses in this exercise.
Source
Gender
Smoke
Interaction
Error
Total
df
1
1
1
12
15
SS
0.005625
0.950625
0.275625
3.787500
5.019375
MS
0.005625
0.950625
0.275625
0.315625
F
0.0178
3.0119
0.8733
Review Exercises
275
Ho: there is no gender-smoking interaction
H1: there is gender-smoking interaction
α = 0.05 and dfnum = 1, dfden = 12
C.V. F = Fα = F0.05 = 4.7472
calculations:
F = MSInt/MSErr
0.05
= 0.275625/0.315625
= 0.8733
F
1
4.7472
P-value = Fcdf(0.8733,99,1,12) = 0.3685
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that there is no
interaction between gender and smoking. There is not sufficient evidence to reject the
claim that a person’s body temperature is not affected by an interaction between gender
and smoking habits.
1
12
Ho: μ1 = μ2 [there is no gender effect]
H1: at least one μi is different
α = 0.05 and dfnum = 1, dfden = 12
C.V. F = Fα = F0.05 = 4.7472
calculations:
F = MSGen/MSErr
0.05
= 0.005625/0.315625
= 0.0178
F
1
4.7472
P-value = Fcdf(0.0178,99,1,12) = 0.8960
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that there is no gender
effect. There is not sufficient evidence to reject the claim that a person’s body temperature
is unaffected by that person’s gender.
1
12
Ho: μ1 = μ2 [there is no smoking effect]
H1: at least one μi is different
α = 0.05 and dfnum = 1, dfden = 12
C.V. F = Fα = F0.05 = 4.7472
calculations:
F = MSSmo/MSErr
0.05
= 0.950625/0.315625
= 3.0119
F
1
4.7472
P-value = Fcdf(3.0119,99,1,12) = 0.1082
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that there is no smoking
effect. There is not sufficient evidence to reject the claim that a person’s body temperature
is unaffected by that person’s smoking habits.
1
12
8. The following preliminary values are identified.
n
Σx
Σx2
x
s2
presidents
38
589
12627
15.500
94.5270
popes
24
315
5981
13.125
80.2880
monarchs
14
318
11722
22.714
346.0659
276
CHAPTER 12 Analysis of Variance
x = (Σni x i )/Σni = [38(15.500) + 24(13.125) + 14(22.714)]/[38+24+14] = 1222/76 = 16.079
Σni( x i – x )2 = 38(15.500-16.079)2 + 24(13.125-16.079)2 + 14(22.714-16.079)2 = 838.54
2
Σdfi s i = 37(94.5270) + 23(80.2880) + 13(346.0659) = 9842.98
s 2B = Σni( x i – x )2/(k-1) = 838.54/2 = 419.27
s 2p = (Σdfi s i2 )/Σdfi = 9842.98/73 = 134.84
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 [assumed] and dfnum = 2, dfden = 73
C.V. F = Fα = 3.1504
calculations:
F = s 2B /s 2p
0.05
= 419.27/134.84
F
1
3.1504
= 3.1095
P-value = Fcdf(3.1095,99,2,73) = 0.0506
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that μ1 = μ2 = μ3. There
is not sufficient evidence to reject the claim that the three groups have the same mean
longevity. No, the survival times for the three groups do not appear to differ significantly.
2
73
Cumulative Review Exercises
The following summary information applies to exercises 1-4.
n
Σx
Σx2
x
s2
presidents
38
589
12627
15.500
94.5270
popes
24
315
5981
13.125
80.2880
monarchs
14
318
11722
22.714
346.0659
1. a. The means for the three groups are also follows.
pres: 15.5 years
popes: 13.1 years
mon: 22.7 years
b. The standard deviations for the three groups are as follows.
