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Ask Dr. Mikel Harry The Sigma Conversion Table Displays Yields That Don’t Match My Understanding of USL/LSL for the Process. Could You Please Explain Why There Is a Discrepancy?

The Sigma Conversion Table Displays Yields That Don’t Match My Understanding of USL/LSL for the Process. Could You Please Explain Why There Is a Discrepancy?

To facilitate an answer to your question, consider the following table. The first column is the given measure of process capability (provided in the form of Z).  The second column is the corresponding short-term yield of the first column (based on a one-sided specification). Of course, the third column is the associated long-term yield, also predicated on a one-sided specification. Naturally, the yield values could be converted to dpmo by the simple relation dpmo = (1 – Yield) * 10^6.  However, for puposes of our discussion, we will remain in the domain of yield.

1.00   0.841344740      0.308537533
2.00   0.977249938      0.691462467
3.00   0.998650033      0.933192771
4.00   0.999968314      0.993790320
5.00   0.999999713      0.999767327
6.00   0.999999999      0.999996599

So as to extend this insight, let us consider a global example. Let us say that a certain normally distributed critical-to-quality characteristic (CTQ) of some product was assigned a unilateral (one-sided) specification during the course of design. Related to this CTQ, it will be known that the performance requirement was provided in the form of an upper specification limit (USL). Furthermore, no target was specified. In other words, a nominal specification was not given. Based on these understandings, the case facts are presented as follows:

USL = upper specification limit = 100.0 grams.
M = process mean = process center = 90.1 grams. = short-term process standard deviation = 5.0 grams. = | USL – M | / = | 100.0 – 90.1 | / 5 = 1.98.

Owing to a dearth of empirical data (of a longitudinal nature), we must rationally approximate the long-term capability ( To do this, we elect to apply the proverbial 1.5 sigma shift factor (Z.shift = 1.5). Of course, this compensatory measure provides us with a rational means to analytically emulate or otherwise postulate the influence of unknown long-term reproducibility error (of a random nature). Thus, we project the long-term capability as = – Z.shift = 1.98 – 1.50 = .48. Again, we stress that the provision of such an index is merely a rational approximation designed to support an on-going regiment of quality reporting. Such an approximation should only be made in the light of empirical ambiguity, not as a routine substitute for measured experience.

Continuing our reasoning, we logically recognize the quantity of .48 as an equivalent standard normal deviate (of long-term origin). So as to ascertain the equivalent one-tailed probability of this metric, we could employ the benefits of an Excel spreadsheet.  For example, we would compute “=Normsdist( = .6843863, or about 68.44 percent. Thus, we can now better understand some of the critical connections between tail area probabilities and indices of capability, as well as the various statistical constructs underlying various types of Six Sigma conversion tables.

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