Acceptance Sampling

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Acceptance Sampling Dec 10, 2008

Presented by: Girish Harsha AIM, Jodhpur Management Trainee

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• Statistical quality control technique, where a random sample is taken from a lot, and upon the results of the sample taken the lot will either be rejected or accepted • Acceptance sampling is a quality control procedure used when a decision on the acceptability of a batch has to be made from tests done on a sample of items from the batch • The main purpose of acceptance sampling is to decide whether or not the lot is likely to be acceptable, not to estimate the quality of the lot.

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• Accept Lot – Ready for customers • Reject Lot – Not suitable for customers • Statistical Process Control (SPC) – Sample and determine if in acceptable limits • Purposes Determine the quality level of an incoming shipment or, at the end production – Ensure that the quality level is within the level that has been predetermined • Inspection • The observation and measurement of inputs and outputs Accept without inspection 100% inspection Acceptance sampling

Advantages and Disadvantages of Acceptance sampling

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Dec 10, 2008

Advantages

• It is usually less expensive because there is less inspection (less personal, less inspection time, less inspection error) • There is less handling of product, hence reduced damage • Rejection of entire lots as opposed to the simple return of defectives often provides a stronger motivation to the vendor for quality improvement.

Disadvantages

• Risks of accepting “bad” lots and rejecting “good” lots. • Less information is generated • Need planning and documentation

Sampling Plan

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• A plan for acceptance sampling specifying the number of units to sample and the number of sample units that must conform to specifications if the shipment is to be accepted • Type of sampling plan ▫ Single sampling plan The lot disposition is determined by one single sample.

▫ Double-sampling plan The decision from the first sample is to (1) accept the lot, (2) reject the lot, (3) take second sample. If we need to take the second sample, the lot disposition is determined by both the first and the second sample.

▫ Multiple-sampling plan It is an extension of double-sampling plan. Acceptance sampling based on many small samples.

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Acceptance Number

Dec 10, 2008

• The number of sample units specified in a sampling plan that must conform to specifications if the shipment is to be accepted • sample size n and acceptance number c

• Single sampling If the lot size is N=10000, then the sampling plan n=89, c=2 means that from a lot of size 10000 a random sample of n=89 units is inspected and the number of nonconforming or defective items d observed. If d is less or equal to c = 2, the lot will be accepted. If d is larger than 2, the lot will be rejected

Rule: If d ≤ c , accept lot, else reject the lot

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Type of Risks

Dec 10, 2008

• Producer’s Risk ▫ Type I Error (α ) ▫ The risk or probability of incorrectly concluding that the conversion process is out of control ▫ Reduced by using wide control limits ▫ Increased by using narrow control limits

• Consumer’s Risk ▫ Type II Error (β ) ▫ The risk or probability of incorrectly concluding that the conversion process is in control ▫ Increased by using wide control limits ▫ Decreased by using narrow control limits

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Operating characteristic curve

Dec 10, 2008

• The graph of probability of accepting a shipment as a function of the quality of the shipment ,for a given sampling plan ▫ If a shipment is of high quality (low % defective), good sampling yields a high probability of accepting the shipment ▫ If a shipment is of poor quality (high % defective), the plan yields a low probability of accepting the shipment

• Helps to keep the high cost of inspection down • Aids in selection of plans that are effective in reducing risk

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Ideal OC curve

1.0

Accept all lots 1.5% defective or less, and reject all lots having quality level more than 1.5%

Pa

0 p

1.5%

OC for Single sampling alpha

Beta

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How to get OC curve?

Dec 10, 2008

n! P{d defectives} = f ( d ) = p d (1 − p ) n −d d ! ( n − d )! c

n! Pa = p{d ≤ c} = ∑ p d (1 − p ) n −d d =0 d ! ( n − d )! Larger the sample size the steeper the graph. i.e. the larger the sample size, better the plan discriminates b/w good batches and bad batches. Note: Provided the batch should be large enough

Double sampling

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• Application of double sampling requires that a first sample of size n1 is taken at random from the (large) lot. The number of defectives is then counted and compared to the first sample's acceptance number c1and rejection number r1. Denote the number of defectives in sample 1 by d1 and in sample 2 by d2, then: ▫ If d1≤ c1, the lot is accepted. ▫ If d1 ≥r1, the lot is rejected. ▫ If c1 < d1 < r1, a second sample is taken.

