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9.6 Cohen’s Kappa Coefficient 197
p
o
=
0 + 0
1000
κ =
p
o
− p
e
1 − p
e
=
0.00 − 0.905
1.00 − 0.905
=−9.53.
The sample size does not affect the calculation of kappa coefficient and increasing
or decreasing the sample size does not change it. Because the kappa coefficient uses
relative frequencies. For example, if in Table
9.12, all cells are reduced 10 times, i.e.,
d = 0, c = 95 instead of 950, b = 5 instead of 50, a = 0, the kappa coefficient will
still be κ =− 9.53.
Example 9 If in example 6, the sample size is 100 and d = 0, c = 90, b = 10, 0 =
a (Table
9.13). Find the kappa coefficient.
Solution: The purpose of this example is not to show the effect of increasing or
decreasing the sample size on the kappa coefficient, but we want to express the effect
of changes in the cells within the table or the probability of observations (prevalence).
The probability of observed agreement, agreement by chance, and kappa
coefficient is:
p
e
=
(
10 × 10
)
+
(
90 × 90
)
100
2
= 0.82
p
o
=
0 + 0
100
= 0.00
κ =
p
o
− p
e
1 − p
e
=
0.00 − 0.82
1.00 − 0.82
=−4.56.
Comparison of examples 8 and 9 shows that when the cells in the table change,
the probability of chance and random agreement decreases from 0.905 to 0.82, and
as a result, the kappa coefficient will change from − 9.53 to − 4.56.
Example 10
If in Example 9, the sample size is 100 and d = 0, c = 50, b = 50, 0
= a (Table
9.14). Find the kappa coefficient.
Table 9.13 Results of clinical and laboratory diagnosis with different agreement
Expert1
Yes No
Expert 2 Yes 0 10 5
No 90 0 95
90 10 100
https://t.me/med1917
198 9 Health and Disease Measures
Table 9.14 Results of clinical and laboratory diagnosis with different agreement
Expert1
Yes No
Expert 2 Yes 0 50 50
No 50 0 50
50 50 100
Solution: The purpose of this example is to show the kappa coefficient is equal − 1.
The probability of observed agreement, agreement by chance, and kappa
coefficient is:
p
e
=
(
50 × 50
)
+
(
50 × 50
)
100
2
= 0.50
p
o
=
50 + 50
100
= 1.00
κ =
p
o
− p
e
1 − p
e
=
1.00 − 0.50
1.00 − 0.50
=−1.0.
Understanding the concept of zero is very important for the Kappa coefficient.
The zero coefficient occurs when one of the experts assigns all observations to
only two cells in a row (or column) of a table and does not assign anything to two
cells in another row (or column).
Tip
The kappa coefficient will be zero when all observations are assigned to only
two cells of one row (column) of the table and nothing is assigned to the other
two cells of the row (column).
Example 11 If in Example 10, the sample size is 100 and the number of items in
the cells is equal to d = 50, c = , b = 50, a = 0 (Table
9.15), the kappa coefficient
will be equal to zero.
Table 9.15 Hypothetical results for zero agreement
Expert1
Yes No
Expert 2 Yes 0 50 50
No 0 50 50
0 100 100
https://t.me/med1917
9.6 Cohen’s Kappa Coefficient 199
p
e
=
(
0 × 50
)
+
(
100 × 50
)
100
2
= 0.50
p
o
=
0 + 50
100
= 0.50
κ =
p
o
− p
e
1 − p
e
=
0.50 − 0.50
1.00 − 0.50
= 0.0.
Also, if the number of items in the row cells is equal to in another example are
equal to a = 85, b = 15, c = 0, d = 0, (Table
9.16), the kappa coefficient will be
equal to zero.
p
e
=
(
100 × 85
)
+
(
0 × 1550
)
100
2
= 0.85
p
o
=
0 + 85
100
= 0.85
κ =
p
o
− p
e
1 − p
e
=
0.85 − 0.85
1.00 − 0.85
= 0.0.
