6_Attribute Control Charts

6_Attribute Control Charts - Spring 2008 Lecture Notes -...

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Spring 2008 Lecture Notes - EMSE 182, 282, Professor Harris, Attribute Control Charts - 1 1 CONTROL CHARTS FOR ATTRIBUTES CONTROL CHATS FOR FRACTION NONCONFORMING p Chart Preliminaries • Fraction Nonconforming œ # nonconforming #items • Conformance defined as a set of characteristics, must conform to all. • Could look at opposite which is process yield • Statistical model used is binomial distribution Pr{D d|n,p} p (1 p) , d 0, . .., n œœ œ ˆ‰ n d dn d • Assuming - production process is stable -Pr{an item is nonconforming} p œ E[D] np, VAR[D] np(1 p) npq œ • Sample fraction nonconforming P and the distribution of P ^^ œ D n is obtained as Pr{P p} Pr{D np | n,p} with ^ œ E[P] p and VAR[P] œ œ œ œ E[D] VAR[D] nn n np npq pq ##
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Spring 2008 Lecture Notes - EMSE 182, 282, Professor Harris, Attribute Control Charts - 2 2 p-chart • with Standards (p* known) plot p against CL p* 3 ^ i d nn p*(1 p*) œœ 3 É • with estimates select m 20 to 25 samples of size n and estimate œ p ( )/m ( ) ±± 3œ" 3œ" mm d nm p ^ 3 i plot p against CL p 3 ^ i d p (1 p ) 3 É N o t e s 3-sigma limits often used in practice Ä sometimes CL rounded to integer values Ä if estimated, the CL are trial limits and may have to be adjusted Ä if using target make sure that the target and true process value of p Ä are close pay attention to cvalues below LCL, they are usually signs of Ä recording error but consistancy may be siigns of process improvement
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Spring 2008 Lecture Notes - EMSE 182, 282, Professor Harris, Attribute Control Charts - 3 3 • Design approaches n should be large for the chart to be effective Ä when p is small, avoid 1 defect causing an out of control signal Ä if =Pr{D 1} n /p where is obtained from the #- -  Ê œ Poisson Tables as the smallest value of such that - Pr{D 0} 1 œŸ # Example p .001, .75 n 1.4/.001=1400 œœ Ê œ # if LCL > 0 p L 0 (L is usually 3) ÄÊ É p(1 p) n n L Ê 1p p 2 to detect shift p p ÄÄ $ p p L ( t h e U C L ) Ê $ É p(1 p) n L or n p(1 p) Ð Ñ $ É p(1 p) n L # $ Example if L 3 and p .02, then to detect a .01 shift in p œ¸ n p(1 p) n .02(.98) 1764 Ð Ñ œ œ L3 .01
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6_Attribute Control Charts - Spring 2008 Lecture Notes -...

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