Statman
@StatmanMember since October 6, 2003
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April 9, 2004 at 1:48 am #98293
Hi Malcolm,
I do not tell someone something that I don’t believe or that I can’t justify.
I tell them that a six sigma process does not have 3.4 PPM defective it has .001 PPM. But I also tell them that a six sigma process will be held in statistical control for the long term. We all know, however, that a process held in statistical control for the long term is very rare and therefore, we will most likely see more defects than 0.001 PPM.
Knowing that a process will drift and shift over time and the amount of drift and shift will depend on the level, method and economics of control, some have attempted to create a fudge factor (a swag), to estimate the lack of statistical control using a 1.5 shift to the Ztable. There is no generally accepted mathematical, statistical, or empirical justification for this swag.
The confusion comes from a rather bazaar convention in six sigma to report the process as if it were held in statistical control for the long term, or in other words, the short term performance is equal to the long term performance. After all, if the process is going to produce 3.4 PPM it is a 4.5 sigma process not a 6 sigma process. When we add to this convention the 1.5 sigma swag, we get this disconnect between the z table and “6 sigma quality level = 3.4 defects per million opportunities.
My hope is that this disconnect will eventually die out and the George Group and other growing contributors to six sigma can help by refusing to continue teaching this mendacity. After all, Six Sigma draws its strengths from being based in data and sound theory; we should practice what we preach.
Cheers,
Statman0April 9, 2004 at 12:02 am #98284Please find a different screen name. Statman is already taken.
You ask a good question however, and I think that Stan has given you very sound advice. Adding a constant to make all values positive before you transform is usually the safest method.
Cheers,
Statman (the original)0April 8, 2004 at 7:49 pm #98277Thanks Gabriel, this is a big help
0April 8, 2004 at 3:51 pm #98224It is a real problem, with real data
0April 8, 2004 at 1:16 am #98198Malcolm,
You can try to run from this but you can not hide.
You say that you have provided an explanation from a credible source. However, any one that has read Bothe’s book would know that he has no intention to “explain” the shift. If you notice, this quote that you posted is in the last chapter (14th) of a book on process capability measurement. It is on the 832 page of a 900 page book. He has spent 13 chapters and 800+ pages explaining how one should go about measuring and using process capability metrics in just about every imaginable situation. This chapter is almost like an after thought. I would think that if this so called short coming of SPC that would leave one to suspect a error of such magnitude, he would have addressed it much sooner than page 832.
Also, this chapter does not provide an explanation nor does it provide a justification but only explains why the shift is used by some six sigma practitioners. You will also notice that he never makes a judgment as to it being correct or incorrect. Considering the timing of the publication of Bothe’s book, (1997) one can only speculate on why he included this add on chapter.
You said that “a copy of that chapter is what I give Black Belts in training who ask the same question and want more detail than I’ve provided here.” Maybe you should think about giving them the other 13 chapters and ignore this one because all you are doing is perpetuating this misinformation and propaganda. It is disingenuous of you to then carp about it not being practically defensible or just a academic construct.
Rather than continuing this perpetuation, maybe you should sit down and try to prove it yourself. What I think that you will find is even if the source of the shift is transient and not constant (will come and go). An inflation of the long term variation relative to the short term variation would have to see a rate of transient disturbances in an order of magnitude that would contribute over 55% to the total variation to justify a 1.5 shift and a Shewhart control chart will detect this with an ARL of 2. Another way to look at is a capability analysis of 25 subgroups of 4 will have a probability of not detecting a special cause using rule 1 of the western electric rules will be 1.59×10^53 for a process that has a 1.5 shift.
As much as you are trying to show your outstanding wisdom of six sigma on this site lately, I think I will stick to the tried and true advice of Stan, Darth, and Gabriel.
Statman0March 8, 2004 at 3:44 pm #96571SSBB,
You are using a normal approxamation to calculate the confidence intervals for a binomial distribution. The published article has used the exact calculations from a binomial.
