AQL and Confidence Level

Six Sigma – iSixSigma Forums Old Forums General AQL and Confidence Level

Viewing 8 posts - 1 through 8 (of 8 total)
  • Author
  • #32321


    Not sure if I got it right, but let me give it a shot.
    I believe you are referring to (1-alpha risk (type I error)) or the probability of accepting the lot when its quality level is equal to the AQL. To compute the alpha risk (and thereby, 1-alpha or “confidence level”) for any given sampling plan (sample size and Acceptance Number) with a specified AQL, you can make use of the Hypergeometric Distribution. Use the MS EXCEL, vary the sample size until you get the desired alpha. If you are getting an error on the probability, your lot size maybe too large (Excel can not factorial (!) of very large numbers). In this case, make use of the Poisson distribution.



    Rams –
    Excelleant summary except the Binomial is usually the approximating distribution to the Hypergepmetric when the ratio of sample to lot size (n/N) is <= 0.1.
    Poisson approximates the Binomial and by default the Hypergeometric when(n/N ) >= 0.1 and np <5. If np greater than five use normal.
    All of these are easy to set up in Excel and produce OC curves except the Hypergeometric as you observed.



    Thanks Rams – u got it right – that was exactly what I wanted – another requirement will be to determine the associated consumer risk as well. will revert for any further help after trying it out.
    I shud have used the word risk instead of confidence level though they are complementary



    Hi rams,
    Here’s my specific problem.
    1. 1.5 AQL
    2. Lot size 151 to 280
    3. Sample size is 32 ( Normal inspection )
    4. From Table 5A of ISO 2859-1 : 1999  I can infer that the Producer’s risk ( probability that good pices are rejected ) is 8.42 % by Poisson and 8.30 % by Binommial distrn.
     5. Can you pls guide me as to exactly how these figures were arrived at ?



    Go learn about OC curves – any good SQC book will have the topic addressed.



    First, a correction. You said:
    “the Producer’s risk ( probability that good pices are rejected )”
    In fact, the producer’s risk is the chance that a lot with a quality level = AQL is rejected just by chance (such a lot would contain bad parts!). Tha is (takeing the binomial distribution) that the % non conforming in the lot = 1.5% (what would be “acceptable” according to the “acceptable quality level” or AQL) but the lot is rejected because, just by chance, it does not meet the accepting criteria based on sampling.
    In your post, you don’t mention the accept/reject criteria, but I guess  from the figures (you will see why) that it is 1 non conforming = accept, 2 non conforming = reject. In such a case, the chances to accept a lot would be = to the chances to find 0 or 1 or, what is the same, to find up to 1 non conforming in the sample. Then the probability to reject a lot would be = 1-probability to accept the lot = 1-probability to find up to 1 non conforming in the sample.
    Now, let’s see. If you take samples of size 32 from a batch containing 1.5% of deffectives, then (I’m using BINOMDIST in Excel) the probability to find 0 non conforming in the sample is 61.7%, to find 1 it’s 30%, to find 2 it’s 7.1% and so on. If we take the cummulative, the probability to find up to 0 is 61.7%, to find up to 1 it’s 91.7%, to find up to 2 it’s 98.8% and so on. Because you said that the AQL was 1.5, this batch would be in the limit of “acceptable”.
    As you see, the chances to accept such a lot (criteria 1=accept, 2=reject) is 91.7%, then the chances to reject it is 8.3%. Because in this lot the quality level = AQL, this is by definition the producer’s risk.
    Now, are you the producer or the consumer? Specially if you are the customer you might be interested in the consumer risk, which is the chances to accept a lot with the worst quality level the consumer is willing to accept. We have already seen that the chances to reject a lot with 1.5% non conforming is 8.3%. A too low chance if it was the worst quality you are willing to accept. If it was, then the consumer risk would be 91.7%. Big risk! So, what is the worst quality level you are willing to accept? We could reverse the question, and see which quality level would give us a consumer risk of about 8% (just to take about the same than the consumer risk). Well, you need 12.5%!!! of defectives, then the chances to fail to reject it would be about 8%.
    A final comment. Note that the lot size was not mentioned. Both binomial and poisson assumes that the sample is very small compared to the lot.



    Hi Gabriel,
    Bingo ! Pls accept my sincere thanks for your explanation.
    1. Your guess abt accept /reject criteria was correct.
    2. Your final comment … lot size is not known
    lot size is 151 to 280 ( I had mentioned it in my posting ). At 1.5 AQL the chance of defective pieces in the lot is 0.015. np is 0.015*32=0.48
    a) random sampling from the lot
    b) samples are not replaced after checking
    c) np< 5
    d) n is not greater than 0.1 N ( 32 > 0.1 *280 )
    Is this a binomial distrn ?



    Hi Prasad,
    I didn’t mean that you didn’t mention the lot size, I meant that all the analysis (about probabilities, risks, etc) was made without mentioning the lot size, and that that was because the binomial distribution consideres infinite population.
    Replacing is a way of simulating an infinite population. If the population is not infinite and replacing is not done, then the distribution is not binomial but hypergeometrical. If the lot is much larger than the sample, then the hyper closely matches the binomial. In your case, you are not replacing and the lot does not seems to be that large. Yet, if this sampling plan is in the standard I guess it should work. I really do not know how much would it affect, so let’s try:
    Cases (c)              Cummulative probability P(X = c or less) [%]
                              Binomial      Hyper (lot=280)   Hyper (lot=151)
        0                      61.6                 61.4                     62.0
        1                      91.7                 93.4                     95.6
        2                      98.8                 99.5                    100.0
    – AQL = 1.5 was used for the binomial.
    – An AQL of 1.5 would imply 4.2 deffectives in 280 and 2.3 in 151. Because “deffectives” has to be integer (no decimales), 4 and 2 deffectives in the lot were used (that explains the 100% chances to find up to 2 in a sample of 32 from 151: there are only 2 deffectives in all the lot!). That number of defectives in those lots would equal a non conforming fraction of 1.4% and 1.3% respectivelly (lower than the 1.5% used for the binomial). This “better” quality expalins, in part, the increased cummulated probabilities. For example, in a population of 267 with 4 deffectives (that would be 1.498%, very very close to 1.5%) the chances to find up to 1 deffective in a sample of 32 would be 92.8% (closer from the binomial than the values for lots of 280 and 151).
    – BINOMDIST and HYPGEOMDIST was used in Excel.
    – My opinion? This analysis is splitting hairs. Anyway, the power of the acceptance sampling to detect bad lots is so poor that is typically no longer used in companies that speak in terms of “zero deffects”, “six sigma” or “PPM” (remember from the previos post: To have a good chance to reject a lot with this sampling plan it has to have 12.5% of non conforming units!!! And you still would be accepting such a horrible lot 8% of the times!!!) So if you are still going to use it, then use it and that’s it. After all, you are already assuring to your customet that you will accept lots with 1.5% of deffectives more than 90% of the times. Does it really matters if this “more than 90%” is actually 92 or 95%?

Viewing 8 posts - 1 through 8 (of 8 total)

The forum ‘General’ is closed to new topics and replies.