Mr. Tom McCraw September 22, 1976 Page 5. B. The question of whether to cleanup an island or part of- an island can be put in a hypothesis testing framework. In particular, what is known as "acceptance sampling" appears to be a useful approach since there is no need to make any assumptions (normal, lognormal, etc.) about-the statistical distribution of the data. The basic idea is to specify (1) an activity level, say L, above which cleanup is indicated, (2) a propersion (p,) of samples with activities greater than L that is acceptable, (3) a proportion (po) of samples with activities greater than L that is not (3) oa ere (4) the allowable risk (a) of concluding that cleanup is necessary when it really isn't, and (5) the risk (8) of concluding that cleanup is not necessary when in fact cleanup is necessary. Once these quantities have been specified we can determine (i)the number of samples n required in order: to meet these specifications, and (ii) the rejection number r. If r or more of the n samples have activities greater than L, then cleanup is required. Note that this approach assumes we are willing to tolerate a certain proportion (pj) of samples with activities greater than L without cleaning up the area. Of course, P, can be specified to be as smal] as we choose. The risk 8 should be specified as a smal] quantity since the consequences of not cleaning up a contaminated area could be considerable to the inhabitants of the area. 1-8 is known as the “power" of the design, i.e. the probability that the area is cleaned up when the actual proportion is po. On the other hand we would also like a to be near zero so as to avoid unnecessary cleanup operations. In the following table we give values of n and r for various values of py, Pg, a, and 8. These were obtained using Table 13 in Burstein, H., 1971. Attribute Sampling;Tables and Explanations, McGraw-Hill, 464 pp. These values of n and r are for a non-sequential sampling plan. A sequen- tial plan would probably require fewer samples. From the results in TABLE 1 we note that: a) As a gets larger the number of samples (n) required decreases when P1» P2. and B remain constant. Hence, if we are willing to risk Spending more money on cleanup, the number of samples we need to -collect decreases. b) As B increases (power decreases) the number of samples n also decreases when Pj, pz, and & remain constant. Hence, if we are willing to take a higher risk of missing some areas needing cleanup, we won't need to take as many samples. c) As po increases, the number of samples (n) decreases. If our cleanup criterion is that 10% rather than 2% of the samples must be greater than L before cleanup is started, then only 113 rather than 3063 samples need be taken (assuming p] = a = B=.01). That is it will take many fewer samples to detect a difference between Py = .O] and po = .10 than to detect a difference between p and pp = .02. Hence, as py and po are placed closer together( for given a and g), the number of samples (n) increases.