When np and n(1p) are both greater than or equal to five and n is greater than 30, then the normal approximation is pretty good. If not, as in your case (n(1p) =0.987), the approxamation is not so good.
Minitab will calculate the exact binomial CIs for you in the 1proportions test.
Statman0March 8, 2004 at 3:32 pm #96567Hi SSBB,
That should have been: BINOMDIST(1,20,0.25,TRUE) = 0.0243.
We are looking at the probability of 1 success out of 20 when the true proportion is 25%. where a success is a mismatch. The Confidence intervals were determined in Minitab using the 1proportions test and exact calculations.
Sorry for the error. I had the right calculation and logic just the wrong formulas in the post.
Cheers,
Statman0March 5, 2004 at 9:29 pm #96511I gave you an answer. If you don’t like it follow the other advise that you got. It appears that you do not know enough about what you are doing to follow a lecture.
What is your Pchart telling you? How are you using a Pchart if you have varing sample size? Shouldn’t you use an npchart for varing sample size?
Statman0March 5, 2004 at 8:09 pm #96500Gabriel,
Yes, the control limits are conservative and biased towards preventing a type I error. This of course makes sense we have a situation where the cost of adjustment to a process is high relative to the cost of small shifts in the process. We tend to treat these alpha risks as if they were handed down by God and surely, I think we could agree that God likes 0.05 as much .0027.
Your suggestion of increasing the alpha risk to a 2 sigma decision line is one option. One way to accomplish this is using EWMA or CUSUM type of charts. Another option is to increase the number of data points by decreasing the area of opportunity. In other words, look at defects per week rather than per month. This will give us 52 data points and higher confidence in detecting a shift when using the run rules. If the defect rate is very small than using an exponential control chart and assessing units between occurrences is another option.
My objection is in using traditional hypothesis testing (an enumerative method) in an analytic application. In hypothesis testing, we decide whether the unusual values are simply different because of random sampling error or they because they are truly different from others. Reference distributions have been developed that tell us what the probability of this sampling error is in a random sample obtained from a population. In order to apply this principle to a sequential sample from an ongoing process, there must be the assurance that the process is in statistical control. In order to assure the process is in statistical control, one must control chart the process. If you are control charting the process, then hypothesis testing is, at best redundant and can be misleading.
Also, how do we determine the before and after sampling frame? We could continue to increase the before by going back in history until we have reached a sample size that will make any change, real or random, significant. This makes the practice irrelevant.
I feel like I have posted this before.
Cheers,
Statman0March 5, 2004 at 5:19 pm #96483Gabriel,
As usual, you make some very good points.
To build on what you are saying, I would rather see the BB close the project when they have reached the point of dimensioning return on the project rather than continue to try to work toward an unrealistic goal. Too often, we start the project by swagging an unrealistic goal. It is not until the BB has reached the analyze phase that a realistic goal can be assessed with data. Yet, we keep the BB on the project even though the return on the investment (BB and team time) is very low.
For a program (Six Sigma) that prides itself on making “data based” decisions, the goal setting in the define phase is anathema. These goals are usually nothing more than a swag or educated guess. Holding hard to these goals without some common sense analysis when data is available is foolhardy and will lead to some potentially negative results such as keeping a BB and team engaged in a project that is getting little return on investment when they could be deployed elsewhere – another project with higher return. Organizational interest in the project will wane after about 4 months and it is not a efficient use of resources.
Cheers
Statman0March 5, 2004 at 4:49 pm #96480Lets see, sequential months of data and I want to know if there has been a shift in the process……I don’t know maybe I could use a
CONTROL CHART
Try either a Cchart or a Uchart.
And fire the trainer that led you to believe that this is an application of hypithesis testing.
Statman0March 5, 2004 at 4:41 pm #96479Joe,
Thank you for the compliment on my post.
