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Process Sigma calculation

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  • August 22, 2002 at 8:36 pm #30167

    Elsa Rodriguez
    ★ 20 Years ★
    Participant

    In the process sigma calculatrion formula why is there a 1.5  sigma shift and what does that mean?
    Process Sigma = NORMSINV(1-((Total Defects) / (Total Opportunities))) + 1.5
    Also there is another formula published: process Sigma = 0.8406 + SQRT(29.37 – 2.221 * (ln(DPMO))).What does each constant mean or where does they come from?
     
     
     
     
     

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    August 22, 2002 at 10:13 pm #78330

    Perryman
    ★ 10 Years ★
    Participant

    Hi Elsa
    The addition of 1.5 in your first formula is to convert your sigma value into short term.  The common understanding is that in the long term, you will have shift and drift in your process, which will therefore produce more defects and consequently lower your overall sigma value.  Since your formula uses discrete data to start (which is considered long term already), you have to at the conventional 1,5 to reflect a short term sigma value.  Many organizations express their sigma values in the short term, by convention.
    You can read more about this by clicking on the Six Sigma Q&A on the blue bar.
    As for the second formula, I haven’t plugged any numbers into it yet but I suspect the natural log is to illustrate the non linear nature of sigma versus defect rate while the sqrt of 29.37 added to the 0.8406 allows for a sigma value in your equation even when you have a DPMO of zero.
    Cheers,
    Patch

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