# Binomial Data Sigma Value

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• #30610

Terry
Member

Hello All
We have been using the standard normal distribution (Z score) to get sigma values for binomial data (I am aware of the limitations of this concept) and I was wondering how others handle this situation: If you have binomial data and have a process where there have been no defects identified, how do you determine a Sigma value? and if all output is classified as a defect? I realize that mathematically this is impossible, and that eventually you would have a defect, but what do you do to attach a value (for the layperson or just as a closeout requirment)?  I am hoping this thread does not go down a rat hole.
Thanks
Terry

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#79883

Jagdish R
Participant

In a process where the defects are measured as Pass/Fail  or you count the total number of defects that is counted as an attribute you can always calculate the PPM of the process where you are capturing the attribute defect. Use the PPM to Sigma table to calculate your Sigma for the process.
Does that make sense or have I given you a very different view ?
Jagdish

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#79884

Terry
Member

Well yes, but say we have a process that has a 20% defective, and we have poke yoked it so afterwards we see 0 PPM defective after being in Control for 3 mos. Now we need to close it ou.

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#79896

Jamie
Participant

I just ran into this while making a spreadsheet to calculate sigma value based on DPMO. What happens when you enter in 0 defects…. well sigma level goes to infinite. What I did to handle this was assume that instead of infinity it approached 6 sigma long term or 7.5 sigma short term. Without an incredibly large sample size I don’t know of much else you can do. Though I think its kind of a mute point, if you truly arent seeing any defects after three months is there really a need to track it? (tracking data is a non-value added activity).
Jamie

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#79897

abasu
Participant

With attribute data, to show true six sigma capability you need a sample size close to 1,000,000.  So the question is what is the sample size examined in the 3 months.

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#79898

denton
Participant

Just run Capability (Binomial) in Minitab.  It reports a Z score for the process.

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#79899

Terry
Member

If you have no errors this message appears:

** Error ** All defectives are 0 – no calculations can be done;
execution aborted.

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#79900

RT
Member

I would close the project with zero defects noted or a Sigma level of infinity and go back to that process in 6 Months or so to see if the process is still defect free.  I had this same problem and that’s what I did.  Unfortunately some of these project trackers require some Sigma level input so the  bean counters can blindly base savings on Sigma levels.  If thats the case just put in Zlt = 6 and move on.

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#79901

Jamie
Participant

It sure does because the tails of the normal curve go to infinite. Remember 6 standard deviations is 0 defects per million (not 3.4, thats quoted with a 1.5 shift). So I’d say again (see my previous post) that that you either need to increase sample size or declare that you have approached 6 sigma long term/7.5 sigma short term. Only other option I see is calculate sigma with only 1 defect, stating you are likely better than this. But is this just a numbers game, have you accomplished what you need to do?… 0 defects for three months (assuming an adequate sample size) sounds like a real victory. Time for a new project.
Jamie

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#79906

Ropp
Participant

Hi Terry,
With 0 defectives, you cannot get a point estimate of the population percentage nonconforming (call it p’).  However, you can compute an upper confidence bound for p’ with the following formula:
UCB (at 1 – alpha confidence) for p’ = 1 – alpha ^ (1/m)
where UCB is the upper confidence bound, alpha is the alpha risk (usually chosen to be .10, .05, or .01), the symbol ^ means “to the power of,” and m is the number of conforming units tested.
For example, if you have tested 2642 units with 0 nonconforming units found, an upper .99 confidence bound (alpha = .01) for the process percentage of nonconforming units is:
UCB (.99 confidence) for p’ = 1 – (.01) ^ (1/2642) = 1 – .998258 = .001742 = 0.1742%
The “Z” value corresponding to this percentage is 2.92.  Thus, you can report, with .99 confidence, that the sigma level for this process is at least 2.92.
I didn’t invent the above formula, I got it from the book, Measuring Process Capability, by DR Bothe.
I hope this helps.

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#79914

Terry
Member

Wow, that is the coolest damn thing I have ever seen. I will go with it. Thank you!

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#79915

Mel Alexander
Participant

If your process has a defective rate of 20% (p=0.20), then the Average Run Length (ARL) is 5 (=1/p=1/0.20), depending on how frequently your sampling process is done (either hourly, daily, weekly, monthly). You didnt say what the sampling frequency was nor did you say how many units constituted the p=20%.
You may want to consider using “inter-event counts” (i.e., number of successive conforming units between nonconforming items) and assess the PPM using control charts specific for PPM processes based on the ARL and the geometric or negative binomial distribution probability model. For more information, read J. C. Bennyanss article “DESIGN AND USE OF g-TYPE CONTROL CHARTS FOR INDUSTRY AND HEALTHCARE: SPC Charts for Hospital Infections and Adverse Events, ” on http://www1.coe.neu.edu/~benneyan/papers/g_chart_overview/
or Lloyd Nelsons article LS (1994): “A Control Chart for Parts-Per-Million Nonconforming Items”, Journal of Quality Technology 26(3), 1994, pp. 239-240.

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#79952

Jamie
Participant

Dave, That was extremely enlightening and an excellent suggestion.
Thanks, Jamie

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