Regarding part of your answer to a post found here, you state:
“The calculation of AQL is not dependent on lot size. In other words, a sample size of 315 gives a minimum AQL of 0.04, so a larger sample is required to estimate an AQL of 0.01.”
Can you explain for the non-statistical folks like me people how that math works? Specifically, I am wondering what the minimum sample size would be for an AQL of 0.25, when using Special Inspection level S2? Would it be a minimum of 50, no mater what the lot size is?
Acceptance sampling procedures were developed during the early 1920s at Western Electric Company and later formalized at Bell Telephone Laboratories where terms like producer’s risk and consumer’s risk were established. Later, during World War II, sampling plans such as MIL-STD-105 were developed by Harold F. Dodge and others working with the Army Quartermaster Corps (Dodge, 1967).
Two special features were employed in order to gain agreement with the large body of military suppliers. One was the use of the acceptable quality limit (AQL) as opposed to the RQL in presenting the plans. The goal at the time was to focus on rewarding suppliers for production whose quality levels were considered good. RQLs were recognized but not often brought to the surface during discussions. Also, at that time, the term “AQL” was deliberately vague or inexact. It was a close approximation, not an exact probability statement.
The other feature was the practice of increasing sample sizes with increased lot sizes. As noted in Section 3, in most situations, the lot size does not factor in plan construction (based on the binomial). For many, however, this lacks intuitive appeal. Therefore, in the development of MIL-STD-105 and its derivatives a deliberate increase in sample sizes for higher lot sizes was introduced, with corresponding increases in acceptance numbers for similar AQLs. Clearly, this practice resulted in over-sampling and consequent increased inspection costs. Government operatives believed that the increased sampling cost was of small consequence relative to the power to persuade.
For the binomial distribution you solve for the AQL that gives a high probability of passing. Usually this probability is set at 95%. For example if you have a sample size of 80 units with an accept/reject of 1, an AQL of 0.65% would have a 90% probability of passing the sampling plan.
You can use Excel to solve this with the function
Hope this helps,