Questioner01 wrote:For a sample of 100 with accept number of 3, the LTPD at 10% probability of acceptance is 5.56% (attribute sampling plan).

I think the LTPD10 is 6.55% for this plan. (n=100,c=3) Do you agree?

I interpret this to mean that there is a 10% chance that up to 5.6% of the population may be defective.

Not exactly. Distinguish between before and after sample is taken/seen/decided. The 10% Pa is a conditional probability. It only applies when the condition (p'=LTPD) is true.

However, once a sample is observed, you can use the sample to estimate the percent of defective of the lot. The proper method is with confidence limits.

For example for the 4 possible samples that produce acceptance, the upper one-sided 90% confidence limits are:

n=100, X=0, Upper CL=2.28 %defective

n=100, X=1, Upper CL=3.83 %defective

n=100, X=2, Upper CL=5.23 %defective

n=100, X=3, Upper CL=6.56 %defective

My interpretation of the confidence limit for the sample: n=100, X=0 is that I am 90% confident that the % of defectives in the lot is

(less than or equal to)=(no greater than) 2.28 % defective.

If the lot size is known, you can express the upper CLs in terms of actual defectives in the lot. For example, if the lot size is 1000 units, then I am 90% confident that no more than 22.8=23 of them are defective.

The balance of the population, 94.4% I consider to be defective free.

Answered.

The sampling plan is sufficient to characterize 94.4% of the population as defective free. It is not sufficient to characterize 5.6% of the population.

Answered.

Is there any meaning or significance to the ratio, 94.4:5.6?

I don't think so.

How does one interpret this ratio, or its inverse, if it has any meaning at all?

I don't think so.

Is this something like an odds ratio?

I don't think so.