If the AQL=0.02, LQL=0.15, alpha=0.1, and beta=0.2 in an acceptance sampling plan for a company, What is the best approximate value of sample size n and acceptance number c

Question

If the AQL=0.02, LQL=0.15, alpha=0.1, and beta=0.2 in an acceptance sampling plan for a company, What is the best approximate value of sample size n and acceptance number c

Answer 1

With c = 0, at n = 5, at AQL = 0.02, (1 - .02)^5 = 0.9039207968 >= 1 - alpha = 1 - .1 = .9 This is the maximum n to meet the condition.

At LQL = .15, .85^5 = 0.4437053125. Unfortunately, this is greater than .2. Thus, c = 0 does not work.

(We could have done this step by noting that log_.98 .9 = ln .9 / ln .98 = 5.21516814563063, and log_..85 .2 =

9.90307970528525. While n <= 5 for the alpha condition, it must be greater than or equal to 10 for the beta condition).

With c = 1, we can have up to n = 26 and meet the requirement that P(0 or 1 failures at AQL = .02 with n trials) >= .9

The smallest n such that P(0 or 1 failures with n trials at LQL = .15) <= Beta = .2 is n = 19. In Excel, binom.dist(1,19,.15,1) = 0.1985

Incidentally, P(0 or 1 failures at AQL = .02 with 19 trials) is binom.dist(1,19,.02,1) = 0.945384049965565 >= .9

Thus, our smallest sample size meeting the conditions is n = 19 with c = 1.

There are other ways to solve this problem. However, with such a small c necessary, this seems like a nice way to go about it.