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13.7 The ANOVA Table 329
SS
B
=
k
j=1
n
j
y
j
− y
2
=
k
j=1
n
j
j=1
y
ij
2
j
− CF,
SS
W
=
k
i=1
n
j
j=1
y
ij
− y
i
2
=
y
2
ij
−
k
j=1
n
j
j=1
y
ij
2
n
j
.
In this table, y
j
and n
j
are the mean and number of observations of jth, y, N are the
grand mean and total number of samples for all groups, K is the number of groups
being compared,
y
ij
is the observed value of the ith sample from jth group, and the
F is test statistic obtained from the proportion of two variances (
F =
MS
B
MS
W
).
In this table, the significance level is determined using the p-value. To obtain it,
the position of the statistic in the distribution to which it belongs must be known. In
other words, there must be a distribution that is a criterion. This distribution is called
the F distribution, the first letter of the famous statistician Ronald Fisher. That is
why the F test is another name for the ANOVA test.
As mentioned earlier, the deviation of each observation from the grand mean is
divided into two parts: “deviation of each observation from the mean of its own group”
and “deviation of the mean of each group from the grand mean” as the following
formula and Figs.
13.1 and 13.2.
y
ij
− y = y
ij
+
−y
i
+ y
i
−
y =
y
ij
− y
i
+
y
i
− y
.
In the right Fig. 13.2, the grand mean of population is known, but it is not enough
to explain all points (values) of the population alone. More accurately, based on
the common characteristics of points with each other, three separate groups can be
Fig. 13.1 Geometric
representation of between
group and within group
deviations
Within G
Between G
Within G
4.5
Group1
Group2
3
3.5
4
5
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330 13 Analysis of Variance
Grand
Mean
Circle
Mean
Plus
Mean
Triangle
Mean
Between
Group
Within
Group
Fig. 13.2 Schematic representation of between and within group deviations
identified to better describe the population. Therefore, as shown in Fig. 13.2 (middle),
the population is partitioned around three means. The whole population can now be
explained in terms of the mean of each group.
In the analysis of variance, the mean deviation of each group from the grand mean
(line arrow) is squared (
y
i
− y
2
) and multiplied by the sample size in each group
(
n
j
) and is divided by the degree of freedom between groups (k − 1) to obtain the
between group variance (MS
B
) and put it in the nominator of the fraction F. Then,
we squared the data deviation of each group from the mean of its own group (dotted
arrow) (
y
ij
− y
i
2
) and divide their sum by the degree of freedom within the group
to obtain the intragroup variance (MS
W
) and we put it in denominator of the fraction
F.
The test statistic obtained from the ratio of between groups variance to within
group variance has an F distribution (Fig.
13.3). If the calculated F is greater than
the critical value of the 0.05 with degree of freedom (
df = (n − k, k − 1)), the null
hypothesis is rejected. That is, at least the mean of one group is different from the
means of the other groups.
Fig. 13.3 Changes in the F
distribution in terms of the
degree of freedom of the
nominator and the
denominator of the fraction
0.2
0.4
0.6
0.8
0
1.0
0
12 3
45
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13.7 The ANOVA Table 331
Example 13.2 A study was designed to compare the effect of three treatments on
the mobility score of physiotherapy patients. In this study, 24 patients were randomly
assigned to three groups: control (A), just physiotherapy (B), and physiotherapy with
counseling (C). The results of this study are presented in Table
13.5. Is the mobility
score of these 3 groups equal?
Solution: In this example n
1
= n
2
= n
3
= 8, n = n
1
+ n
2
+ n
3
= 8 + 8 + 8 = 24
and required calculations for the implementation of ANOVA are given on the right
side of Table
9.5.
First, we obtain the correction factor (CF), which facilitates the rest of the
computational process Table
13.6.
CF =
y
ij
2
n
=
(
1008
)
2
24
= 42336,
SS
T
=
y
2
ij
− CF = 43464 − 42336 = 1128,
SS
B
=
n
j
j=1
y
ij
2
n
i
− CF =
(
280
)
2
8
+
(
360
)
2
8
+
(
368
)
2
8
− 42336 = 592,
SS
W
= SS
T
− SS
B
= 1128 − 592 = 536.