• If a second sample of size n2 is taken, the number of defectives, d2, is counted. • The total number of defectives is D2 = d1 + d2. Now this is compared to the acceptance number c2 and the rejection number r2 of sample 2. In double sampling, r2 = c2 + 1 to ensure a decision on the sample. ▫ If D2 ≤c2, the lot is accepted. ▫ If D2 ≥r2, the lot is rejected.

Multiple sampling

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• It involves inspection of 1 to k successive samples as required to reach an ultimate decision • The procedure commences with taking a random sample of size n1 from a large lot of size N and counting the number of defectives, d1. if d1≤ c1 the lot is accepted. if d1≥ r1 the lot is rejected. if c1 < d1 < r1, another sample is taken

• If subsequent samples are required, the first sample procedure is repeated sample by sample. For each sample, the total number of defectives found at any stage, say stage i,

Contd..

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• Acceptable Quality Level (AQL) ▫ The AQL is a percent defective that is the base line requirement for the quality of the producer's product. The producer would like to design a sampling plan such that there is a high probability of accepting a lot that has a defect level less than or equal to the AQL

• Lot Tolerance Percent Defective (LTPD) ▫ The LTPD is a designated high defect level that would be unacceptable to the consumer. The consumer would like the sampling plan to have a low probability of accepting a lot with a defect level as high as the LTPD , also known as Rejection Quality Level

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Contd..

• Average Outgoing Quality (AOQ)

Dec 10, 2008

▫ In this case, all rejected lots are made perfect and the only defects left are those in lots that were accepted. AOQ's refer to the long term defect level for this combined LASP and 100% inspection of rejected lots process. If all lots come in with a defect level of exactly p, and the OC curve for the chosen (n ,c) LASP indicates a probability pa of accepting such a lot, over the long run the AOQ can easily be shown to be: AOQ=pap(N-n)/N

• Average Outgoing Quality Level (AOQL) ▫ A plot of the AOQ (Y-axis) versus the incoming lot p (Xaxis) will start at 0 for p = 0, and return to 0 for p = 1 (where every lot is 100% inspected and rectified). In between, it will rise to a maximum. This maximum, which is the worst possible long term AOQ, is called the AOQL *LASP- Lot Acceptance Sampling Plans

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Contd.

Dec 10, 2008

• Average Total Inspection (ATI)

▫ When rejected lots are 100% inspected, it is easy to calculate the ATI if lots come consistently with a defect level of p. For a LASP (n , c) with a probability pa of accepting a lot with defect level p, ATI = n + (1 - pa) (N - n) where N is the lot size

• Average Sample Number (ASN)

▫ ASN refers to comparisons b/w average numbers of items inspected in sampling where other characteristics of the plans ,such as their OC curves ▫ A plot of the ASN, versus the incoming defect level p, describes the sampling efficiency of a given LASP scheme ASN= [pa(n1)+pr(n1)]n1 +[pa(n2)+pr(n2)](n1+n2) = n1 + [pa(n2) +pr(n2)]n2 = n1 + n2 p(n2)

where p(n2)= probability of drawing n2

Examples

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• A single sampling plan uses a size of 12 and acceptance number 1. what is probability of acceptance of lots having size=40 and defective 2% Solution: • Given , N=40, n=12, c=1 and p= 1/40 =0.025 • Number of defective items= 40 * 0.025=1 • Number of non defective items= 40-1 =39 • Po=39C12/40C12 =0.7 • P1=39C11/40C12 =0.3 • Hence total probability of acceptance=Po + P1=1.0