1.5 SHIFT SIMULATION

Reigle,
Shouldn’t it be AF1=AVERAGE(A1:AD1)+3*AE1? Whay are you adding the “sampling” standard deviation to the “population” average, which is supposed to be unknwon?
I mean, shouldn’t we compare Xbar+3S (what is the upper limit you can tell from the sample) against µ+3sigma (what is the upper limit of the actual population?

Reigle,
Give it up. Mikel has been exposed for what he is not (a change agent) and your groupie behavior is sad.
Try this. Set up a control chart with a given mean and a given std dev. Shift the mean and run 30 samples of 5. How long is it before the shift is detected? One? Two at most?
Mikel and Reigle have no clothes. (actualy Mikel has no clothes and Reigle pretends not to see that)

Reigle,
 
Congratulations!
 
You have successfully provided a simulation that proves that the upper (1-alpha)*100% confidence interval for the sample standard deviation for n=30 and alpha is small is approximately 1.5.  Gee, I think they are doing a similar proof in green belt training now.  Maybe you should sign up for one so that you can get a grasp on basic statistics.
 
Unfortunately, you have not provided a proof or simulation of the 1.5 sigma shift.  Let’s summarize your simulation:
 
Steps 1-3:  Calculate a series of sample standard deviations (S) from a random sample of n=30, average = 100, s = 10.
 
Step 4: Determine the 3*standard deviation upper limit for the sample based on the population average (100+3*S).  If we take the worst case level of S (upper 99.5 limit), S will be approximately 15 (1.5* s) so this value will be 145.
 
Step 5:  Determine the difference between the upper 3*sigma population limit and the upper 3*S sample limit based on the population average (D): D = (100+3*S) – (100+3*s) = 3*(S- s).  For worse case D = 3*(15-10) = 15.  Essentially, we have calculated the linear distance between the upper 3*S and the upper 3*s limits.
 
Step 6: Divide D by s: shift* = 3*(S- s)/ s = 3*(c-1) where c= S/ s and since c for n=30 and alpha = 0.005 is approximately 1.5, shift* = 1.5.
 
Step 7-8:  Rinse and repeat and graph.
 
So we know that the worse case difference between the upper 3*S and the upper 3*s limits is 1.5* s.  
 
So what!  The whole purpose of calculating the Z value is to determine the probability of the process producing product beyond the spec limit; or the percent of the population out of spec.  Determining the number of standard deviations between the “point of unity” of the short term and long term distribution is a meaningless operation.  The distribution with the true population sigma does not have the same pdf of a distribution using the worst case sampling error estimate of sigma.  Shifting the Zst (Z of the true population) by 1.5 sigma does not produce the same cumulative probability beyond the upper spec limit of the sample distribution except in the unique situation that Zst is 4.5.
 
If, in your example, the upper spec of this process is 145, then
Zst = (145- 100)/10 = 4.5 and Zlt=(145-100)/15 = 3.0 and the Z shift is 1.5. 
 
However, let’s make the upper spec 130. 
Then Zst = (130- 100)/10 = 3.0 and Zlt=(130-100)/15 = 2.0.  Now the shift is 1.0
 
Want another example?  Lets make the upper spec 160:
Then Zst = (160- 100)/10 = 6.0 and Zlt=(160-100)/15 = 4.0.  Now the shift is 2.0
This example shows that using your assumptions about sampling error, a process that is 6 sigma short term is 4 sigma long term not 4.5.
 
Lets make this the general rule and throw out this 1.5 shift:
Let C = SLt/Sst
Then ZLt = Zst/C
 
Or if the goal is to determine the S.qual based on the required S.design, then you can simply divide S.Design by C.  The arguments to use the “shift” is only adding complexity and confusion and limiting the application of C.
 
Statman

Reigle,
 
You said:
 
“By your own math, you are now showing the algebraic equivalency between an inflation of the standard deviation (due to random sampling error) and a linear off-set in the mean (exactly what Dr. Harry said in his book)”.
 
No, I said: “Essentially, we have calculated the linear distance between the upper 3*S and the upper 3*s limits.”  And I go on (for about the fifth time on this forum) to prove that a linear off-set in the mean is not equivalent to an inflation of the standard deviation due to the non-equivalency in the pdfs of the two distributions.
 
 
Give me one example from process study or new product qualification in which there is no consideration of the range of acceptability of the process (ie the specs).  Without the specifications, the capability of the process has no meaningful context.
 
Let me give you this from your example: 
 
Mean short term is 100, standard deviation short term is 10.  We will only launch the product if the PPM long term is less than 400. 
 
It has also been determined (through extensive research) C= S.lt/S.st = 1.5
 
Do we launch the product?
 
