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Control Chart

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Viewing 13 posts - 1 through 13 (of 13 total)
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  • #47583

    I’Anson
    Participant

    I have continuous data that is not normally distributed, which control charts should I consider using and why?

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    #158727

    Omashi Sabachi
    Participant

    It  depends

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    #158740

    GrayR
    Participant

    Use a variables x-bar/R chart.  Your data don’t need to be normally distributed to use it.

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    #158743

    Chris Seider
    Participant

    Be wary of advice from those which seem so confident.  Either use I-MR or Xbar-R charts.  If you use I-MR charts, I’d consider finding the distribution type and find the 3 sigma equivalent since your distribution is non-normal.  Some might say I’m wasting your time by suggesting to find the equivalent but I’d hate to have you respond to “out of control” indications if they weren’t. 
    Xbar-R charts should be used IF you can find a rational reason for subgroups.  However, if little process knowledge is known, I’ve seen some respected individuals use Xbar-R just to get the process of investigating special causes started.
    Good luck.

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    #158746

    Jim Shelor
    Participant

    Daniel,
    If you have Minitab, the I,MR chart has a selection for doing a Box-Cox Transformation on the data as it is analyzed.
    Jim Shelor

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    #158762

    Craig
    Participant

    Daniel,
    It might help if you explain your situation a little more. Why is the data non-normal? Is it intrinsically non-normal where you would generally use a transformation or did you just happen to fail a normality test on a batch of data? (Because of outliers, etc.)
    HACL

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    #158770

    Snow
    Participant

    What are the considerations for determining sample size in variable control charting?  For example, using a larger sample size will result in tighter control limits, but is this always desirable?

    What if you run an IMR and you obtain indications of instability but running an XbarR shows stability?  What does it mean and what is the approach? 
    THANKS

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    #158775

    Ron
    Member

    If you are using a shewhart control chart it does not require data to be normal. That feature is one of the remarkable things about shewhart control charts.
     

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    #158797

    I’Anson
    Participant

    I have non normal service time data. As you can imagine it’s skewed to the left. The data transforms well using the natural log (LN).
    I have a process target/mean that I want to calculate upper and lower spec limits for based on a Cp of 1.33.
    Can I the usl and lsl limit by using the sigma of the transformed data.
    For example use the transformed sigma multiple it by 6, then multiple it by 1.33. Divide the result by 2, add it to the transformed mean value. convert the transformed result back.

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    #158799

    Jim Shelor
    Participant

    Daniel,
    The USL and LSL are limits that should be supplied by the engineer or whoever is the owner of the process.  You should not be calculating these limits from the results of your data.
    There must be a time you do not want to exceed for your service.  That time is your USL.  Your LSL is naturally 0.
    Best regards
    Jim Shelor
     

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    #158801

    Mikel
    Member

    Daniel,Just a nuance, but your data is skewed to the right. Your method will work assuming your lower limit does not go below
    0.You should take Mr. Shelor’s advice about whether it is reasonable to
    set your limits this way however. The upper limit should be VOC; the
    lower limit is not 0, but instead should represent the absolute
    minimum time that the service can be accomplished correctly.

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    #158805

    Chris Seider
    Participant

    I’m not sure from the business point of view why you worry about calculating the Pp but if you want to remember that Minitab will calculate it for you if you know which distribution best fits your process data.  Just be sure to use the capability non-normal command.
    They use the best estimation of the data that represents the 6s equivalent spread as the Pp calculation for normal distributions.  Do all of this on non-transformed data.
    My 3.4 cents.  Good luck on reducing that service time….I”m presuming this.

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    #158843

    Savage
    Participant

    Stan,
    You stated “the lower limit is not 0, but instead should represent the absolute minimum time that the service can be accomplished correctly.” I agree with you but could the lower limit also be defined as missing?
    The reason I ask, if the minimum time that the service can be completed is X (thus LSL=X) and my process average is X+1 upper limit is X+10, then my Cpk/Ppk might be ‘artificially’ low.
    Matt

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