Sampl Grn Blt Crtification Examination Qustions with Answrs (Grn Blt crtification xaminations assum that that th participant has succssfully compltd th Champion crtification xamination at th Univrsity of Miami. This sction only prsnts qustions byond th Champion crtification lvl. Howvr, Grn Blt crtification xaminations ar cumulativ in that thy covr th matrial rquird for both Champion and Grn Blt crtification.) QUESTION: Provid a non-tchnical dfinition for Six Sigma managmnt. It is an organizational initiativ dsignd to crat brakthrough improvmnts in manufacturing, srvic and administrativ procsss. For xampl, Motorola stablishd a goal to rduc dfcts 0-fold with a 50% rduction in cycl tim vry yars. QUESTION: Dfin a procss. Draw a pictur. A procss is th vhicl for transforming inputs into outputs, s figur blow. Fdback loops ar usd to mov data to appropriat points in th procss for dcision making purposs. Inputs Manpowr Srvics Matrials / Goods Mthods Environmnt Procss Transform inputs into outputs by crating tim, plac form valu Outputs Manpowr Srvics Matrials / Goods Mthods Environmnt Fdback Loops QUESTION: Dscrib th thr typs of fdback loops. (Non, Spcial only, Common and Spcial caus) No fdback loop: A procss without a fdback loop will dtriorat and dcay du to ntropy. Spcial caus only fdback loop: A procss in which all fdback is tratd as spcial will xhibit a doubling or xplosion in th variation of its output. Spcial and common caus fdback loop: A procss in which fdback is statistically rcognizd as common or spcial will xprinc improvmnt of its output. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
QUESTION: Explain th origin of th 6 and.4 in Six Sigma managmnt. Us a diagram that includs th Voic of th Customr and th Voic of th Procss. Voic of Customr quals Voic of Procss - Procss is stabl and cntrd on nominal. - 0.0 sigma shift in th man rsults in,700 DPMO. Voic of Customr quals Voic of Procss - Procss is stabl. -.5 sigma shift in th man rsults in 66,807 DPMO. Through th continuous procss improvmnt, th variation of procss is rducd. Thus, th rlationship btwn Voic of Customr and Voic of Procss movs to a bttr lvl, Voic of Procss is half th Voic of Customr, that is --- th procss improvs from a - sigma procss to a 6-sigma procss. Voic of Procss is half Voic of Customr - Procss is stabl and cntrd on nominal. - 0.0 sigma shift in th man rsults in dfcts pr billion opportunitis. Voic of Procss is half Voic of Customr - Procss is stabl. -.5 sigma shift in th man rsults in.4 DPMO. QUESTION: Dscrib th rols and rsponsibilitis of a procss ownr.. A Procss Ownr has th authority to chang a procss.. A Procss Ownr should b idntifid and dsignatd arly in a Six Sigma projct.. A Procss Ownr is rsponsibl for managing and holding th gains for th improvd procss, and for improving and innovating th procss in th futur. 4. A Procss Ownr mpowrs popl in th procss. 5. A Procss Ownr works with th projct tam. (This can also b a champion rol.) 6. A Procss Ownr coordinats tam logistics.. (This can also b a MBB or BB rol.) 7. A Procss Ownr ngotiats rsourcs for tam.. (This can also b a champion rol.) 8. A Procss Ownr links procss and organizational objctivs. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
A Procss Ownr undrstands thir procss capability and its rlationship to th organization. 9. A Procss Ownr nsurs that customr s nds tak priority. (This is also mainly th rsponsibility of th top xcutivs of an organization.) 0. Procss ownr optimizs th ntir procss, not just componnt of th procss. QUESTION: Dscrib th rols and rsponsibilitis of a champion.. A Champion should b a mmbr of th Excutiv Committ, or at last a trustd dirct rport of a mmbr of th Excutiv Committ. Champions tak a vry activ sponsorship and ladrship rol in implmnting Six Sigma projcts.. A Champion translats stratgic masurs into a projct (coordinats with Excutiv Committ).. A Champion slcts th tam ladr. 4. A Champion dvlops and ngotiats th projct chartr. 5. A Champion rmovs obstacls to th projct. 6. A Champion obtains rsourcs and hlps th tam control th budgt. 7. A Champion rviws tam progrss. 8. A Champion hlps kp th tam focusd. 9. A Champion assurs us of Six Sigma mthods and tools. 