Please answer this for me with no additional information
 
Statman

Ok, so maybe this reply won’t add value, but it’s late and I can’t sleep.  Especially after reading Reigle’s last response.  Reigle, this isn’t a jab – just some constructive criticism.  I have yet to read a response from you that hasn’t made my head spin.  I openly admit I’m not deeply involved in math or statistics.  The reason so many people anticipate responses from Statman, Stan, Gabriel, Mike Carnell is because they stress simplicity.  If you feel constant hostility after you post, I thinks it’s because others are left confused by what you’ve said.  I’m pretty sure every post from you is about the 1.5 sigma shift.  Do you have ideas or experiences in other aspects of Six Sigma?  One thing I can count on from your responses is that you’ll use WAY TOO many words and not even answer the question posed.  From someone sitting on the sidelines, I believe it’s really easy to prove a theory when you constantly make up the numbers to support the theory.  The Six Sigma community is large – if anyone has some real data that supports the shift, let’s see it.  Let’s use it. 
Enough said.  I’m off the box now. 

Using subgroups of 4 (to make it simple because sqrt(4)=2) a shift in the process average equal to 1.5 process standard deviations equals a 3 sigma shift in Xbar (remember the centrla limit theorem?). Then your new shifted µ will match the CL for Xbar. That means that 1 out of 2 points will be beyond the control limit. Using only the “beyond control limits” criteria the chcances to detect such a shift are 0.5, 0.75, 0.87, 0.94, 0.97, … , 1-0.5^n in the first n points. Add to that the criteria of 2 out of the last three points in C zone (beyond 2 sigma to the same side) and you are almost certain to detect the shift in the first two points.
Stan is right on this. With subgroups of 5 (probably the most typical subgroup size in the industry practices) a 1.5 sigma shift will be asily detected in the next point, next two points at most.
An you should know that. Or you could have run the simulation proposed by Stan.

thanks Gabriel,
Good to be back and good to hear from you
Statman

Gabriel,
 
You forgot a few more constraints on the use of the 1.5 shift:
 
Ø      Is valid only when the actual difference between short and long term can not be empirically established through components of variation analysis
Ø      Is valid only when the process is free from special causes and only subject to the uncertainty of sampling error
Ø      Is valid only when the underlying distribution of the process is normal
Ø      Is valid only when the metric is continuous
 
This is why I think a better title to Harry’s book is “The incredible Shrinking Shift”. 
 
I also love the argument that this book is not about the justification of the use of the 1.5 sigma shift but about where it comes from.  This is like writing a book about why it was once believed that the earth was flat.  The only difference is that we are still seeing the practice of inappropriate application of the shift.  Only crack-pots of the flat earth society are doing calculations based on a flat earth.
 
Cheers,
 
Statman

Reigle,
 
Obviously, you have no intention to solve the problem that I gave you.  I have posted the solution below.  This is to show the users of this form how misleading the 1.5 shift is and the dangers of using it even when we have established that the ratio of long to short term process standard deviation is 1.5.  After reviewing this solution, it should be clear that the concept of shifting the distribution mean by C  to compensate for an inflation in the standard deviation by C just flat wrong.
 
 
Solution to the example:
 
Mean short term is 100
 
Sigma short term is 10
 
Z.st is 5.0
 
Ratio of the C = S.lt/S.st = 1.5
 
Requirement to launch the new product is that the PPM defective long term is no greater than 400s
 
DO WE LAUNCH THE PRODUCT?
 
Using the formula: Z.lt =  Z.st/C
 
Z.lt = 5.0/1.5 = 3.3333
 
Pr(Z>3.333) = 1-NORMSDIST(3.3333) (in excel) = 0.000429
 
0.000429*1,000,000 = 429 PPM
 
Since 429 > 400, the product does not meet requirement to launch – Do not Launch product
 
Note:  Using the Harry/Reigle method of the 1.5 sigma shift, we would draw the wrong conclusion since:
 
Z.lt = Z.st-1.5 = 5-1.5 = 3.5 and
 
Pr(Z>3.5) = 1-NORMSDIST(3.5) (in excel) = 0.000233
 
0.000233*1,000,000 = 233 PPM
 
Proof:
 
Z.st = (USL-T)/S.st  
 
5.0 = (USL – 100)/10  USL = 150
 
C = S.lt/S.st

Reigle,
If it makes you feel better, I can agree to always disagree with you if you are just spouting Mikel retoric. Much ado about nothing of value.
Got anything worthwhile or original to say?

An animated Shewhart chart with 1, 2 and 3 shift is available at http://www.multiqc.com/shewhart.htm

Reigle,
 
Thank you!!!!
 
This is what I have been saying all along!  There is no linear shift in the numerator of the Z equation that will compensate for an inflation of sigma in the denominator and the “shift” is dependent on the size of Z.st.
 
We are finally in full agreement!!!  Thanks for coming over to my side!
 
