Estimation of the proportion of infected people with the aid two data sets:

repeated q-PCR tests and a cohort survey

Addis Ababa University

The Hebrew University of

Jerusalem

Gertner Institute forEpidemiology and Health

Policy Research

Ezer Miller, Amit Huppert, Ilya Novikov, Laurence Freedman

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.

Background

2.0 mm

Sand fly Sand fly feeding ♀ feeding ♀

bloodblood

2.5 µm2.5 µm

LeishmaniaLeishmania promastigotes in the promastigotes in the sand fly gutsand fly gut

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.

The parasitemia (no. of parasites per ml) level obtained from the cohortstudy in north Ethiopia during spring 2011 via q-PCR. n=4756.

PCR result

0j=1

1-10j=2

11-100j=3

101-1000j=4

>1000j=5

4076P(e1=0,j=1)=0.857

468P(e1=1,j=2)=0.0984

93P(e1=1,j=3)=0.0196

96P(e1=1,j=4)=0.0202

23P(e1=1,j=5)=0.0048

Cohort survey results - table

ei=experimental value in i test = 1 (infected) or 0 (uninfected)j = group index

Cohort survey results – Pie chart

0 1 - 10 11 - 100 101 - 1000 > 1000

The parasitemia (no. of parasites per ml) level obtained from the cohortstudy in north Ethiopia during spring 2011 via q-PCR. n=4756.

The proportion of infected people is~14%.

The q-PCR has a limited accuracy.

The goal of the study is to estimateq = the probability of being infected = the proportionof 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 .

Study goal

The probability of getting false-negative result (FN) and false-positive result (FP) are not zero !

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

Repeated test resultsei = 1 → infected, ei = 0 → uninfected

Statistical approachCalculation 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

Expressing these probabilities as a function of The true probability of being infected = qThe probability of getting false-negative result = pThe probability of getting false-positive result = r

Calculation the variance of q*

Calculation of q*, p* and r*.

1

2

3

4

785.0)0,0()1,0|0()0,0( 11221 jePjeePeeP

04375.0),1(),1|0()0,1( 1

5

21221

kjePkjeePeePk

07208.0)1,0()1,0|1()1,0( 11221 jePjeePeeP

5

211221 0992.0),1(),1|1()1,1(

k

kjePkjeePeeP

Calculation of the estimated probabilities of two separatedmeasurements by using the cohort and the repeated test results:

1

The relationship between the probability of two separated measurements and q, p, and r

q = P(T=1) = the proportion of infected peoplep = P(e=0 | T=1) = probability of getting false-negative result.r = P(e=1 | T=0) = probability of getting false-positive result

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)

T = true statuse = measured status

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

2

P(e1=1,e2=1)= (1-r)2(1-q) = 0.099P(e1=1,e2=0 U e1=0, e2=1) = 2r(1-q)(1-r) = 0.116P(e1=0,e2=0) = r2(1-q)+q = 0.785

p* = the probability of getting false-negative result = 0.369q* = the infection probability = 0.249

Calculation of q*, p* and r*.

For zero false-positive probability, r=0

For zero false-negative probability, p=0

P(e1=1,e2=1)= (1-p)2q = 0.099P(e1=1,e2=0 U e1=0, e2=1) = 2qp(1-p) = 0.116P(e1=0,e2=0) = p2q+1-q = 0.785

r* = the probability of getting false-positive result = 0.06q* = the infection probability = 0.11

3

u = P(e1=0,e2=0)s = P(e2=1,e1=1)

q* is a functions of u* and s*.

*)*,())((2*)()(*)()(*)]*,([ 22 suCovs

f

u

fsVar

s

fuVar

u

fsufVar

q* = f(u*,s*)

Calculation of the variance of q through ad hoc method

4

u = p2q+1-qs = (1-p)2q

1

1,

)1(

122

pu

q

ps

q

*)*,(])1)(1(

1[*)(]

1

1[*)(]

)1(

1[*)(

32

22

2suCov

ppuVar

psVar

pqVar

Example of calculation the variance of q in case where r = FP = 0

q* = 0.249, Var(q*) = 0.0011, σ* = 0.0033

SummaryA cohort survey combined with repeated tests results can be usedtogether in order to estimate the proportion of infected people withinthe population.

Our analysis shows that:

The probability of having false-negative result is high – 37%, when there is no false-positive tests.

Under the assumption of zero false-positive result ,the proportion of infected people among the populationis much higher than recorded in the cohort survey (~14%).

These result have an important implications in the analysisof a dynamic infectious model we are currently developing.

The proportion of infected people is ranged between 11% (FN=0)and 25% (FP=0).

Thank you for listening !

The model development and analysis could not have been performed without the valuable help and support from

Prof. Laurence FreedmanDr. Amit HuppertDr. Ilya NovikovDr. Asrat HailuDr. Ibrahim AbassiProf. Alon Warburg

The study is funded by Bill and Melinda Gates Foundation,(BMGF)

Acknowledgements…