We now substitute the above-calculated values in the ANOVA table as below:
The calculated F statistic (11.597) is greater than the critical value of
F
(
0.95, 2, 23
)
= 4.28 in the F distribution tables, so the null hypothesis that the
mean of the groups is equal is rejected.
Table 13.5 Mobility score in 3 groups
Sample A B C A
2
B
2
C
2
1 35 38 47 1225 1444 2209
2 38 43 53 1444 1849 2809
3 42 45 42 1764 2025 1764
4 34 52 45 1156 2704 2025
5 28 40 46 784 1600 2116
6 39 46 37 1521 2116 1369
7 34 54 48 1156 2916 2304
8 30 42 50 900 1764 2500
Sum
y
i1=280
y
i2=360
y
i3=368
9950 16418 17096
y
ij
= 1008
y
2
ij
= 43464
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332 13 Analysis of Variance
Table 13.6 One-way ANOVA table for mobility score
Source of variation Sum of square (SS) Degree of freedom
(DF)
Mean of square
(MS)
F
Between 592 2 296 11.597
Within 536 21 25.52 –
Total 1128 23 –
Table 13.7 Blood pressure
reduction in 16 subjects after
3 treatments
Medication Exercise Herbal diet
9 6 7
11 8 10
10 7 11
13 4 13
12 5 9
11 – –
Example 13.3 A research project is designed to investigate the effect of 3 different
treatments on reduction of systolic the blood pressure of individuals diagnosed with
high blood pressure. They are randomly allocated to three groups; (1) medication,
(2) exercises, and (3) an herbal diet. After a month, the reduction in each person’s
blood pressure is recorded and is shown in Table
13.7. Is there difference among the
means of 3 treatments?
Solution: In this example
n
1
= 6, n
2
= 5, n
3
= 5, n = n
1
+n
2
+n
3
= 6+ 5 + 5 = 16
and required calculations for the implementation of ANOVA are given on the right
side of Table
13.8.
First, we obtain the correction factor (CF), which facilitates the rest of the
computational process Table
13.9.
Table 13.8 Necessary calculation for ANOVA
Sample Med Exe Herb Med
2
Exe
2
Herb
2
1 9 6 7 81 36 49
2 11 8 10 121 64 100
3 10 7 11 100 49 121
4 13 4 13 69 16 169
5 12 5 9 144 25 81
6 11 – – 121 – –
Sum
y
i1=66
y
i2=30
y
i3=50
736 190 520
y
ij
= 146
y
2
ij
= 1446
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13.8 Running ANOVA on SPSS 333
Table 13.9 One-way ANOVA table for mobility score
Source of variation Sum of square (SS) Degree of freedom
(df)
Mean of square
(MS)
F
Between 73.75 2 36.875 11.984
Within 40 13 3.077 –
Total 113.75 15 –
CF =
y
ij
2
n
=
(
146
)
2
16
= 1332.25,
SS
T
=
y
2
ij
− CF = 1446 − 1332.25 = 113.75,
SS
B
=
n
j
i=1
y
ij
2
n
j
− CF =
(
66
)
2
6
+
(
30
)
2
5
+
(
50
)
2
5
− 1332.25 = 73.75,
SS
W
= SS
T
− SS
B
= 113.75 − 73.75 = 40.
We now substitute the above-calculated values in the ANOVA table as below:
The calculated F statistic (11.984) is greater than the critical value of
F
(
0.95, 2, 13
)
= 6.70 in the F distribution tables, so the null hypothesis that the
mean of the groups is equal is rejected.
13.8 Running ANOVA on SPSS
To run a one-way analysis of variance in SPSS, all quantitative data (dependent
variable) must be entered in one column and the names of groups (independent
variable or factor) in another column. Then select the path of Analyze\Compare
Means\One-way ANOVA to open a windows like Fig.