I have yet to see a good reference for attribute Gage R&R that includes a discussion of Cohen’s kappa coefficient. Although I have not spent anytime looking for one so there may be one out there. A brief but very good discussion of Kappa is in the following Quality Progress article:
D. Futrell (1995). “When Quality is a Matter of Taste, Use Reliability Indexes,” Quality Progress, 28 (5), 8186
I tend to look at an application of a method and try to understand the underlying statistical/mathematical methodology and assumptions then search for the reference source. In most cases, the method is nothing more than the application of a fundamental inferential statistical concept that has been renamed to give the author credit or association to the application. After all, there are only four inferential (analytical) applications; all the thousands of methods are just variations on those four themes for various situations.
Most of the in depth work in qualitative measurement system analysis has been in the field of Psychometrics not industrial statistics.
Cheers,
Statman0February 22, 2004 at 6:30 pm #95902Newby,
Your Quality Guru is full of it.
As is often the case, the Guru is misinterpreting the concept of confidence intervals. All values within a confidence interval do not have equal likelihood of occurrence. There is not a uniform distribution of probability across the CI. A better way to state the question is: given the true proportion of agreement between the appraiser and the standard is 75%, what is the probability that on a given evaluation the appraiser would score 19 or better out of 20? The answer to that from excel is BINOMDIST(19,20,0.25,TRUE)= 0.0243. So there is only a 2.43% chance that the appraiser is at a 75% agreement level and will score 19 out of 20 or better.
If you had run 30 samples, there is still a chance, albeit smaller, that the true proportion of agreement is 75% with 29 out of 30. It is BINOMDIST (29, 30, 0.25, TRUE) = 0.002. The table below lists the probability of the result (19 out of 20, 29 out of 30, and 28 out of 30) for various assumed true proportions of agreement (P.null). I have also listed the resulting 95% lower CI for each result. If you were to run 30 trails rather than 20 with 29 matches, the lower limit would only increase to 82%. If there was one additional failure, the lower limit would have been 78%. I don’t know what your gurus interpretation of an acceptable lower limit would be but if you have one mismatch, you would have to have 55 trials for the 95% lower CI to be greater than 90%. This, of course, makes the whole intent of the study trivial. As soon as you have one mismatch you might as well stop and declare the inspection system flawed.
P.null Matches Trials Probability 95% lower CI
0.75 19 20 0.0243 0.751
0.80 19 20 0.0692 0.751
0.85 19 20 0.1756 0.751
0.90 19 20 0.3917 0.751
0.95 19 20 0.7358 0.751
0.75 29 30 0.0020 0.823
0.80 29 30 0.0105 0.823
0.85 29 30 0.0480 0.823
0.90 29 30 0.1837 0.823
0.95 29 30 0.5535 0.823
0.75 28 30 0.0020 0.779
0.80 28 30 0.0442 0.779
0.85 28 30 0.1514 0.779
0.90 28 30 0.4114 0.779
0.95 28 30 0.8122 0.779
A better measure of the reliability of the inspection system is the Kappa coefficient. The Kappa coefficient is the proportion of agreement between raters and the standard after agreement by chance has been removed. Minitab has the option to give you Kappa values. As a general rule, if Kappa is lower than 0.7, the measurement system needs attention. The problems are almost always caused by either an ambiguous operational definition or a poorly trained rater. Kappa coefficients above 0.9 are considered excellent, and there is rarely a need to try to improve beyond this level. I have reevaluated your data based on the information you have provided and got the following results for Kappa:
Within Appraiser
Appraiser Response Kappa SE Kappa Z P(vs > 0)
1 f 1.0000 0.2236 4.4721 0.000
2 f 0.8746 0.2236 3.9114 0.000
3 f 0.8746 0.2236 3.9114 0.000
Each Appraiser vs Standard
Appraiser Response Kappa SE Kappa Z P(vs > 0)
1 f 1.0000 0.1581 6.3246 0.000
2 f 0.9373 0.1581 5.9280 0.000
3 f 0.9373 0.1581 5.9280 0.000
Between Appraisers
Response Kappa SE Kappa Z P(vs > 0)
f 0.9179 0.0577 15.8989 0.000
All Appraisers vs Standard
Response Kappa SE Kappa Z P(vs > 0)
f 0.9582 0.0913 10.4966 0.000
As you can see, the only concern is the within appraiser Kappa level.