When raters classify observations into n classes instead of 2, the kappa coefficient
is calculated from the following formula
κ =
O
kk
−
O
k
.×O
.k
n
n −
O
k
.×O
.k
n
,
where O
kk
is the cell frequency in the kth row and the kth column (diagonal element)
of the K × K table, n is the grand total,
O
k
is the marginal total in the kth row, O
k
is
the marginal total in the kth column. Obviously, this formula can also be used for 2
× 2 tables.
Table 9.16 Another hypothetical results for zero agreement
Expert1
Yes No
Expert 2 Yes 85 15 100
No 0 0 0
85 15 100
https://t.me/med1917
200 9 Health and Disease Measures
Table 9.17 Hypothetical assessment of intrathecal synthesis by two laboratories
Laboratory 1 Laboratory 2
Positive Doubtful Negative To t a l
Positive 40 6 4 50
Doubtful 7 10 8 25
Negative 3 4 58 65
Total 50 20 70 140
Example 12 In the hypothetical example, following the intrathecal assessment of
130 patients have sent to two laboratories and the results were reported positive,
negative, and doubtful in Table
9.17. Calculate the kappa agreement coefficient.
Solution: In this example, O
11
= 40, O
22
= 10, O
33
= 58 and n = 140.By
substituting them in the formula, we have:
κ =
(
40 + 10 + 58
)
−
50×50
130
+
20×25
130
+
70×65
130
140 −
50×50
130
+
20×25
130
+
70×65
130
=
49.92
81.92
= 0.609.
This kappa coefficient means that the agreement between two laboratories is
moderate.
9.7 Relative Risk (RR)
When the population is divided into two groups: those exposed to a factor (E) and
those not exposed to that factor (
E), and the occurrence of a certain disease in them
is denoted by D, the relative risk indicates how much more likely it is for the disease
to occur among the exposed (E) group compared to the unexposed individuals.
D
E
In epidemiology, the relative risk (RR) provides insights into the increased
likelihood of an exposed group developing a specific event (such as a disease,
injury, risk factor, or death) compared to a non-exposed group. By understanding
the exposure and disease status of a research population, the corresponding
numbers can be filled in a table, such as Table
9.18.
https://t.me/med1917
9.7 Relative Risk (RR) 201
Table 9.18 Disease status in terms of exposure status
Disease
Yes No
Exposure Yes Diseased + Exposed Healthy + Exposed
No Diseased + Unexposed Healthy + Unexposed
Thus, the probability or risk of disease among the exposed (p
1
= p(D|E)) is equal
to:
probability
(
risk
)
of disease
=
No. of diseased & Exposed
No. of diseased & Exposed + No. of Healthy & Exposed
.
And the probability or risk of disease among the unexposed (p
0
= p(D|E))is
equal to:
probability
(
risk
)
disease
=
No. of diseased & Unexposed
No. of diseased & Unexposed + No. of Healthy & Unexposed
.
The relative risk (RR) is obtained by dividing the probability of disease among
the exposed by the probability of disease among unexposed.
RR =
probability of getting disease in exposed group
probability of getting disease in unexposed group
RR =
p
1
p
0
=
p(D|E)
p(D|E)
.
The relative risk points out that the occurrence of disease (D) in the presence and
exposure to a factor (E) is how many times greater or lesser than the occurrence
of disease (D) in the absence of factor E. This measure is also called the incidence
relative risk or “risk ratio” of the disease (mostly related to prospective and cohort
studies). Relative risk is compared with 1, because the expected value of relative risk
(RR) under the null hypothesis is 1 and means there is no difference between groups.
Tip
The relative risk or risk ratio (RR) of an event is the probability of its
occurrence in exposed group to a risk factor as compared with the probability
of its occurrence in an unexposed or control or reference group. The RR is
estimated as the absolute risk with the risk factor divided by the absolute risk
in the unexposed group.
https://t.me/med1917
202 9 Health and Disease Measures
We must remind that in a prospective or cohort study, groups of subjects are
followed up for predetermined time to observe whether the outcome of interest
occurs; we can use the relative risk (RR).
