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32 2 Tables and Charts
Fig. 2.8 Boxplot for the length of individuals’ ears by gender
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2.6 Exercises 33
Interperate the Boxplot
1. When the median is near the center of the box (approximately halfway
between the first and third quartiles), that is, the lines are about the same
length and the two whiskers are approximately equal in length, the
distribution is approximately symmetric.
2. If the median falls to the left of the center of the box (the median is closer
to the first quartile than to the third quartile), that is, the right line is larger
than the left line or the upper whisker is longer than the lower whisker, the
distribution is positively skewed or skewed to the right.
3. When the median falls to the right of the center (the median is closer to the
third quartile than to the first quartile), that is, the left line is larger than
the right line or the lower whisker is longer than the upper whisker, the
distribution is negatively skewed or skewed to the left.
A boxplot is often employed to discern the skewness of a dataset’s distribution.
Skewness refers to a lack of symmetry in the distribution, where one tail extends
farther than the other. In a right-skewed distribution (also known as positively
skewed), the boxplot displays a longer right whisker, indicating that relatively few
data values have high values, with the majority located toward the left. Conversely,
in a left-skewed distribution (or negatively skewed), the boxplot exhibits a longer
left whisker, suggesting that most data values are situated toward the right.
Therefore, by observing the lengths of the whiskers relative to each other, one can
infer the skewness of the distribution.
2.6 Exercises
1.
The following data is the blood glucose level (in millimoles per liter) of 40
first-year students.
4.1– 4.0– 3.6– 3.8– 4.6– 4.4– 3.3– 3.6– 3.8– 4.1– 3.9– 4.7– 3.4– 3.7– 2.2– 5.0– 4.7– 3.3–
4.2– 3.6– 4.1– 3.8– 4.8– 3.3– 4.7– 4.4– 4.9– 4.5– 3.7– 4.0– 4.9– 3.4– 2.9– 6.0– 4.0– 3.4–
5.1– 4.3– 4.8– 5.0
A. Draw a stem-and-leaf diagram for this data.
B. Tabulate this data and make its frequency distribution.
C. Draw a histogram for this data.
2. The method of maternal delivery of 600 mothers in a hospital is as follows. Draw
a suitable chart for how these children are born.
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34 2 Tables and Charts
Type of delivery No.
Cesarean 57
Normal 476
With aid 67
Total 600
3. Floating particles in the air at a sample of 49 cities in micrograms per cubic meter
are as follows:
A. Tabulate them using the Fisher suggestion.
B. Draw a stem-and-leaf diagram for it.
22– 42– 43– 49– 50– 27– 22– 23– 24– 25– 44– 63– 57– 27– 30– 27– 45– 49– 46– 12– 42–
21– 32– 27– 16– 23– 74– 25– 32– 79– 51– 31– 33– 23– 19– 12– 38– 38– 24– 65– 51– 47–
25– 36– 28– 42– 28– 30
4. In a sample of 50, the size of the arm circumference in centimeters and the weight
in kilograms are as described in Table
2.9.
A. Tabulate the above data in two-dimensional table (e.g., weights in rows and
arm circumference in columns).
B. Plot a scattergram for weights and arm circumference.
Table 2.9 Arm circumference (cm) and weight (kg) of 50 samples
Weight Arm Weight Arm Weight Arm
78 29 66 26 71 30
68 25 68 26 66 26
72 26 73 29 78 31
61 24 66 26 58 24
79 33 60 25 76 24
88 31 78 29 58 25
69 26 82 31 59 28
71 25 66 26 60 27
68 29 64 25 55 23
73 28 64 28 79 28
53 31 74 27 65 26
53 23 62 25 68 25
46 22 82 33 61 27
87 34 72 31 80 27
52 23 95 31 71 26
82 29 74 28 80 26
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2.6 Exercises 35
C. Draw a stem and leaf for each variable separately.
D. Interpret the box-and-whiskers plot for these variables.
5. Draw a suitable chart for the number of medications prescribed in 1000
prescriptions of patients referred to a pharmacy shown in Table
2.10.