Since you arrived at 3.3333 (Z.lt) by multiplying Z.st by (1-K) and K = 1-1/c, it is quite easy to show that your equation is equal to my equation:
 
(1-k) = 1/C
 
So the correct calculation is Z.lt = Z.st/C          not Z.lt = Z.st – Z.shift
 
Of course, we always have an estimate of C don’t we?  It has been shown in your simulation that it is the upper 99.5% CI on the sample standard deviation.
 
Therefore, we need to change all of those conversion tables don’t we?
 
No more shift – YES
 
Statistics is saved!
 
Reigle, Please tell Dr. Harry for me that he has been wrong all this time in his advice to adjust Z.st by subtracting 1.5 to get long term.  He may take it better from you.
 
Cheers and once again thank you,
 
Statman
 
Ding Dong the shift is dead the wicked shift is dead.

Got me?  What do you mean you got me?
I couldn’t be happer!  More evidence that the Shift is dead!
Ding Dong the shift is dead……

Reigle,
 
 
“You talk-the- talk, but will you be able to walk-the-walk?”
 
What are we going to do arm wrestle over this?
 
I hope that the competition is better than what I have seen so far.
 
Statman
 

Go create a data set with a 1.5 sigma shift and apply SPC. One maybe two samples are all that is needed to detect and if the shift occurs concurrent with some process change (most do) and we follow simple rules of how to turn on a process after a change, you will never turn on a 1.5 shift. Period. Go try it instead of coming back with something that is a quote.
Reigle – go try it and report back the results.

Aside from silly labels, there is nothing the least bit original about Dr. Mikel.
Go read Managerial Breakthrough by Dr. Juran from 1964 if you want to see where Mikel got the idea for the Breakthrough Strategy. Mikel read it in 1983. Strange but true.
 

Yes,
I’m done.  The topic is now dead….like the shift
Cheers,
Statman
PS what should we discuss?

Reigle,
Thank your for your efforts  in raising these issues. I appreciate your dielectic. (Sometimes I suspect that you are playing the devil’s advocate!)
Without wishing to offend Dr. Wheeler, may I suggest that he study Shainin’s (or whoever invented it) randomised sequence as I’ve found this helpful in isolating temporal variation – even when the objective is only achieved after several processing steps. Another method that uses randomisation to detect temporal variation is the ‘split lot.’
As has been pointed out previously, X-bar and R charts are very senstive to small deviations of the process mean. More important though, in many processes the batch uniformity is a much larger source of variation than any variation in the mean, which was totally missed by corporate Motorola, but not by SPS. And it is truly sad to learn that SPS is to be sold off this first quarter.
Cheers,
Andy
 

Reigle,
If you are really setting up a debate (I think you are just pretending – like playing house or playing cowboy, something you and Mikel should know about), I will cover any of Statman’s logistics including any and all costs. Statman – do you want to stay and eat at the Phoenician?
I will also make sure everyone interested in such a debate is given the simplest of simulations that shows that basing this agrument on SPC is silly. A 1.5 shift simply would come and immediately be dealt with. So simple – not worthy of all the smoke and mirrors you and Mikel like so much.

I concur .. the purpose of a control chart is to find sources of variation and eliminate them – not to guardband against them.
Even if there is a ‘mathematical’ type of variation in a ‘theoretical’ process, which I wouldn’t descibe as a shift; it is possible to use both warning limits (2 sigma) and control  limits on a Shewhart Chart to detect smaller ‘transients’ in the mean, using the rules suggested by the Western Electric Company.
·             2 of 3 consecutive points fall outside warning (2-sigma) limits, but within control (3-sigma) limits.
·             4 of 5 consecutive points fall beyond 1-sigma limits, but within control limits.
·     8 consecutive points fall on one side of the centre line.
I agree with previous comments  – the 1.5 sigma shift is unnecessary and confusing and detracts from the ‘real’ issues affecting western industry  – rolled yield.

No you told me that k = 1 – (1/c)  then you multiplied Z.st by (1-K)
I told you that 1-K = 1/c and that you have done the same calculations that I have done to get the correct answer:
Z.lt = Z.st/c
 and now we are in agreement that the formula
Z.lt = Z.st – Z.shift   is BS
We have declared this string dead until after the debate.  But if you would like to post a comment to it each day, I am all right with that because I like seeing the topic “Death blow to the 1.5 sigma shift” show up each day.
Statman

Never heard of him. 
But tell him Hi for me and I look forward to meeting him.  I did visit his   —- website which I had also never heard of.  Tell him that he looked like a deer caught in the headlights on his CNBC appearance.  I guess he is not use to being on tv.
I hope that you will have more than just gear heads – Mech-Es –  at this debate.
Give me some possible dates.

