Chapter 5 A case study in Pulse Polio Immunisation 5.1 Introduction This chapter features a worked example of Bayesian graphical modelling. Our data for this exercise are (i) the antibody measurements after oral polio vaccination (OPV) obtained from the publication of W.H.O. (ii) The statistics of Pulse Polio immunization held in Kannur district of Kerala State. We begin our analysis by finding a relation ship between antibody and time after Trivalent Oral Polio Vaccination (TOPV). Also, we shall formulate graphical models of the variables involved using relevant Probability distribution. More over, the success and efficacy of pulse polio immunization programme in Kannur district also be analysed. 5.2 Back ground Poliomyelitis is an acute infectious disease mainly which affects the central nervous system. It is caused by polio virus with three sub types 1, 2 and 3. This disease is associated with poor environment hygiene especially the lack of safe water and poor 94
Chapter 5 95 sanitation. (see Figure 5.1). The clinical picture may manifest in the form of minor illness, non - paralytic polio myelitis or paralytic poliomyelitis. Figure 5.1: Figure 5.2: A study on this reveal that around 2, 50000 new cases of poliomyelitis occurs in each year over the world. Of these fatality is 10% while 25% become severely impaired and another 50% have some residual weakness and 15% recovered by treatment. This distribution exhibited in the Figure 5.2. 5.3 Public health importantance The public health importance of poliomyelitis in India is discussed in the following four dimension and they are the following. 1. The epidemiological picture
Chapter 5 96 2. The social dimension 3. The density of population 4. The economic aspects. This classification is presented graphically below (Figure 5.3) Figure 5.3: 5.4 Epidemiological picture Poliomyelitis is one of the major health problems in India. A study on this shows that the prevalence of paralytic poliomyelitis in India per 1000 children aged 5-9 years 6.2 in rural and 6.5 in Urban. While the incidence per 1000 children aged 0 to 4 years is 1.6 rural and 1.7 Urban. On an average of 17433 cases 323 deaths have been reported every year in India, ofcourse the death cases is 18% of affected cases which is very high compared to global rate of fatality. However, the under reporting cases has been a serious problem for us and probably only 10% of the total cases have been reported. Considering this factor it is estimated that around 700 children per
Chapter 5 97 day become paralysed in India due to polio and nearly 125 to 140 loss their lives. Remarkably India seems to have more polio cases than the rest of the world. 5.5 Social Dimensions The poliomyelitis almost invariably affects the under-privileged overworked and socially backward sections of the population. 5.5.1 Density of population India is having large number of regions having very high density of population. In these regions people are facing many problems like poor sanitation, shortage of pure drinking water etc. So there is greater chances for affecting this disease in thickly populated regions. 5.5.2 Economic Consequences The economic loss due this disease in very heavy interms of labour, cost of treatment and preventive programmes. It is estimated that global eradication of polio would save billions of dollars every year. 5.6 Options for Prevention The are several options for the prevention of poliomyelitis. The first option is providing pure water and sanitation facilities. The second option is immunization programme.this is the most feasible and effective method of control.
