iSixSigma

Justifying outlier

Six Sigma – iSixSigma Forums Old Forums General Justifying outlier

Viewing 6 posts - 1 through 6 (of 6 total)
  • Author
    Posts
  • #28554

    woey
    Member

    In process monitoring, if I found and proved that an out of specification is an outlier, what tools shall I use to justify that outlier? Any ideas? Thanks!

    0
    #71338

    Laszlo Gunderud
    Participant

    Refer to the National Bureau of Standards Handbook 91 “Experimental Statistics” Section 17-3.1. By selecting your alpha risk, say 0.05, and with your sample size known, you calculate an alpha prime equal to 1-(1-alpha)^(1/sample size). You then look up the z-score for 1-(alpha prime/2). With this z-score, you can reasonable expect to find your data with +/- (z-score)*(population standard deviation) from your population mean. Points beyond this interval are assumed outliers based on the alpha risk your selected. If you do not know your population standard deviation and/or your population mean, the Handbook provides formulas and tables for the calculation.
    Example: alpha = 0.05, sample size equals 30
    alpha prime becomes 1 – (1 – 0.05) ^ (1/30) = 0.0017
    z-value lookup becomes 1 – (0.0017)/2 = 0.9991, thus you find z to be around 3.12
    Thus any value beyond +/- 3.12*(pop. std. dev.) from the pop. mean is considered an outlier.
     

    0
    #71360

    Sridhar
    Member

    Hi,
            I think the simplest way of looking for outlier is Box-Plots. The points which are falling outside the whiskers of box plot are called as outliers.
    hope this helps you
    sridhar
     

    0
    #71409

    Tom Black
    Member

    A true “outlier” is a point that was caused by a special cause variation.  You should know beyond a reasonable doubt the cause and know that it is something that can only happen once (or was caused by something you can fix and poke yoke).  Once you have done that you can eliminate that point from your capability calcualations.
    If you do not know the cause; if you only know that the point is outside the three sigma control chart limits, then you have to consider that part of the normal process.  You cannot eliminate it.
    Tom Black, MBB

    0
    #71414

    Rick S
    Member

    An easy method of using a box and whisker approximation is to calculate a maximum point for comparison to the data point in question.  The formula is:  mean + or – (0.67*process standard deviation) + or – (1.5 (1.34*process standard deviation)).  The mean + or – (0.67*process standard deviation) gives the maximum / minimum boundry of the interquartile range.  The mean is added to or subtracted from depending on whether the point in question is above or below the specification limits.  The (1.5(1.34*process standard deviation)) is added or subtracted based again of whether the point in question is above or below specification limits.  If the point in question is beyond the range of the calculated points it can be considered an outlier.  You may also substitue the 1.5 with 3 to estimate an extreme outlier.

    0
    #71453

    Helpful
    Participant

    Don’t have blinders on and get totally consumed by the statistics.  Keep two other thoughts in mind:  1.  there are only 2.7 chances in a thousand that the result is part of your process… it may never be explained and could have been a measurement error. It might not be real.  If you are going to question this one point, why don’t you question the other 997?  2. How important is the flyer?  If your customer couldn’t care less, or the result is within engineering spec (but outside SPC control limits), or the damages incured is 10 cents, or it is going to consume tons of labor and process shut down time… do you want to do it?  The odds are 997 to 3 that this point is not real.  I’m not saying get sloppy… but question what the impact is and what your response should be.

    0
Viewing 6 posts - 1 through 6 (of 6 total)

The forum ‘General’ is closed to new topics and replies.