SAMPLING PLANS G Venu Gopal Department Of Statistics Avanthi Degree & PG College

1. SAMPLING PLANS G VenuGopal Department Of Statistics Avanthi Degree & PG College Hyderabad

2. STATISTICAL QUALITY CONTROL • It is a powerful productivity technique for effective diagnosis of lack of quality in any of the materials, process, machines or end-products. The Quality Control can be done in two stages: • 1.Process Control • 2.Product Control

3. SQC PROCESS CONTROL PRODUCT CONTROL • For process control the tools are control charts • For the construction of control charts we use hypothesis testing i.e., H₀: θ = vs H₁ : θ≠ • where is the target value. • For product control the tools are acceptance sampling plans. • For the construction of acceptance sampling plans • H₀: θ < vs H₁ : θ>

4. (Acceptance Quality Level ) AQL: A lot with small fraction defective p (let) that the producer does not wish to be rejected more often than a small proportion of time. p is the acceptance quality level and a lot of p is satisfactory to the consumer. P[Rejecting a lot of quality p] = 0.05 Pa =P[Accepting a lot of quality p]=0.95 ( Lot Tolerance Proportion or percent defective ) LTPD: It is the worst quality of the lot which the consumer is willing to accept in an individual lot.p2 is called Lot Tolerance Percentage Defective.

5. Producer’s Risk: It is the probability that a process producing acceptable values of a particular quality characteristic will be rejected as unsatisfactory. i.e., Pp=P [ rejecting a lot of quality p1 ] = α=P[Type I error]. Hence, Pa =P[Accepting a product of quality p1 ]=0.95 =1- α Consumer’s Risk: It is the probability of allowing a process that is operating unsatisfactorily relative to some p quality characteristic to continue in operation. i.e., Pc=P [ accepting a lot of quality p2 ] =β = P[Type II error]

6. Average sample number (ASN):Average number of sampling units required to be inspected before passing a judgment on the lot. • Operating Characteristic Curve (OC- Curve): • It is a measure of efficiency of a sampling plan. This curve plots the probability of accepting the lot corresponding to the lot fraction defective. • Pa = P[accepting the lot of quality ‘p’];where, p is the fraction defective. • By increasing the sample size, ideal OC-Curve can be achieved i.e., the precision with which the plan differentiates between good and bad quality lots increases with n, if c is kept proportional to n. • As the acceptance number decreases, the plan tightens, hence the OC-Curve shifts to the left. • There are two types of OC Curves. They are Type-A OC- Curve and Type-B OC- Curve. Pa Pa Lot quality product quality

7. SAMPLING PLANS: These are developed by “Harold F.Dodge “and Harry.G.Romig at the bell telephone laboratories and used before world war-II. • SINGLE SAMPLING PLAN : • In this case, the decision of acceptance or rejection of a lot is taken based on one sample only. • Let N be a lot size from which a random sample of size ‘n’ is selected. • Let ‘X’ be the number of defectives in the sample and • let C be the maximum allowable acceptance no. of defectives in the sample. • If X ≤C then we accept the lot . • If X > C then we reject the lot.

8. Inspect n X Compare x c Accept lot X≤ c X>c Reject lot

9. DESIGNING OF A SINGLE SAMPLING PLAN DETERMINATION OF ‘n’ AND ‘c’ : In a lot of incoming quality ‘p’, the probability of obtaining ‘x’ defectives in a sample of size ‘n’ follows Binomial distribution. If p is the fraction defective operating at the time of observation and is the fraction defective given by the producer such that if the quality is better than i.e., p< , the producer expects his product to be accepted with a probability > 1- α.

10. Hence, the total probability of accepting a product with the fraction defective and with the number of defectives less than or equal to c is given by = > 1- α ….. (1) • If the sample is large (n>20) then it may be approximated by Poisson distribution. • If the sample is very small (n<5) then it may be approximated by Hyper geometric distribution.

11. Similarly , if p2 is the fraction defective given by the consumer such that if the quality is worse than p2 i.e., p> p2, the consumer wants to reject the product . Hence, the total probability of accepting a product with the fraction defective p2 and with the number of defectives less than or equal to c is given by Pc=P {Accepting a lot of quality p2} = ; p2 >p. P should always lie between p1 and p2. ..(2)

12. Example: If p1 = 0.01, p2 = 0.08, α= 0.05 and β= 0.10. From Molina's tables calculate n p1 and n p2 for various values of c and obtain the ratios of np2 to n p1 . Let us denote it by R. Also obtain the ratio of p2 to p1, which will be 8 . Check between what values of R the above 8 lies. C1 corresponds to the upper value of R and C2 to the lower value of R. For the above data 8 lies between C1 =1 and C2=2. The consumer can select c1=1 , if heis very particular about the quality of the product or he can take c2=2 if he is not very particular about the quality. If the consumer selects c1=1 then n can be obtained by dividing n p1 for the selected c by 0.01. i.e., n=0.35 /0.01 which is approximately 36. If he selects C2=2 then n=0.81/0.01 which is approximately 82.

