Sampling Plan for Destructive Testing - Personal Body Armor Plates

[elamo]

Big first time request follows. Here goes…….

I am looking for a sampling plan that has a valid statistical basis that can be used to make a decision about the acceptability of batches of personal body armor plates where high user confidence is required by testing a very small sample due to the high cost of destructive testing.

As the customer I will expect the manufacturer to prove that their manufacturing process is controlled and repeatable and all plates meet all of the non destructive requirements for the plates before being sent for destructive testing to prove they meet the ballistic performance requirements.

I would prefer that no matter what quantity makes up a homogenous batch size for the manufacturer that the Acceptance number for the sample size always remains at zero. The only way I can find to achieve this is to reduce the AQL as the batch size increases as shown below.

Batch size 1-150, AQL 2.5%, Sample size 5, ACC 0 REJ 1
Batch size 151-500, AQL 1.5%, Sample size 8, ACC 0 REJ 1
Batch size 501-1200, AQL 1.0%, Sample size 13, ACC 0 REJ 1

Based on Special Inspection Level S-4 in AS1199.1:2003 Sampling procedures for inspection by attributes Part 1: Sampling schemes indexed by AQL for lot-by-lot inspection (identical to ISO 2859-1:1999).

Can anybody tell me if the above plan is reasonable and if so how I can figure out what confidence level will be achieved by using the sampling plan above? If not, any other suggestions for sampling to achieve a confidence level of say, 90%, 95% or 97.5% that the quality of the batch will be no worse than the AQL?

That’s all I ask...... Any assistance will be sincerely appreciated.

 

[Marc]

I don't see anything wrong with your sampling plan as long as your qualification testing went OK.

 

[Mike S.]

Take a peek at "Zero Acceptance Number Sampling Plans" by Nicholas L. Squeglia. It is well written. I think it might help you.

Good luck -- let us know what you decide on!

PS -- Now that would be some fun testing...

 

[elamo]

Thanks Mike

I'm pursuing getting a copy of your recommended book and look forward to the answers it may provide.

Regards

Tracey

 

[wmarhel]

If you'd like, just send me half a dozen plates and I'll sample them for you.

Wayne

 

[Tim Folkerts]

If you want to acheive a specified confidence level, CL, for a sampling plan with c=0 where the defect rate is x, then the formula (unless I messed up the math somewhere) for the required sample size, n, is

n = ln(CL) / ln(1-x)

For example, of x = 0.01 and CL = 0.95, you need

n = ln(0.95) / ln(0.99) = 5.1

which should be rounded XXup to 6XX down to 5. In other words, if only 1% of the products are truly bad, then you should expect to find 1 bad item in a sample of XX6XX 5 less than 1-CL = 5% of the time. Thus finding a bad item is unusual and would be grounds to question the quality of the lot.


On the other hand, with a sample size of 5, it works of that even a lot with 45% defective would be accepted about 5% of the time! Thus you could only be 95% confident of catching lots that are more than 45% defective. That is the weakness of c=0 sampling plans - they are very poor at rejecting lots that are actually quite poor quality.


Tim F


P.S. These calculations assume that the lot size is much larger than the sample size. If this is not true, then the results improve slightly, so your confidence would be slightly higher than estimated here.

 

[Mike S.]

Tim,

Interesting stuff. Do you have a source for your equation?

Mike

 

[Tim Folkerts]

Do you have a source for your equation?

I just made it up.


Well, actually I just derived it. If x is the proportion of bad parts, then 1-x is odds of drawing a good part. If you draw n parts, the odds that they are all good is (1-x)^n. As long as this number stays above your desired CL, then the test is reasonable. So the cut-off point is

(1-x)^n = CL

ln( (1-x)^n) = ln(CL) ....... take ln of both sides

n ln(1-x) = ln(CL) ............ a basic rule for logs

n = ln(CL) / ln(1-x) .......... solve for n

And there you go...


ONE CORRECTION. In my previous post I rounded up, but I should have rounded down. Using the previous example with CL = 0.95 and x = 0.01, the answer was 5.1. For samples sizes up to 5, finding a defect is highly unlikely, so finding a defect would be reason to suspect that something is happening. For samples of 6 or more, finding a defect happens at least 1-CL = 5% of the time, so it isn't significant for the confidence level we had chosen. I'll make that change to the earlier post.

Tim F


P.S. It works just the same with log rather than ln. So if you are an engineer rather than a mathematician, you can use n = log(CL) / log(1-x).

 

[mdurivage]

I have a question about the math and the rationale. I do not understand that when the CL decreases, the sample size increases. It just seems that it should be the opposite.

For example CL 95%, defect rate 0.01 n =5
CL 90%, defect rate 0.01 n = 10

Thanks!