Question The 1.5 SD shift in mean
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 This topic has 11 replies, 8 voices, and was last updated 19 years, 2 months ago by Gabriel.

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July 11, 2003 at 8:57 pm #32769
Tom ZachariahMember@TomZachariah Include @TomZachariah in your post and this person will
be notified via email.Having just completed my Six Sigma:SPC course, I have a question about the ‘famous’ 1.5 standard deviation shift in the process mean. According to Montgomery (Introduction to Statistical Quality Control), this is a ‘logical inconsistancy’. As process analysis is based on analysing a stable process (even if it has high variation), the process mean should be stable. If it shifts or keeps shifting then the analysis becomes impossible. Can anyone clarify?
Tom0July 12, 2003 at 12:45 am #87887What the heck is a Six Sigma SPC course?
On the question of the mean shifting, go get some real data from a process and see what it is doing. Montgomery’s theory is good for books, but short on reality.
How would you ever verify that a mean is not shifting anyway?0July 12, 2003 at 2:45 am #87888
marklamfuParticipant@marklamfu Include @marklamfu in your post and this person will
be notified via email.Firstly, the most important condition of a stable process is to have small std, the high variation means process data can not follow statistical rules, at that time, we can not to precisely control process & predict what will be happen for the process because the population of process can not determined through samples analysis.
Secondly, if process have a small std, the mean will be limited as small swing range even have shift comparing to normal value.
A noshift mean can not indicate process is stable, “mean shift” also do not equal to “mean unstable”.0July 12, 2003 at 2:55 am #87889
Chris ButterworthParticipant@ChrisButterworth Include @ChrisButterworth in your post and this person will
be notified via email.If the process mean shifts – and they all do – the analysis isn’t impossible, it’s just not exact. That’s OK. You can still get some good work done. Using the word ‘impossible’ implies that any effort is a waste of time.
The goal isn’t to get a process stable so that you can get accurate Cpk indices. The goal is to provide results.
0July 14, 2003 at 3:28 am #87907Hi Tom,
Much discussion has taken place over the 1.5 sigma shift, both pro and con. Perhaps the best explanation I have come across that gives a real reason for including such a shift in our capability calculations is in an article by Davis Bothe in Quality Engineering (Volume 14, Number 3, 2002, pp. 479487), called “A Statistical Reason for the 1.5sigma Shift.”
I hope this article helps you better understand this concept.0July 14, 2003 at 11:44 am #87914With the exception of pedantic discussions on various theories the concept of the 1.5 sigma shift is simply “Shift Happens”. The purpose for this information is to force the observer of a process to continue pursuit of perfection.
I would not spend any more time on the topic than that…0July 14, 2003 at 1:58 pm #87921
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.“Firstly, the most important condition of a stable process is to have small std”
That makes no sense. Small compared to what? Does ot means that a turning opereation can not be as stable as a grinding operation because its variation is larger?
“A noshift mean can not indicate process is stable, “mean shift” also do not equal to “mean unstable””
Unless the shif of the mean is predictable and part of the normal process behavior and due to common causes only (pretty unlikely), a shif in the mean is unstability by definition.
The key thing is: real processes are never fully stable, and hence tend to show more variation in the long term than in the short term. That’s unstability. That’s reality.0July 14, 2003 at 2:42 pm #87923Well said and much nicer that what I thought about when I saw the statement about the small std deviation – how dumb is that?
0July 14, 2003 at 3:24 pm #87924
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Thanks. I also typed it much nicer than what I thought.
0July 14, 2003 at 4:28 pm #87927
Dr DarthParticipant@DrDarth Include @DrDarth in your post and this person will
be notified via email.Let’s not forget that the 1.5 shift was empirically determined after observing Motorola manufacturing processes over the course of better than a decade. As stated, shift does happen over a long enough period of time. For the most part, ignore the 1.5 and determine what your true process shift is between long term and short term variation.
0July 15, 2003 at 1:24 pm #87946
marklamfuParticipant@marklamfu Include @marklamfu in your post and this person will
be notified via email.Answer your question:
1. Small std. compare to product spec. and expected process capability , if expected CPk>1, small std means that 6std 1.33, small std means that 8std<(USLLSL).
2. For grinding process you mentioned, if machine setting have limited change, the parameter mean of product will be shifted, but, the std. may be no change. in other words, CPK is lower, CP is not change, if CPK is not lower than your expected value, the process still stable.
For a six sigma process, even process mean shifts 1.5 sigma , the process capability still enough (CPK = 1.5, CP=2, dppm=3.4), at that time, the process still is stable!
3. Any process have variation , “variation” is not “unstable”.
I am very gold to discuss this issue with you!0July 15, 2003 at 3:36 pm #87958
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Marklamfu,
My point of view is that you are just wrong, and that you are indeed making a couple of big conceptual errors: 1) You are mixing “stability” with “capability”. 2) You are failing to distinguish “variation due to common causes” and “variation due to special cuases”.
“if CPK is not lower than your expected value, the process still stable.”
Stability means equality of behavior over the time. Capability means ability to comply with specifications now and in the future. A process can be fully stable and yet have a Cpk of 0.5. Further more, a process can be fully stable and there could be no specification at all. For example, you could controlchart your body temeperature on a dayly basis and you will find it is consitently somewhere at random between 35 and 37ºC (it is stable), except in specific circumstances when there is a special cause of variation such as an infection and it goes to 39ºC (unstablity). Now tell me, what is the “specification” of your body temperature? What would the Cpk be?
“Any process have variation , “variation” is not “unstable””
That’s true, all processes have variation and that’s why we study them in a statistical way. And it is also true that “variation” is not “unstable”. Not necesary, but can be. Random variation is never unstability. Nonrandom variation is allways unstability (because you have a change in the behavior). I can not imagine a 1.5 sigmas shift in the process mean and call it “random varriation”. If the process mean shifted, then the process distribution changd, then the process behavior changed, then there you have unstability by definition.
“For a six sigma process, even process mean shifts 1.5 sigma , the process capability still enough (CPK = 1.5, CP=2, dppm=3.4), at that time, the process still is stable!”
Once again you are mixing capability (Cp, Cpk, sigma level) with stability. The reason why a 6 sigma process is “corrected” with a 1.5 sigma mean shift is just a kind of “safety margin” based on experience. The fact is: Real processes are NOT fully stable. Real processes DO shift. So let’s take that fact into account. There you have the1.5 sigmas shift. It is a measure of the unstability that there allways is and we cannot avoid.
Your simpistic reasoning that “good Cpk” = “stable” and bad “Cpk = unstable” is just meaningless. You just can’t put the “=” sign betewwn Cpk and Stability because they are not comparable. You are mixing apples with oranges. The right way is:
Good Cpk = Capbale; Bad Cpk = Uncapable
Random variation = Stable; Nonrandom variation = Unstable
That’s my view. I can be wrong, but I really feel I am not so wrong this time.0 
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