Outbound Shipping Quality - Package Contents Sampling Plan - 1,500 Packages an Hour

In our shipping operation, we send out about 1,500 packages an hour. I'd like to determine an hourly sampling plan that will tell me, with 95% confidence, that our outbound quality is 99% -- that is, the contents in the packages are 99% correct.

How do I go about establishing the plan? We do not have an quality department to address this type of program, so the request has defaulted to me!

Thanks for the help.

I believe you are missing one more piece of information, and that is what is the variability of your quality before packing.

If you have a big defect rate that is steady all over the day (say 12%), then two packages will have the enough power to confirm the alternate hypothesis that you have a defect level greater than 1%.

So, you need to know what is your defect rate and how does it distribute over the day. Then you can calculate the standard deviation of that sample and use it to calculate the sample size that will give you enough power.

What about the sampling plans explained in MIL-STD-1916? I am attaching a copy if you like to review it.

The binomial distribution should give you the information you need. It allows you to predict the odds of find c defects in a sample of N items when the probabilty of any item being bad are P. In Excel, you can find the odds of upto A defects using

=BINOMDIST(c, N, P, TRUE)

If we assume you use a c=0 plan, then that would be

=BINOMDIST(0, N, 0.01, TRUE)

Then you adjust N to give the results you want.



You also need to determine just what you mean by 95% confident that 99% are good. Do you want to accept the lot only when you are 95% sure it is good, or do you only want to reject the lot if you are 95% sure it is bad.

To be sure the lot is good, the function should return a value of 0.05 (=1-0.95) or less. Playing with N indicates this takes 299 samples! If you draw less than 299 pieces, there is still at least a 5% chance that a lot with just over 1% defective could still show no defects, so you can't be 95% sure it is 1% or lower.

On the other hand, is takes a much smaller sample to show the lot is bad. If you draw up to 5 pieces and find even a single bad part, that would be extremely unlikely for 1% defective, so you are 95%+ certain the lot is above 1% bad. At a sample size of 6, there is at least a 5% chance of getting 1 defect even for 1% bad, so getting one defect is not 95% sure evidence the lot is bad. For 6 (and up to 35 samples) you would need to 2 defects to ensure the lot is bad.

Note that for a sample of 300, 0 defects will ensure you it is good, 7 defects will ensure you it is bad, but anything in between is indeterminant.

You will find it challenging to show the boxes are good. It will be much easier to show the boxes are bad.


Tim F

P.S. I'm assuming that the orders are considered either right or wrong - for example a box with 1000 parts and 999 correct would still be considered a single defect, not a 99.9% success. Also, these numbers assume the sample is much smaller than the total, which is only sort of true for 300 out of 1500.

Here is a very good link.

http://www.joma.org/mathDL/4/?pa=content&sa=viewDocument&nodeId=428&pf=1

you could also use the poisson instead of the binomial - it will give you the same sample size for this case. you must accept on 0 and reject (revert to 100% sampling) on 1 bad box with this formula:

n = -natural log(1-confidence) / fraction defective (p)

in your case the math = -natural log(1-.95) / .01

pabloquintana has an important caveat: if you have widely varying quality (really good with small bursts of bad) then you will have less 95% probability of detecting the bad in teh short run. If you truly randomply sample you will have 95% protection in the long run, but be prepared to miss the goal.

Unfortunately with categorical sampling, there is no 100% protection of zero defects. (unless of course you use 100% sampling with a true poke yoke device)