UNIT 7 SINGLE SAMPLING PLANS

UNIT 7 SINGLE SAMPLING PLANS Structure 7. Introduction Objectives 7. Single Smpling Pln 7.3 Operting Chrcteristics (OC) Curve 7.4 Producer s Risk nd Consumer s Risk 7.5 Averge Outgoing Qulity (AOQ) 7.6 Averge Smple Number (ASN) nd Averge Totl Inspection (ATI) 7.7 Design of Single Smpling Plns Stipulted Producer s Risk Stipulted Consumer s Risk Stipulted Producer s Risk nd Consumer s Risk Lrson Binomil Nomogrph 7.8 Summry 7.9 Solutions/Answers 7. INTRODUCTION In Units 5 nd 6, you hve lernt the vrious fetures of smpling inspection pln such s AQL, LTPD, producer s risk, consumer s risk, OC curve, ASN, ATI, etc. The sme fetures chrcterise different types of smpling plns. In Unit 5, you hve lso lernt tht the min types of cceptnce smpling plns for ttributes re: i) Single smpling pln, ii) Double smpling pln, iii) Multiple smpling pln, nd iv) Sequentil smpling pln In this unit, we focus on the single smpling plns for ttributes. In Sec. 7., we explin the single smpling pln nd its implementtion. We describe vrious fetures of the single smpling pln such s the operting chrcteristic (OC) curve, producer s risk, consumer s risk, verge smple number (ASN) nd verge totl inspection (ATI) in Secs. 7.3 to 7.6. Finlly, we describe the design of the single smpling plns in Sec. 7.7. In the next unit, we shll discuss the double smpling plns for ttributes. Objectives After studying this unit, you should be ble to: describe single smpling pln; compute the probbility of ccepting or rejecting lot of given incoming qulity in single smpling pln; construct the operting chrcteristics (OC) curve of single smpling pln; compute the consumer s risk nd producer s risk in single smpling pln;

Product Control compute the verge smple number (ASN) nd the verge totl inspection (ATI) for single smpling pln; nd design single smpling plns. 7. SINGLE SAMPLING PLAN A smpling pln in which decision bout the cceptnce or rejection of lot is bsed on single smple tht hs been inspected is known s single smpling pln. For exmple, suppose buyer purchses cricket blls in lots of 500 from compny mnufcturing cricket blls. To check the qulity of the lots, the buyer drws rndom smple of size 0 from ech lot nd tkes decision bout ccepting or rejecting of the lot on the bsis of the informtion provided by this smple. Since the buyer tkes the decision bout the lot on the bsis of single smple, this smpling pln is single smpling pln. A single smpling pln requires the specifiction of two quntities which re known s prmeters of the single smpling pln. These prmeters re n size of the smple, nd c cceptnce number for the smple. Let us suppose tht the lots re of the sme size (N) nd re submitted for inspection one t time. The procedure for implementing the single smpling pln to rrive t decision bout the lot is described in the following steps: Step : We drw rndom smple of size n from the lot received from the supplier or the finl ssembly. Step : We inspect ech nd every unit of the smple nd clssify it s defective or non-defective. At the end of the inspection, we count the number of defective units found in the smple. Suppose the number of defective units found in the smple is d. Step 3: We compre the number of defective units (d) found in the smple with the stted cceptnce number (c). Step 4: We tke the decision of cceptnce or rejection of the lot on the bsis of the smple s follows: Under cceptnce smpling pln If the number of defective units (d) in the smple is less thn or equl to the stted cceptnce number (c), i.e., if d c, we ccept the lot nd if d > c, we reject the lot. Under rectifying smpling pln If d c, we ccept the lot nd replce ll defective units found in the smple by non-defective units nd if d > c, we ccept the lot fter inspecting the entire lot nd replcing ll defective units in the lot by non-defective units. The steps described bove re shown in Fig. 7..

Single Smpling Plns Fig. 7.: Procedure for implementing single smpling pln. Let us explin these steps further with the help of n exmple. Exmple : Suppose mobile phone compny produces mobiles phones in lots of 00 phones. To check the qulity of the lots, the qulity inspector of the compny uses single smpling pln with n = 5 nd c =. Explin the procedure for implementing it. Solution: For implementing the single smpling pln, the qulity inspector of the compny rndomly drws smple of 5 mobile phones from ech lot nd clssifies ech mobile of the smple s defective or non-defective. At the end of the inspection, he/she counts the number of defective mobiles (d) found in the smple nd compres it with the cceptnce number (c). If d c (= ), he/she ccepts the lot nd if d > c (= ), he/she rejects the lot under the cceptnce smpling pln. Under rectifying smpling pln, if d c (= ), he/she ccepts the lot by replcing ll defective mobiles found in the smple by non-defective mobiles nd if d > c, he/she ccepts the lot by inspecting the entire lot nd replcing ll defective mobiles in the lot by non-defective mobiles. You my like to explin the procedure for implementing single smpling pln yourself. Try the following exercise. E) A mnufcturer of silicon chip produces lots of 000 chips for shipment. A buyer uses single smpling pln with n = 50 nd c = to test for bd outgoing lots. Explin the procedure for implementing it under cceptnce smpling pln. So fr you hve lernt bout the single smpling pln nd how it is implemented. We describe vrious fetures of the single smpling pln in Secs.7.3 to 7.6. 7.3 OPERATING CHARACTERISTIC (OC) CURVE

