This document discusses various measures of disease frequency used in epidemiology, including count, ratio, proportion, rate, and prevalence. It provides definitions and examples of each measure. Key measures discussed in depth include incidence rate, cumulative incidence rate, attack rate, and person-time rate. Public health applications and calculations using these measures are presented.

1. Andamlak Dendir
Assistant professor of epidemiology
Wolkite University, Ethiopia
Measures of disease
frequency
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 1

2. Measures of disease frequency
 Count
 Ratio
 Proportion
 Rate
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 2

3. Count
Simplest & most basic measure
Absolute number of persons who have disease or
characteristic of interest.
Useful for health planners & administrators: for
allocation of resources (e.g. quantity of ORS needed by
diarrheal cases)
Also used for surveillance of infectious disease for
early detection of outbreaks.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 3

4. Ratio
 Ratio is the relative size of two quantities
 It quantifies the magnitude of occurrence of something in
relation to another.
 One number divided by another number
 The magnitude of one character divided by another
Example: sex ratio
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 4

5. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 5

6. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 6

7. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 7

8. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 8

9. Proportion
 It is a ratio of two quantity in
which Numerator included
in the denominator
 Its result ranges between 0
and 1 or (0–100%)
 Parcentage = Proportion x
100
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 9

10. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 10

11. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 11

12.  This line listing in Table
presents some of the
information collected on
infants born at General
Hospital with neonatal
listeriosis.
a. What is the ratio of
males to females?
b. What proportion of
infants lived?
c. What proportion of
infants were delivered in
a delivery room?
d. What is the ratio of
operating room
deliveries to delivery
room deliveries?
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 12

13. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 13

14. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 14

15. Frequently used measures of
morbidity
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 15

16. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 16

17. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 17

18. General formula
Incidence rate
= new cases occurring during a given time period × 10n
population at risk/person time at risk during the same time period
 The numerator of an incidence should reflect new cases of disease
which occurred or were diagnosed during the specified period.
 The numerator should not include cases which occurred or were
diagnosed earlier.
 Persons who are included in the denominator should be able to
the disease that is being described during the time period covered.
 Denominator is can be average /midpoint population/person time at
risk / population at risk
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 18

19. Cumulative Incidence rate (CIR)
= new cases occurring during a given time period × 10n
population at risk at start of time period/study
It is a measure of the probability or risk of
getting disease
 A measure of “average risk”
 CIR answers the question: “what is the probability or chance that an
individual develops the outcome over time”
Answer the question what proportion of the
population will develop illness during the
specified time period
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 19

20. Cumulative Incidence Rate(CIR)
Denominator is the size of the population at the start of
the time period/study
 it is a proportion, because all persons in the numerator
are also in the denominator.
It is a rate because it includes time period and population-
at-risk
Its synonyms are incidence proportion ,attack rate
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 20

21. Cumulative Incidence Rate(CIR)
The probability that an event will occur (if we calculated it during
outbreak it is called Attack rate)
Number of new cases with disease in a specified time period
CI = ----------------------------------------------------------
Number of disease-free people at the start of the time period
Disease-free persons are Population at risk
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 21

22. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 22

23. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 23

24. Class activity
 In a study of oral contraceptive (OC) use and bacteruria , a total of
2390 women aged 16 to 49 years were identified who were free from
bacteruria . Of these, 482 were OC users at initial survey in1973. at
second survey in 1976 , 27 of the OC user had developed bacteruria .
Q. calculate the cumulative incidence of bacteruria among OC user
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 24

25. Answer
Given
 population at risk at start of study=482
 population who developed bacteruria=27
Required
 CI=?
Solution
CI= total pupulation who develop bacterurai in a given
period/population at risk
27/482 x100= 5.6%
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 25

26. Example
A. the following figure
showing episodes of an
illness X (a disease that
develops lifelong
immunity), in a
population of 200 people
followed for 3
consecutive months.
What is the cumulative
incidence of illness X
among the population in
the month Sept 1 to Oct
1?
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 26

27. Attack Rate
 An attack rate is a variant of an incidence rate, applied to a
narrowly defined population observed for a limited time, such as
during an epidemic.
 expressed as a percent, so 10n equals 100.
Attack rate = Number of new cases among the population during the
period × 100
Population at risk at the beginning of the period
 It is a proportion—the persons in the numerator are also in the
denominator.
 It is a measure of the probability or risk of becoming a case.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 27