Chapter 5 98 5.6.1 Oral polio vaccine and inactivated polio vaccine Two types of polio vaccine are available. One is oral polio vaccine (OPV - Sabin) and the other is inactivated polio vaccine (IPV - Sark). Both are effective in controlling and eliminating poliomyelitis. In India OPV is used in the national immunization programme. Hence, we shall concentrate our investigation only on OPV. Our investigation mainly contains two sections. In the first section, we study the persistence of of antibody after booster vaccination on a data obtained from WHO with the help of graphical models. In the later section we shall investigate on the effectiveness of immunerisation programme launched in Kannur District of Kerala state on the basis of the data collected from this region. 5.7 Effect of vaccination and Analysis on antibody measurements using Beyesean networks We have collected a raw data from the publication of WHO related to the persistence of antibody after vaccination and the response to booster vaccination with Trivalent OPV (T.O.P.V). A study was performed on this with graphical models and conclusions were drawn accordingly. 5.7.1 Methods The study population consists of 175 children ranging from age 2 to 17 years. The age distribution of children receiving polio vaccination is given in Table I. Table I Age Distribution of Children Receiving Polio Booster Vaccination
Chapter 5 99 Age (Years) Frequency (N=175) 2 1 3 3 4 9 5 18 6 13 7 11 8 21 9 21 10 22 11 18 12 13 13 16 14 4 16 4 17 2 Table II Gives the number and frequency distribution of TOPV administered to children. The Table I, shows that all but four children had received TOPV. The four children who had not received TOPV had received types I, II and III MOPV. Table II Number and Frequency of TOPV Administered
Chapter 5 100 No. Frequency (N=175) None 4 1 x 17 2 x 27 3 x 31 4 x 57 5 x 33 6 x 6 * TOPV indicates trivalent oral polio vaccine. Table III Gives the distribution of neutralized antibody titre with respect to various intervals of time on different polio virus since last immunization. Table III. Polio Neutralizing Antibody in Pre-Booster Vaccination Sera
Chapter 5 101 Polio 1 lees than 6 mo 21 0 3 1 4 3 3 4 1 2 19.7 6.12 mo 4 0 0 1 1 1 1 0 0 0 11.3 1-2 yr 49 1 4 10 14 10 6 2 1 1 9.8 3-4 yr 42 4 6 7 10 7 5 3 0 0 7.5 5-6 yr 19 2 6 4 4 2 1 0 0 0 4.3 7-8 yr 23 3 4 8 2 5 1 0 0 0 4.5 greater than 9yr 17 2 5 5 2 2 0 1 0 0 3.2 Total 175 12 28 36 37 30 17 10 2 3 Polio 2 less than 6 mo 21 1 2 2 3 4 3 4 0 2 17.1 6-12 mo 4 0 1 0 2 0 1 0 0 0 8.0 1-2 yr 49 2 10 10 8 6 8 3 2 0 8.8 3-4 yr 42 1 4 12 11 10 2 0 1 1 8.0 5-6 yr 19 4 4 2 4 5 0 0 0 0 4.3 7-8 yr 23 2 5 8 3 3 3 1 0 0 8.1 greater than 9 yr 17 5 3 4 4 1 0 0 0 0 3.0 Total 175 15 29 36 35 29 17 6 3 3 Polio3 less than 6mo 21 4 2 3 3 5 2 1 1 0 7.5 6.12 mo 4 0 0 2 1 0 1 0 0 0 8.0 1-2 yr 49 8 15 9 8 7 2 0 0 0 3.7 3-4 yr 42 9 11 8 9 3 1 0 0 1 3.7 5-6 yr 19 7 5 3 1 2 1 0 0 0 2.6 7-8 yr 23 7 4 6 5 1 0 0 0 0 2.8 greater than 9 yr 17 9 4 1 1 1 1 0 0 0 2.1 Total 175 44 41 32 28 19 8 1 1 1
Chapter 5 102 5.8 POLIO TYPE I From table III we obtain the antibody measurements of children measured on various intervals in Pre booster vaccination. From the data it is noted that the antibody of children dicreases rapidly as time increases since last vaccination. The table shows that geometric mean of antibody measured in six months period since the last vaccination is 19.7 and the same measured after 9 years or greater since last vaccination is 3.2. Hence, we conclude that there is a rapid decrease in antibody as time increases. The bivariate graph drawn (See Figure 5.4) shows that there exists an exponential relationship between antibody and time and hence we shall fit an expression of the form y = AB t (43) to the data given in table III, where y denote the reciprocal of antibody and t denote the time since the last vaccination. Figure 5.4: Using logarithmic transformation of (1) we have logy = loga + tlogb (44) Now we shall find the values of A and B. Let logy = Y, LogA = a and logb = b.