0. Champion is th liaison btwn xcutiv managmnt and th projct ladr. QUESTION: Dscrib th rols and rsponsibilitis of a black blt.. A Black Blt is a full-tim chang agnt who may not b an xprt in th procss undr study.. A Black Blt srvs as tam ladr.. A Black Blt is a mastr of th DMAIC modl. 4. A Black Blt provids guidanc and training to tam mmbrs. 5. A Black Blt hlps th tam rfin a projct chartr (assums Champion / Procss Ownr draftd th initial chartr) 6. A Black Blt intrfacs with th champion and MBB. 7. A Black Blt lads th projct tam. 8. A Black Blt hlps th tam analyz data and dsign xprimnts. 9. A Black Blt provids training in tools and tam functions. 0. A Black Blt hlps th tam prpar for rviws.. A Black Blt rcommnds stratgic Six Sigma projcts.. A Black Blt lads and coachs Grn Blts lading lowr-lvl tams. QUESTION: Dscrib th rols and rsponsibilitis of a Grn Blt.. A Grn Blt is a part-tim (5%) projct ladr or mmbr and provids most of th functions of a Black Blt (tam ladr) for lowr-lvl projct tams.. Grn Blts ar th work horss of stratgic Six Sigma Managmnt fforts. Most managrs should b or bcom Grn Blts.. For lowr lvl tams, a Grn Blt Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
4 Rfins a draft projct chartr (procss ownrs and champions should coauthor th first draft) Rviws th draft chartr with Champion Slcts tam mmbrs Facilitats th tam Communicats with th Champion and othr stakholdrs of th procss Analyzs data Provids training in basic tools Coordinats tam fforts with highr lvl tams Complts documntation Complts th control plan Sprads th lssons larnd QUESTION: Dscrib th rols and rsponsibilitis of a Mastr black blt A Mastr Black Blt is a provn ladr and chang agnt for Six Sigma managmnt. A Mastr Black Blt provids tchnical xprtis in Six Sigma managmnt.. Tachs Black Blts and Grn Blts.. Mntors Black Blts and Grn Blts.. Coordinats svral Black Blt projcts simultanously. 4. Improvs and innovats th Six Sigma procss. 5. Counsls top managmnt on Six Sigma managmnt. QUESTION: Crat a srvic xampl to xplain rolld throughput yild, DPO, DPMO, procss sigma. A srvic has 4 stps and ach stp has only on opportunity for a dfct. Th yild of stp is 0.99, th yild of stp is 0.95, th yild of stp is 0.99, and th yild of stp 4 is 0.97. Th four stps ar indpndnt of ach othr. What is th rolld throughput yild (RTY), th DPO, th DPMO, and th procss sigma? RTY=.99x.95x.99x.97=.90 DPO=.0-RTY=.097 DPMO=97,000 Procss Sigma=approx.8 QUESTION: Construct a tabl that shows how to prioritiz potntial Stratgic Six Sigma projcts basd on thir rlationships with businss objctivs. Explain how th tabl functions to accomplish its aim. Six Sigma Projct B O Projct Projct Projct BO W E W BO I G H W BOm T S Wm =wak =modrat Gitlow Lvin Wightd A Cours Avrag in Six of Sigma CTQs Managmnt4/9/007 9=strong
5 W i s ar dvlopd by th Financ Dpartmnt. Th sum of W i =.0. Cll valus ar dtrmind by tam mmbrs with th strong guidanc of th financ dpartmnt. Th wightd avrags that ar shown in th last row of th columns ar rankd from smallst to largst. Th largst avrag is considrd th highst priority projct. Altrnativly, a control chart could b usd to find a projct with a wightd avrag that is out of control on th high sid from th othr projct s avrags. QUESTION: Giv an xampl of a projct chartr. Labl ach of th 5 parts of th projct chartr. Dcras (dirction) of th numbr of customr complaints (CTQ masur) causd by at-hom rpairs (procss) from 0 pr day to 0 pr day (CTQ targt) by March, 004 (dadlin). QUESTION: Explain th trm SIPOC analysis. Construct a chart to illustrat your xplanation. A SIPOC analysis is a simpl tool for idntifying th Supplirs and thir Inputs into a Procss, th high lvl stps of a procss, th Outputs of th procss, and th Customrs (markt) sgmnts intrstd in th outputs. QUESTION: Dfin CTQ. Rlat your answr to a SIPOC analysis. CTQ stands for Critical to Quality charactristic. It is a charactristic of a procss, product or srvic that is critical to th satisfaction of a stakholdr. QUESTION: Dfin X in rspct to a SIPOC analysis. Xs ar th inputs and procss stps that caus variation in th outputs of a procss (CTQs). QUESTION: What is th tactic of Six Sigma managmnt in rspct to Xs? - Dfin vital fw Xs - Idntify lvl of critical Xs - Slct bst