6.) When testing donated blood for infectious diseases, each sample can either b

Question

6.) When testing donated blood for infectious diseases, each sample can either be tested individ-ually or in groups of b samples. When pooled samples are tested, a negative test indicates that all of the donors contributing to the group are disease-free. On the other hand, if a group test is positive, then each donor contributing to the group must be individually tested to determine which are infected. If there are D donors and the prevalence of the disease in the population is p, nd conditions on p, b and D such that the expected number of tests performed using groups is less than the number required without grouping. For convenience, you may assume that D is an integer multiple of b.

Explanation / Answer

So in that vein, let's say that p = 2% = 0.02, D = 100, and b = 5. In each group of 5 samples, you can calculate the probability of whether a group has a positive test (one or more) or not. The probability that all the tests in a group will be negative is 98%5. In terms of p, D, and b, that probability is (1-p)b. The probability that one or more donors in a sample is positive is therefore 1 - 98%5 or 1 - (1-p)b.

The expected number of tests for the one sample, therefore, is 98%5 1 + [1 - 98%5] 6 (because if we have to test all 5 donors individually, that's 6 tests, including the test on the first pooled sample). This works out to about 1.48 (which is definitely better than doing the 5 tests we'd have to do without grouping). In terms of p, D, and b, that's (1-p)b 1 + [1 - (1-p)b] (b+1).

For the whole group of D donors, let's first take the assumption that D is an integer multiple of b, and call g = D/b. In our example with specific numbers, g = D/b = 100/5 = 20. Now, if you don't group, you have to run 100 tests. But if you divide the 100 donors into 20 groups with 5 donors in each group, your expected total number of tests is:

20 {98%5 1 + [1 - 98%5] 6} 29.6

Definitely better than 100 tests.

So, in terms of p, D, and b, your expected number of tests is

D/b {(1-p)5 1 + [1 - (1-p)5] (b+1)}

Now, The condition required is

D/b {(1-p)5 1 + [1 - (1-p)5] (b+1)} < D

You could factor out the D and end up with

1/b {(1-p)5 1 + [1 - (1-p)5] (b+1)} < 1

And you could multiply both sides by b and end up with

{(1-p)5 1 + [1 - (1-p)5] (b+1)} < b

You could probably multiply out the left side, and then rearrange things to get all the p's on one side of the inequality and all the b's on the other side. Then you'd have a condition that relates p to b; that is, if you know p, you can then choose a value for b that not only meets the condition (to save on the expected number of tests), but also optimally meets the condition (in this case, to make the inequality as "big" as possible, to make the left side as much smaller than the right side as possible).

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