Chapter 5 103 So that (44) becomes Normal equations of (45) are Y = a + bt (45) Y = na + b t (46) And ty = a t + b t 2 (47) From the table values of III we can formulate table IV and values of y, t, ty, t 2 can be obtained. Table IV No Time since last Mean Time (t). G.M. of Y= logy Yt t 2 vaccination antibody 1 less than 6 mo 3 mo (0.25 years) 19.7 1.2833 0.3208 0.0625 2 6-12 mo 9 mo (0.75 years) 11.3 1.0530 0.7898 0.5625 3 1-2 years 1.5 years 9.8 0.9912 1.4868 2.25 4 3-4 years 3.5 7.5 0.8750 3.0625 12.25 5 5-6 years 5.5 4.3 0.6334 3.4837 30.25 6 7-8 years 7.5 4.6 0.6627 4.97036 56.25 7 greater than 9 years 9.5 3.2 0.5051 4.7985 90.25 Total 27.5 6.0037 18.9124 191.875 Substituting the values obtained from table IV in equations (45) and (46) we have the following equations 6.0037 = 7a + (27.5)b. and 18.9124 = 67.5a + 191.875b. Solving the equations we get a = 1.0765 and b = 1.9443 Hence A = 11.926 and B = 0.8796
Chapter 5 104 Therefore the exponential relation between y and t is y = (11.926) (0.8796)t. From this relation when value of one variable is known the other variable can be estimated. Moreover we can formulate the frequency of distribution of antibody measurements of 175 children as it is follows. Table V Frequency distribution of antibody measurement of 175 children Antibody y i Frequency f i 1 12 2 28 4 36 8 37 16 30 32 17 64 10 128 2 256 3 Total 175 For further analysis of the data we shall present the data graphically. From the graph (See Figure 5.5) We infer that the frequency distribution of antibody can be approximated to normal probability distribution. If y denote the antibody measurement then y N(µ, σ 2 ), where estimates of µ and σ are X and s 2 computed from the sample. Now = yi f i y = N 1.12 + 2.28 + 4.36 + 8.37 + 16.30 + 32.17 + 64.10 + 128.2 + 256.3 175 y = 18.263 and s 2 = 1372.132
Chapter 5 105 Figure 5.5: Hence the graphical model of the antibody can be presented as follows. Figure 5.6: To interpret the graph let v be a nod in a graph. Let V be the set of all nodes. We define parent of v to be any node with an arrow emanating from it pointing v and descendant v of be any node on a directed path staring from v. Hence µ and σ are parents of y. The joint distribution of all random quantities is
Chapter 5 106 fully, specified in terms of the conditional distribution of each node given its parents. P (V ) = P (v)/(v/parentv) v V The likelihood terms and its models are as follows y = AB t y N(µ, σ 2 ) where µ = y and σ 2 = s 2 which are estimated from the sample data 5.9 Study on polio 2 From table 3 we can obtain the distribution of antibody on polio 2 with respect to time and is presented in the table 6 given below. As in polio I we can obtain relationship between antibody y and time t in the form y = AB t. Table VI No Time since last Mean Time (t). G.M. of Y= logy Yt t 2 vaccination antibody 1 less than 6 mo 0.25 17.1 1.2330 0.3082 0.0625 2 6-12 mo 0.75 8.0 0.9031 0.6773 0.5625 3 1-2 years 1.5 8.6 0.9345 1.4018 2.25 4 3-4 years 3.5 8.0 0.9031 3.1609 12.25 5 5-6 years 5.5 4.3 0.6335 3.4843 30.25 6 7-8 years 7.5 6.1 0.7993 5.9948 56.25 7 greater than 9 yrs 9.5 3.0 0.4771 4.5327 90.25 Total 27.5 5.8836 19.56 191.875 Taking logarithmic transformation of the equation y = AB t and substituting the values from the table we get the values of A and B as A = 10.1648 and B = 0.907
