59 is correct sample for the given condition of 95% Reliability with 95% Confidence. This is NOT an AQL plan.
A Reliability of 95% means that you want to reject the lot if you have 5% defective or more. Confidence means that you want to reject that defective rate 95% of the time.
the formula is:
n= LN(1-Confidence)/LN(Reliability)
so, n = LN(1-.95)/LN(.95) = 58.40397
sample sizes should be rounded UP, so n = 59.
However this is an acceptance sampling plan and for OQ validation the sample size should be calculated differently as you are trying to estimate the actual defect rate...acceptance sampling only provides some assurance that the defect rate is probably not greater than some value for 'most' lots. That isn't a good approach for OQ validation.
you can use the acceptance sampling plan approach for your PQ validation lots.
A Reliability of 95% means that you want to reject the lot if you have 5% defective or more. Confidence means that you want to reject that defective rate 95% of the time.
the formula is:
n= LN(1-Confidence)/LN(Reliability)
so, n = LN(1-.95)/LN(.95) = 58.40397
sample sizes should be rounded UP, so n = 59.
However this is an acceptance sampling plan and for OQ validation the sample size should be calculated differently as you are trying to estimate the actual defect rate...acceptance sampling only provides some assurance that the defect rate is probably not greater than some value for 'most' lots. That isn't a good approach for OQ validation.
you can use the acceptance sampling plan approach for your PQ validation lots.