Product Control You hve studied in Unit 3 of MST-003 tht if A nd B re mutully exclusive events then P AorB P A P B The nottion Np C cn lso be repersened s Np. x 4 x You hve studied in Unit 6 tht the operting chrcteristics (OC) curve is n importnt spect of n cceptnce smpling pln. This curve displys the discrimintory power of the smpling pln. Tht is, it shows the probbility tht lot submitted with certin frction defective will be either ccepted or rejected. In this section, we discuss how to construct the OC curve for single smpling pln. You hve lernt tht for constructing n OC curve, we require the probbilities of ccepting lot corresponding to different qulity levels. Therefore, we first compute the probbility of ccepting lot of incoming qulity p for single smpling pln. You hve studied in Sec. 7. tht in single smple pln, we ccept the lot if the number of defective units (d) in the smple is less thn or equl to the cceptnce number (c). It mens tht if X represents the number of defective units in the smple, we ccept the lot if X c, i.e., X = 0 or or,, or c. Therefore, the probbility of ccepting the lot of incoming qulity p is given by P p P X c P X 0 or,..., or c X 0,,,..., c re PX 0 PX... PX c mutully exclusive c PX x () x0 We cn clculte this probbility if we know the distribution of X. Generlly, in qulity control, rndom smple is drwn from lot of finite size without replcement. So in such situtions, the number of defective units (X) in the smple follows hypergeometric distribution. In lot of size N nd incoming qulity p, the number of defective units is Np nd the number of non-defective units is N Np. Therefore, the probbility of getting exctly x defective units in smple of size n from this lot is given by Np NNp Cx Cnx PX x ; x 0,,..., min N Np,n () C x Thus, we cn obtin the probbility of ccepting lot of qulity p in single smpling pln by putting the vlue of P[X = x] in eqution () s follows: c c Np C NNp x nx (3) N P p P X x C x0 x0 x We know from industril experience tht n is usully smll for ny economiclly worthwhile production process. Therefore, when smple size n is smll compred to lot size (N), i.e., when N 0n, we know tht the hypergeometric distribution is pproximted by the binomil distribution with prmeters n nd p where p is the lot qulity. It is fr esier to clculte the probbilities with the help of the binomil distribution in comprison with the hypergeometric distribution. Therefore, we cn tke P X x C p p x Thus, the probbility of ccepting lot of qulity p using binomil pproximtion is given by C

c c x (4) P p P X x C p p x0 x0 However, for rpid clcultion, we cn use Tble I entitled Cumultive Binomil Probbility Distribution which is given t the end of this block. We cn lso pproximte the binomil distribution to nother distribution. When p is smll nd n is lrge such tht np is finite, we know tht the binomil distribution pproches the Poisson distribution with prmeter λ = np. Therefore, the probbility of ccepting lot of qulity p using the Poisson pproximtion is given by P p c x e (5) x x0 We cn use Tble II entitled Cumultive Poisson Probbility Distribution given t the end of this block for clculting this probbility. Note: Tble I nd Tble II do not list tbulted vlue for ech vlue of p nd λ, respectively. In such cses, we interpolte it s we hve discussed in Unit 4 of MST-004 or use scientific clcultor to clculte the probbility of ccepting lot. We now illustrte how to compute the probbility of ccepting lot in different situtions with the help of n exmple. Exmple : A mnufcturer of silicon chip produces chips in lots of 000. A single smpling pln is used to test for bd outgoing lots. If the qulity of incoming lot is 0.0, clculte the probbility of ccepting the lot in the following cses: i) n = nd c =, nd ii) n = 60 nd c =. Solution: i) It is given tht N 000, n, c nd p 0.0 If X represents the number of defective chips in the smple, the lot is ccepted if X c, i.e., X. Therefore, the probbility of ccepting the lot is given by P p P X c P X P X 0 P X Single Smpling Plns For Poisson distribution x e PX x x Np NNp x N x0 x0 Cx PX x Since N 0n, we cn use the binomil distribution (with prmeters n nd p where p is the lot qulity) s the pproximtion of the hypergeometric distribution. Therefore, the probbility of ccepting the lot of qulity p is given by x P p P X C p p x0 For rpid clcultion, we cn use Tble I for obtining this probbility. C C nx From Tble I, for n =, x = c = nd p = 0.0, we hve 5

Product Control nx Cx p p 0.9938 x0 Therefore, the probbility of ccepting the lot of qulity p = 0.0 is given by ii) It is given tht x x0 n x P p P X C p p 0.9938 N 000, n 60,c nd p 0.0 Since p is smll, n is lrge nd np = 0.65 is finite, we cn use the Poisson distribution (with prmeters λ = np) s the pproximtion of the hypergeometric distribution. Therefore, the probbility of ccepting the lot of qulity p is given by e P p PX x x0 where np 60 0.0 0.60. x For rpid clcultion, we cn use Tble II for obtining this probbility. From Tble II, for λ = 0. 60 nd x = c =, we hve x0 e x x 0.9769 Hence, the probbility of ccepting lot of qulity p = 0.0 is given by x e P p PX 0.9769 x x 0 You cn clculte the probbility of cceptnce of lot for different lot qulities in the sme wy. We now tke up the construction of the OC curve for single smpling pln. As you know, the OC curve is constructed by tking the qulity level (proportion defective) on the X-xis nd the probbility of ccepting the lot on the Y-xis. So for construction of the OC curve, we first consider different qulity levels such s p = 0.0, 0.0, 0.03 nd then clculte the corresponding probbility of cceptning lot s discussed in Exmple. Let us consider n exmple to demonstrte the construction of the OC curve. Exmple 3: Suppose consumer receives lots of 500 cndles from new supplier. To check the qulity of the lot, the consumer drws one smple of size 0 nd ccepts the lot if the inspected smple contins t most one defective cndle. Otherwise, he/she rejects the lot. Construct the OC curve for this pln. Solution: It is given tht N 500, n 0, c For constructing the OC curve, we hve to clculte the probbilities of ccepting the lot corresponding to different qulity levels. 6

If X represents the number of defective cndles in the smple, the consumer ccepts the lot if X c, i.e., X. Therefore, the probbility of ccepting the lot is given by Single Smpling Plns P p P X c P X P X x x0 Since N 0n, we cn use binomil distribution. Therefore, x P p P X C p p x0 We use Tble I for clculting the probbilities of ccepting the lot corresponding to different qulity levels such s p = 0.0, 0.0, 0.03 s we hve discussed in Exmple. These probbilities re shown in the tble given below: Incoming Lot Qulity Probbility of Accepting the Lot 0 0.0 0.983 0.0 0.940 0.04 0.803 0.06 0.6605 0.08 0.569 0.0 0.397 0. 0.89 0.4 0.084 0.6 0.47 0.8 0.08 0.0 0.069 We now construct the OC curve by tking the qulity level (proportion defective) on the X-xis nd the probbility of ccepting the lot on the Y-xis s shown in Fig. 7.. Probbility of Accepting the lot P (p) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0.00 0.0 0.04 0.06 0.08 0.0 0. 0.4 0.6 0.8 0.0 Proportion Defective (p) Fig. 7.: The OC curve for the single smpling pln. You my now like to construct the OC curve for the following exercise.