actions ndd to implmnt lvls of critical Xs - Implmnt critical Xs and tst for bst lvls of critical Xs that minimiz variation in th CTQs \QUESTION: Dfin must-b, on way, and attractiv quality that can rsult from a Kano survy. Mak sur your answr is basd on th dimnsions of prformanc and satisfaction. Draw a pictur to illustrat your xplanation. Must-b- Usr satisfaction is not proportional to th prformanc of th fatur. Th lowr th prformanc, th lowr th usr satisfaction, but high prformanc crats flings of indiffrnc to th fatur. On-way - Usr satisfaction is proportional to th prformanc of th fatur; th lss prformanc th lss usr satisfaction, and th mor prformanc, th mor usr satisfaction. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
6 Attractiv - Usr satisfaction is not proportional to th prformanc of th fatur; low lvls of prformanc crat flings of indiffrnc to th fatur, but high lvls of prformanc crat flings of dlight to th fatur. High Satisfaction Attractiv quality On-way quality Low Prformanc High Prformanc Must-b quality Low Satisfaction QUESTION: Provid an xampl of a qustion on a Kano survy. Column Column Column Column 4 CTQs Study Cntr in th dormitory How would you fl if th following CTQ wr prsnt? Dlightd [ ] Expct it and lik it [ ] No fling [ ] Liv with it [ ] Do not lik it [ ] Othr [ ] How would you fl if th CTQ wr not prsnt? Dlightd [ ] Expct it and lik it [ ] No fling [ ] Liv with it [ ] Do not lik it [ ] Othr [ ] What prcntag cost incras, ovr currnt costs, would you b willing to pay for this CTQ? 0% 0% 0% 0% 40% or mor QUESTION: Giv xampls of classification attribut data. Explain why ach xampl is classification attribut data. Exampls of classification attribut data: classifying mploys by dpartmnt, classifying cars by makrs, and so on. Each xampl is classification attribut data bcaus itms ar classifid into on of two or mor catgoris. QUESTION: Giv xampls of count (ara of opportunity) attribut data. Explain why ach xampl is ara of opportunity attribut data. Exampls of count attribut data: numbr of union grivancs pr wk, numbr of typ rrors in a pag, numbr of chocolat chips in a cooki. Each xampl is count attribut data bcaus it is from counts of th numbr of occurrncs pr unit. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
7 QUESTION: Giv xampls of masurmnt data. Explain why ach xampl is masurmnt data. Exampls of masurmnt data: cycl tim, wight, and tmpratur. Masurmnt data ar continuous data. QUESTION: What is th purpos of an oprational dfinition. An oprational dfinition promots ffctiv communication btwn popl by putting communicabl maning into an adjctiv. QUESTION: Crat an oprational dfinition for " pound bar" that rsults in attribut data. Critria: Slct a bar from invntory. Plac it on a digital scal. If th digital radout is btwn.999 and.00, inclusiv, thn th bar is classifid as pounds. If th digital radout is not btwn.999 and.00, inclusiv, thn th bar is classifid as not pounds. Tst: Slct a particular bar and put it on th scal. Rcord th digital radout. Dcision: If.999 digital radout.00, thn bar = pounds. If digital radout <.999 or digital radout >.00, thn bar pounds. QUESTION: Crat an oprational dfinition for " pound bar" that rsults in masurmnt data. Critria: Us a digital scal to wigh a bar. Tst: Slct a particular bar and put it on th digital scal. Rcord th digital radout. Dcision: Us bar basd on digital radout. QUESTION: A customr wants to buy chocolat bars with a nominal wight of 6.0 ouncs and will accpt a tolranc of 0.05 ouncs ithr sid of nominal. As a supplir of chocolat bars you want to undrstand your procss. Your procss producs chocolat rctangls that ar cut from largr blocks of chocolat and thn packagd as six-ounc bars. Evry 5 minuts, thr chocolat bars ar wighd, prior to packaging. Th chart blow shows th wights for ach bar xamind in a svn-hour day. Tim Obsrvation # Wight (oz) 9:5 9:0 9:45 6.0 5.99 6.0 5.98 5.99 6.0 6.0 6.0 6.0 6.0 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
8 0:00 6.0 6.0 0:5 6.00 5.99 6.0 0:0 5.99 6.00 6.00 0:45 6.0 6.0 6.00 :00 6.0 6.0 6.0 :5 6.0 6.0 6.00 :0 6.00 6.0 6.0 :45 6.0 6.0 :00 6.0 6.0 6.00 :5 6.0 6.0 :0 6.0 6.0 6.0 :45 6.0 6.0 6.0 :00 6.0 6.00 6.0 :5 6.0 6.0 :0 6.05 6.0 :45 6.0 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