13.4.
Select the name of the dependent variable (in this example Mobility) and move
it to the right box titled Dependent List. We move the name of the multilevel group
or factor (say, treatment) that is qualitative (nominal or ordinal) to the Factor box.
Then by selecting options, another window will open. In this window, by selecting
the option of “Homogeneity of variance test”, Levene statistic is calculated in order
to test the equality of the variance of the groups. Also, by selecting the “Fixed and
random effects” option, it calculates the standard error, standard deviation, 95%
confidence interval for the fixed effects model.
The most important output of SPSS software for ANOVA is the following two
tables.
https://t.me/med1917
334 13 Analysis of Variance
Fig. 13.4 Path and main windows of one-way ANOVA in SPSS
Test of homogeneity of variances
Levene statistic df1 df2 Sig.
Mobility score Based on mean 0.161 2 21 0.852
Based on median 0.152 2 21 0.860
Basedonmedianand with
adjusted df
0.152 2 20.444 0.860
Based on trimmed mean 0.161 2 21 0.852
The p-value in this test was p = 0.852 and is more than the significance level of
0.05. Therefore, the hypothesis of the p-value in this test was 0.852 p = 0.852 and
more than the significance level of 0.05. Therefore, the hypothesis of equal group
variance is not rejected and the condition of homogeneity of variances for ANOVA
is established.
The main table of ANOVA will be as follows
ANOVA
Mobility score
Sum of squares df Mean square F Sig.
Between groups 592.000 2 296.000 11.597 .000
Within groups 536.000 21 25.524
Total 1128.000 23
The calculations and conclusions are exactly the s ame as the manual calculations
performed above.
The main table of ANOVA for Example 13.2 is as follows
https://t.me/med1917
13.9 Identifying Which Means is Different 335
ANOVA
Reduction
Sum of squares df Mean square F Sig.
Between groups 73.750 2 36.875 11.984 .001
Within groups 40.000 13 3.077
Total 113.750 15
13.9 Identifying Which Means is Different
Analysis of variance (ANOVA) is the most common method of comparing the means
of several groups. After conducting ANOVA, if there is no reason to reject the null
hypothesis, there will be no more questions and the research question will be fully
answered. However, if the rejection of the null hypothesis and the significance of
the mean of the groups are concluded, according to ANOVA, it is not possible to
say which particular groups are different from each other. In order to reach the full
answer to the question, the researcher needs to know which pair of groups led to
such a result. Is only one group different from the other groups or are the groups
significantly different from each other?
For this purpose, empirical (informal) methods such as constructing confidence
interval and boxplots or argumentative (formal) methods such as multiple comparison
tests and range tests can be used.
Calculate confidence interval estimates of the means for each of the different
groups, and then check if one or more of them does overlap with the others.
Construct boxplots for different groups, and then examine if one or more of them
does overlap with the others.
Range tests allow user to identify subsets of means that are not significantly
different from each other.
Multiple comparison tests (such as Bonferroni, Duncan, Tukey, Scheffe, Dunnet,
Fisher LSD, and Student–Newman–Keuls) using pairs of means make adjustments
to overcome the problem of increasing a significance level when the number of tests
increases. These tests are usually selected based on whether or not the variance of
the groups compared is equal, whether the sample size of the groups is equal or not,
and whether the number of comparisons is low or high.
Example 13.4
Subjects have to press a button when they hear a signal. Three
treatment signals with different intensities (low, moderate, and severe) are given to
hearing problem patients. The time elapsed between presentation of the signal and
button is pushed by patients in hundredths of a second are recorded in the table
below. Are these three treatments different? If so, which groups are significantly
different from the others?
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336 13 Analysis of Variance
Signal Observations
Low 27 24 28 21 32
Moderate 19 14 15 13 –
Severe 21 18 22 14 16
Solution: In this example n
1
= 5, n
2
= 4, n
3
= 5, n = n
1
+n
2
+n
3
= 5+ 4 + 5 = 14
and required calculations for the implementation of ANOVA are given on the right
side of Table
13.10.