Product Control E) A hospitl receives disposble injection syringes in lots of 000. A single smpling pln with n = 5 nd c = is being used for inspection by the qulity inspector of the hospitl. Construct the OC curve for this pln. You hve lernt in Unit 5 tht the cceptnce or rejection of the entire lot depends on the conclusions drwn from the smple. Thus, there is lwys chnce of mking wrong decision. It mens tht lot of good qulity my be rejected nd lot of poor qulity my be ccepted. This leds to two kinds of risks:. Producer s risk, nd. Consumer s risk. We now discuss these risks for the single smpling pln. 7.4 PRODUCER S RISK AND CONSUMER S RISK In Unit 5, we hve defined the producer s risk s follows: The probbility of rejecting lot of cceptnce qulity level (AQL) p is known s the producer s risk. Therefore, the producer s risk for single smpling pln is given by P P rejecting lot of cceptnce qulity level p p P ccepting lot of cceptnce qulity level p P p (6) We cn compute P p from eqution (3) by replcing the qulity level p with p s follows: c c Np NNp x N x0 x0 Cx P p P X x C C nx Therefore, from eqution (6), the producer s risk is given by C C c Np NNp x nx p (7) N x0 Cx P P p For rpid clcultion of the producer s risk for single smpling pln, we cn lso use pproximtions s we hve discussed in Sec. 7.4. Therefore, if we use the pproximtion of the hypergeometric distribution to the binomil distribution with prmeters n nd p, the producer s risk is given by c nx p x (8) x0 P P p C p p We now explin the Consumer s Risk for single smpling pln: By definition, the probbility of ccepting lot of unstisfctory qulity (LTPD) p is known s the consumer s risk. Therefore, the consumer s risk for single smpling pln is given by P P ccepting lot of qulityp c 8

We cn compute the consumer s risk for the single smpling pln from eqution (4) by replcing the qulity level p with p s follows: Single Smpling Plns c c Np NNp x nx (9) P P p P X x c N x0 x0 Cx If we pproximte the hypergeometric distribution to the binomil distribution with prmeters n nd p, the consumer s risk is given by C c nx c x (0) x0 P P p C p p Let us tke up n exmple from rel life to explin these concepts. Exmple 4: Suppose tyre supplier ships tyres in lots of size 400 to the buyer. A single smpling pln with n = 5 nd c = 0 is being used for the lot inspection. The supplier nd the buyer s qulity control inspector decide tht AQL = 0.0 nd LTPD = 0.0. Compute the producer s risk nd consumer s risk for this single smpling pln. Solution: It is given tht N 400, n 5,c 0, AQL(p ) 0.0 nd LTPD(p ) 0.0 Since N 0n, we cn use the binomil distribution. Therefore, we cn use eqution (8) to clculte the producer s risk for the single smpling pln. We first clculte the probbility of ccepting lot of qulity p = p = 0.0 using Tble I. From Tble I, for n = 5, x = c = 0 nd p = p = 0.0, we hve 0 x x0 P p P X 0 C p p 0.860 Therefore, from eqution (8), we clculte the producer s risk s follows: p P P p 0.860 0.399 It mens tht if there re severl lots of the sme qulity p = 0.0, bout 3.99% of these will be rejected. This is obviously risk for the supplier becuse it ws greed upon by both tht lots of qulity 0.0 will be ccepted wheres the qulity inspector is rejecting 3.99% of those. Similrly, we cn clculte the consumer s risk using eqution (0). We first clculte the probbility of ccepting lot of qulity p = p = 0.0 using Tble I. From Tble I, for n = 5, x = c = 0 nd p = 0.0, we hve x x0 0 n x P p P X 0 C p p 0.059 Therefore, from eqution (0), the consumer s risk is given by P P p 0.059 c It mens tht if there re severl lots of the sme qulity p = 0.0, bout 0.59% of these will be ccepted by the qulity inspector even though this qulity is unstisfctory. This is obviously the buyer s risk. C 9

Product Control For prctice you cn lso compute the producer s risk nd consumer s risk in the following exercise. E3) Suppose in E, the cceptnce qulity level (AQL) nd lot tolernce percent defective (LTPD) re 0.04 nd 0.0, respectively. Clculte the producer s risk nd consumer s risk for this pln. 7.5 AVERAGE OUTGOING QUALITY (AOQ) You hve studied in Unit 6 tht the concept of verge outgoing qulity is prticulrly useful in the rectifying smpling pln where the rejected lots re inspected 00% nd ll defective units re replced by non-defective units. The AOQ is defined s follows: The expected qulity of the lots fter the pplictions of smpling inspection is clled the verge outgoing qulity (AOQ). It is clculted from the formul given below: Number of defective units in the lot fter the inspection AOQ () Totl number of units in the lot So in single smple pln, we cn obtin the formul for verge outgoing qulity by considering the following situtions: i) If the lot of size N is ccepted on the bsis of smple of size n, (N n) units remin un-inspected. If the incoming qulity of the lot is p, we expect tht p(n n) defective units re left in the lot fter the inspection. However, the probbility tht the lot will be ccepted by the smpling pln is P. Therefore, the expected number of defective units per lot in the outgoing stge is p(n n) P. ii) If the lot is rejected, ll units of the lot go for 00% inspection nd ll defective units found in the lot re replced by non-defective units. So there is no defective unit t the outgoing stge. The probbility tht the lot will be rejected is ( P ). Therefore, the expected number of defective units per 0 P 0. lot t the outgoing stge is Thus, the expected number of defective units per lot fter smpling inspection p N n P 0 p N n P. is Hence, the verge proportion defective in the outgoing stge or verge outgoing qulity (AOQ) is given by or Number of defective units in the lot fter the inspection AOQ Totl number of units in the lot p N n P AOQ () N If the smple size n is very smll in proportion to the lot size N, i.e., n / N 0, eqution () for AOQ becomes n AOQ p P pp N (3) 0

