Forum for Sampling Practices

Sir:

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

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

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

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.

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

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

Is this something like an odds ratio?

Thank you.

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.

Do you agree?

Thanx for the prompt reply.

Yes, I mistyped. It should have been 6.56%, not 5.56%.

I will read the remainder of your reply and reply to you tomorrow.

As I said earlier, I erred typing 5.56%; it should have been 6.56% as you stated.

I should have been more careful and wrote: "I interpret this to mean that with a sample of 100 with three defectives observed, the LTPD at 10% probability of acceptance is 6.56% (attribute sampling plan)." Then the statement I made,"I interpret this to mean that there is a 10% chance that up to 5.6% of the population may be defective" is appropriate.

Regarding confidence limits, the LTPD value is the upper bound of the % defective for all sampling plans. And 100 - LTPD is the lower confidence bound of the % conforming. Do you agree?

Please pardon me for the delay -- I was fighting some computer problems.

Questioner01 wrote:As I said earlier, I erred typing 5.56%; it should have been 6.56% as you stated.

I should have been more careful and wrote: "I interpret this to mean that with a sample of 100 with three defectives observed, the LTPD at 10% probability of acceptance is 6.56% (attribute sampling plan)." Then the statement I made,"I interpret this to mean that there is a 10% chance that up to 5.6% of the population may be defective" is appropriate.

I agree (using 6.56%)

Regarding confidence limits, the LTPD value is the upper bound of the % defective for all sampling plans.

Yes, I agree. Specifically, the LTPD value is equal to the upper confidence limit when the number of defectives in the sample are equal to the acceptance number.

And 100 - LTPD is the lower confidence bound of the % conforming. Do you agree?

I agree. One can be 90% confident that the % conforming, when n=100 & X=3, is equal to or above (no lower than) 100-6.56=93.44%.

An assumption of this is that for a sequence of lots sampled is that most lots are acceptable.

Questidoner01 wrote:I agree. One can be 90% confident that the % conforming, when n=100 & X=3, is equal to or above (no lower than) 100-6.56=93.44%.

An assumption of this is that for a sequence of lots sampled is that most lots are acceptable.

That example you posted explains a lot of stuff for me. Thanks Stan.