28. Example
 Of 75 persons who attended the dinner of get together
program , 46 subsequently developed gastroenteritis.
Calculate attack rate
x = Cases of GI= 46
y = Number of persons at the program 75
Then, the attack rate for gastroenteritis is = 46/75 × 100 =
61%
In the example above, among persons who attended the
program, the probability of developing gastroenteritis was
61%, or the risk of developing gastroenteritis was 61%.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 28

29. Secondary Attack Rate
 A secondary attack rate is a measure of the frequency of new
cases of a disease among the contacts of known cases.
 The formula is as follows:
SAR= Number of cases among contacts of primary cases during the period
total number of contacts
 To calculate the total number of household contacts, we usually
subtract the number of primary cases from the total number of
people residing in those households.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 29

30. Example
 Seven cases of hepatitis A
occurred among 70 children
attending a child care center.
Each infected child came from
a different family. The total
number of persons in the 7
affected families was 32. One
incubation period later, 5
family members of the 7
infected children also
developed hepatitis A. We will
calculate the attack rate in the
child care center and the
secondary attack rate among
family contacts of those cases.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 30

31. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 31

32. Attack rate in child care center=
x = cases of hepatitis A among children in child care center = 7
y = number of children enrolled in the child care center = 70
Attack rate =x/y× 100 =7/70× 100 = 10%
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 32

33. The secondary attack rate among family contacts of those cases
was
x = cases of hepatitis A among family contacts of children
with hepatitis = 5
y = number of persons at risk in the families (total number
of family members—children already infected) = 32 − 7 =
25
Secondary attack rate =
x/ y× 100 = 25
5/25 × 100 = 20%
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 33

34. Example 3
 In a particular community, 115 persons in a population of
4,399 became ill with a disease of unknown etiology. The 115
cases occurred in 77 households and each household has one
index case The total number of persons living in these 77
households was 424.
 Calculate
 Attack rate of the illness in the community
 SAR of the illness in the affected household
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 34

35. Solution
 AR= 115/4,399 = 26/1,000 or 2.6%
 SAR= (115-77)/424-77)=38/347 = 11.0%
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 35

36. Example
 Seven cases of hepatitis A occurred among 70 children attending a
child care center. Each infected child came from a different family.
The total number of persons in the 7 affected families was 32. One
incubation period later, 5 family members of the 7 infected
children also developed hepatitis A.
 We will calculate the attack rate in the child care center and the
secondary attack rate among family contacts of those cases.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 36

37. Solution
 x = cases of hepatitis A among
family contacts of children with
hepatitis A = 5
 y = number of persons at risk in the
families (total number of family
members—children already
infected) = 32 − 7 = 25
 Secondary Attack
Rate
=5/25%100=20%
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 37

38. Person-time Rate/incidence density Rate
= Number of new cases during observation period × 10n
Person-time at risk
 Its measures the speed of disease occurrence in population at risk
of developing the disease.
 IDR answers the question: “At what rate are new cases of disease occurring in
the population”
 The numerator is still the number of new cases, but the
denominator is a little different.
 The denominator is the sum of the time each person is observed
 each person is observed from a set beginning point to an
established end point (onset of disease, death, migration out of
the study, or end of the study).
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 38

39. The Concept of “person-time”
 Person- time: is sum of length of time period passed free of
illness (at risk) by each individual member of study
 For example
 a person enrolled in a study who develops the disease of
interest 5 years later contributes 5 person-years to the
denominator
 An individual followed for 10 years with out becoming a case,
contributes 10 person-years
 A person who is disease-free at one year and who is then
lost to follow-up contributes just that 1 person-year to the
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 39

40. Cont…
Person-time rates are often used in cohort (follow-up)
studies of diseases with long incubation or latency periods,
such as some occupationally related diseases, AIDS, and
chronic diseases.
IDR ranges from 0 to infinity (it is not a proportion!)
Dimension = per unit time or the reciprocal of time (time-
1)
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 40

41. Example
In a cohort study, 600 people were followed over the course of
3 years as follows:
 100 persons for 1 year
 200 persons for 2 years
 300 persons for 3 years
The no of person years of observation in this study is
calculated as:
 100 persons*1 year= 100 person-years
 200 persons*2 years= 400 person-years
 300 persons*3 years= 900 person-years
 Total = 1400 person-years
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 41

42.  calcua
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 42

43. Special types of incidence rate
 Usually calculated to know the
morbidity rate of case in
total population at risk of
developing the disease in a
given specific time period .
 For example the morbidity
rate of TB in Ethiopia in 1992
can be calculated by dividing
the number of new TB case
reported for the year by
total Ethiopia mid –year
population in 1992
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 43