pres: 94.5270 = 9.7 years popes: 80.2880 = 9.0 years mon: 346.0659 = 18.6 years
2. Let the presidents be group 1.
x1 -x 2 = 15.500 – 22.714 =
original claim: μ1 – μ2 ≠ 0 mg
Ho: μ1 – μ2 = 0 years
H1: μ1 – μ2 ≠ 0 years
α = 0.05 [assumed] and df = 13
C.V. t = ±tα/2 = t0.025 = ±2.160
calculations:
t x1 -x 2 = (x1 -x 2 - μ x1 -x 2 )/s x1 -x 2
= (-7.214 – 0)/ 94.5270/38 +346.0659/14
= -7.214/5.2160 = -1.383
P-value = 2∙tcdf(-99,-1.383,13) = 0.1899
0.025
0.025
-2.160
0
0
_ _
x 1- x2
2.160
t13
Cumulative Review Exercises
277
conclusion:
Do not reject Ho; there is not sufficient evidence to conclude that μ1 – μ2 ≠ 0. There
is not sufficient evidence to support a claim that there is a difference between the mean for
presidents and the mean for British monarchs.
3. Yes, the longevity times for presidents appear to come from a population that is approximately
normal. The histogram is approximately bell-shaped and the points on the normal probability
plot approximate a straight line. Because of the truncation of the times at 0, there is not the
opportunity for unusually low scores to symmetrically balance unusually high scores – and this
slight departure form normality is evident in both the histogram and the normal probability
plot.
U.S. Presidents
Normal Probability Plot
9
2.0
8
1.5
1.0
6
z score
Frequency
7
5
4
3
0
-0.5
-1.0
2
-1.5
1
0
0.5
-2.0
0
5
10
15
20
25
30
35
longevity following inauguration (years)
0
10
20
30
40
longevity following inauguration (years)
4. n = 39 and x = 15.000 and s = 94.5270 = 9.7224
σ unknown, n > 30: use t with df=37
α = 0.05, tdf, α/2 = t37,0.025 = 2.026
x  tα/2∙s/ n
15.500  2.026(9.7224)/ 38
15.500  3.195
12.3 < μ < 18.7 (years)
5. Listed in order, the n=9 values are: 96.5 97.3 97.6 97.6 98.0 98.2 98.3 98.4 98.9
Σx = 880.8
Σx2 = 86204.96
a. Temperatures are measurements at the interval level, because differences are meaningful but
ratios are not (since there is not a meaningful zero).
b. Body temperatures are continuous data. Even though they are typically reported to the
nearest tenth of a degree, they actually can take on any values on a continuum.
c. x = (Σx)/n = 880.8/9 = 97.8667, rounded to 97.87 °F
d. x = x5 = 98.0 °F
e. R = 98.9 – 96.5 = 2.4 °F
f. s2 = [n(Σx2) – (Σx)2]/[n(n-1)]
= [9(86204.96) – (880.8)2]/[9(8)] = 36.00/72 = 0.5000
rating
frequency
s = 0.5000 = 0.7071, rounded to 0.71 °F
1.0 – 1.9
1
2
2
g. s = 0.5000 (°F)
2.0 – 2.9
4
6. The requested frequency distribution is given
In the box at the right
3.0 – 3.9
4.0 – 4.9
5.0 – 5.9
6.0 – 6.9
7.0 – 7.9
8.0 – 8.9
0
1
4
9
9
7
35
278
CHAPTER 12 Analysis of Variance
7. The histogram is given below. The actual cut points for the bars are 0.95, 1.95, 2.95, etc.
9
8
Frequency
7
6
5
4
3
2
1
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
movie viewer rating
No; it does not appear that the ratings are from a population with a normal distribution. The
distribution appears to be negatively skewed, and not bell-shaped.
8. This exercise requires the following summary statistics.
n
Σx
Σx2
x
s2
PG
3
19.9
133.89
6.633
0.9433
PG-13
20
128.6
904.88
6.430
4.1043
R
12
74.1
510.05
6.175
4.7711
informal comparison of the three sample means:
The means of 6.63, 6.43, and 6.18 do not appear to be very different – especially
considering the fairly wide range of ratings that extend from 1.9 to 8.6.