You asked, “Is he yanking our chain or does he know what he is talking about?” In my opinion he is doing worse than yanking your chain. He is acting in the worse way that a coach/expert can act; be it that the coach is a MBB, BB, or Guru. He is playing data/method policeman and not providing any value added help to your project. Disregarding the work that you have done just because it does not meet some semiarbitrary sample size requirement is not value added. In one book on attribute MSA I looked some time ago had a stated 30 part requirement and then proceeded to show an example with 20 parts. I guess the rules only apply to the practitioners.
The questions he should be asking of these results and having you investigate are:
Ø What is the inspection used for?
Ø How critical is a failed good part?
Ø How critical is a pass bad part?
Ø Were the mismatches that 2 of the 3 appraisers experienced on the same sample?
Ø Were the mismatches passed bad parts or failed good parts?
Ø If they were passed bad parts, were the parts close to standard?
Ø If they were failed good parts, were the parts borderline acceptable?
Statman0February 21, 2004 at 5:20 am #95878PA,
You have received some interesting advice on your question. I am not sure if you feel better or worse about your situation so let me see if I can help. Actually, Tim Folkerts’ strategy was on the right track so I am just going to build off of what he was doing.
What you want to do is test the minimum number of parts that will tell you that you have reduced the number of leaks. There is no reason to continue to test if the number of failures in the test reaches a level that will force you to an excessively large sample size. Also, you do not want to continue to test if no or few failures have occurred. Therefore, you will want to take a sequential approach.
First, the hypothesis you are testing is Ho: P = Pnull vs. Ha: P < Pnull
I will also assume that you will accept a .05 alpha risk.
You must establish the current proportion leaking (this is Pnull). In one of your posts you stated that it was 20%, in another you said it was 30%. The current level of leaks will determine the sample size; the smaller this proportion, the more samples that will be required.
Below is a table of alpha and beta risks for various sample sizes, failure rates, and null proportions. Basically what this table is doing is determining the sample size required to control the alpha risk level at about 5% when failures of zero to five are detected in the test. The beta risk is the risk of not detecting a 50% improvement when 1 to 4 more failures are detected. In other words, if 1 failure is detected in 14 samples with p.null at 0.2 and you determined that there was no improvement, then you have a 41.5% chance of not detecting an improvement of 50%. I will explain how to use this table below:
P.null Failures Sample Alpha +1 Beta +2 Beta +3 Beta +4 Beta
0.2 0 14 0.0440 0.415 0.158 0.044 0.009
0.2 1 22 0.0480 0.380 0.172 0.062 0.018
0.2 2 30 0.0442 0.353 0.175 0.073 0.026
0.2 3 37 0.0450 0.309 0.160 0.071 0.027
0.2 4 44 0.0440 0.274 0.146 0.068 0.028
0.2 5 50 0.0480 0.230 0.122 0.058 0.025
0.3 0 9 0.0404 0.401 0.141 0.034 0.006
0.3 1 14 0.0475 0.352 0.147 0.047 0.012
0.3 2 19 0.0462 0.316 0.144 0.054 0.016
0.3 3 24 0.0424 0.287 0.139 0.057 0.020
0.3 4 28 0.0474 0.235 0.115 0.049 0.018
0.3 5 33 0.0414 0.217 0.111 0.049 0.019
Procedure: (I will assume that P.null is .30)
1. Begin the test
Ø If 9 parts are tested without a failure, conclude that there has been a significant improvement at an alpha = .0404
Ø If 1 failure is detected, continue testing up to 14 parts, if no more failures then conclude that there has been a significant improvement at an alpha = .0475
Ø If 2 failures are detected before you reach 9, (beta risk is .141), this is a judgment call. You can stop testing if the beta risk is acceptable and conclude that you could not detect an improvement or continue to test up to 19 parts. However, this does not allow any more failures before the end of the test. If no more failures occur, conclude that there has been a significant improvement at an alpha = .0462
Ø If 1 failure occurs before you reach 9 and one after you have tested 9 but before you reach 14, continue testing up to 19 parts, if no more failures then conclude that there has been a significant improvement at an alpha = .0462
Ø If 3 failures are detected before you reach 9 (beta risk is .034), stop the test and conclude the no improvement could be detected.