6. Draw a suitable chart for the reasons why people did not get the Covid-19 vaccine.
Reason Freq
Religious believes 5
Fear of vaccine side effects 6
No time 18
Crowded queues for vaccination 24
Lack of approved vaccine by WHO 30
Propaganda of traditional anti-vaccine groups 29
Total 112
7. Draw a box plot for the number of painkiller pills consumed by rural households
in the last month. Is the stem-and-leaf plot suitable for the data which are mostly
one digit?
6– 11– 2– 3– 11– 8– 9– 11– 4– 3– 7– 12– 4– 3– 0– 1– 8– 2– 8– 4– 9– 4– 6– 9– 5– 4– 4– 5– 3–
2– 4– 8– 0– 6– 8– 9– 3– 6– 5– 8–
8. Table 2.11 shows the calcium level of two groups of 75 married women by age.
The first group used contraceptive pills and the other group did not use the pill.
A. Classified the calcium levels in two-dimensional table (e.g., consuming pill
in rows and calcium level in columns).
B. Classified by age and calcium level.
Table 2.10 Number of medications prescribed in 1000 prescriptions of patients
No. of medications 1 2 3 4 5 6 7 Sum
Frequency 37 137 218 340 192 56 20 1000
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36 2 Tables and Charts
9. From the database of an insurance company, we have selected a sample of 40
patients who have undergone diagnostic imaging tests. The samples of patients
from three types of CT scan, MRI, and radiology (R) are listed below. Categorize
this data and draw a suitable chart for it.
CT, R, MRI, R, R, R, R, R, CT, MRI, R,
CT, MRI, MRI, R, CT, CT , R, MRI, CT,
R, R, MRI, R, CT, MRI, CT, CT, CT, R,
MRI, CT, MRI, R, CT, R, R, R, CT.
Table 2.11 Calcium levels in two groups of 75 women by age (group 1: taking contraceptive pill,
group 2: not taking pills)
Age Group 1 Group 2 Age Group 1 Group 2 Age Group 1 Group 2
25 96 104 27 94 101 30 96 99
19 105 99 27 96 97 30 99 104
20 93 95 27 103 98 31 91 98
21 106 101 28 105 106 31 99 111
21 106 98 28 93 100 32 108 103
21 95 108 28 101 97 32 97 99
21 99 102 29 98 99 32 90 94
21 105 96 29 99 94 54 104 108
22 97 98 30 98 107 33 99 100
22 98 98 30 99 99 33 95 106
22 105 102 30 96 104 33 104 101
22 95 100 30 99 95 34 95 93
22 106 104 31 91 98 35 100 91
23 100 92 31 99 111 35 105 107
23 101 103 31 95 96 36 103 90
24 103 106 32 99 103 36 98 104
24 104 108 32 97 99 37 99 105
24 102 96 32 90 94 37 91 100
25 93 101 54 106 86 38 94 102
25 103 101 26 99 103 39 90 95
54 106 86 27 98 95 39 94 108
27 101 106 28 105 106 39 99 91
26 99 103 29 98 99 40 99 97
26 95 96 29 99 94 40 100 100
27 98 98 30 98 107 40 96 91
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Chapter 3
Summarizing Data
3.1 Introduction
A large amount of data without processing and summarizing is not understandable
and sensible and even is ambiguous and confusing to be. In order to make sense and
tangible the set of observations, it is necessary to summarize them with statistical
indicators in order to provide a clear picture of the entire dataset. These measures
also condense the dataset down to a few representative indexes and are usually
brought into two general headings, “central tendency” and “dispersion indicators”,
to represent the entire data distribution and form the basis of descriptive statistics.
One reason we explain central tendency before dispersion indices is that many
measures of dispersion involve measures of central tendency and they cannot be
computed and understood without central tendency. Another reason, in descriptive
statistics, measures of central tendency come up early on before measures of
dispersion.