Let us now discuss the construction of the AOQ curve for single smpling pln. As you know, the AOQ curve is constructed by tking the qulity level (proportion defective) on the X-xis nd the AOQ on the Y-xis. So for constructing the AOQ curve, we first consider different qulity levels such s p = 0.0, 0.0, 0.03 nd then clculte the corresponding AOQ. Let us consider some exmples for clculting AOQ nd constructing the AOQ curve. Exmple 5: Suppose in Exmple 3 the submitted lot qulity is p = 0.0. The rejected lots re screened nd ll defective cndles re replced by the non-defective cndles. Clculte the verge outgoing qulity (AOQ) for this pln. Solution: The submitted lot qulity is p = 0.0 nd we hve to clculte the AOQ for this single smpling pln. It is given tht N 500, n 0, c nd p 0.0 From eqution (), the AOQ for the single smpling pln is AOQ p N n P N where P is the probbility of ccepting the lot of qulity p. Therefore, for clculting AOQ, we hve to clculte P. We hve lredy clculted this probbility in Exmple 3. So we directly use the result: P 0.940 On putting the vlues of N = 500, n = 0, p = 0.0 nd P = 0.940in eqution (), we get p N n P 0.0 500 0 0.940 AOQ 0.08 N 500 In the sme wy, you cn clculte AOQ for different submitted lot qulities. Exmple 6: Suppose in Exmple 4, the rejected lots re screened nd ll defective tyres re replced by non-defective tyres. Construct the AQO curve for this pln. Solution: It is given tht N 400, n 5, c 0 For constructing the AOQ curve, we first clculte the probbilities of ccepting the lot corresponding to different qulity levels using Tble I. Then we clculte the AOQ for ech qulity level using eqution (). The probbilities of ccepting the lot nd the AOQs corresponding to different qulity levels re given in the following tble: Incoming Lot Qulity Probbility of Accepting the Lot AOQ 0 0 0.0 0.860 0.0083 0.0 0.7386 0.04 Single Smpling Plns

Product Control 0.04 0.54 0.009 0.06 0.3953 0.08 0.08 0.863 0.00 0.0 0.059 0.098 0. 0.470 0.070 0.4 0.04 0.0099 0.6 0.073 0.0079 0.8 0.050 0.006 0.0 0.035 0.0000 We now construct the AOQ curve by tking the qulity level (proportion defective) on the X-xis nd the corresponding AOQ vlues on the Y-xis s shown in Fig.7.3. Averge Outgoing Qulity(AOQ) 0.05 0.0 0.05 0.0 0.005 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Fig. 7.3: The AOQ curve for Exmple 5. You my now like to clculte the AOQ nd construct the AOQ curve. Try the following exercises. E4) Assuming tht the lot size is lrge reltive to the smple size, clculte the pproximte verge outgoing qulity (AOQ) for the single smpling pln with n = 0 nd c = 0 contining 0% defective units. E5) A computer mnufcturer purchses computer chips from compny in lots of 00. Twelve computer chips re smpled t rndom nd inspected for defects. The computer mnufcturer ccepts the lot if the inspected smple contins t most one defective chip. Otherwise, he/she rejects the lot. If the rejected lots re screened nd ll defective computer chips re replced by non-defective chips. Construct the AOQ curve for this pln. You hve lernt bout the OC curve, producer s risk, consumer s risk nd AOQ for single smpling pln. We now discuss ASN nd ATI of the pln. 7.6 AVERAGE SAMPLE NUMBER (ASN) AND AVERAGE TOTAL INSPECTION (ATI) Two other fetures tht re lso useful for ny smpling pln re the verge smple number (ASN) nd the verge totl inspection (ATI). We now discuss these for single smpling pln in some detil. Averge Smple Number (ASN) Proportion Defective (p)

You hve studied in Unit 6 tht the verge smple number is the expected number of smple units per lot, which is required to rrive t decision bout the cceptnce or rejection of the lot under the cceptnce smpling pln. In cceptnce single smpling pln, the decision bout the cceptnce or rejection of lot is tken on the bsis of single smple tht hs been inspected. Therefore, the ASN in single smpling pln is simply the smple size n. It mens tht ASN is constnt in single smpling pln. Therefore, ASN n (4) The curve drwn between the ASN nd the lot qulity (p) is known s the ASN curve. The ASN curve for single smpling pln is stright line s shown in Fig. 7.4. Single Smpling Plns Fig. 7.4: The ASN curve for single smpling pln. Let us consider Exmple 4. In this exmple, the qulity control inspector tkes the decision of cceptnce or rejection of the lot on the bsis of the single smple of size n = 5. So the ASN for this single smpling pln is 5. Averge Totl Inspection (ATI) You know tht the concept of verge totl inspection (ATI) is considered under rectifying smpling pln in which rejected lots under go 00% inspection. It is defined s follows: The verge number of units inspected per lot under the rectifying smpling pln is clled the verge totl inspection (ATI). So in rectifying single smpling pln, the number of units to be inspected will depend on two situtions given below: i) If the lot of size N is ccepted on the bsis of smple of size n, the number of units inspected is n nd the probbility of the ccepting the lot is P. ii) If the lot is rejected on the bsis of smple, we inspect the entire lot of size N nd the probbility of rejecting the lot is ( P ). Therefore, we cn compute the ATI for single smpling pln s follows: ATI Averge number of units inspected per lot

Product Control (inspected number of units probbility of tking decision) ATI np N P This cn lso be written s ATI n P N n (5) The curve drwn between ATI nd lot qulity (p) is known s the ATI curve. Let us tke up n exmple to illustrte this concept. Exmple 7: Clculte the ASN for the pln given in Exmple 4. If the rejected lots re screened nd ll defective tyres re replced by non-defective tyres, construct the ATI curve for this pln. Solution: It is given tht N 400, n 5, c 0 In the first cse, the decision of cceptnce or rejection of the lot is tken only on single smple of size n = 5. Therefore, the ASN for this pln is simply the smple size, i.e., ASN = n = 5. For construction of the ATI curve, we first clculte the probbilities of ccepting the lot corresponding to different qulity levels using Tble I. Then we clculte the ATI for ech qulity level using eqution (5). We hve lredy clculted these probbilities in Exmple 6 nd we use those results. Substituting the vlues of N, n, p nd P in eqution (5) we cn clculte ATI. The probbilities of ccepting the lot nd the ATIs corresponding to different qulity levels re given in the following tble: Incoming Lot Qulity Probbility of Accepting the Lot ATI 0 5.00 0.0 0.860 68.86 0.0 0.7386 5.64 0.04 0.54 9.9 0.06 0.3953 47.8 0.08 0.863 89.77 0.0 0.059 30.73 0. 0.470 343.4 0.4 0.04 359.9 0.6 0.073 37.86 0.8 0.050 380.37 0.0 0.035 386.45 We now construct the ATI curve by tking the qulity level (proportion defective) on the X-xis nd the corresponding ATI vlues on the Y-xis s shown in Fig.7.5. 4