Ø If 3 failures are detected before you reach 14, (beta risk is .147), this is, once again, a judgment call. You can stop testing if the beta risk is acceptable and conclude that you could not detect an improvement or continue to test up to 24 parts. However, this does not allow any more failures before the end of the test. If no more failures occur, conclude that there has been a significant improvement at an alpha = .0424
Ø If 1 failure occurs before you reach 9, one after you have tested 9, but before you reach 14, and one after 14 but before you reach 19, continue testing up to 24 parts, if no more failures conclude that there has been a significant improvement at an alpha = .0424
Ø I would recommend stopping if the number of failures reaches 4 unless the 4th occurs after 19. In this case, you will want to continue to 28 since the beta risk is still 28.7% with 4 at 24.
The same type of strategy can be employed if the P.null is .20.
I hope this helps. It is better than testing 247 samples as some one has suggested.
Statman0February 19, 2004 at 5:16 pm #95789Who’s mathematics? Do you really believe that parameters estimated from a process in statistical control will be off by that much?
You have been drinkin too much from the well of Dr Harry
So what is Ppk in worse case?
Statman0February 19, 2004 at 4:30 pm #95782Interesting set of articles.
You not have noticed the little caveat in their discussion of the dangers of Cpk. If not I pasted it below:
“(Yes, for this example, we have ignored the first cardinal rule: Before one looks at Cpk, the process must be in control.)”
This of course invalids any thing the author is saying about Cpk. Makes you kind of wonder what his point is.
Also, I can’t find any thing in the articles that tells me that a larger subgroup size will inflate (over estimate) the capability or that a 1.33 Cpk will have .65% defective.
Statman0February 19, 2004 at 3:06 pm #95776You said:
“cpk is not a good measure because it is always inflated(more sub group size more better cpk)……. with cpk 1.33 you can have 0.65% out of spec parts”
What in the ____ are you talking about?
Statman0February 19, 2004 at 5:36 am #95751You have done successful Six Sigma Projects both at Stategic and Operations level and 36 years of business and academic experience plus a phd and yet you were still compelled to take this exam.
Why? Do you have some sort of neurotic need for approval?
What if you would have failed? Would you have considered your whole life to be a lie?
Maybe you just like to have a lot of letters after your name.
Statman0February 19, 2004 at 3:31 am #95747Now your talkin’ my language!
Throw in a few Belgium beers and we can have an interesting debate
Cheers,
Statman0February 19, 2004 at 12:00 am #95738You know Darth,
I wish I was as optimistic as you about the discriminating powers of the market. I can’t help but think about the scene in the movie “Raiders of Silicone Valley” where Bill Gates tells Steve Jobs that it is not about who has the best equipment or who has the best software, it is about who can sell it (or something like that).
I actually have less concern about the quality and competency of the Six Sigma training providers. At the end of the day, arguing about which one is better is like arguing about whether Bud light, Coors light, or Miller Lite are better or worse than each other. They all are about the same: bland, unsatisfying, and indistinguishable.
