Assignment Questions Acceptance Sampling

Utkarsh Shrivastava (N18802109)

10. Compare and contrast chain sampling and sequential sampling plans. When are

they used?

Sol. Acceptance sampling is the methodology that deals with procedures by which decisions to

accept or not accept lots of items are based on the results of the inspection of samples. Special

purpose acceptance sampling inspection plans (abbreviated to special purpose plans) are tailored

for special applications as against general or universal use.Chain sampling can be viewed as a

plan based on a cumulative results criterion (CRC), where related batch information is chained or

cumulated. The phrase chain sampling is also used in sample surveys to imply snowball

sampling for collection of data. It should be noted that this phrase was originally coined in the

acceptance sampling literature, and should be distinguished from its usage in other areas.

Chain sampling is extended to two or more stages of cumulation of inspection results with

appropriate acceptance criteria for each stage. The theory of chain sampling is also closely

related to the various other methods of sampling inspection such as dependent-deferred

sentencing, tightenednormaltightened (TNT) sampling, quick-switching inspection etc.

15. Consider a single sampling plan with a lot size of 1500, sample size of 150, and acceptance

number of 3. Construct the OC curve. If the acceptable quality level is 0.05% nonconforming

and the limiting quality level is 6% nonconforming, describe the protection offered by the plan at

these quality levels.

Sol.

0.05% nonconforming, they will be accepted just about all the time thus acceptance rate is 100%, while lots that are 6% nonconforming will be accepted about 2.1% of the time,

using the sampling plan N = 1500, n = 150, c = 3.

16. The calculations for the OC curve are shown in Table 10-2. For lots that are

0.05%nonconforming, they will be accepted just about all the time; while lots that are 6%

nonconforming will be accepted about 0.2% of the time. For the same level of p, the probability

of lot acceptance is lower for this plan compared to that in Problem 15.

Sol.

OC curve calculation above the graph. For lots that are 0.05% nonconforming, they will

be accepted 100%; while lots that are 6% nonconforming will be accepted about 0.2% of

the time. Thus for the same probability the lot acceptance is lower as compared to

previous one.

20. For the double sampling plan N=2000, n1 =80, c1 = 1, r1 = 3 , n2 = 100, c2 = 2, r2 = 3,

construct and interpret the ASN curve. Suppose that process average nonconforming rate is

1.5%. Would you prefer the stated double sampling plan or a single sampling plan with n = 100,

c = 2 in order to minimize ASN?

Sol.

The ASN curve is calculated for the values in the table. Since for p = .015 ASN is 101.6 whereas for a single sampling plan with n = 100, c = 2 is 100. Therefore in order to

minimize ASN, single sampling plan would be preferred.

22. A computer monitor manufacturer subcontracts its major parts to four vendors: A, B, C, and

D. Past records show that vendors A and C provide 30% of the requirements each, vendor B

provides 25%, and vendor D provides 15%. In a random sample of 10 parts, 4 were found to be

nonconforming. It is known that vendors A and C operate at a level of 5% nonconformance rate,

while vendors B and D are at 10% and 15%, respectively. What is the probability that the

sample came from vendor A? From vendor B? From vendor C?

Sol. Given, Probabilites from Vendor P(A) = .30, P(B) = .25, P(C) = .30, P(D) = .15

Let the probability of coming from different vendors be G, having 4 non-conforming parts and

having a sample size of 10.

P(G|A) = (10 4) (.05)4 (.95)6 = .00096

P(G|B) = (10 4) (.10)4 (.90)6 = .0112

P(G|C) = (10 4) (.05)4 (.95)6 = .00096

P(G|D) = (10 4) (.15)4 (.85)6 = .0401

Further, Probablities of coming from different vendors is given by

P(A|G) = (.0096 *.30) / (.00096)(.30) + (.0112)(.25) + (.00096)(.30) + (.0401)(.15) = .0370

P(B|G) = (.0112 *.30) / (.00096)(.30) + (.0112)(.25) + (.00096)(.30) + (.0401)(.15) = .2982

P(C|G) = (.00096 *.30) / (.00096)(.30) + (.0112)(.25) + (.00096)(.30) + (.0401)(.15) = .0307

Thus we can analyze the various probabilities and make interpretations based on that such as,

probability of B has gone up and probability of A and C have gone down.

30. Consider a double sampling plan given by the following parameters: N = 1200, n1 = 50, c1 =

1, r1 = 4, n2 = 110, c2 = 5, r2 = 6. Find the probability of accepting lots that are 4%

nonconforming. What is the probability of rejecting a lot on the first sample?

Sol. The probabilities for accepting 4 % lots that are non-conforming is given as

P(First) = P( x1