9 :00 :5 :0 :45 :00 :5 :0 :45 4:00 6.0 6.0 6.0 6.0 6.05 6.0 6.06 6.0 6.05 6.0 6.05 6.0 6.06 6.05 6.05 6.0 6.0 6.0 6.06 6.06 6.05 Us Minitab to construct a run chart and a histogram from this data st. Run Chart of Wight 6.06 6.05 6.0 Wight 6.0 6.0 6.00 5.99 5.98 5.97 4 6 8 0 4 6 Sampl 8 0 4 6 8 Numbr of runs about mdian: 0 Expctd numbr of runs: 5.00000 Longst run about mdian: 0 Approx P-Valu for Clustring: 0.0706 Approx P-Valu for Mixturs: 0.9794 Numbr of runs up or down: 8 Expctd numbr of runs: 8. Longst run up or down: Approx P-Valu for Trnds: 0.486 Approx P-Valu for Oscillation: 0.569 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
0 Histogram of Wight 0 5 Frquncy 0 5 0 5.98 6.00 6.0 Wight Intrprt th charts. What conclusions about th distribution of th wights can you rach basd on ths charts? Th histogram shows that th most of chocolat bars wigh mor than 6.00 ouncs. Th run chart shows that th obsrvd wights plottd ovr tim hav upward drift in th wights of chocolat bars throughout th day. Th run chart also indicats th nd for action on th procss. Us Minitab to construct an X and R-chart from this data st. Is th procss stabl? If no, at what tims is it unstabl? Xbar-R Chart of Wight 6.06 6.060 Sampl Man 5 6.00 6.05 6.000 6 9 5 Sampl 8 4 7 UC L=90 _ X=7 LC L=6.00444 0.048 UC L=0.049646 Sampl Rang 0.06 0.04 0.0 _ R=0.0986 0.000 6 9 5 Sampl No, whn 9:0am, 0:5am, 0:0am, :5pm and 4:00pm. What is th ovrall procss man? What is th stimat of th procss standard dviation? (0.0986/d ) = (0.0986/.69) = 0.09. Show th formula you usd to stimat th procss standard dviation. ( R d ) Why did you us this formula vrsus anothr formula? ( R d ) was usd instad of th formula for th sampl standard dviation bcaus ( R d ) considrs only short trm variation, whil th formula for th sampl standard considrs long trm variation. ( R d ) assums that th procss is stabl. If th procss 8 4 7 LC L=0 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
is not stabl, th out of control points will jump out of th control limits basd on ( R d ). Th ( R d ) limits ar tightr than limits basd on th sampl standard dviation. Stat th LSL and USL for th abov procss. LSL = 5.95 and USL = 6.05 Calculat th actual procss yild using th mpirical data. 80 84 = 0.95 Calculat th thortical procss yild using th normal distribution. Is this a rasonabl calculation? If ys, why? If no, why not? Z LSL = (5.95 )/ 0.09 = -6.4969, hnc, P( X < -6.4969) = 0.0000 Z USL = (6.05 )/ 0.09 =.87, hnc, P( X > -.87) = 0.0 P (5.95 X 6.05) = 0.0000 + 0.0 = 0.0 Th abov calculation is not rasonabl bcaus th procss is not in statistical control. Comput th actual DPMO from th mpirical data. DPO = RTY = 0.95 = 0.048 DPMO =,000,000 x DPO =,000,000 x 0.048 = 48,000 DPMO Comput th thortical DPMO using th normal distribution. Is this a rasonabl calculation? If ys, why? If no, why not? DPO = 0.0 DPMO =,000,000 x 0.0 =,00 Th DPMO calculation is not rasonabl bcaus th procss is not stabl at this tim. Comput th Procss Sigma. Which DPMO should you us? Why? Procss sigma assuming a.5 sigma shift in th man for th mpirical DPMO of,00 is. to.. Procss sigma assuming a.5 sigma shift in th man for th thortical DPMO of 48,000 is.7 to.8. Nithr DPMO should b usd bcaus th procss is not stabl. A procss sigma calculation is not appropriat for this procss at this tim. QUESTION: Construct a control chart from th following data. Day Numbr of Entris Inspctd Numbr of Dfctiv Entris Fraction of Dfctiv Entris 00 6 0.0 00 6 0.0 00 6 0.0 4 00 5 0.05 5 00 0 0 6 00 0 0 7 00 6 0.0 8 00 4 0.07 9 00 4 0.0 0 00 0 0 00 0.005 00 8 0.04 00 0.0 4 00 4 0.0 5 00 7 0.05 6 00 0.005 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