x = (Σni x i )/Σni = [3(6.633) + 20(6.430) + 12(6.175)]/[3+20+12] = 222.6/35 = 6.360
Σni( x i – x )2 = 3(6.633-6.360)2 + 20(6.430-6.360)2 + 12(6.175-6.360)2 = 0.7328
2
Σdfi s i = 2(0.9433) + 19(4.1043) + 11(4.7711) = 132.3504
s 2B = Σni( x i – x )2/(k-1) = 0.7328/2 = 0.3664
s 2p = (Σdfi s i2 )/Σdfi = 132.3504/32 = 4.13595
Ho: μ1 = μ2 = μ3
H1: at least one μi is different
α = 0.05 and dfnum = 2, dfden = 32
C.V. F = Fα = F0.05 = 3.3158
calculations:
F = s 2B /s 2p
0.05
= 0.36644.13595
F
1
3.3158
= 0.0886
P-value = Fcdf(0.0886,99,2,32) = 0.9154
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that μ1 = μ2 = μ3. There
is not sufficient evidence to reject the claim that PG, PG-13 and R movies have the same
mean viewer rating.
2
32
9. These are the necessary summary statistics.
n=6
Σxy = 110.9247 n(Σxy) – (Σx)(Σy) = 5(110.9247) – (52.41)(10.40) = 9.5595
Σx = 52.41 Σx2 = 554.0767 n(Σx2) – (Σx)2 = 5(554.0767) – (52.41)2 = 23.5754
Σy = 10.40 Σy2 = 23.7642
n(Σy2) – (Σy)2 = 5(23.7642) – (10.40)2 = 10.6610
Cumulative Review Exercises
279
a. r = [n(Σxy) - (Σx)(Σy)]/[ n(Σx 2 ) - (Σx) 2 n(Σy 2 ) - (y) 2 ]
= 9.5595/ [ 23.5754 10.6610]
= 0.6030
Ho: ρ = 0
H1: ρ ≠ 0
α = 0.05 [assumed] and df = 3
C.V. t = ±tα/2 = ±t0.025 = ±3.182 [or r = ±0.878]
calculations:
tr = (r – μr)/sr
0.025
0.025
= (0.6030 – 0)/ [1 - (0.6030)2 ]/3
-0.878
0
0.878
r
= 0.6030/0.4605
-3.182
0
3.182
t
= 1.309
P-value = 2∙tcdf(1.309,99,3) = 0.2817
conclusion:
Do not reject Ho; there is not sufficient evidence to conclude that ρ ≠ 0. No; there is not
sufficient evidence to support the claim of a linear correlation between the amounts of
discarded paper and discarded plastic.
b. x = (Σx)/n = 52.41/5 = 10.482
y = (Σy)/n = 10.40/5 = 2.080
2
b1 = [n(Σxy) – (Σx)(Σy)]/[n(Σx ) – (Σx)2]
= 9.5595/23.5754 = 0.4055
bo = y – b1 x
= 2.080 – (0.4055)(10.482) = -2.1703
ŷ = bo + b1x
= -2.1703 + 0.4055x
c. No. Since the correlation in part (a) was not significant, the regression line in part (b)
should not be used for making predictions. The best predictive equation is simply ŷ = y
for all values of x.
3
10. Ho: gun type and circumstances are independent
H1: gun type and circumstances are related
α = 0.05 and df = 2
2
C.V. χ2 = χ 2 = χ 0.05
= 5.991
calculations:
CIRCUMSTANCES
TYPE
handgun
rifle
accident
31
(31.85)
13
(12.15)
44
self-inflict
35
(30.40)
7
(11.60)
42
assault
162
(165.75)
67
(63.25)
229
228
87
0.05
2
5.991
315
χ2 = Σ[(O – E)2/E]
= 0.023 + 0.696 + 0.085
0.059 + 1.824 + 0.223 = 2.909
P-value = χ2cdf (2.909,99,2) = 0.2335
conclusion:
Do not reject Ho; there is not sufficient evidence to reject the claim that gun type and
circumstances are independent. For firearm injuries, there is not sufficient evidence to
reject the claim that the type (handgun or rifle/shotgun) of gun involved and the
circumstances of the event (unintentional, self-inflicted or assault) are independent.
2
2