In prospective and cohort studies, we look at the table in this (horizontal) way
(Table
9.19).
And then:
RR =
pr
(
D
|
E
)
pr
D
E
=
a
(
a + c
)
b
(
b + d
)
.
When
pr
(
D
|
E
)
pr
(
D
|
E
)
is interpreted as the ratio of the prevalence of the disease in the
two groups, it is the “relative risk of disease” and shows that the prevalence of a
particular disease in exposed group is much higher than the prevalence of the same
disease in unexposed people.
Example 13 Out of 951 mothers who drank alcohol at least once a day during
pregnancy, 7 babies were born with impairment. Of the 31,530 mothers who drank
less than once a day during pregnancy, 86 had birth impairment babies. Does drinking
alcohol during pregnancy increase the risk of birth impairment in the newborns?
Solution: In this example, event I = impairment and event E = daily consumption
of alcohol at least once during pregnancy is defined. The conditional probability of
having a impairment is
pr
(
I
|
E
)
=
7
951
= 0.00736 provided the pregnant mother
has consumed alcohol at least once a day equals and the conditional probability of
having a impairment is
pr
I
E
=
86
31530
= 0.00273 given the pregnant mother has
consumed alcohol less than once a day. Then according Table
9.20, a = 7, b = 944,
c = 86, d = 31,444, a + c = 93, and b + d = 32,388.
Dividing these two probabilities gives the relative risk, and we have to compare
RR with 1.
RR =
pr
(
I
|
E
)
pr
I
E
=
0.00736
0.00273
= 2.70.
Table 9.19 Disease status and exposure in cohort studies
https://t.me/med1917
9.7 Relative Risk (RR) 203
Table 9.20 Impairment of
newborns in terms of mother
drinking
Cohort study Impairment
I I
Alcohol drinking E 7 944
E 86 31,444
Total 93 32,388
This means that impairment in newborns who’s their mothers drank alcohol at
least once a day during pregnancy is 2.70 times more likely than newborns whose
mothers drank less than once a day.
Confidence interval for relative risk.
Like other statistical estimations, we can confidence interval for RR. But because its
logarithm follows the normal distribution, we first find the standard error of
log
e
RR
using the following formula:
Se
log
e
RR
=
1
a
+
1
b
−
1
a + c
+
1
b + d
.
So, the lower and upper boundaries of
(
1 − α
)
%
confidence interval for log
e
RR
are:
log
e
RR − Z
1−
α
2
× Se
log
e
E
to log
e
RR + Z
1−
α
2
× Se
log
e
RR
,
where Z
1−
α
2
is corresponding value obtained from standard normal distribution.
Finally, using back transformation (antilogarithm) from these boundaries the
confidence interval for relative risk (RR) is obtained.
Example 14 Find the confidence interval of relative risk for data in example 13.
Solution: In Example 13, the values were a = 7, b = 944, c = 86, d = 31,444 and
RR = 2.70. To calculate the confidence interval, we must first calculate the standard
error of the logarithm of relative risk (
log
e
RR) as follows:
Se
log
e
RR
=
1
a
+
1
b
−
1
a + c
+
1
b + d
=
1
7
+
1
944
−
1
93
+
1
32388
= 0.3649.
https://t.me/med1917
204 9 Health and Disease Measures
The corresponding value of the standard normal distribution table for 95%
confidence level is 1.96 and
log
e
2.70 = 0.993.
log
e
RR − Z
1−
α
2
× SE
log
e
RR
to log
e
RR + Z
1−
α
2
× Se
log
e
RR
=
(
0.993 − 1.96 × 0.3649 to 0.993 + 1.96 × 0.3649
)
=
(
0.993 − 0.7152 to 0.993 to 0.7152
)
=
(
0.2778 to 1.7082
)
.