Hypothesis:Let S.lt be any real number.Let c be any real number other than zero.Let’s define S.st = S.lt/c.Let’s define k = 1-1/c
Thesis:There exist S.shift that belongs to the real numbers such as:S.lt = S.st – S.shift
Demonstration:k = 1-1/c ==> c = 1/(1-k)S.st = S.lt/c = S.lt/[1/(1-k)] = S.lt*(1-k) = S.lt – k*S.ltBecause of the property of the addition and division of being closed operations in the real numbers, since c is a real number other than zero then k = 1-1/c is a real number.Because of the same property of the multiplication, S.shift = k*S.lt is also a real number.
QED
So yes, S.st = S.lt – S.shift.
However, this is as trivial as saying that a number can be expressed as the sum of other two numbers. The question is wether there exist ONE S.shift or not, because we don’t need Dr Harry to tell us that there always exists some number S.shift such as S.lt = S.st + S.shift. This number just happen to be S.lt – S.st.
And note that I dind’t declare what S.st, S.lt, S.shift, c, or k are, and I didn’t even mention standard deviations, chi-square distributions, or any other statistical concept, as I didn’t mentioned processes, specifications, subgroups, etc. It’s just abstract algebra.
“The shift exist, but we never said it was a constant” Gee, thanks for the new and revolutionary info.

Maybe I am missing something but is the argument whether a shift exists or whether it is exactly 1.5?  I have been debating the notion of the shift for the past decade but never from the perspective of whether processes can shift over time.  My issue is whether the absolute value of 1.5 applies to every organization.  It has been my impression that each organization needs to assess its long term process variation and compare it to the short term variation to determine what is an appropriate number.  Most of the rangling in this thread has been to either prove or disprove the 1.5.  I have been using control charts on a number of processes for 5 years.  If I compare the variation over that time period and look at it on a composite month to month basis I find that my process shifts only about .83 sigma.  Shift happens but why would I be inclined to use 1.5 rather than .83?

Praveen,
you said:
“Another scenario, if everyone uses a different level of shift, and every one would be calling their processes as six sigma. Now, there could be a race, hey, my six sigma is better than yours because of smaller shift!”
That is just flat silly.
The size of shift is not of critical importance.  what is important is that the correct compenstation for long term inflation in the variation is used to precisely estimate the defect rate of the process.

Stop your naked promotion of your group and it’s strategic partners!!!!
And I am not interested in your recommended reading list
I will ask isixsigma to remove this and any other future business promotions.

It is interesting that YOU are accusing someone of lacking integrity.  Don’t people with integrity do what they say they are going to do?  I thought I remembered a string of comments from you saying that you were not going to participate in this forum anymore…  And yet, here you are again pushing your 1.5 shift / 3.4 dpmo / Harry is the man agenda.  If you had integrity, you would do what you said and stop taking up space.
I think you also said that there would be some great debate on this topic.  Haven’t seen that yet either… 

Dear Gentlemen,
 
It seems that a public debate should be a good opportunity to escalate this mental combat you’ve been fighting at this forum.
I, myself, am based in Europe and know that there is a Six Sigma summit 27th through 30th of April in London. I suppose that in the US there must also be such a venue. I also assume that a company organizing such a venue would be more than willing to provide a stage for the above mentioned debate. You might probably be at such a summit anyway. I also suppose that from within the adience (or indeed the adience itsself) proper judges can be appointed.
So in true mathematical ways: the problem can be solved, and is therefore no longer interessting as it only requires basic hard work to get to the solution.
I  implore all parties involved (Reigle Stewart, Statman and the party organizing  a Six Sigma Summit) to take perform the basic hard work to have this debate realized.
 
 
Kind Regards,
 
Hein-Jan

Hi Darth,
 
I am responding to your post so that the title of this string will show up again today – Just kiddingJ.
 
I am working on a theory about the ratio of long term to short term variation of a process and the number for the difference short to long (0.83 sigma) that you said you have observed in your process actually fits a hypothesis of mine.  But I am curious about a couple of things.
 
Ø      How did you calculate this – did you determine it by subtracting Z.lt from Z.st?
Ø      Is your process centered at target?  And if not, did you use |SL – Xbar| or |SL – T| in the numerator of the Z value?
Ø      Is your process in (or nearly in) statistical control?
Ø      What is your Z.st?
Ø      What is the ratio of S.lt/S.st (S is the standard deviation)?
 
Just want to check a few things out.  I am assuming the this process performance is measured with a variables output.
 
 
As far as your question about the debate, I don’t think that anyone will argue that the long term variation will not be greater than short term even when the process demonstrates statistical control.  This has been recognized for years.  In fact, it is the concept behind Cp vs Pp calculations that have been around much longer than six sigma.  The question is about the 1.5 shift and it’s appropriateness in estimating long term capability.
 
Thanks,
 
Statman