7 00 0.05 8 00 0.005 9 00 4 0.0 0 00 0 0 00 4 0.0 00 5 0.075 00 4 0.0 4 00 0.005 Totals 4800 0 What typ of data is in th abov matrix? Classification typ attribut data What typ of control chart should b usd to study th abov data? p chart with constant subgroup siz Us Minitab to construct a control chart for th abov data. 0.08 0.07 0.06 P Chart of Df. Entris Proportion 0.05 0.04 0.0 0.0 +SL=0.0584 +SL=0.04645 +SL=0.0448 _ P=0.05 0.0 0.00 -SL=0.005 -SL=0 -SL=0.000855 4 6 8 0 4 Sampl Is th procss stabl? If not, on what days is it not stabl? Days 8 and ar byond th uppr control limit. QUESTION: Analyz th following data st using a control chart. Monthly accidnt data is listd across th rows for 6 months 5 4 4 9 7 8 6 6 5 6 6 9 6 7 4 4 4 5 7 7 4 4 What typ of data is in th abov matrix? Count typ attribut data What typ of control chart should b usd to study th abov data? c-chart Us Minitab to construct a control chart for th abov data. 6 8 0 4 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
C Chart of Accidnts 5 +SL=5.045 0 +SL=0.457 Sampl Count 5 0 5 +SL=5.890 _ C=. -SL=6.655 0 -SL=.0087 -SL=7.4095 4 8 6 0 Sampl Is th procss stabl? If not, in what months is it not stabl? Ys QUESTION: Construct a control chart for th following yars of wkly sals data for th Latin Amrican Division of a hospital supply company. Wkly Hospital Supply data is listd across th rows for yars 0,594.0 9,447.7 4,884.9 0,895.5 4,40.5 5,774.4,847.8,095.5 7,8. 6,05.5 4,475.,59.9 9,7.0 6,508.4,97.7 0,496.,970.9,406. 0,.7 5,977.4,057. 9,58.7 7,054.7 9,044. 9,096.,69.,780. 8,6.7 9,876. 0,6.4 6,456.7 0,099.8 0,65. 0,9. 8,6.7 0,6.9,898.7,67.,889.,94. 6,445.9 7,67.,689.7,86. 7,44. 40,09.7,0. 4,86. 6,859.6,640.8,579.6 6,55.0 9,44.0 9,994. 7,0.9 6,67. 4,04.6 0,08.6,597.9,.8 5,77.4 5,07. 6,550.8 9,70.4,08.6,5.0 9,79.0 7,870.4 5,90.8 6,550.6,94.4 9,9.0 9,478.6,89.4 6,.7 9,647.5 4,99.8 4,959.7 6,594.5 7,086. 4,945.,.7 7,87.,874. 8,8.8 6,0.9 8,595.0 8,770. 8,607.7 8,645.,746.6,09.9,058.7,578.4 9,64.7 6,87.8,886.7 8,049.8,45. 6,07.7 7,8.6 4,88.4 7,77. 0,5.,49.6 5,899. 7,8.8 0,. 6,6.4 9,99. 5,09.5,004.0,8. 8,74. 7,70. 5,96.0,95.7 5,46.9,448.6 7,494.,00.7,65.4 5,49.0 7,897.6 9,6. 4,90. 0,608.8 4,694.7 9,046.7 6,5.4 4,64.4 4,97. 8,996. 5,99. 9,056.,705.0,959.,8.6 4,567.5,97.9,5.7 9,655. 7,8.7 9,9.4,899.,66.4 8,906.8 4,7.5 9,80.8 5,484.7,547. 5,99.4 4,76.7 8,666.0 6,977.7 7,805.9 What typ of data is in th abov matrix? Masurmnt data What typ of control chart should b usd to study th abov data? 4 8 6 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
4 An I-MR chart should b usd dus to masurmnt data and a subgroup siz of on. Us Minitab to construct a control chart for th abov data. I-MR Chart of HospitalSupply +SL=4659. Individual Valu 40000 0000 0000 0000 +SL=986.8 +SL=980.5 _ X=4674. -SL=767.8 -SL=006.5 0 6 48 64 80 Obsrvation 96 8 44 -SL=755.8 0000 +SL=697.5 Moving Rang 0000 0000 +SL=0698.8 +SL=4470. MR=84.54 0 6 48 64 80 96 Obsrvation Is th procss stabl? If not, in what wks is it not stabl? Th procss is stabl. 8 44 -SL=0.89 -S -S L=0 QUESTION: Explain th purpos of a Gag R&R study. Th purpos of a Gag R&R study is to dfin th validity of a masurmnt systm, spcifically, to stimat th proportion of obsrvd variation du to unit-to-unit variation and masurmnt variation. QUESTION: Construct a Gag R&R run chart from th following data. Inspctor Rading CTQ 0 0 9 0 8 7 8 8 9 9 9 0 5 5 5 6 7 7 7 7 7 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
5 7 8 7 7 8 8 8 6 6 6 6 RunChart of CTQ by Rading, Inspctor Gag nam: Dat of study: Rportd by : Tolranc: Misc: 0 Inspctor 9 CTQ 8 7 Man 6 5 Panl variabl: Rading Inspctor QUESTION: Explain how flowcharts can b constructd to idntify and highlight th non-valu addd stps in a procss. Valu Addd Stps or Xs Non Valu Addd Stps or Xs NOTE: Show cycl tims for ach stp or X. QUESTION: Explain th rlationship btwn Y = f(x) and a flowchart in rspct to Six Sigma managmnt. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
6 As you can s from th following figur, CTQ = f(x, X, X, X 4, X 5 ) and CTQ = f(x, X, X, X 4, X 5 ) CTQ(s) () -------------------------- () -------------------------- Start X X X No Ys X 4 X 5 Stop QUESTION: Explain how Failur Mods and Effcts Analysis (FMEA) is usd to idntify th Xs that caus CTQs to b out of spcification. Construct a tabl to illustrat your xplanation. Failur Mod and Effcts Analysis (FMEA) is usd to idntify, stimat, prioritiz, and rduc th risk of failur in CTQs through th dvlopmnt of countrmasurs basd on Xs. Thr ar 0 stps to conducting a FMEA. First, tam mmbrs idntify th critical paramtrs and thir potntial failur mods through brainstorming or othr tools, that is, ways in which th dsign might fail (columns and of th tabl blow). Scond, tam mmbrs idntify th potntial ffct of ach failur (consquncs of that failur) and rat its svrity (columns and 4 of th tabl blow). Th dfinition of th svrity scal is shown blow. Third, tam mmbrs idntify causs of th ffcts and rat thir liklihood of occurrnc (columns 5 and 6 of th tabl blow). Th dfinition of th liklihood of occurrnc scal is shown blow. Fifth, tam mmbrs Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
