Estimation of the proportion of infected
people with the aid two data sets:
repeated q-PCR tests and
a cohort survey
Ezer Miller, Amit Huppert, Ilya Novikov, Laurence Freedman
The Hebrew
University of
Jerusalem
Addis Ababa
University
Gertner Institute for
Epidemiology and Health
Policy Research
Background
Visceral Leishmaniasis (VL) or Kala Azar disease is a vector borne
infectious disease that is responsible for 500,000 infections annually.
In the last 5 years there is a disease outbreak in northern Ethiopia
The statistical model presented here is a part of a large study,
that aims to explore the disease transmission dynamics
in northern Ethiopia.
Leishmania promastigotes in the
sand fly gut
Sand fly ♀ feeding blood
2.0 mm
2.5 µm
Leishmania parasites are transmitted by blood-sucking
phlebotomine sand flies. Female sand flies take blood
which constitutes a source of protein for maturing their
eggs. Leishmania amastigotes that are ingested with the
blood, transform into promastigotes and multiply in the
gut of the infected sand fly.
Cohort survey results - table
The parasitemia (no. of parasites per ml) level obtained from the cohort
study in north Ethiopia during spring 2011 via q-PCR. n=4756.
ei=experimental value in i test = 1 (infected) or 0 (uninfected)
j = group index
PCR result
0
j=1
1-10
j=2
11-100
j=3
101-1000
j=4
>1000
j=5
4076
468
93
96
23
P(e1=0,j=1)=0.857
P(e1=1,j=2)=0.0984 P(e1=1,j=3)=0.0196 P(e1=1,j=4)=0.0202 P(e1=1,j=5)=0.0048
Cohort survey results – Pie chart
The parasitemia (no. of parasites per ml) level obtained from the cohort
study in north Ethiopia during spring 2011 via q-PCR. n=4756.
0
1 - 10
11 - 100
101 - 1000
> 1000
The proportion of infected people is
~14%.
Study goal
The q-PCR has a limited accuracy.
The probability of getting false-negative result (FN) and
false-positive result (FP) are not zero !
The goal of the study is to estimate
q = the probability of being infected = the proportion
of infected people within the population
To achieve this, the cohort participants were divided into five
groups according to their parasitemia level, and a repeated assay
performed on selected members of each group.
Repeated test results
ei = 1 → infected, ei = 0 → uninfected
Group index
1st
measurement
n
2nd measurement:
The number
who got
positive result
in the 2st test
Conditional
probabilities
calculated according to
the repeated test
results.
j=1
0
107
9
P(e2=1|e1=0,j=1)=0.084
j=2
1 - 10
108
64
P(e2=1|e1=1,j=2)=0.503
j=3
11 - 100
48
41
P(e2=1|e1=1, j=3)=0.854
J=4
101 -1000
24
23 P(e2=1|e1=1,j=4)=0.9583
j=5
> 1000
19
19
P(e2=1|e1=1,j=5)=1
Statistical approach
1
Calculation of the estimated probabilities of two separated measurements:
P(e1=0,e2=0) = ?
P(e1=0,e2=1 U e1=1, e2=0) = ?
P(e1=1,e2=1) = ?
ei = experimental measurements
2
Expressing these probabilities as a function of
The true probability of being infected = q
The probability of getting false-negative result = p
The probability of getting false-positive result = r
3
Calculation of q*, p* and r*.
4
Calculation the variance of q*
1
Calculation of the estimated probabilities of two separated
measurements by using the cohort and the repeated test results:
P ( e1  0 , e 2  0 )  P ( e 2  0 | e1  0 , j  1) P ( e1  0 , j  0 )  0 . 785
P ( e 1  0 , e 2  1)  P ( e 2  1 | e 1  0 , j  1) P ( e 1  0 , j  1)  0 . 07208
5
P ( e1  1, e 2  0 ) 
 P (e
 0 | e1  1, j  k ) P ( e1  1, j  k )  0 . 04375
2
k 2
5
P ( e1  1, e 2  1) 
 P (e
k 2
2
 1 | e1  1, j  k ) P ( e1  1, j  k )  0 . 0992
2 The relationship between the probability
of two separated measurements and q, p, and r
q = P(T=1) = the proportion of infected people
p = P(e=0 | T=1) = probability of getting false-negative result.
r = P(e=1 | T=0) = probability of getting false-positive result
T = true status
e = measured status
P(e1=0, e2=0) = P(e1=0, e2=0| T=1)P(T=1)+P(e1=0,e2=0|T=0)P(T=0) =
p2q+(1-r)2 (1-q)
P(e1=0,e2=1U e1=1, e2=0)=P(~|T=1)P(T=1)+P(~|T=0)P(T=0) = 2r(1-q)(1-r)+2qp(1-p)
P(e1=1,e2=1) = P( (~|T=0)P(T=0)+p(~|T=1)P(T=1) = r2(1-q)+q(1-p)2
3
Calculation of q*, p* and r*.
For zero false-positive probability, r=0
P(e1=1,e2=1)= (1-p)2q = 0.099
P(e1=1,e2=0 U e1=0, e2=1) = 2qp(1-p) = 0.116
P(e1=0,e2=0) = p2q+1-q = 0.785
p* = the probability of getting false-negative result = 0.369
q* = the infection probability = 0.249
For zero false-negative probability, p=0
P(e1=1,e2=1)= (1-r)2(1-q) = 0.099
P(e1=1,e2=0 U e1=0, e2=1) = 2r(1-q)(1-r) = 0.116
P(e1=0,e2=0) = r2(1-q)+q = 0.785
r* = the probability of getting false-positive result = 0.06
q* = the infection probability = 0.11
4 Calculation of the variance of q
through ad hoc method
q* is a functions of u* and s*.
u = P(e1=0,e2=0)
s = P(e2=1,e1=1)
q* = f(u*,s*)
Var [ f ( u *, s *)]  (
f
u
) Var ( u *)  (
2
f
s
) Var ( s *)  2 (
2
f
u
)(
f
s
) Cov ( u *, s *)
Example of calculation the variance of q in case where r = FP = 0
u = p2q+1-q
s = (1-p)2q
q
s
Var ( q *)  [
1
(1  p )
] Var ( s *)  [
2
2

1
(1  p )
1
p 1
2
2
,
q
u

1
p 1
2
] Var ( u *)  [
2
1
( p  1)( p  1)
q* = 0.249, Var(q*) = 0.0011, σ* = 0.0033
3
]Cov ( u *, s *)
Summary
A cohort survey combined with repeated tests results can be used
together in order to estimate the proportion of infected people within
the population.
Our analysis shows that:
The probability of having false-negative result is high –
37%, when there is no false-positive tests.
The proportion of infected people is ranged between 11% (FN=0)
and 25% (FP=0).
Under the assumption of zero false-positive result,
the proportion of infected people among the population
is much higher than recorded in the cohort survey (~14%).
These result have an important implications in the analysis
of a dynamic infectious model we are currently developing.
Thank you for listening !
Acknowledgements…
The model development and analysis could not have been
performed without the valuable help and support from
Prof. Laurence Freedman
Dr. Amit Huppert
Dr. Ilya Novikov
Dr. Asrat Hailu
Dr. Ibrahim Abassi
Prof. Alon Warburg
The study is funded by Bill and Melinda Gates Foundation,
(BMGF)