400 Single Smpling Plns Averge Totl Inspection (ATI) 350 300 50 00 50 00 50 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Proportion Defective (p) Fig. 7.5: The ATI curve for Exmple 6. You cn try the following exercise bsed on ASN nd ATI for prctice. E6) Clculte the ASN nd construct the ATI curve for the pln given in E5. So fr you hve lernt the vrious fetures of the single smpling pln. We now discuss how to design single smpling pln. 7.7 DESIGN OF SINGLE SAMPLING PLANS The design of single smpling pln implies the determintion of the prmeters of the pln, i.e., the smple size n nd the cceptnce number c. These numbers hve to be decided in dvnce before pplying the single smpling pln technique. There re severl pproches for determining the prmeters n nd c for the single smpling. Sometimes, more thn one pln will stisfy the criteri, but the best pln is the one with the lrgest smple size. It provides dequte protection to both producer nd consumer. We now discuss some of the pproches. 7.7. Stipulted Producer s Risk In this pproch, the producer s risk α nd its corresponding cceptnce qulity level (AQL) p re specified. We would like to design single smpling pln in such wy tht the lots of qulity level p re ccepted 00( α)% of the time. According to this pproch, we first select n cceptnce number (c) nd then find the vlue of np corresponding to c nd α with the help of Tble III given t the end of this block. Then the vlue of n is obtined by dividing np by p (= AQL) s follows: n np p (6) If the computed vlue of n is frction, it is rounded off to the next integer. For different vlues of the cceptnce number (c), we my get different vlues of smple size (n) (s you will see in Exmple 8). This mens tht for the sme

Product Control 6 producer s risk nd cceptnce qulity level, we hve different single smpling plns. If we drw the OC curve for ech smpling pln, we will see tht ech pln hs different consumer s risk. So out of these, we choose the smpling pln which gives the best protection to the consumer ginst cceptnce of poor qulity lots. We now describe this procedure with the help of n exmple. Exmple 8: Suppose tyre supplier ships tyres in lots of size 400 to the buyer. The supplier nd qulity control inspector of the buyer decide the cceptnce qulity level (AQL) to be %. Design smpling pln which ensures tht the lots of qulity % will be rejected 5% of the time. Solution: The supplier nd the qulity control inspector desire the smpling pln for which AQL (p ) = % nd the producer s risk (α) = 5%. To find the desired smpling pln, first of ll, we choose the cceptnce number c s c =. Then we look up the vlue of np corresponding to c = nd α = 0.05 from Tble III. We hve np 0.355 Therefore, we cn obtin the smple size s follows: np 0.355 p 0.0 n 7.75 8 Hence, the required single smpling pln is n 8 nd c Similrly, for c = nd α = 0.05, the vlue of np is 0.88. Therefore, the smple size is np 0.88 p 0.0 n 40.9 4 Hence, the required single smpling pln is n 4 nd c Similrly, for c = 5 nd α = 0.05, the vlue of np is.63. Therefore, np.63 p 0.0 n 30.65 3 Hence, the required single smpling pln is n 3 nd c 5 The OC curves for the three smpling plns (s discussed in Sec. 7.4) re shown in Fig. 7.6. Incoming Lot Probbility of Accepting the Lot Qulity n = 8 nd c = n = 4 nd c = n = 3 nd c = 5 0.0000.0000.0000 0.0 0.986 0.990 0.9978 0.0 0.9505 0.954 0.953 0.04 0.8393 0.7750 0.573 0.06 0.7055 0.5505 0.959

0.08 0.579 0.356 0.0445 0.0 0.4503 0.086 0.0075 0. 0.3460 0.56 0.000 0.4 0.60 0.0607 0.000 0.6 0.90 0.0303 0.0000 0.8 0.39 0.045 0.0000 0.0 0.099 0.0066 0.0000 Single Smpling Plns Probbility of Accepting the lot P (p) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Proportion Defective(p) Fig. 7.6: The OC curves for Exmple 8. From Fig. 7.6, we conclude tht out of the three smpling plns, the smpling pln with n = 3, c = 5 provides the best protection to the consumer becuse it hs the lowest probbility of ccepting poor qulity lots. However, this smpling pln hs the lrgest smple size (n = 3), which increses the inspection cost. 7.7. Stipulted Consumer s Risk According to this pproch, the consumer risk β nd its corresponding lot tolernce percent defective (LTPD) p re specified. We desire to determine single smpling pln in such wy tht we will be ccepted lots of qulity level p 00β% of the time. To design the pln using this pproch, we first select n cceptnce number (c) nd then find the vlue of np corresponding to c nd β with the help of Tble III. Then the vlue of n cn be obtined by dividing np by p LTPD s follows: n np p (7) For different vlues of c, number of smpling plns my stisfy this criterion. If we drw the OC curve for ech smpling pln, we will see tht ech pln hs different producer s risk. So out of these, we choose the smpling pln which gives the best protection to the producer ginst the rejection of good qulity lots. We now describe this procedure with the help of n exmple.