Now, using back transformation (antilogarithm from the upper and lower
boundaries), a 95% confidence interval is obtained for the relative risk.
(e
0.2778
to e
1.7082)
= (1.320 to 5.519).
The confidence interval is not wide because the sample size in both groups is
large.
9.8 Odds Ratio (OR)
Before starting about odds ratio, we note that “odds” is a singular word, not the plural
of “odd” and has advantages for some types of analysis, as it is not constrained to lie
between 0 and 1 but can take any value from zero to infinity (Bland 2015).
There is a main difference between the retrospective or case control studies,
with prospective studies. In retrospective study, the selection of subjects is based on
outcome (disease) not on exposure to a factor. Then it looks backwards to see/
examine if subjects have been exposed to a characteristic factor (such as a risk
factor or a protection factor) in relation to the outcome that is established at the
start of the study.
https://t.me/med1917
9.8 Odds Ratio (OR) 205
Tips
The odds of an event are a ratio of the frequency (or probability) of its
occurrence to the frequency (or probability) of its non-occurrence.
The OR is a comparison of the odds of an event after exposure to a risk factor
with the odds of that event in a control or reference situation. The OR is
estimated as the odds of an event in the exposure group divided by the odds of
that event in the control or reference group. An OR of 1.00 means that there is
no increase or decrease in risk.
If the probability of occurrence of an event is p and the probability of its non-
occurrence is 1 − p, the odds of that event are defined as
p
1−p
. Now, if the probability
of an event occurring in the two groups of patients and non-patients is
p
1
and p
2
,
respectively, the odds of the event in the two groups will be
p
1
1−p
1
and
p
2
1−p
2
respectively,
and the odds ratio in both groups is expressed as follows:
OR =
p
1
1−p
1
p
2
1−p
2
.
The odds ratio illustrates how strongly the presence or absence of a certain
characteristic (disease) relates to the presence or absence of another characteristic
(Exposure). We can use the odds ratio to observe if a certain outcome (e.g.,
developing the disease) is associated with exposure to a hypothesized risk factor
(e.g., smoking). With an odds ratio, the outcome can be the starting point with
which we can determine the relative odds of someone having been exposed to a risk
factor. Alternatively, we can also use it to describe the ratio of disease odds given
the exposure status. Once we know the exposure and disease status of a research
population, we can fill in their corresponding numbers in Table
9.21.
In retrospective and case control studies, we look at the table from vertical
direction (Table 9.22).
Thus, the odds of the exposed are:
Odds for exposed =
No. of diseased & Exposed
No. of Healthy & Exposed
=
a
b
.
Table 9.21 Disease status and exposure in retrospective (case control) studies
https://t.me/med1917
206 9 Health and Disease Measures
Table 9.22 Disease status and exposure in case control studies
And the odds of unexposed are:
Odds for Unexposed =
No. of Diseased & Unexposed
No. of Healthy & Unexposed
=
c
d
.
The odds ratio (OR) is obtained by dividing the probability of disease among the
exposed by the probability of disease among unexposed.
OR =
Diseased & exposed
Healthy & Exposed
Diseased & Unexposed
Healthy & Unexposed
==
a
c
b
d
=
a × d
b × c
.
We can define as:
OR =
Diseased & exposed
Diseased & Unexposed
Healthy & Exposed
Healthy & Unexposed
=
a
b
c
d
=
a × d
b × c
.
And then:
OR =
p
1
1−p
1
p
0
1−p
0
.
Example 15
In a case control study of people aged 45–54, they developed diabetes
by residential areas (rural, urban) according to Table
9.23:
Solution: In t his example, a = 269, b = 504, c = 331 and d = 496, and the odds
ratio is equal to:
OR =
a × d
b × c
=
269 × 496
504 × 331
= 0.80.
Table 9.23 Diabetic patients
by residential area
Diabetes
Case Control
Residential area Rural 269 504
Urban 331 496
https://t.me/med1917
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