7 idntify th currnt controls for dtcting ach failur mod and rat th organization's ability to dtct ach failur mod (columns 7 and 8 of th tabl blow). Th dfinition of th dtction scal is shown blow. Sixth, tam mmbrs calculat th RPN (Risk Priority Numbr) for ach failur mod by multiplying th valus in columns 4, 6 and 8 (column 9 of th tabl blow). Svnth, tam mmbrs idntify rcommndd actions and contingncy plans, prsons rsponsibl, and targt compltion dats for rducing or liminating ach failur mod (columns 0 and of th tabl blow). Eight, tam mmbrs idntify th dat th action was takn to rduc or liminat ach failur mod (column of th tabl blow). Ninth, tam mmbrs rank th svrity (column of th tabl blow), occurrnc (column of th tabl blow) and dtction (column 4 of th tabl blow) of ach failur mod aftr th rcommndd action (column 0 of th tabl blow) has bn put into motion. Tnth, tam mmbrs multipl th valus in columns, 4 and 5 of th tabl blow to calculat th RPN (Risk Priority Numbr) for ach failur mod aftr th rcommndd action (column 6 of th tabl blow) has bn put into motion. Format for a FMEA 4 5 6 7 8 9 Critical Para- Mtr Potntia l Failur mod Potntia l Failur Effct S v r i t y Potnti al Causs O c c u r r n c C u r r n t C o n t r o l s D t c t i o n 0 R R P N c o m m n d d A c t i o n Rsponsibility and Targt Dat A c t i o n T a k n S v r i t y 4 O c c u r r n c 5 D t c t i o n 6 R P N Bfor RPN = Aftr RPN = Dfinition of svrity scal = likly impact of failur Impact Rating Critria: A failur could Bad 0 Injur a customr or mploy \/ 9 B illgal \/ 8 Rndr th unit unfit for us Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
8 \/ 7 Caus xtrm customr dissatisfaction \/ 6 Rsult in partial malfunction \/ 5 Caus a loss of prformanc likly to rsult in a complaint \/ 4 Caus minor prformanc loss \/ Caus a minor nuisanc; can b ovrcom with no loss \/ B unnoticd; minor ffct on prformanc Good B unnoticd and not ffct th prformanc Dfinition of Occurrnc scal = frquncy of failur Impact Rating Tim Priod Probability of occurrnc Bad 0 Mor than onc pr day > 0% \/ 9 Onc vry -4 days < = 0% \/ 8 Onc pr wk < = 5% \/ 7 Onc pr month < = % \/ 6 Onc vry months < =. pr,000 \/ 5 Onc vry 6 months < = pr 0,000 \/ 4 Onc pr yar < = 6 pr 00,00 \/ Onc vry - yars < = 6 pr million (approx. Six Sigma) \/ Onc vry -6 yars < = pr tn million Good Onc vry 6-00 yars < = pr billion Dfinition of Dtction scal = ability to dtct failur Impact Rating Dfinition Bad 0 Dfct causd by failur is not dtctabl \/ 9 Occasional units ar chckd for dfcts \/ 8 Units ar systmatically sampld and inspctd \/ 7 All units ar manually inspctd \/ 6 manual inspction with mistak proofing modifications \/ 5 procss in monitord with control charts and manually inspctd \/ 4 control charts usd with an immdiat raction to out of control condition \/ control charts usd as abov with 00% inspction surrounding out of control condition \/ all units automatically inspctd or control charts usd to improv th procss Good dfct is obvious and can b kpt from th customr or control charts ar usd for procss improvmnt to yild a no inspction systm with routin monitoring QUESTION: Dfin capability of th procss in statistical trms. Capability is a masur of th rlativ rlationship btwn th Voic of th Procss and th Voic of th Customr. This rlationship considrs th diffrntial btwn th man and nominal of th procss. Th capability of a stabl and normally distributd procss is dfind as 99.7% of its output will b in th intrval btwn LNL Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
