US Public Health Service (PS-1)

Group 8, Section C

Background Information:

The existing procedure for detecting syphilitic men was to subject each person to the Wasserman-type blood test. The alternative method suggests blood sera be drawn from groups of people, and the groups should be subjected to chemical analysis. If the group test reveals the presence of antigen, the individual blood samples would be tested to identify the person.

Assumptions

1. The event of one person testing positive is independent of any other person testing positive.

2. Every person testing positive for the disease is equally likely.

Solution

• Let N be the total population, r be the number of infected people in the population. Hence, N-r represents the number of people in the population without the disease.

• Let n denote the number of samples in a group. Therefore, [pic] be the number of groups.

Let ‘p’ represent the probability that a group tests negative for the disease, that is, none of the group members carry the antigen.

Probability Distribution

In a population of N, there are r people who carry the disease and N-r people who are free from the disease. Samples of size n are chosen from this population. Let X denote the number of people who test positive for the disease in the sample n. Then, p follows hypergeometric distribution.

[pic]

For the problem at hand, k = 0.

[pic]

Therefore, 1-p will represent the probability that the group tests positive.

Number of tests to be performed through sampling = Number of groups + Testing each person in the groups testing positive

If the proposed technique is to require lesser number of chemical analyses than those required by the existing procedure

[pic]

[pic]

[pic]

f(x) was plotted on MS Excel for different values of N, n and r, as shown here:

|N |r |n |f(n) |
|1000 |10 |500 |0.001067 |
|1000 |10 |250 |-0.05147 |
|1000 |10 |200 |-0.10116 |
|1000 |10 |100 |-0.33693 |
|1000 |10 |50 |-0.57731 |
|1000 |10 |25 |-0.73543 |
|1000 |10 |20 |-0.76632 |
|1000 |10 |5 |-0.75089 |
|1000 |10 |2 |-0.48009 |

Table 1: N = 1000; r = 10

|N |r |n |f(n) |
|1000 |20 |500 |0.001999 |
|1000 |20 |250 |0.001026 |
|1000 |20 |200 |-0.00599 |
|1000 |20 |100 |-0.109 |
|1000 |20 |50 |-0.33487 |
|1000 |20 |25 |-0.55972 |
|1000 |20 |20 |-0.61499 |
|1000 |20 |5 |-0.70374 |
|1000 |20 |2 |-0.46038 |
|1000 |20 |1 |0.02 |

Table 2: N = 1000; r = 20

|N |r |n |f(n) |
|1000 |5 |500 |-0.02894 |
|1000 |5 |250 |-0.23251 |
|1000 |5 |200 |-0.32186 |
|1000 |5 |100 |-0.57983 |
|1000 |5 |50 |-0.75337 |
|1000 |5 |25 |-0.84087 |
|1000 |5 |20 |-0.85374 |
|1000 |5 |5 |-0.7752 |
|1000 |5 |2 |-0.49002 |
|1000 |5 |1 |0.005 |

Table 3: N = 1000; r = 5

[pic]Figure 1: Proposed vs Existing Procedure

Answers

1. Yes, the proposed technique requires lesser number of chemical analyses than those required by the existing procedure. This is true at all values of negative f(n) in the above graph.

2. For each of the above cases, the n-value (group size) corresponding to the smallest f(n) value represents the most efficient group size. This value has been highlighted in Tables 1, 2 and 3.

Alternate Solution

Assumptions

1. The event of one person testing positive is independent of any other person testing positive.

Every person testing positive for the disease is equally likely.

Solution

Once again, let N be the total population and r be the number of infected people in the population. Let ‘p’ be the probability that a person chosen at random will test positive for the disease.

Let X be a random variable representing the number of people in a group testing positive. It follows a binomial distribution Bin(N, k) such that

[pic]

Now, the probability for no person in a group testing positive is,

[pic]

So, the probability that atleast one person in the group tests positive is

[pic]

Number of tests to be performed through sampling = Number of groups + Testing each person in the groups testing positive. For the proposed technique to be cost effective, this figure should be less than the total population (when you test each person).

[pic]

Simplifying, we get

[pic]

[pic]

f(x) was plotted on MS Excel for different values of N, n and r, as shown here:

|N |r |n |f(n) |
|1000 |10 |500 |-0.00457 |
|1000 |10 |250 |-0.07706 |
|1000 |10 |200 |-0.12898 |
|1000 |10 |100 |-0.35603 |
|1000 |10 |50 |-0.58501 |
|1000 |10 |25 |-0.73782 |
|1000 |10 |20 |-0.76791 |
|1000 |10 |10 |-0.80438 |
|1000 |10 |5 |-0.75099 |
|1000 |10 |2 |-0.4801 |

Table 4: N = 1000; r = 10

|N |r |n |f(n) |
|1000 |5 |500 |-0.07957 |
|1000 |5 |250 |-0.28161 |
|1000 |5 |200 |-0.36196 |
|1000 |5 |100 |-0.59577 |
|1000 |5 |50 |-0.75831 |
|1000 |5 |25 |-0.84222 |
|1000 |5 |20 |-0.85461 |
|1000 |5 |10 |-0.85111 |
|1000 |5 |5 |-0.77525 |
|1000 |5 |2 |-0.49003 |

Table 5: N = 1000; r = 5

|N |r |n |f(n) |
|1000 |20 |500 |0.001959 |
|1000 |20 |250 |-0.0024 |
|1000 |20 |200 |-0.01259 |
|1000 |20 |100 |-0.12262 |
|1000 |20 |50 |-0.34417 |
|1000 |20 |25 |-0.56346 |
|1000 |20 |20 |-0.61761 |
|1000 |20 |10 |-0.71707 |
|1000 |20 |5 |-0.70392 |
|1000 |20 |2 |-0.4604 |

Table 6: N = 1000; r = 20

[pic]
Figure 2: Proposed vs Existing Procedure (Alternate Solution)

Answers

1. Yes, it is again observed that the proposed technique requires lesser number of chemical analyses than those required by the existing procedure. This is true at all values of negative f(n) in the above graph.

2. Again, the n-value (group size) corresponding to the smallest f(n) value represents the most efficient group size. This value has been highlighted in Tables 4, 5 and 6.

Comparison of Solutions

|N |r (disease occurance in N)|Efficient Group Size (Hypergeometric) |Efficient Group Size (Binomial) |
|1000 |5 |20 |20 |
|1000 |10 |20 |10 |
|1000 |20 |5 |10 |

Table 7: Comparison of Solutions