Chapter 5 107 Hence relation between y and t takes the form Y = (10.1648)(0.907) t From this relation when value of one variable is known then the other variable can be estimated. Now forming the frequency distribution of antibody of 175 we obtain the table as follows. From the table Frequency distribution of antibody of polio type 2 Antibody (y) Frequency (f) y.f y2 f. less than 2 (1) 15 15 15 2 29 58 116 4 36 144 576 8 35 280 2240 16 29 464 7424 32 17 544 17400 64 8 512 32768 128 3 384 49152 256 3 768 196608 Total 175 3169 326299 yi f i y = N = 3169 175 = 18.1086 S 2 = 1 y 2 f = 1 [(326298)] (18.1086)2 N 175 = 1864.5657-327.9214 = 1536.6443 Hence s = 39.2001 Thus the graphical model of antibody is presented as follows :
Chapter 5 108 Figure 5.7: y = AB t, That is, logy = loga + tlogb y N(µ, σ 2 ) where µ = y and σ 2 = s 2 5.10 Study on polio type III Again from table III we can again obtain the antibody measurements of polio type 3 with respect to time and the exponential relation y = AB t can be fitted to the data. Table VIII
Chapter 5 109 No Time since last Mean Time (t). G.M. of Y= logy Yt t 2 vaccination antibody 1 less than 6 mo 0.25 7.5 0.8751 0.2588 0.0625 2 6-12 mo 0.75 3.0 04771 0.3578 0.5625 3 1-2 yr 1.5 3.7 0.5682 0.8523 2.25 4 3-4 yr 3.5 3.7 0.5682 1.9887 12.25 5 5-6 yr 5.5 2.6 0.4150 2.2825 30.25 6 7-8 yr 7.5 2.8 0.4472 3.354 56.25 7 greater than 9 yr 9.5 2.1 0.3222 3.0609 90.25 Total 27.5 3.673 12.115 191.875 Taking logarithmic transformation of the equation y = AB t and solving for A and B we obtain. A = 4.2964 and B = 0.9384 Hence the relation between y and t is Y = (4.2964)(0.9384) t From this relation when one variable is known the other variable can be estimated.
Chapter 5 110 5.11 Frequency distribution of Antibody of polio type III Antibody (y) Frequency (f) y.f y2 f. less than 2 (1) 44 44 44 2 41 82 164 4 32 128 512 8 28 224 1792 16 19 304 4864 32 8 256 8192 64 1 64 4096 128 1 128 16384 256 1 256 65536 Total 175 1486 101584 From the table we obtain y as And Hence s = 22.5471 y = 1 1 yf = 1486 = 8.4914 175 175 s 2 = 1 175 Y 2 f (y) 2 = 1 175 [101584] (8.4914)2 = 508.3761
Chapter 5 111 Thus the graphical model of antibody is presented on follows. Figure 5.8: y = AB t That is, logy = loga + tlogb also y N(µ, σ 2 ) where µ = y and σ 2 = s 2 5.11.1 The following table gives the persistence of Booster response in polio neutralizing antibody. Table 10 Persistence of Booster response in polio neutralizing antibody
Chapter 5 112 No.change in for fold Sustained for Sustained for titer at 3 Booster at 3 fold Booster at fold Booster at weeks No. (%) weeks No. (%) 6 to 10 weeks No. (%) 6 months No. (%) Polio 1 12/26 (46) 14/26 (54) 14/14 (100) 13/14 (93) Polio 2 10/26 (36) 16/26 (62) No sample available 13/16 (81) Polio 3 14/26 (64) 12/26 (48) 12/12 (100) 11/12 (92) Total 36/78 (46) 42/778 (54) 37/42 (88) The data in Table 10 shows that at six months period in polio I, 93% of children were sustaining four fold or greater antibody; in polio 2 it is 81% and in Polio 3 it is 92%. Hence a total of 88% of children were sustaining four fold or greater antibody when it in measured at 6 months period. A diagramatic representation of the distribution is shown in Figure 5.9. Figure 5.9: 5.12 Summary and Conclusions In summary our observation lead to the following conclusions.