9 (man + [ R /d ]) and UNL (man - [ R /d ]), givn masurmnt data and a subgroup siz btwn and 0 inclusiv. QUESTION: Construct a dot plot for th CTQ. Construct dot plots to study th CTQ for th diffrnt lvls of X and X. Construct main ffcts plots and intraction plots for th following data st. X X CTQ - - 8 + - 9 - + 4 + + - - 7 + - 8 - + 5 + + 4 Dotplot of CTQ 8 0 CTQ 4 Dotplot of CTQ vs X X - 8 0 CTQ 4 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
0 Main Effcts Plot (data mans) for CTQ 7.5 5.0 Man of CTQ.5 0.0 7.5 5.0 - X Dotplot of CTQ vs X X - 8 0 CTQ 4 Main Effcts Plot (data mans) for CTQ 4 Man of CTQ 0 9 8 - X Intraction Plot (data mans) for CTQ 5 4 X - Man 0 9 8 7 - Thr is a significant intraction btwn X and X that will affct th CTQ. Th intraction is sn in th crossd lins. X Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
QUESTION: Explain th purpos of a k full factorial dsign. Th purpos of a k full factorial dsign is to undrstand th rlationships btwn a st of Xs ach with only two lvls, and th intractions of th X s, on th CTQ s. QUESTION: Crat th standard ordr matrix for a full factorial dsign. Tst X X X - - - + - - - + - 4 + + - 5 - - + 6 + - + 7 - + + 8 + + + QUESTION: Explain how th standard ordr matrix for a full factorial dsign is usd to stimat th ffcts of on main factor on th CTQ. Add up th valus of th CTQ for which th X is +, thn add up th valus of th CTQ for which X is -, finally, divid th diffrnc of th sums by th numbr of plus signs. QUESTION: Explain how th cofficint pattrn for a full factorial dsign is usd to stimat th ffcts of on intraction factor (X i X j ) on th CTQ. Multiply th signs of X i and X j to crat a nw variabl, X i X j. Nxt, add up th valus of th CTQ for which th intraction X i X j is +, thn add up th valus of th CTQ for which X i X j is -, finally, divid th diffrnc of th sums by th numbr of plus signs. QUESTION: Why randomiz runs in an xprimntal dsign. Randomization is don in xprimntal dsigns to rmov th ffcts of lurking variabls. MULTIPLE CHOICE QUESTIONS (corrct answr is in boldfac typ): Answrs:. A full factorial dsign with rplications provids information about: a) main factor ffcts b) two way intractions c) thr way intraction d) all of th abov. A full factorial dsign attmpts to prvnt th ffct of lurking variabls by a) rplication b) randomization c) intraction d) non of th abov Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
) all of th abov. A full factorial dsign with rplications pr cll rquirs runs. a) b) c) 4 d) 8 ) 6 4. A full factorial dsign can b physically rprsntd by which shap? a) squar b) cub c) rctangl d) circl ) non of th abov 5. Th purpos of a full factorial dsign is to: a) provid information on th ffcts of main factors b) provid information on th ffcts of intractions c) if possibl, rmov th ffcts of lurking d) all of th abov ) non of th abov 6. Us th following dsign matrix display to answr th following qustions. Tst X X X CTQ - - - 6 + - - 7 - + - 8 4 + + - 7 5 - - + 6 + - + 4 7 - + + 4 8 + + + 5 7. Th abov dsign matrix layout is calld: a) random ordr b) run ordr c) standard ordr d) non of th abov ) all of th abov 8. Comput th avrag ffct of X: a) + b) - c) -4 d) +5 ) non of th abov [+4+4+5]/4-{6+7+8+7]/4= 4-7 = - 9. Comput th avrag ffct of th intraction btwn X and X. a) - b) +/ c) + Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
d) / ) non of th abov [6+8+4+5]/4-[7+7++4]/4=/4-/4=/ QUESTION: Explain th purpos of a pilot tst. A pilot tst srvs four purposs. First, it validats a rvisd bst practics mthod. Scond, it highlights th risks (.g., FMEA) involvd in using th rvisd bst practic mthod. Third, it promots a smooth implmntation of th rvisd bst practic mthod. Finally, it facilitats buy-in by th stakholdrs of rvisd bst practic mthod. QUESTION: Analyz th following full factorial dsign with rplications data using Minitab. StdOrdr RunOrdr CntrPt Blocks A B C CTQ 5-0 - - - 8 00 4-9 5 - - 6 6 - - 7 - - 9 6 8 99 7 9-0 - - 4-4 0 - - 0 4-4 6 4-4 9 5 - - - 8 5 6 - - 0 Analysis of Varianc for CTQ (codd units) Sourc DF Sq SS Adj SS Adj MS F P Main Effcts 4884.7 4884.7 68.5 76.8 0.000 -Way Intractions 555.0 555.0 85.00 55.44 0.000 -Way Intractions 870. 870. 870.5 40.7 0.000 Rsidual Error 8 7.0 7.0.7 Pur Error 8 7.0 7.0.8 Total 5 948.0 Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