Product Control Exmple 9: Suppose, in Exmple 8, the supplier nd the qulity control inspector decide the lot tolernce percent defective (LTPD) to be 5%. Determine the single smpling plns using c =, nd 8 which ensure tht the lots of qulity 5% will be ccepted 0% of the time. Solution: It is given tht LTPD (p ) = 5% nd the consumer risk (β) = 0% To find the desired smpling pln, we first look up the vlue of np corresponding to c = nd β = 0.0 from Tble III. We hve np 3.890 Therefore, the smple size is np 3.890 p 0.0 n 38.90 39 Hence, the required single smpling pln is n 39, c Similrly, for c = nd β = 0.0, the vlue of np is 5.3. Therefore, the smple size is np 5.3 p 0.0 n 53. 54 Hence, the required single smpling pln is n 54, c Similrly, for c = 8 nd β = 0.0 the vlue of np is.995. Therefore, np.955 p 0.0 n 9.55 30 Hence, the required single smpling pln is n 30, c 8 7.7.3 Stipulted Producer s Risk nd Consumer s Risk According to this pproch, the producer s risk with its corresponding cceptnce qulity level (AQL) nd consumer s risk with its corresponding lot tolernce percent defective (LTPD) re specified. We hve to design single smpling plns which stisfy both producer s nd consumer s risks, such tht the lots of AQL re to be rejected no more thn 00α% of the time nd lots of LTPD re to be ccepted no more thn 00β% of the time. Here the criteri re more stringent thn the previous pproches nd we my not hve much flexibility in choosing the cceptnce number nd the ssocited smpling pln. So it my be difficult to design smpling pln tht will stisfy both producer s nd consumer s risks. In this pproch, we first find the operting rtio R s follows: R p p (8) 8

The vlues of R corresponding to vrious cceptnce number (c) nd α nd β re lso listed in Tble III. We choose vlue of R from Tble III which is exctly equl to the desired vlue of R corresponding to the desired α nd β. Generlly, the tbulted vlue of R is not equl to the desired vlue of R. So in such situtions, we tke the tbulted vlues of R between which the desired vlue of R lies. Then we look up the corresponding vlues of cceptnce number (c) in Tble III nd find the vlue of smple size (n). The following two pproches re used to find n:. Stisfy Producer s Risk Stipultion Exctly nd come Close to Consumer s Risk According to this pproch, we find the vlues of n s we hve discussed in Sec. 7.7.. It mens tht we first find the vlue of np corresponding to c nd α with the help of Tble III. Then we find n from eqution (6). Thus, np n p The vlues of n re obtined for both vlues of c. In this wy, we get two smpling plns which stisfy producer s risk exctly. Out of these, we choose the smpling pln which is close to stisfying the consumer s risk. For tht we find the vlue of p for ech pln. We find the vlues of np corresponding to the desired β nd ech c from Tble III. Then we find the vlue of p for ech pln s follows: p np n (9) We choose the smpling pln for which the clculted p is closer to the desired p. Another criterion for choosing the smpling pln is tht we select the smpling pln which hs the smllest smple size in order to minimize inspection costs. Alterntively, we cn choose the smpling pln which hs the lrgest smple size in order to get mximum informtion.. Stisfy Consumer s Risk Stipultion Exctly nd come Close to Producer s Risk According to this pproch, we find the vlues of n s we hve discussed in Sec. 7.7.. It mens tht we first find the vlue of np corresponding to c nd β with the help of Tble III. Then we find n from eqution (7). Thus, np n p The vlues of n re obtined for both vlues of c. In this wy, we get two smpling plns which stisfy consumer s risk exctly. Out of these, we choose the smpling pln which is close to stisfying the producer s risk. For tht we find the vlue of p for ech pln. We find the vlues of np corresponding to the desired α nd ech c from Tble III. Then we find the vlue of p for ech pln s follows: p np n (0) Single Smpling Plns 9

Product Control 0 We choose the smpling pln for which the clculted p is closer to the desired p. Another criterion for choosing the smpling pln is tht we select the smpling pln which hs the smllest smple size in order to minimize inspection costs. Alterntively, we cn choose the smpling pln which hs the lrgest smple size in order to get mximum informtion. We now describe this procedure with the help of n exmple. Exmple 0: Suppose, in Exmple 8, the supplier nd the qulity control inspector decide the cceptnce qulity level (AQL) to be % nd the lot tolernce percent defective (LTPD) to be 8%. Design smpling pln which ensures tht lots of qulity % will be rejected 5% of the time nd lots of qulity 8% will be ccepted 5% of the time. Solution: It is given tht AQL p % 0.0 nd 5% 0.05 LTPD p 8% 0.08 nd 5% 0.05 To design the desired smpling pln, we first clculte the operting rtio (R) from eqution (8) s follows: p 0.08 p 0.0 R 4.0 From Tble III, we see tht the desired vlue of R = 4.0 lies between 4.03 nd 3.604 for α = 0.05 nd β = 0.05. From Tble III, the corresponding cceptnce numbers (c) re 5 nd 6. We hve to find the vlue of smple size (n). We first find the plns which stisfy the desired producer s risk exctly. For this, we find the vlue of np corresponding to c = 5 nd α = 0.05 with the help of Tble III. From Tble III, for c = 5 nd α = 0.05, we hve np =.63. Therefore, from eqution (6), we hve np.63 p 0.0 n 30.65 3 Similrly, for c = 6 nd α = 0.05, the vlue of np is 3.86. Therefore, np 3.86 p 0.0 n 64.3 65 So both the plns with n = 3, c = 5 nd n = 65, c = 6 stisfy the producer s risk exctly. Out of these plns, we hve to find the pln which is closer to stisfying the desired consumer s risk. For this we find np corresponding to c nd β. From Tble III, for c = 5, β = 0.05, we hve np = 0.53. Therefore, from eqution (9), we hve np 0.53 n 3 p 0.08 Similrly, for c = 6, β = 0.05, we hve np =.84. Therefore,