14. O.C.CURVE The o. c. curve for the incoming quality ‘p ‘ is given by when p<0.10,a good approximation can be given by binomial expansion[(1-n/N)+n/N]Np i.e., = when p<0.10 and also n/N<0.10,a good approximation is given by the Poisson distribution

15. In the above example when c=1 and n=49 the oc curve can be drawn in the following way :

16. Graph: ASN: For a single sampling plan ASN is a constant i.e., the sample size ‘n’

17. DOUBLE SAMPLING PLAN Inthis method, if the decision about accepting or rejecting a lot based upon the first sample is non-conclusive then we can draw a second sample and based upon the second sample a decision is taken. Let N be the Lot size. Let n1 be the size of the first sample with c1 as the acceptance number of defectives . Let X1 be the number of defective items in the first sample. Similarly, let n2 be the size of the second sample with c2 as the acceptance number of defectives for the combined sample. Let X2 be the number of defective items in the second sample.

18. PROCEDURE: • Take a sample of size n1 from the lot of size N. • If X1≤c1,accept the lot. • If X1>c2,reject the whole lot. • If c1+1<X1≤c2 , take a second sample of size n2 from the remaining lot. • If X1 + X2≤c2 , accept the lot. • If X1 + X2 >c2, reject the whole lot.

19. Inspect n1 X1 Compare X1 c1,c2 X1>C1 X1≤C1 C1+1<X1≤C2 Inspect n2 Accept lot Reject lot X=X1+X2 compareX c2 X ≤C2 X>C2

20. DESIGNING OF A DOUBLE SAMPLING PLAN DETERMINATION OF n1, n2,c1,c2 and c3: Let n1 be the size of the first sample. Accept the first sample if X1≤c1 Reject the first sample if X1>c2 . Take a second sample of size n2 ; ifc1< X1 ≤c2 Accept if X1+X2 ≤c3 Reject if X1+X2 > c3 We have five variables which are to be found from two equations . ASN: In an acceptance rejection double sampling plan , the number of items inspected for a lot is either n1, when a lot is accepted or rejected on the basis of the first sample.

21. EXAMPLE • Let p1=0.01, p2=0.08,α=0.05 and β=0.10 • STEPS: • Calculate R= / = 0.08 / 0.01 =8 • From the available tables with n2=kn1 k= 1 or 2 , read the corresponding values of c1 and c2 =2. • If we decided to hold ( ,β) point then n1 =(n ) 0.10/ p2 • =2.42/0.08 • =30 • If k=2, then n2=2*30=60 • ASN=30*1.511=45, where 1.511 is the ASN divided by n 1value obtained from the tables corresponding to the selected plan. p2 p1 p2 p1

22. O.C.CURVE OF DOUBLE SAMPLING PLAN: The lot will be accepted under the following mutually exclusive ways: • 0≤ d1≤c1; • X1=c1+1; X2≤c2- (c1+1) • X1=c1+2; X2≤c2- (c1+2) • . • . • . • X1=c2,X2=0 • By the addition theorem of probability , the probability of acceptance for a lot of incoming quality ‘p’ is given by • Pa(p)= • where g(x, p) is the probability of finding x defectives in the firstsample and h(y, p/x) is the conditional probability of finding y defectives in the second sample under the condition that x defectives have already appeared in the first sample. Thus, • g(x,p)= • h(y,p/x)=

23. Pa(p)= + ..(a1) =Pa1(p)+ Pa2(p), (say) Where Pa1 and Pa2 , are the probabilities of acceptance on the basis of first and second sample respectively.

24. III II pa I p • Curve –I represents the probability of acceptance on first sample. • Curve –II represents the probability of acceptance based on combined samples. • Curve –III represents the probability of Rejection on first sample

25. Example Let n1=50 , n2=100 , c = 1, c2=3 and p=0.05. Let pai andpaii denoted the probabilities of acceptance on the first and second samples respectively. Hence the probability of acceptance on the combined samples is pa= pai + paii Pai = P[accepting X1≤ c1] or the given p=0.05 pai = 0.279 We draw a second sample if c1 < X1 ≤ c2. The lot is accepted based on the second sample if ; =

26. We draw a second sample if c1 < X1 ≤ c2. The lot is accepted based on the second sample if ; Case-(i):- X1=2 and X2=o or 1. i.e., p[X1=2,X2≤1] = p[X1=2] p[X2 ≤ 1] = [ ] [ ] =0.009 CASE-(ii): X1=3 and X2=0 On similar lines as above we get , P[X1=3 , X2=0]= 0.001 Thus, the probability of acceptance based n the second sample is paii=p[X1=2 , X2≤ 1] + p[X1=3 , X2=0]= 0.010

27. The probability of acceptance of this lot which has a fraction defective 0.05 is • pa= pai + paii • =0.279+0.010 • =0.289 • Similarly the other points on the OC- curve can be calculated. • Advantages of double Sampling plan over single sampling plan: • Psychological advantage of giving the lot a second chance. • Sometimes t is also possible to reject the lot without inspecting the entire second sample.

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