4 Main Effcts Plot (data mans) for CTQ A B 45 40 5 Man of CTQ 0 5 45 - C - 40 5 0 5 - Intraction Plot (data mans) for CTQ - - 60 A - A 40 0 60 B - B 40 0 C Cub Plot (data mans) for CTQ 0.5 99.5 9.0.5 B 4.5 9.0-4.5 - A Th high-high-high -way intraction ruls th xprimnt. To maximiz CTQ, st all factors at th high lvl..5 - C QUESTION: Explain th purpos of risk managmnt. Th purpos of risk managmnt is to maintain control and minimiz th risk of failur of product, srvic or procss by using risk abatmnt plans. QUESTION: List th stps involvd in dvloping a risk abatmnt plan. Tam mmbrs construct risk abatmnt plans for risk lmnts with high and mdium risk lmnts; that is, a risk lmnt scor of 6 to 5. Tam mmbrs idntify procss, product or srvic changs to rduc th risk for ach high and mdium risk lmnt (Xs). Nxt, tam mmbrs stimat th risk lmnt scor aftr th risk abatmnt plan is st into motion. Tam mmbrs idntify th risk lmnt ownr and st Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
5 a compltion dat for th risk abatmnt plan to bcom oprational. Finally, tam mmbrs documnt th risk abatmnt plans. A format for a risk abatmnt plan is shown blow. Potntial Risk Elmnt Potntial Harm Masur for Risk Risk Elmnt Scor Elmnt Bfor Aftr Countrmasur Risk Ownr Compltion Dat for Countrmasur Tam mmbrs carry out all risk abatmnt plans. Thy documnt th lssons larnd for ach risk lmnt and transfr th knowldg to othr rlvant risk lmnts. Nxt, tam mmbrs incorporat th risk abatmnt plans into a control plan for th procss ownr. Tam mmbrs turn th procss control plan ovr to th procss ownr. Finally, th procss ownr continuously turns th PDSA cycl to continu improving th procss, product, or srvic. QUESTION: Explain th purpos of mistak proofing. Explain th functioning of mistak proofing. Mistak proofing is usd to crat robust stps in a procss or srvic, or componnts of a product, dnotd as Xs, which ar not suscptibl to human rror. It is usd on th critical paramtrs shown in column in th tabl blow and on th potntial failur mods shown in column of th tabl blow. Column in th tabl blow is usd to list a mistak proofing solution to potntial failur mods (column ). Thr ar many typs of mistak proofing solutions to prvnt failur mods, for xampl, alarms that indicat dangr to an oprator, color coding mdical rcords by typ (grn for pdiatric and orang for griatric), or a hand harnss for a prss to prvnt an oprator from gtting his/hr hands crushd in th prss. Tam mmbrs can brainstorm for mistak proofing solutions to potntial failur mods, or thy can rviw th litratur on mistak proofing tchniqus. Format for a FMEA 4 5 6 7 8 9 0 4 5 6 Critical Paramtr Potntial Failur mod Potntial Failur Effct Svrity Potntial Causs Occurrnc Currnt Controls Dtction RPN Rcommndd Action Rsponsibility and Targt Dat Action Takn Svrity Occurrnc Dtction RPN Bfor RPN = Aftr RPN = QUESTION: Explain why standardization is such an important part of Six Sigma. Standardization of a procss crats a known bst practic flowchart that can b improvd by on or mor mploys, prhaps a Six Sigma projct tam. Thr ar multipl variations of a procss without standardization. Th lack of standardization Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007
6 maks procss improvmnt vry difficult bcaus thr is not on procss to b improvd, rathr, thr ar many vrsions of th sam procss ndd improvmnt. QUESTION: Explain th function of control charts in th Control phas of th DMAIC modl. Control charts ar important in th Control Phas of th DMAIC modl bcaus thy can b usd to monitor stability of th Xs and th CTQs. Also, control charts promot procss monitoring and work towards th limination of mass inspction mthods. QUESTION: Explain th purpos of a QC procss chart. Draw a QC chart to illustrat your xplanation. Th purpos of a QC procss chart is to monitor implmntation of rvisd bst practic mthod. QUESTION: Explain th purpos of documntation in Six Sigma managmnt. Documntation has at last 4 purposs in Six Sigma. First, it provids a prmannt rcord of th projct and dtails for th futur ownr of th procss. Scond, it nsurs that th improvmnt and rational is documntd so that th problm dos not com back. Third, it nsurs that th knowldg gaind by th tam is savd and shard. Finally, it documnts vital information about th projct from which othrs can bnfit. Gitlow Lvin A Cours in Six Sigma Managmnt4/9/007