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Sigma level from process capability

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  • #45478

    Struk
    Member

    HiI have a problem…
    I need to estimate sigma level of a process
    given is output from Minitab Process capabilty data window:
    defects PPM – 80000
    cp 2.09
    cpl 2.10
    cpu 2.09
    cpk 2.09USL =70
    LSL = 40
    Sample mean = 55.0263
    Sample N =500
    Std Dev(Within) = 2.38962
    Std Dev(Overall) = 8.13597I need the Sigma Level?
    Also, please tell me the mathematical way of doing it…as i need to understand how we arrive at it?

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    #148735

    Edwards
    Participant

    Read past posts. 
    Sigma levels are MEANINGLESS.

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    #148742

    Al
    Participant

    According to the book: Basic Statistics Tools for Continuous Improvement, by  Kiemele, Schmidt, and Berdine (Air Academy Associates airacad.com). Pages 9-11-915.
     
    Sigma level = number of std deviations between the center of the process and the nearest spec (this is the Z value);  
    Where Z value = Minimum ( (USL-xbar)/sdev, ((xbar-LSL))/sdev))
     
    Cpk = Sigma Level/3
     
    Sigma Capability = 0.8406 + SQRoot(29.37- 2.221 + ln(dpm))
     
    Sigma Capability ¡Ö Sigma Level + 1.5.
     
    Sigma Capability vs dpm
    (Motorola¡¯s sigma capability index and dpm)
     

    Sigma Capability

    dpm

    6.0

    3.4

    5.0

    2.33

    4.0

    6,210

    3.0

    66,807

    2.0

    308,538
     
    I hope it helps.
     
    Al

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    #148743

    Struk
    Member

    Hi Al
    Actually I have got 2 types of StdDev from Minitab output. One is StdDv(Within)=2.3, other is StDev(Overall)=8.6
    Which one should I use for the formula u told:
    Where Z value = Minimum ( (USL-xbar)/sdev, ((xbar-LSL))/sdev))
    I calculated Z with StDev(overall)….. it is 6.28815
    Now, how to get Sigma level from there, mathematically?
     

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    #148752

    Prof. Paul Swanson
    Participant

    This table is completely incorrect. 
    Do not use it !

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    #148754

    Tierradentro
    Participant

    Please  explain  why?

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    #148761

    Struk
    Member

    Thanks guys for your replies. Unfortunately I am still unable to calculate the sigma level mathematically. According to MINITAB output, in our observed process, DPMO is 80,000. How to get sigma level based on this the above info mathematically (I have already used several online calculators; sigma level is ~2.9. I want to arrive it mathematically)
    Thanks in advance

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    #148762

    Adrian P. Smith
    Participant

    If you are using Excel you can use the following formula:
    =NORMINV(1-DPMO/1000000; 0; 1)
    This will give you Sigma WITHOUT 1.5 shift (if you want the 1.5 shift, of course just add 1.5 to the result). (Obviously replace DPMO with the cell where your DPMO value is stored.)
    Alternatively, without 1.5 shift:
    =-(NORMSINV(DPMO/1000000))
    With 1.5 shift:
    =-(NORMSINV(DPMO/1000000)-1.5)
     
    If you not using Excel, try the following:
    Sigma =SQRT(LN(1/(1-EXP(-DPMO/1000000))^2))-(2.55155+0.802853*SQRT(LN(1/(1-EXP(-DPMO/1000000))^2))+0.010328*SQRT(LN(1/(1-EXP(-DPMO/1000000))^2))^2)/(1+1.432788*SQRT(LN(1/(1-EXP(-DPMO/1000000 ))^2))+0.189269*SQRT(LN(1/(1-EXP(-DPMO/1000000))^2))^2+0.001308*SQRT(LN(1/(1-EXP(-DPMO/1000000))^2))^3)
    Again add 1.5 if required.
    Adrian
     

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    #148763

    Prof. Paul Swanson
    Participant

    The origins of six sigma’s numbers will help people understand how the many error’s have been created. A brief summary follows:

