Attribute sampling plan
- Thread starter Flyn17
- Start date
Mustapha
Starting to get Involved
What are you inspecting and what are you trying to accomplish? I can give you a basic run down, but it would help if you gave more info on what you're trying to do. Both sampling plans below have a confidence level of 95%
You can either use one of the standard sampling plans such as ANSI Z1.4 or Squeglia c=0. Most companies are moving towards Squeglia's c=0 sampling plans because it technically gives you the same result of the Z1.4 plans with less samples taken for measurement.
In your example, under Z1.4, under general inspection level 2 and an AQL of 1.0, your sampling code would be P and the number of samples you take would be 800. You could accept up to 14 defects within that 800 and only reject lot once you hit 15 defects. For c=0, you would take 90 samples and reject the lot on the first defect.
You can either use one of the standard sampling plans such as ANSI Z1.4 or Squeglia c=0. Most companies are moving towards Squeglia's c=0 sampling plans because it technically gives you the same result of the Z1.4 plans with less samples taken for measurement.
In your example, under Z1.4, under general inspection level 2 and an AQL of 1.0, your sampling code would be P and the number of samples you take would be 800. You could accept up to 14 defects within that 800 and only reject lot once you hit 15 defects. For c=0, you would take 90 samples and reject the lot on the first defect.
You must state a RQL and/or AQL level in order to establish a c number.
BUT can you clarify your situation: you state “lot size of 350k(all of them). Do you mean you will inspect all 350k parts? Or are you going to sample randomly from a lot of 350k.
by the way, lot size is actually irrelevant despite some plans that index n based on sample size.
Are you looking for the formulas to use?
BUT can you clarify your situation: you state “lot size of 350k(all of them). Do you mean you will inspect all 350k parts? Or are you going to sample randomly from a lot of 350k.
by the way, lot size is actually irrelevant despite some plans that index n based on sample size.
Are you looking for the formulas to use?
I am inspecting cracks on a product. lets say, I have not done a IMV for this inspection and sold over 350k products, and found that we have 3 complaints due to the cracks. how do I support the statement, we are doing better in terms of Conf/reliability even if we were to validate the inspections.
AQL=.25% RQL=5%
for validation AQL=.25% RQL=5%
Yes, inspect all of them
lot size is actually irrelevant despite some plans that index n based on sample size - did not understand this
formulas would help
Could you help me with formula, on how to proceed to further
Before formulas we need to understand the situation. The correct math matches the situation.
a few questions:
a few questions:
- Do you inspect all parts for cracks? Or only a sample. If a sample what is the sample size and how did you decide on it?
- How is the inspection done?
- Have you performed an MSA on the inspection? If so what method did you use and what were the results?
- And the most important question of all: what is the timing of the complaints? Are they randomly scattered throughout the 350k parts or are all they all from the same time frame / shipment & Customer / Lot & Customer? How many ‘good’ parts between teh complaints? There is a huge difference between 3 complaints over 350k parts and 3 complaints from the last 10 units shipped. Any math must take that into account.
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