The concern, in my opinion, is what passes for “master” and expert internally in an organization; or what external consultation they are receiving on an on going basis. These people will have the most influence on the performance of a Black Belt, The training, whether it is 2, 4 or 6 weeks, will only provide an awareness level of Six Sigma not expertise.
I have always been under the opinion that the title of “Master” was a title of respect that one earned through years of demonstrated accomplishment and expertise in ones field. Now, you can go to a one or two week course to become a “master”. Worse, in most companies, it is an appointed position. Imagine if your organization got its legal advice from someone from engineering that was appointed chief counsel and sent to a 2 week training class in business law.
Statman0February 18, 2004 at 7:08 pm #95712As long as we are quoting Dr. Deming, how about this one from Out of the Crisis:
“…American management have resorted to mass assemblies for crash courses in statistical methods, employing hacks for teachers, being unable to discriminate between competence and ignorance. The result is that hundreds of people are learning what is wrong. No one should teach the theory and use of control charts without knowledge of statistical theory through at least the master’s level, supplemented by experience under a master. I make this statement on the basis of experience, seeing every day the devastating effects of incompetent teaching and faulty application.”
As relevant today as it was in 1982 when he wrote it!
Statman0February 9, 2004 at 3:20 am #95204Reigle,
Do you understand the difference between alpha risk and beta risk? Do you know how the power of a test is defined? How about an error of type I vs. an error of type II?
Maybe you have not been trained in the concept that there is a difference between RejectSupport significance tests and AcceptSupport significance tests.
Let me give you a little help in this area. A while ago you proposed a simulation by creating thousands of random normal samples with mean = 100 and standard deviation = 10. You showed that that about 0.5 percent of the standard deviations will be 14.9 form the random sample. You concluded that the worse case sampling error would cause you to underestimate the true standard deviation because it can be as big as 14.9.
However, THE TRUE POPULATION STANARD DEVIATION IS 10. Therefore, if you conclude that it is greater than 10 you are making an error of type I – the population variation has increased when it actually has not.
Acting in a way that might be construed as highly virtuous in the rejectsupport situation, for example, maintaining a very low alpha level like .005, is actually “stacking the deck” in favor of the design engineers theory that the variation has not increased ( an AcceptSupport situation). I do understand how easy it is for the novice to sometimes get confused and retains the conventions applicable to RS testing.
Statman
0February 8, 2004 at 10:51 pm #95201Reigle,
Let me get this straight. An engineer has just finished configuring a certain assembly. He has already determined the specifications on each of the 5 components without the determination of the requirements on the CTQ (which is the gap between the 4 parts and the envelope). The process is already spec’ed and up and running without consideration of the tolerances to meet the design requirement. He has already taken a sample which he has arbitrarily determined 30 is sufficient with no consideration of the power of the sample. And now he wants to know the “probability of interference fit?”
Talk about a contrived example. Or maybe is this what you consider DfSS and what you are teaching as DfSS. I can’t think about a situation that could be any more antiDfSS. Look at your situation: Arbitrary specifications on the components without consideration of the requirements on the CTQ, CTQ requirements (range of acceptability) not established prior to configuration (don’t you think that there may be a upper requirement on the gap – why not design the parts at 1.23 nominal so that there is no possible way to have “interference fit”), acceptable risk of violating the CTQ requirements not established but estimated after the fact, qualification sample set at 30 random parts with no consideration in the power in estimating process parameters or ability to assess process stability due to sampling at random.
This isn’t DfSS this is what has always been done. The only thing that you have added is an overly conservative fudge factor due to sampling error. And you couldn’t even do that correctly.
