US Public Health Service

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US Public Health Service
Section B, Group 6 Probability & Statistics

Problem Statement
‡ The U.S. army subjected each prospective inductee to a blood test that detected syphilitic antigen. Number of analyses required = Number of persons examined ‡ However, syphilitic antigen can be detected at great levels of dilution also.

Problem Statement Contd.
‡ So, statisticians proposed an alternate solution
± Club the individual blood sera into groups ± Test the sera for a group as a whole, instead of individuals ± If a group s blood sera tests positive for the antigen, each member of the group would be retested.

Probability Analysis
‡ Let p = probability that a random selection would yield an infected individual ‡ Let N = Total people to be examined ‡ Let n = size of group for testing blood sera ‡ Number of groups thus formed = N/n ‡ Let X = Number of people infected in a group of size n.

Probability Analysis Contd.
‡ The binomial probablity of obtaining zero infected persons in a group of n is P(X=0) = (1 p)n ‡ Probability of atleast one person being infected in a group of n is P(X>0) = 1 P(X=0) = 1 (1 p)n ‡ Let Y = Number of groups tested positive, i.e. the number of failures in (N/n) trials

Probability Analysis Contd.
‡ As Y is a binomial random variable, E(Y) = Number of trials × Probability that a group tests positive = (N/n) × Probability of at least one infected = (N/n)[1 (1 p)n] ‡ The number of chemical analyses required by the grouping procedure = Number of groups + n×(Number of groups that test positive) = N/n + n×(N/n)[1 (1 p)n]

Probability Analysis Contd.
‡ Ratio of number of tests required now to those required originally (N) = 1/n + 1 (1 p)n = (n+1)/n (1 p)n ‡ This ratio can now be used to determine whether the proposed grouping technique requires lesser number of chemical analyses than the existing procedure ‡ If ratio < 1 for all values of p then the new technique is better

Excel Analysis
‡ For various values of p ranging from 0.01 to 0.5, the ratio of tests required currently to those required originally were plotted ‡ The group sizes were varied from 2 to 20, to check the optimal group size ‡ The graph thus obtained is demonstrated on the next slide

Graph
1.4 1.2

1

p = 0.35 p = 0.3 p = 0.25

Relative Ratio

p = 0.2
0.8

p = 0.15 p = 0.1 p = 0.06 p = 0.05 p = 0.03 p = 0.02
0.2

0.6

0.4

p = 0.01

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Group Size

Analyzing Results
‡ From the graph we can notice that for p 0.15 and group size varying from 2 to 20, the relative ratio is < 1 ‡ But, if more than 15% of people are syphilitic, and the number of people in each group is large, then the original method of testing is preferable ‡ As the prevalence of syphilitis increases, more individuals may be re-tested thereby worsening the relative ratio

Efficient Group Sizes for p < 0.3
p 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.15 0.2 0.3 Optimum Size of Group (Graph Minima) 11 8 6 6 5 5 4 4 4 4 3 3 3 Relative Ratio 0.195 0.274 0.333 0.383 0.426 0.466 0.502 0.533 0.564 0.593 0.719 0.821 0.990

Conclusion
‡ The proposed technique will require lesser number of chemical analyses if prevelance of infection is less than 15% ‡ The efficient group size gradually decreases from 11 for p = 0.01 to 3 for p = 0.3.

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