Chapter 5 113 1. There is loss in antibody with respect to time since last vaccination and an exponential relation exist between antibody and time. 2. The distribution of antibody (y) can be approximated to normal probability distribution and the likelihood functions and models can be derived from this. 3. Antibody is not related to sex and age of children at the time of immunization 4. Four fold a greater antibody prevails at six months period after Booster vaccination. 5. From the analysis we suggest that for persons traveling to countries where poliomyelitis to prevalent is advisable to take booster TOPV as a preventive measure irrespective of age. 5.13 A study on the effectiveness of pulse polio immunization conducted in Kannur District of Kerala state. A study on Poliomyelitis in India give us the following information. About 50% Polio myletis cases occurring in children of 12 months of age; 75 to 80% cases in 24 months and 90-95 in 36 months of age. If the target age of immunization is set at 0-36 months and the coverage about 90 95% will not achieve polio under control. Considering this factor the principle adopted in the programme is over immunization is safer then under immunization. Hence under this programme all children best the age of 5 are given two doses of OPV every year and the goal to eradicate poliomyelitis from this region can be made a reality.
Chapter 5 114 5.13.1 Distribution of Polio affected cases among different age groups In India a study on this shows that 90 95% of polio affected cases occurs among children of 3 years age and only 5% cases occurs among children of age between 3-5 years. this distribution is presented graphically in the diagram given bellow. Figure 5.10: 5.13.2 The Polio immunization programme The immunization programme (giving OPV) has commenced in India in 1998. As a part of the programme the same started in Kannur District in the same year. Since then the Programme is continuing every year. Under this programme all children below 5 years age will be given two doses OPV every year and as a result on completion of 5 years of age each children will be receiving 8-10 doses of OPV. The statistics collected from the medical authorities of Kannur shows that an average of 206011. Children were given 2 doses of OPV every year during the period 2005 to 2009. (See table below), Also average of the target achieved is 99.16
Chapter 5 115 5.13.3 Kerala Health Services District: Kannur Sl. NO. Year Target No. No. of children % of the target of children Vaccinated achieved (p ) 1 2005 208554 206567 99.05 2 2006 208554 206936 99.22 3 2007 207226 205068 98.96 4 2008 207226 205794 99.31 5 2009 207226 20560 99.26 Total 1030055 Average of Children Vaccinated during the year 2005-09 = 206011 Average % of the target achieved (P ) = 99.16 Moreover the data declare that since 2001 no polio affected case was reported in Kannur which shows the total success of the programme. Now we shall formulate a statistical procedure to test the truth of the claim made by the District medical authorities with the help of graphical Models. From the table we see that an average of 206011children were vaccinated during the period 2005 to 2009. For a period of 5 years also the data claims that since 2001 no polio case was reported in the District of Kannur. Let P denote the proportion of success of the vaccination programme. From the data published by Dr. Sabin, he claims 98% efficiency of the programme. So we shall choose the null hypothesis as Ho : P = P 0 = 0.98; Since in Kannur District it has produced in 100% Success we shall choose H 1 : P = P = 1 (where P is the value of P in Kannur District). We shall test H 0 against H 1.
Chapter 5 116 According to statistical theory P N(P 0, p 0 q 0 ). Consequently we have n Z = P P 0 p0 q 0, n where Z is the standard normal variable and n is the average of the children vaccinated From the sample study Z = 1 0.98 0.98 0.02 206011 where n denote the average number of children vaccinated during 2005 to 2009 Thus Calculated value of Z is 64.84 This Value of Z falls in the critical region at 05% level of significance. Hence we reject the Hypothesis P = 0.98 which establishes the acceptance of the hypothesis H 1 : P = 1. Therefore, the study establishes 100% success of the programme in Kannur District. The likelihood function and the graphical model of P (the proportion of success) are given as follows. P N(µ, σ 2 ) where µ = P 0 = 0.98 and σ = where P denote the proportion of target achieved in the immunization programme., 0.98 0.02 n, Figure 5.11:
Chapter 5 117 5.14 Summary and Conclusion. In Summary our observation warrant the following conclusions. 1. The process of giving OPV to all children of age 0-5 years shall be continued without drop outs. 2. Providing 8-10 doses of OPV to children up to the age of 5 years results in 100% control of the disease. 3. Steps shall be taken to achieve 100% immunization coverage. By implementing this kind of immunization programme all over India we can eradicate the acute infectious disease poliomyelitis from our country.