np.84 n 65 p 0.07 Since the vlue of p = 0.08 corresponding to the pln n = 3, c = 5 is equl to the desired vlue 0.08, the pln with n = 3, c = 5 is the best single smpling pln. We now find the plns which stisfy the desired consumer s risk exctly. From Tble III, for c = 5 nd β = 0.05, we hve np = 0.53. Therefore, from eqution (7), we hve np 0.53 p 0.08 n 3.4 3 Similrly, for c = 6 nd β = 0.05, we hve np =.84. Therefore, np.84 p 0.08 n 48.0 49 So both the plns with n = 3, c = 5 nd n = 49, c = 6 re stisfied the consumer s risk exctly. Out of these plns, we hve to find the pln which is closer to the desired producer s risk. For this we find np corresponding to c nd α. From Tble III, for c = 5, α = 0.05, we hve np =.63 Therefore, np.63 n 3 p 0.098 Similrly, for c = 6, α = 0.05, we hve np = 3.86. Therefore, np 3.86 n 49 p 0.0 Since the vlue of p = 0.098 corresponding to the pln n = 3, c = 5 is pproximte equl to the desired vlue 0.0, the pln n = 3, c = 5 is the best single smpling pln. 7.7.4 Lrson Binomil Nomogrph There is lso grphicl method for designing the single smpling plns when the producer s risk with corresponding cceptnce qulity level (AQL) nd consumer s risk with corresponding lot tolernce percent defective (LTPD) re specified. This method is bsed on Lrson binomil nomogrph nd is used when we tke the binomil pproximte to the hypergeometric distribution, i.e., when N 0n. The Lrson binomil nomogrph (shown in Fig. 7.7) is grph of the cumultive binomil distribution. It hs two scles. On the left scle, proportion defective (p) is shown s probbility of occurences on single tril (p). This scle is known s p-scle. On the right scle, the probbility of cceptnce (P ) is shown s probbility of less thn or equl to c occurences in n trils (p). This scle is known s P -scle. Single Smpling Plns

Product Control Fig. 7.7: Lrson binomil nomogrph. The procedure of designing single smpling pln by using the nomogrph is quite simple. We plot the vlues of AQL (p ) nd LTPD (p ) on the left scle nd the corresponding vlues of ( α) nd β on the right scle. Then we join p with ( α) nd p with β by stright lines. We red the vlues of smple size (n) nd cceptnce number (c) t the intersection of the lines on the grid. To demonstrte this method, we consider Exmple 0. It is given tht AQL p % 0.0 nd 5% 0.05 LTPD p 8% 0.08 nd 5% 0.05 For designing single smpling pln by using Lrson nomogrph, we first plot p = 0.0 nd p = 0.08 on the p-scle on the nomogrph. Then we plot α (= 0.95) nd β = 0.05 on the P -scle on the nomogrph. After plotting these points, we drw stright line joining p (= 0.0) nd α (= 0.95) nd nother stright line joining p ( = 0.08) nd β ( = 0.05) s shown in Fig. 7.8. At the intersection of the two lines, we red the vlues of n nd c.

Single Smpling Plns Fig. 7.8 From Fig. 7.8 we hve, n = 35 nd c = 5. You cn now check your understnding of how to design single smpling pln by nswering the following exercises. E7) A bll bering supplier nd n utomobile compny hve decided to check the qulity of bll berings in lots of size 000 with cceptnce qulity level (AQL) s %. Design the single smpling plns using c =, 4 nd 6 which ensure tht lots of qulity % will be rejected % of the time. E8) A consumer receives lots of 5000 cndles from new supplier. To check the qulity of lots, the consumer nd supplier wnt to use the single smpling pln which stisfies consumer s risk of 5% for lots of qulity 5%. Determine smpling plns for the specified consumer s risk nd LTPD for cceptnce number c = 3 nd 6. We end this unit by giving summry of wht we hve covered in it. 7.8 SUMMARY. The min cceptnce smpling plns for ttributes re: i) Single smpling pln, ii) Double smpling pln,

Product Control iii) Multiple smpling pln, nd iv) Sequentil Smpling Pln.. A smpling pln in which decision bout the cceptnce or rejection of lot is bsed on single smple tht hs been inspected is known s single smpling pln. There re two prmeters of single smpling pln: n size of the smple, nd c cceptnce number for the smple. 3. In single smpling pln, if number of defective units (d) in the smple is less thn or equl to the stted cceptnce number (c), i.e., if d c, we ccept the lot nd if d > c, we reject the lot under cceptnce smpling pln. 4. In single smpling pln, if d c, we ccept the lot nd replce ll defective units found in the smple by non-defective units nd if d > c, we ccept the lot by inspecting the entire lot nd replcing ll defective units in the lot by non-defective units under rectifying smpling pln. 5. The probbility of ccepting lot of qulity p for single smpling pln is given by x P p P X c C p p c x0 6. The produce s risk nd consumer s risk for single smpling pln re given by c p x x x0 P P p C p p nd c nx c x x 0 P P p C p p 7. The AOQ for single smpling pln is AOQ p N n P 8. The ASN nd ATI for single smpling pln re N ASN n nd ATI n P Nn 9. Designing single smpling pln implies determining the smple size (n) nd cceptnce number (c). 7.9 SOLUTIONS/ANSWERS E) To check the qulity of the lots, the buyer rndomly drws 50 silicon chips from ech lot. After tht he/she inspects ech nd every chip drwn from the lot for certin defects nd clssifies ech chip of the smple s defective or non-defective. At the end of the inspection, he/she counts the number of defective chips (d) found in the smple nd then compres the number of defective chips (d) with the cceptnce number (c). If d c =, he/she ccepts the lot nd if d > c =, he/she rejects the lot on the bsis of the inspected smple. It mens tht if the 4

buyer finds 0 or or defective chips in the smple, he/she ccepts the lot. Otherwise, he/she rejects the lot. E) It is given tht N 000, n 5, c For constructing the OC curve, we hve to clculte the probbilities of ccepting the lot corresponding to different qulity levels. If X represents the number of defective syringes in the smple, the qulity inspector ccepts the lot if X c =. Therefore, the probbility of ccepting the lot is given by Single Smpling Plns P p P X c P X P X x Since N 0n, we use the binomil distribution. We cn use Tble I for clculting the probbilities of ccepting the lot corresponding to different qulity levels such s p = 0.0, 0.0, 0.03.These probbilities re shown in the tble given below: Incoming Lot Qulity x0 Probbility of Accepting the Lot 0 0.0 0.9980 0.0 0.9868 0.04 0.935 0.06 0.89 0.08 0.6768 0.0 0.537 0. 0.4088 0.4 0.3000 0.6 0.30 0.8 0.467 0.0 0.098 We construct the OC curve by tking the qulity level (proportion defective) on the X-xis nd the probbility of ccepting the lot on the Y-xis s shown in Fig. 7.9. 5

