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© 2007 Pearson Education   AQL LTPD Acceptance Sampling Plans Supplement I.

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Presentation on theme: "© 2007 Pearson Education   AQL LTPD Acceptance Sampling Plans Supplement I."— Presentation transcript:

1 © 2007 Pearson Education   AQL LTPD Acceptance Sampling Plans Supplement I

2 © 2007 Pearson Education Sequential Sampling Chart 8 8 – 7 7 – 6 6 – 5 5 – 4 4 – 3 3 – 2 2 – 1 1 – 0 0 – Reject Continue sampling Accept Cumulative sample size ||||||| Number of defectives

3 © 2007 Pearson Education Sequential Sampling Chart 8 8 – 7 7 – 6 6 – 5 5 – 4 4 – 3 3 – 2 2 – 1 1 – 0 0 – Reject Decision to reject Continue sampling Accept Cumulative sample size ||||||| Number of defectives

4 © 2007 Pearson Education Operating Characteristic Curve Ideal OC curve Probability of acceptance Proportion defective

5 © 2007 Pearson Education Operating Characteristic Curve Ideal OC curve Typical OC curve Probability of acceptance Proportion defective

6 © 2007 Pearson Education Operating Characteristic Curve  1.0  Ideal OC curve Typical OC curve AQL LTPD Probability of acceptance Proportion defective

7 © 2007 Pearson Education Constructing an OC Curve Example I – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance

8 © 2007 Pearson Education Probability Proportionof c or less defectivedefects (p)np(P a )Comments n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Constructing an OC Curve Example I.1

9 © 2007 Pearson Education Probability Proportionof c or less defectivedefects (p)np(P a )Comments n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance np Constructing an OC Curve Example I.1

10 © 2007 Pearson Education Probability Proportionof c or less Defectivedefects (p)np(P a )Comments 0.01 (AQL)0.6 n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance np Constructing an OC Curve Example I.1

11 © 2007 Pearson Education Probability Proportionof c or less defectivedefects (p)np(P a )Comments 0.01 (AQL) n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance np Constructing an OC Curve Example I.1

12 © 2007 Pearson Education Probability Proportionof c or less defectivedefects (p)np(P a )Comments 0.01 (AQL) n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance np Constructing an OC Curve Example I.1

13 © 2007 Pearson Education Probability Proportionof c or less defectivedefects (p)np(P a )Comments 0.01 (AQL) n = 60 c = – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Constructing an OC Curve Example I.1

14 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Probability Proportionof c or less defectivedefects (p)np(P a )Comments 0.01 (AQL)  = – = (LTPD)  = n = 60 c = 1 Constructing an OC Curve Example I.1

15 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Constructing an OC Curve Example I.1

16 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – ||||||||||  = (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance  = Constructing an OC Curve Example I.1

17 © 2007 Pearson Education Drawing the OC Curve Application I.1

18 © 2007 Pearson Education Finding  (probability of rejecting AQL quality: p =.03 np =5.79 Pa =Pa =  = 1 –.965 = Drawing the OC Curve Application I.1 Cumulative Poisson Probabilities

19 © 2007 Pearson Education Finding  (probability of accepting LTPD quality: p =.08 np =15.44 Pa =Pa = 0.10  = P a = 0.10 Drawing the OC Curve Application I.1 Cumulative Poisson Probabilities

20 © 2007 Pearson Education Drawing the OC Curve Application I.1

21 © 2007 Pearson Education Drawing the OC Curve Application I.1

22 © 2007 Pearson Education Understanding Changes in the OC Curve (with c = 1) 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk n (p = AQL)(p = LTPD)

23 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk n (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with c = 1)

24 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk n (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with c = 1)

25 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk n (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with c = 1)

26 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk n (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with c = 1)

27 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance n = 60, c = 1 n = 80, c = 1 n = 100, c = 1 n = 120, c = 1 Operating Characteristic Curves (with c = 1)

28 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk c (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with n = 60)

29 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk c (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with n = 60)

30 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk c (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with n = 60)

31 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk c (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with n = 60)

32 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance Producer’sConsumer’sRisk c (p = AQL)(p = LTPD) Understanding Changes in the OC Curve (with n = 60)

33 © 2007 Pearson Education 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance n = 60, c = 1 n = 60, c = 2 n = 60, c = 3 n = 60, c = 4 Operating Characteristic Curves (with n = 60)

34 © 2007 Pearson Education Acceptance Sampling Plan Data AQL BasedLTPD Based AcceptanceExpectedSampleExpectedSample NumberDefectivesSizeDefectivesSize

35 © 2007 Pearson Education Average Outgoing Quality Example I – 1.2 – 0.8 – 0.4 – 0 – |||||||| |||||||| Defectives in lot (percent) Average outgoing quality (percent) ProportionProbability Defectiveof Acceptance (p)np(P a ) = ( )/ = ( )/ = ( )/ For p = 0.01:0.01(0.974)(1000 – 110)/1000 = For p = 0.02:0.02(0.819)(1000 – 110)/1000 = For p = 0.03:0.03(0.581)(1000 – 110)/1000 = For p = 0.04:0.04(0.359)(1000 – 110)/1000 = For p = 0.05:0.05(0.202)(1000 – 110)/1000 = For p = 0.06:0.06(0.105)(1000 – 110)/1000 = For p = 0.07:0.07(0.052)(1000 – 110)/1000 = For p = 0.08:0.08(0.024)(1000 – 110)/1000 = Noise King uses rectified inspection for its single-sampling plan with n = 110, c = 3, N = 1000 Estimate the probabilities of acceptance for portion defective values from 0.01 to 0.08 in steps of 0.01

36 © 2007 Pearson Education Average Outgoing Quality Example I – 1.2 – 0.8 – 0.4 – 0 – |||||||| |||||||| Defectives in lot (percent) Average outgoing quality (percent) ProportionProbability Defectiveof Acceptance (p)np(P a ) = ( )/ = ( )/ = ( )/

37 © 2007 Pearson Education 1.6 – 1.2 – 0.8 – 0.4 – 0 – |||||||| |||||||| Defectives in lot (percent) Average outgoing quality (percent) Average Outgoing Quality Example I.2

38 © 2007 Pearson Education AOQL 1.6 – 1.2 – 0.8 – 0.4 – 0 – |||||||| |||||||| Defectives in lot (percent) Average outgoing quality (percent) Average Outgoing Quality Example I.2

39 © 2007 Pearson Education AOQ Calculations Application I.2 Management has selected the following parameters:  

40 © 2007 Pearson Education AOQ Calculations Application I.2

41 © 2007 Pearson Education Solved Problem 1.0 — 0.9 — 0.8 — 0.7 — 0.6 — 0.5 — 0.4 — 0.3 — 0.2 — 0.1 — 0 — |||||||||| Proportion defective (hundredths)(p) Probability of acceptance (P a ) (AQL)(LTPD)  =  = 0.049


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