    Bill Smith, a Motorola engineer claims that for uncontrolled processes “batch to batch variation can be as much as +/-1.5 sigma off target.” He gives no references or justification for this. In reality there is no limit to how much that uncontrolled processes may vary.
    At that time Motorola used Cp=1. Bill Smith suggested “Another way to improve yield is to increase the design specification width.” He broadens specification limits to Cp=2.
    Mikel Harry derives +/-1.5 as a theoretical “shift” in the process mean, based on tolerances in stacks of disks. Stacks of disks obviously bear no relation to process.
    Harry names his Z shift. The Z shift makes a number of additional errors: his derivation dispenses with time yet he refers to time periods; he claims a “short term” and “long term” yet data for both are taken over the same time period.
    Harry realises his error in the 1.5 and says it “is not needed”.
    Harry in about 2003 makes a new derivation of 1.5 based on errors in the estimation of sigma from sample standard deviations. For a special case of 30 points, p=.95 he multiplies Chi square factor by 3, subtracts 3 and gets “1.5”. The actual value ranges from 0 to 50+. He calls this a “correction”, not a shift.
    Harry’s partner Reigle Stewart adds a new calculation he calls a “dynamic mean off-set.”: 3 / sqrt( n ) where 3 is the value for control limits and n is the subgroup size. For n=4 he gets “1.5”. Reigle says “This means that the classic Xbar chart can only detect a 1.5 sigma shift (or larger) in the process mean when subgroup size is 4″Reigle is quite incorrect. Such data is readily available from ARL (Average Run Length) plots.
    Furthermore six sigma tables are based on the assumption of normal distributions.  No data will ever exactly follow a normal distribution to the extreme six sigma tail. 
    In summary, the 1.5 does not exist, despite the many attempts to prop it up. Calculations involving 1.5 are hence meaningless. That is, 3.4 dpmo is meaningless. Six sigma tables are meaningless. Sigma levels are meaningless.
    DO NOT USE SIX SIGMA TABLES.

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    #148765

    Ken Feldman
    Participant

    Come on Dr. Paul, tell us how you really feel!!!!!

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    #148774

    Green Belt
    Participant

    This is supposed to be a professional forum. The professor has told us why he thinks the 1.5 shift is rubbish.
    Hacl and Darth should tell us why they think the 1.5 is not rubbish, rather than making foolish remarks.

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    #148776

    Anonymous
    Participant

    It is a professional forum. It’s degree of professionalism isn’t measured by a single response. Now if you choose to behave as a professional you can do a search and find more posts than should be necessary to explain peoples position concerning the shift. Being redundant certainly isn’t professional neither is trying to beintelligent when you actually are posting rubbish.

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    #148784

    Chen
    Participant

    A more important question is do you believe each defect is independent?

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    #148791

    Ken Feldman
    Participant

    Lighten up Green Belt.  And when did you ever conclude that this is supposed to be a professional forum?  Anyone can post and it is obvious that many haven’t been in the profession very long. 
    If you had read some of my previous posts you would have ascertained that I too am not a believer in the 1.5 shift as a universal truth.  I am not quite as emotional as the Prof.  Dr. Harry has put severe constraints around his derivation of the 1.46 shift and has never assumed it to be a constant as it is often used.  My advice has always been to determine the true shift based on historical data and then use that as appropriate.  Having a table with a constant 1.5 shift built in has too much potential error to be of much use.

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    #148793

    Savage
    Participant

    Well why don’t you answer his question?  Can you?  I’ve yet to see anything of substance from you. 

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    #148798

    Simon Hartwell
    Member

    Good afternoon all
           going back to the original question.
     As I understand it, the Sigma level  = 3x CP
    So with a CP of 2.10  you would get a Sigma Level of 6.30

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    #148806

    O’Connell
    Participant

    That is pretty stupid but as good an answer as any of Harry and Reigle’s !!!!

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    #148807

    Markert
    Participant

    “If you had read some of my previous posts you would have ascertained that I too am not a believer in the 1.5 shift as a universal truth.”
    You should tell your crazy friend Praveen Gupta.  He claims that all processes in all industries are out of control “13-14% of the time because of the 1.5 shift”
     
     

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    #148809

    BB
    Participant

    RE: “Dr. Harry has put severe constraints around his derivation of the 1.46 shift and has never assumed it to be a constant as it is often used.”

    What are the constraints ?
    How are the constraint levels calculated ?
    What is the derivation or proof of these constraint levels ?
    Do you have a link to the proof of the constraint levels ?

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    #148810

    BB
    Participant

    I should have also asked:
    5. Do the constraints apply to Harry’s 1.5 “drift” as well as Harry’s 1.5 “correction” and Reigle’s “dynamic” 1.5 ?

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    #148812

    Ashman
    Member

    If the 1.5 doesn’t exist, how can it have constraints ?I’d love to see how Darth , hacl and the SS cronies answer BB.Crap + crap = more crap

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    #148814

    Mikel
    Member

    Praveen is not a real smart guy.