3.2 Measures of Central Tendency
Central tendencies are the most common statistical indicators that measure and
determine the location central, midpoint, and center of gravity of a dataset. Mean,
median, mood, and quartiles are the most important central indices.
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2024
S. H. Saneii and H. Doosti, Practical Biostatistics for Medical and Health Sciences,
https://doi.org/10.1007/978-981-97-3083-4_3
37
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38 3 Summarizing Data
Tip
When the data includes the whole population, the mean is denoted by the
symbol
μ and N is used in the denominator of its fraction. But when data is
taken from a sample, the mean is represented by
x and n is used instead of N
in the denominator of its fraction.
Mean
The most common indicator for summarizing data is the mean, also known as the
average, center of gravity or midpoint. There are different types of mean such as
arithmetic mean, geometric mean and harmonic mean with various uses. This book
deals only with arithmetic mean weight (Fig.
3.1).
The mean of a population is denoted by μ and is obtained from the sum of
observations divided by a total number of data and is equal to:
μ =
x
1
+ x
2
+ x
3
+ ··· + x
N
N
=
N
i=1
x
i
N
,
where the
N
i=1
x
i
is the sum of the observations from 1 to N . We will round the
mean to one more decimal place than the raw data.
The formula for the mean in the frequency tables, where the observations of each
class are given with frequency, is as follows:
μ =
N
i=1
f
i
x
i
N
i=1
f
i
,
where f
i
and x
i
are the absolute frequency and the midpoint of each class. The first
column of each table shows the class width. Calculate the average of lower and upper
limits of each class to get the “midpoint of the class” and replace it with the above
formula.
To facilitate the calculation of the mean, assign a column for f
i
x
i
(product of the
frequency and the midpoint of the each class). Then divide the sum of this column
by the total number of observations to get the average.
Fig. 3.1 Center of gravity of
a dataset
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3.2 Measures of Central Tendency 39
Table 3.1 Calculation for
hemoglobin levels in 70
women
Hemoglobin level x
i
f
i
f
i
.x
i
8.5–9.4 9.0 3 27.0
9.5–10.4 10.0 6 66.0
10.5–11.4 11.0 19 209.0
11.5–12.4 12.0 16 192.0
12.5–13.4 13.0 15 195.0
13.5–14.4 14.0 6 84.0
14.5–15.4 15.0 5 75.0
Total – 70 842.0
Tips
Rounding Rule for the Central Tendency:
Central tendency indexes should be rounded to one more decimal place than
occurs in the raw data. For example, if the raw data are given in whole numbers
(such as family size of 3, 4, 2, … the mean should be rounded to the nearest tenth
(
μ = 3.6). If the data are given in tenths (such as exam weight of children in Kg
17.5, 13.0, 16.5, 18.6), the mean should be rounded to the nearest hundredth
(
μ = 14.65), and so on.
The mean for a sample with size of n is shown with x and equals to:
x =
x
1
+ x
2
+ x
3
+ ··· + x
N
n
=
n
i=1
x
i
n
.
Example 1 For the information in Table 2.5, we assign a column for x
i
and a column
for
f
i
.x
i
. The mean of hemoglobin level of 70 women will be equal to
842
70
= 12.03.
These calculations are shown in Table
3.1.
When we merge two populations with N
1
and N
2
and mean of μ
1
and μ
2
,the
mean of the whole merged population is
μ =
N
1
μ
1
+N
2
μ
2
N
1
+N
2
. It can easily be generalized
into the K populations with
N
1
, N
2
, ···and N
k
and means of μ
1
,μ
2
, ··· and μ
k
.
The means of pooled K populations is
μ =
N
1
μ
1
+ N
2
μ
2
+ ··· + N
K
μ
K
N
1
+ N
2
+ ··· + N
K
.
Example 2
The mean of birth weight of 14 infants whose mothers smoked during
pregnancy is 3.200 and the mean of birth weight of 15 infants whose mothers did not
smoke is 3.590. The grand mean of birth weight of these 29 babies is:
μ =
N
1
μ
1
+ N
2
μ
2
N
1
+ N
2
=
14 × 3.200 + 15 × 3.590
14 + 15
= 3.402.