44. Example -I
 In 1989, 733,151 new cases of gonorrhea were reported among the
Ethiopian population. The 1989 mid-year Ethiopian population was
estimated to be 246,552,000. calculate the 1989 gonorrhea incidence
rate.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 44

45. Solution
Given
 no of new case= 733,151
 Estimated middle year population =246,552,000
 10n = 100,000
Required ? incidence rate
Solution
Incidence rate = (733,151/ 246,552,000)x(105)= 297.4 per
100,000per year or
or
Approximately 3 reported cases per 1,000 population in
1989.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 45

46. Prevalence
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 46

47. Prevalence
Definition: proportion of persons with a particular disease at a
specified point or period of time
Form: (x / y) x 10n, where
x = # new and pre-existing cases at point or period of
time
y = average or mid-point population /total population
at a point in time
10n = depends on how common
 It is expressed as a percentage, per thousand, per hundred
thousand
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 47

48. General formula
Prevalence =
all new and pre - existing cases during a given time period ×
10n
population during the same time period/at a point in time
 The value of 10n is usually 1 or 100 for common attributes. The
value of 10n may be 1,000, 100,000, or even 1,000,000 for rare
traits
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 48

49. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 49

50. Point prevalence
 It is proportion of a population that is affected by disease at a
given point in time.
 The numerator in point prevalence is the number of persons
with a particular disease or attribute on a particular time
 It answers how much of a particular disease is present in a
population at a single point in time—to get a kind of “stop
action” or “snapshot” look at the population with regard to that
disease. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 50

51. Point prevalence
Is not an incidence rate, because the numerator
includes pre-existing cases.
it is a proportion, because the persons in the
numerator are also in the denominator.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 51

52. Period prevalence
 Answer the question how much of a particular disease is
present in a population over a longer period of time/point in
time.
 The numerator in period prevalence is the number of persons
who had a particular disease or attribute at any time during
a particular interval.
 The interval can be a week, month, year, decade, or any other
specified time period.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 52

53. Example1
 In a survey of patients at a sexually transmitted disease clinic
in Hawassa referral hospital, 180 of 300 patients interviewed
reported use of a condom at least once during the 2 months
before the interview (1). The period prevalence of condom use
in this population over the last 2 months is
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 53

54. Solution
1. Identify x and y:
x = condom users = 180
y = total = 300
2. Calculate (x/y) × 10n: 180/300 × 100 = 60.0%.
Thus, the prevalence of condom use in the 2 months before the
study was 60% in this population of patients.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 54

55. Example
 Two surveys were done of the same community 12 months
apart. Of 5,000 people surveyed the first time, 25 had
antibodies to histoplasmosis. Twelve months later, 35 had
antibodies, including the original 25. We will calculate the
prevalence at the first and second survey, and compare the
prevalence with the 1-year cumulative incidence.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 55

56. Answers
1. Point Prevalence at the first survey:
x = antibody positive at first survey = 25
y = population = 5,000
x/y × 10n = 25/5,000 × 1,000 = 1.4 per 1,000
2. Period Prevalence at the second survey:
x = antibody positive at second survey = 35
y = population = 5,000
x/y × 10n = 35/5,000 × 1,000 = 7 per 1,000
2. Incidence during the 12-month period:
x = number of new positives during the 12-month period = 35 − 25 = 10
y = population at risk = 5,000 − 25 = 4,975
x/y × 10n = 10/4,975 × 1,000 = 2 per 1,000
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 56

57. Example
A. the following figure
showing episodes of an
illness X (a disease that
develops lifelong
immunity), in a
population of 200 people
followed for 3
consecutive months.
1. What is the cumulative
incidence of illness X
among the population in
the month Sept 1 to Oct
1?
2. Calculate the prevalence
of ilness X from sept
01-Nov 30
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 57

58. Relationship between Prevalence and Incidence
 Prevalence is a function of:
 the incidence of the condition, and
 the average duration of the condition
 duration is influenced in turn by the recovery rate and mortality
rate
Prev = Incidence x Duration
 This relationship explains why….
 Arthritis is common (“prevalent”) in the elderly
 Rabies is rare.
 Influenza is only common during epidemics.
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 58

59. Concept of the Prevalence
“Pool”
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 59
New cases
(Incidence)
Death
rate
Recovery
rate

60. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 60

61. Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 61

62. Q1
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 62

63. Many thanks!!
Andamlak Dendir (Assistant professor of
Epidemiology ), wolkite university , Ethiopia 63