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5.15 sigma to 6 sigma in gage R R

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  • #39815

    Ryan
    Member

    All experts,
    I would like to know from the experts of this forum on how to arrive 99.73 % when we use 6*sigma. ( 99% for 5.15sigma)
    a reply with an equation would be very helpful
    thanks,
     
    ryan

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    #122119

    Mikel
    Member

    There is no deriviation, the 5.15 goes back to 1962 and a discussion among a bunch of engineers at GM.
    Use 6.

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    #122126

    walden
    Participant

    Look up the area on a standard normal curve between z = -3.0 and z = +3.0 for 6 sigma, and between z = -2.575 and z = +2.575 for 5.15 sigma.

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    #122140

    BBMole
    Participant

    If you are referring to the P/T ratio, the answer I was given was that 5.15 equates to 99% of the variation as stated by Chris

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    #122204

    Ryan
    Member

    Thank you very much on clearing this up to me Stan, Chris, and BBMole.
    I have a follow-up question: I already understand if we try to compute these on z-table, we get 99% for 5.15 and 99.73% for 6, but i am confused on how 99.73 as AIAG says relates with what we say as 99.99966 on 6-sigma value?
    Hope for an answer… thanks again in advance!
    Ryan

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    #122205

    A.S.
    Participant

    99.73 % is with +/- 3 sigma.
    99.99966 % is with +/- 6 sigma.

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    #122210

    Mikel
    Member

    Simple – it does not.

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    #122211

    Mikel
    Member

    Wrong on the +/- 6.

    99.99999990098780%
    is correct

     

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    #122219

    walden
    Participant

    Ryan,
    The 99.73% is the area under the standard normal curve between z = +/-3. To calculate the percentage in “Six Sigma language”, you need to consider the 1.5 sigma shift, which gives you 6 – 1.5 = 4.5. The area under the standard normal curve to the left of z = 4.5 is 99.99966% (or 3.4 PPM).
    Hope this clears up your confusion.
    Chris

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    #122265

    Donwark
    Participant

    As normal you are an idiot Stan.  Go back and take calculus.  Do the math and you will discover your error.  Just more bad advise from another pretender.

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    #122266

    A.S.
    Participant

    My understanding is at +/- 6 sigma ,
    Area under the curve is 99.9999998 % without considering the shift.
    Area under the curve is 99.99966 considering 1.5 sigma shift from the target.
    request some experts to clarify my doubt :whether this 1.5 sigma shift is +/- 1.5 sigma shift or totally 1.5 sigma shift?
    Waiting for doctors reply
     
     

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    #122267

    J. RAVICHANDRAN
    Participant

    Dear Anbu.
    When shift exists in a process, it can be either on RHS  or on LHS of the target.  Therefore if the shift is on RHS +1.5 is effected and if the shift is on LHS -1.5 is effected.  Therefore in a six sigma process +/- 1.5 shift is allowed.

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    #122268

    BTDT
    Participant

    Ryan:There are two Excel functions you can experiment with, NORMSINV gives you the Z value for a corresponding error rate, and NORMDIST allows you to calculate the error rate for a given Z value.”=normsinv(0.005)” gives you the Z value for 1/2 of the 1% defect rate = 2.575835Twice that value gives you the +-range of 5.15167. This is the total range of Z values that will contain 99% of the data. 0.5% will be out of limits on the low side and 0.5% will be out of limits on the high side. When people asked the question about how ‘wide’ a process is, the decision was made to choose the interval that contains 99% of the data and call that the ‘width’.If you want to calculate the error rate that corresponds to +-3 Z values, you can use the function “=1-normdist(3,0,1,TRUE)” this gives you 0.1350% on the high side and “=normdist(-3,0,1,TRUE)” gives you 0.1350% on the low side for a total of 0.2700%If you use these functions to calculate the error for +-6 Z values you get 9.9012E-10 on the low side and 9.9012E-10 on the high side for a total of 1.9802E-09.It is common that people would like to express the TOTAL error rate in terms of a single Z value that you can look up in a table and convert back and forth. I don’t like to use this “Sigma” value to express a DPMO, because people usually think of this as the +- value of Z values – it is not!If I say my DPMO is 1.2E-4, then you can use “=1-NORMSINV(0.00012)” to calculate 4.6729. This is NOT the +- range defining this defect rate, it is the single, one sided function. This is the convention throughout Six Sigma. 0.99 PPB is the error rate for 6 Sigma, not +- Six Sigma.A debatable point(!) is why do you have to use “=normdist(4.5,0,1,TRUE) to get the 3.4 PPM value quoted all the time. You should see that the statement that “6 Sigma = 3.4 DPMO” is only true when you shift it to “4.5 Sigma = 3.4 DPMO”. The shift factor is not applicable for the calculation using continuous data here in these examples.BTDT

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    #122269

    BTDT
    Participant

    Proofreading helpsA DPMO of 120 is an error rate of 1.2E-4.BTDT

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    #122270

    A.S.
    Participant

    Thanks Ravi,
    My basic doubt is 99.99966% was arrived in what basis.Whether it is considering 12 -3 (1.5 +1.5) sigma or considering 12 -1.5 sigma.
    Subsequent post from BTDT confirms that it is consiodering 4. 5 sigma on both side.
    Thanks to BTDT. 
     

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    #122274

    Mikel
    Member

    Wrong

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    #122364

    hmm
    Participant

    Stan my friend,
    Isn’t +/- 6 sigma give a ppm non-defective of 99.99999980268?  Or is my calculator off?
     

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    #122365

    hm
    Participant

    Stan my old friend,
    Is my calculator stuck or isn’t +/- 6 sigma have a nondefective rate of 99.999999802685?

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    #122366

    hm
    Participant

    Ooops….looks like my enter key was stuck  LOL!

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    #122369

    Dr. Scott
    Participant

    BTDT,
    I am glad to see that someone finally got it right. It is the whole 1 versus 2 tail calculation, and the idiocy of the 1.5 shift.
    Take Care,
    Dr. Scott

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    #122378

    Ryan
    Member

    Thank you very much BTDT. It is very well understood.
    ryan

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