First, we obtain the correction factor (CF), which facilitates the rest of the
computational process.
CF =
y
ij
2
n
=
(
284
)
2
14
= 5761.143,
SS
T
=
y
2
ij
− CF = 6206 − 5761.143 = 444.857,
SS
B
=
n
j
i=1
y
ij
2
n
i
− CF
=
(
132
)
2
5
+
(
61
)
2
4
+
(
91
)
2
5
− 5801.786
= 310.107,
SS
W
= SS
T
− SS
B
= 444.857 − 310.107 = 134.75.
We now substitute the above-calculated values in the ANOVA table as below
(Table
13.11):
The calculated F statistic (12.657) is greater than the critical value of
F
(
0.95, 2, 11
)
= 7.21 in the F distribution tables, so the null hypothesis that the
mean of the groups is equal is rejected Table
13.11.
Table 13.10 Necessary calculations for ANOVA
Signal Observations Sum
Low 27 24 28 21 32
y
i1=132
y
ij
= 284
Moderate 19 14 15 13 –
y
i2=61
Severe 21 18 22 14 16
y
i3=91
(Low)
2
729 576 784 441 1024 3554
y
2
ij
= 6206
(Moderate)
2
361 196 225 169 – 951
(Severe)
2
441 324 484 196 256 1701
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13.9 Identifying Which Means is Different 337
Table 13.11 One-way ANOVA table for mobility score
Source of variation Sum of square (SS) Degree of freedom
(df)
Mean of square
(MS)
F
Between 310.107 2 155.054 12.657
Within 134.750 11 12.250 –
Total 444.857 13 –
The SPSS output for this comparison is shown below:
ANOVA
MlSec
Sum of squares df Mean square F Sig.
Between groups 310.107 2 155.054 12.657 0.001
Within groups 134.750 11 12.250
Total 444.857 13
In order to identify which groups’ mean was different from the others, we first
draw the boxplot for all three groups.
The 95% confidence interval for these three groups is:
https://t.me/med1917
338 13 Analysis of Variance
Descriptives
Signal Lower Upper
MlSec Low 95% confidence interval for mean 21.24 31.56
Moderate 95% confidence interval for mean 11.07 19.43
Severe 95% confidence interval for mean 14.04 22.36
i
Moderate
Low
10
14
18
22
26
30
Severe
As can be seen from the diagrams above, the difference between low signal and
medium signal and with high signal is significant.
Tukey, Fisher LSD, and Bonferroni post hoc tests also show differences.
Multiple comparisons
Dependent variable: MlSec
(I) Signal (J) Signal Mean
difference
(I−J)
Std.
error
Sig. 95% confidence
interval
Lower Upper
Tukey HSD Low Moderate 11.150
*
2.348 .002 4.81 17.49
Severe 8.200
*
2.214 .009 2.22 14.18
Moderate Low −11.150
*
2.348 .002 −17.49 −4.81
Severe −2.950 2.348 .447 −9.29 3.39
Severe Low −8.200
*
2.214 .009 −14.18 −2.22
Moderate 2.950 2.348 .447 −3.39 9.29
LSD Low Moderate 11.150
*
2.348 .001 5.98 16.32
Severe 8.200
*
2.214 .003 3.33 13.07
Moderate Low −11.150
*
2.348 .001 −16.32 −5.98
Severe −2.950 2.348 .235 −8.12 2.22
Severe Low −8.200
*
2.214 .003 −13.07 −3.33
Moderate 2.950 2.348 .235 −2.22 8.12
Bonferroni Low Moderate 11.150
*
2.348 .002 4.53 17.77
Severe 8.200
*
2.214 .010 1.96 14.44
Moderate Low −11.150
*
2.348 .002 −17.77 −4.53
Severe −2.950 2.348 .705 −9.57 3.67
Severe Low −8.200
*
2.214 .010 −14.44 −1.96
Moderate 2.950 2.348 .705 −3.67 9.57
*
. The mean difference is significant at the 0.05 level.
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