Let’s look at the proper sequence:
Determine the CTQ (the gap between the 4 parts and the envelope)
Determine the range of acceptability of the CTQ (I will assume since the nominal is .016 that the range is (0, 0.032)
Determine the acceptable risk of violating the requirements on the CTQ (3.4 ppm)
Determine the maximum acceptable level of variation in the CTQ
Ø S.gap = .016/4.65 = 0.00344 where 4.65 is the Zvalue that cuts of 1.7*10^6 in each tailEstablish the relationship between the components and the CTQ
Ø Gap = E – (P1 + P2 + P3 + P4)
Ø S.gap = sqrt((S.e)^2+ (S.p1)^2+(S.p2)^2+(S.p3)^2+(S.p4)^2)Determine the allocation of the variation in gap across the components – since the parts will all be made on the same process use equal allocation:
Ø S.gap = sqrt(5*S.c^2)Determine maximum acceptable level of the variation in the components
Ø .00344 = sqrt(5*S.c^2) – S.c = 0.001538Establish a validation sampling plan to assure that the components will meet requirements. This should be based on the minimum detectable bias in both the mean and variation. For example, a validation plan that calls for 25 subgroups of 4 will detect an inflation in S.c of 20% (c=1.2) with a power of 0.9895, a sample of 30 will detect an inflation of 30% with a power of 0.964
Determine the S.design as S.c/c where C is the minimum detectable inflation. (with a sample size of 25 subgroups of 4, c= 1.2 so S.design = 0.00128
Develop or determine the process for the components that will meet requirements on the components (nominal and variation (S.design)).
Verify the process using validation plan
Establish specifications and controls on the component processes.
In your example, the 30 part sample has estimated S.components at .001. With a sampling plan of 30 random samples This is less than what is required for qualification and less than what would be detected to reject the null (0.001538/1.3 = 0.001183). Therefore, you would qualify the process. The estimated worse case due to sampling error sigma level would be
Ø S.component = .001*1.3 = 0.0013
Ø S.gap = sqrt(5*(0.0013)^2) = 0.0029
Ø Z = .016/.0029 = 5.50
Ø PPM = .02
You know of course that you are stacking the deck in setting the alpha risk as low as .005 when the critical risk is the beta risk since we are concerned with S.qual being greater than what is estimated. In other words, we are concerned that when we fail to reject the null that the actual S.components is significantly greater than what is estimated.
But even when you have used the .005 “worse case inflation” of c=1.4867, you have shown once again that a shift in the numerator does nothing but confuse and add complexity. If c=1.4867, then the worse case due to sampling error is:
Ø S.component = .001*1.4867 = .0014867
Ø S.gap = sqrt(5*(0.0014867)^2) = 0.003324
Ø Z = .016/.003324 = 4.81
Ø PPM = .75
Your “equivalent “static” worst case condition to be Z.gap.static = 3.89 (due to equivalent offset in the nominals)” lacks any validity. You can not assess the amount of bias in the mean by assessing the sampling error in the variance.
Your client’s will realize the additional cost of over design due to the voodoo statistics your are promoting.
Statman0January 30, 2004 at 5:08 am #94840Book ain’t sellin is it?
Reigle,
I have been very busy this week and haven’t nor have I time to respond to your assertations about my claims in full I will try to get back to you this week end.
Mean time, I think that you need to go back and reread my poistion. I stated them very clearly in 5 points and a proof in at least two posts. Please reread the points and you will find that this socalled technical review has not addressed any of them the way that I have stated them. I will state them again.
The “mysteries that Dr Harry has (knowingly or otherwise) uncovered:
1. There is no constant shift of 1.5 (or any other number)
2. The shift from short to long term depends on the Zst ratio and the sigma inflation coefficient
3. A six sigma process short term will be 4sigma long term (owing to the worse case sampling error c=1.5) not a 4.5 sigma process long term (since Z.lt = Z.st/c)
4. Because of point #3, a six sigma process will have a DPMO of 31.69 not 3.4 long term.
5. A sigma shift of K in the numerator of the Z equation does not equate to a Z with a K*sigma inflation.
I will debate any one that thinks that they can support a statistical rational for the 1.5 shift in the applications recommended by Dr. Harry. (ie Z.lt = Z.st – 1.5). Including Douglas.