Product Control Probbility of Accepting the lot P (p) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Proportion Defective (p) E3) It is given tht Fig. 7.9: The OC curve for E. N 000, n 5, c, AOL(p ) 0.04 nd LTPD(p ) 0.0 Since N 0n, we use the binomil distribution. Therefore, we use eqution (8) to clculte the producer s risk for the single smpling pln. We first clculte the probbility of ccepting the lot of qulity p = p = AQL = 0.04 using Tble I. From Tble I, for n = 5, x = c = nd p = p = 0.04, we hve x x0 n x P p P X C p p 0.935 Therefore, from eqution (8), we hve p P P p 0.935 0.0765 It mens tht if there re severl lots of the sme qulity p = 0.04, bout 7.65% of these will be rejected. This is obviously risk for the supplier becuse it ws greed upon by both tht lots of qulity 0.04 will be ccepted wheres the qulity inspector is rejecting 7.65% of them. Similrly, we cn clculte the consumer s risk using eqution (0). We first clculte the probbility of ccepting the lot of qulity p = p = LTPD = 0.0 using Tble I. From Tble I, for n = 5, x = c = nd p = p = 0.0, we hve x x0 n x P p P X C p p 0.537 Therefore, from eqution (0), we hve the consumer s risk P P p 0.537 c

It mens tht if there re severl lots of the sme qulity p = 0.0, bout 53.7% out of these will be ccepted by the qulity inspector even though this qulity is unstisfctory. This is obviously risk for the qulity inspector. E4) It is given tht n 0, c 0, p 0% 0.0 Since the lot size is lrge reltive to the smple size, we cn clculte the verge outgoing qulity (AOQ) for the single smpling pln using eqution (3). For clculting AOQ, we hve to clculte the probbility of ccepting the lot corresponding to p = 0.0. If X represents the number of defective units in the smple, the lot is ccepted if X c = 0. Therefore, the probbility of ccepting the lot is given by P p P P X 0 Since the lot size is lrge reltive to the smple size, we cn clculte this probbility by using Tble I. From Tble I, for n = 0, x = c = 0 nd p = 0.0, we hve 0 x0 x n x P P X 0 C p p 0.074 On putting the vlues of p nd P in eqution (3), we get E5) It is given tht AOQ pp 0.0 0.074 0.05.5% N 00, n, c Since N 0n, we cn use the binomil distribution. Therefore, we use eqution () to clculte the AOQ. For constructing the AOQ curve, we hve to clculte the probbilities of ccepting the lot corresponding to different qulity levels. If X represents the number of defective chips in the smple, the mnufcturer ccepts the lot if X c =. Therefore, the probbility of ccepting the lot is given by P p P X P X x x 0 We cn use Tble I for clculting the probbilities of ccepting the lot corresponding to different qulity levels such s p = 0.0, 0.0, 0.03 Then we clculte AOQ for ech qulity level by using eqution (). The probbilities of ccepting the lot nd AOQs corresponding to different qulity levels re given in the following tble: Incoming Lot Qulity Probbilty of Accepting the Lot AOQ 0.0000 0 0.0 0.9938 0.0093 0.0 0.9769 0.084 Single Smpling Plns 7

Product Control 0.04 0.99 0.0346 0.06 0.8405 0.0474 0.08 0.753 0.0565 0.0 0.6590 0.069 0. 0.5686 0.064 0.4 0.4834 0.0636 0.6 0.4055 0.060 0.8 0.3359 0.0568 0.0 0.749 0.057 We construct the AOQ curve by tking the qulity level (proportion defective) on the X-xis nd the corresponding AOQ vlues on the Y-xis s shown in Fig.7.0. Averge Outgoing Qulity(AOQ) 0.06 0.05 0.04 0.03 0.0 0.0 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Proportion Defective(p) E6) It is given tht N 00, n, c Fig. 7.0: The AOQ curve for E5. In the first cse, the decision of cceptnce or rejection of the lot is tken only on single smple of size n =. Therefore, the ASN for this pln is simply the smple size, i.e., ASN = n =. For construction of the ATI curve, we first clculte the probbilities of ccepting the lot corresponding to different qulity levels using Tble I. Then we clculte the ATI for ech qulity level using eqution (5). We hve lredy clculted these probbilities in E5 nd we those results. Substituting the vlues of N, n, p nd P in eqution (5) we cn clculte ATI. The probbilities of ccepting the lot nd the ATIs corresponding to different qulity levels re given in the following tble: Incoming Probbility of ATI Lot Qulity Accepting the Lot 0.0000.00 0.0 0.9938 3.7 0.0 0.9769 6.34 0.04 0.99 7. 0.06 0.8405 4.99

0.08 0.753 58.76 0.0 0.6590 76. 0. 0.5686 93.0 0.4 0.4834 09. 0.6 0.4055 3.77 0.8 0.3359 36.85 0.0 0.749 48.3 Single Smpling Plns We now construct the ATI curve by tking the qulity level (proportion defective) on the X-xis nd the corresponding ATI vlues on the Y-xis s shown in Fig.7.. Averge Totl Inspection (ATI) 40 0 00 80 60 40 0 0 0 0.0 0.04 0.06 0.08 0. 0. 0.4 0.6 0.8 0. Proportion Defective (p) E7) We hve Fig. 7.: The ATI curve for E6. AQL % 0.0 nd % 0.0 To design the desired smpling pln, we first look up the vlue of np corresponding to c = nd α = 0.0 from Tble III. We hve np 0.436 Therefore, the smple size is np 0.436 p 0.0 n 43.6 44 Hence, the required single smpling pln is n 44 nd c Similrly, for c = 4 nd α = 0.0, the vlue of np is.79. Therefore, the smple size is np.79 p 0.0 n 7.9 8 Hence, the required single smpling pln is n 8 nd c 4 Similrly, for c = 6 nd α = 0.0, the vlue of np is.330.

Product Control Therefore, np.330 p 0.0 n 33 Hence, the required single smpling pln is E8) We hve n 33 nd c 6 LTPD p 5% 0.05 nd 5% 0.05 To design the desired smpling pln, we look up the vlue of np corresponding to c = 3 nd β = 0.05 from Tble III. We hve np 7.754 Therefore, the smple size is np 7.754 p 0.05 n 55.08 56 Hence, the required single smpling pln is n 56 nd c 3 Similrly, for c = 6 nd β = 0.05, the vlue of np is.84. Therefore, the smple size is np.84 p 0.05 n 36.84 37 Hence, the required single smpling pln is n 37 nd c 6 30