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40 3 Summarizing Data
Median
The median, denoted by Med, is widely used in survival studies and splits the number
of observations into two equal parts. Thus, half of the observations are smaller and
the other half are larger than median. The median is the midpoint of the sorted and
does not deal directly with the size of the observations, but only with their order and
ranking. Therefore, first all observations should be sorted in ascending order, and the
middle number should be considered as the median. The procedure for calculating
the median depends on whether the number of observations in the dataset is even or
odd.
When the number of dataset is odd, the median is equal to the value of the middle
number (number in row
N +1
2
), but if there are an even number of values in the dataset,
the median will fall between two given values (
N
2
and
N
2
+ 1), that is the average
values of the middle two numbers.
Example 3 Birth weights of 9 babies are as follows:
3.05, 3.50, 2.90, 3.20, 3.75, 3.80, 4.10, 3.90, 3.30
After sorting them in ascending order, we have:
2.70, 2.90, 3.05, 3.20, 3.30, 3.50, 3.75, 3.80 3.90, 4.10
In this example, the number of subject N = 9 is odd. Therefore, the value of the fifth
observation
(
N +1
2
=
9+1
2
= 5) is the median (3.50) in this dataset.
Example 4 Birth weights of 10 babies are as follows:
3.05, 3.50, 2.90, 3.30, 3.75, 3.80, 4.10, 3.90, 3.20, 2.70
After sorting them in ascending order, we have:
2.70, 2.90, 3.05, 3.20, 3.30, 3.50, 3.75, 3.80 3.90, 4.10
Because the number of observations (N = 10) is even, the average value of the fifth
observation (
N
2
=
10
2
= 5) and the sixth observation (
N
2
+ 1 =
10
2
+ 1 = 6), that is,
3.30+3.50
2
= 3.40 will be the median of this data.
Med =
3.30+3.50
2
= 3.40.
Calculating the median in frequency tables is a little more difficult. Since the
nature of these tables is sorted, you don’t need to sort them. Then we form the
cumulative frequency column to identify the class that contains the median. The
class containing median is that one includes the
N +1
2
th observation. That is, the class
whose cumulative frequency is immediately equal to or greater than
N +1
2
. We then
assume that the values in that class is 0 uniform distributed and we interpolate. We
specify the following components in this table.
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3.2 Measures of Central Tendency 41
Tip
The median class is a class whose cumulative frequency is immediately equal
or greater than
N +1
2
.
Med = L
i
+
N +1
2
− F
i−1
f
i
× w
i
,
where:
L
i
is the lower class boundary of the class containing the median.
N is the total number of data.
F
i−1
is the cumulative frequency of precede class of the median class that should
be excluded.
f
i
is the frequency of the median class.
w
i
is the class width of the median class.
Example 5 Table 3.2 determines the albumin level of 185 women based on the
contraceptive pill. Calculate the median albumin level of these 185 women.
Solution: According to the number of samples and cumulative frequency, the median
class is a group whose cumulative frequency is equal to or greater than
N +1
2
=
185+1
2
= 93 (the gray row). Then we substitute the values L
i
= 39.5, f
i
= 43, w
i
=
3 and F
i−1
= 69 in the above formula:
Med = L
i
+
N +1
2
− F
i−1
f
i
× w
i
= 39.5 +
185+1
2
− 69
43
× 3 = 41.17.
Therefore, the exact value of the median is 41.17.
Table 3.2 Frequency distribution of albumin level of 185 females in terms of pill consuming
Albumin level Consumed Not consumed To t a l (f
i
) F
i
31–34 3 1 4 4
35–36 8 5 13 17
37–39 34 18 52 69
40–42 20 23 43 112
43–45 25 23 48 160
46–48 4 19 23 183
49–51 0 2 2 185
Sum 94 91 195 –
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