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    #148818

    Markert
    Participant

    “Praveen is not a real smart guy.”
    Understatement of the week

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    #148820

    Ken Feldman
    Participant

    All is revealed in his online book entitled “Mysteries of Six Sigma”.  I am not endorsing just informing.

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    #148824

    BB
    Participant

    Darth,
    Could you please share with us how this book answers these basic questions.

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    #148826

    Mikel
    Member

    It doesn’t. It is as bad of a joke as the cartoon books.

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    #148829

    BB
    Participant

    Darth claims:  ‘All is revealed in his online book entitled “Mysteries of Six Sigma”.  “
    Where are you Darth ?  Is Stan correct ?

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    #148833

    M G
    Participant

    Interesting how the link to this e-book doesn’t work:RESOLVING THE MYSTERIES OF SIX SIGMAhttps://www.isixsigma.com/newsletter/archive/newsletter.asp?id=310#section6_title

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    #148845

    Ken Feldman
    Participant

    Sorry, don’t intend to quote all the passages.  If you have a burning desire, order the book.  It will answer where he got the 1.46 rounded to 1.5, what caveats he puts around it, his assumptions and his feelings about it’s use as a universal truth.  Of course, you can argue with the math and assumptions but he is clear about not using it as a constant.  That seems to have come a long way from building tables with the shift built in.

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    #148846

    Ken Feldman
    Participant

    You won’t get me arguing with my good friend Stan.  He has some issues with the book and it’s contents.  An enlightened researcher will read the source him/herself and come to their own conclusion.

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    #148848

    Mikel
    Member

    Of course Stan is correct. What a silly question.

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    #148849

    Ken Feldman
    Participant

    Even Amazon doesn’t carry it.  But here is a link to many many more SS books.  What a joke!!!
    http://amazon.com/s/ref=sr_pg_1/103-6053946-2647860?ie=UTF8&keywords=Mysteries%20of%20Six%20Sigma&rh=i%3Aaps%2Ck%3AMysteries%20of%20Six%20Sigma&page=1

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    #148852

    Mikel
    Member

    Yes, it seems that every joker with 6 months experience and a dumb notion that books make big money is writing a Six Sigma book.
    I’ll bet those Six Sigma mysteries have a wide target audiance.
    I have a rare copy of mysteries I would be willing to part with for $100.00 – should I put it on ebay?

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    #148864

    Simon Hartwell
    Member

    The formula using the Z transform to predict yield:
    Z= (x-u)/std dev where, x represents LSL or USL , x is the mean
    For your data:
    Z = 70-55.0263/8.13597 i.e 1.84 using the USL
    P(Z1.84)= 1- 0.03288 = 0.9671 YIELD = 96.71% GIVES A SIGMA VALUE OF 3.3 (Short term) i.e 32,800 defects to the right of the curve after 70
     
    Z= 40-55.0263/8.13597 i.e -1.84 using the LSL
    P(Z-1.84) = 1 – 0.03288 = 0.9671 Yiled = 96.71 gives a sigma value of 3.3 (short terms) i,e 32,800 defects to the left of the curve before 40
    DPMO : 65,600 translates to a short term sigma level of 2.97 ( approx) but this us based on using overall stnd dev.)

    Calculations for the C’s are as follows:
    Cp= 70-40/6 x 2.38962 = 2.09
    Cpk usl = usl-avg/3 std dev = 70-55.0263/3x 2.38962 = 2.08
    Cpk lsl = avg- lsl/3 stnd dev = 55.0263-40/3 x 2.38962 = 2.09
    Cpk=min (Cpk usl, Cpk lsl) = 2.08
    The short term capability indices Cp and Cpk are cal using short term stnd dev – because short term process variation is used these measures are free of subgroup shifts in data and take into account only variation within subgroup

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    #154046

    QA Guy
    Participant

    My software lists Sigma Level & Sigma Capability.  Although I know this stuff well, I’m not sure what these mean.  Please help.
    As far as professionalism goes, we work a lot of hours, and do not need the mental stress of Truman Show behavior.
    Love, qa guy
     

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    #154055

    Jim Shelor
    Participant

    Generally, the Sigma Level includes sigma-common cause and sigma-special cause and the 1.5 sigma shift.
    Sigma Capability is sigma-common cause, that sigma estimated for use in calculating Cp and Cpk.
    Your software should have a help function where the sigmas are defined.  Some programmers use terms in a non-standard way and we need to look the terms up in the help function to be sure what the software is actually calculating.
    Hope this helps.

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