Statman0January 28, 2004 at 12:30 am #94715Stan,
Minitab will identify outliers on a Box Plot using the following rules:
“A line is drawn across the box at the median. By default, the bottom of the box is at the first quartile (Q1), and the top is at the third quartile (Q3) value. The whiskers are the lines that extend from the top and bottom of the box to the adjacent values. The adjacent values are the lowest and highest observations that are still inside the region defined by the following limits:Lower Limit: Q1 – 1.5 (Q3 – Q1)
Upper Limit: Q3 + 1.5 (Q3 – Q1)
Outliers are points outside of the lower and upper limits and are plotted with asterisks (*)”
But the original poster needs to be more specific about what is meant by an outlier. Is it data points that do not fit the probability of a reference distribution? Is it a unusual observation that does not agree with the predicted value from an emperical model?
I don’t know
Cheers,
Statman0January 23, 2004 at 4:16 pm #94570Hi 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,
Statman0January 23, 2004 at 12:28 am #94538Stop 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.0January 22, 2004 at 11:06 pm #94533Praveen,
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.0January 22, 2004 at 11:03 pm #94532What do you think?
Do I sound like I am insecure?
I certainly haven’t spent the last 20 years of my life trying to justify a miscalculation … nor would I.0January 22, 2004 at 10:10 pm #94528Yes Gabriel,
“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.”
You can always explain things clearer than I can and I am the one that English is my first language.
So we go back to my Points from my first post on this topic in November:
The “mysteries that Dr Harry has (knowingly or otherwise) uncovered:
1. There is no constant shift of 1.5 (or any other number)
2. The shift from short to long term depends on the Zst ratio and the sigma inflation coefficient
3. A six sigma process short term will be 4sigma long term (owing to the worse case sampling error c=1.5) not a 4.5 sigma process long term
4. Because of point #3, a six sigma process will have a DPMO of 31.69 not 3.4 long term.
5. A sigma shift of K in the numerator of the Z equation does not equate to a Z with a K*sigma inflation.
That’s all I am trying to say.
Thanks and Best Regards,
Statman0January 22, 2004 at 8:23 pm #94522Ah….That’s what I just said….
At least I can spell algebra0January 22, 2004 at 7:50 pm #94520Never 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 – MechEs – at this debate.
Give me some possible dates.0January 22, 2004 at 7:20 pm #94515No you told me that k = 1 – (1/c) then you multiplied Z.st by (1K)
I told you that 1K = 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.
Statman0January 22, 2004 at 6:57 pm #94513Reigle,
I said I was done commenting on this string but this one I can’t resist.
First of all, Stan is right and now you agree with him. Any sustained shift of 1.5 sigma will be detected quite rapidly with a common Shewhart control chart with n= 4. Using the 8 Nelson/Western Electric rules, the shift will be detected in fewer than 5 subgroups.
So to wiggle out of this, you have had to further constrain the application of the 1.5 shift and suggest that we are not talking about sustained shifts but transient shifts in the subgroup averages. So we have now eliminated a good 95% of the types of process shifts that would inflate the long term sigma of the process.
This leaves us only with those random transient shifts that can go undetected in a common Shewhart control chart; those fleeting buggers that can’t be detected in trend analysis. We are no longer talking about a shift in the numerator of Z but inflation of the standard deviation in the denominator (back to the same old argument). As you and Dr. H have shown us, this inflation factor to have an alignment of the 3 sigma tails by shifting the short term distribution 1.5 sigma is in the situation that:
C = S.lt/S.st = 1.5
So the question is: will an inflation of 1.5*S.lt due to transient random shifts between the subgroups go undetected in a common Shewhart XbarR chart with subgroup size of 4?
The answer is no, they will be detected quite easily.
First, let’s define S.shift as the standard deviation of the transient random shifts Then:
S.lt = (S.st^2 + S.shift^2)^0.5
Since S.lt = S.st*c and C = 1.5,
(1.5*S.st)^2 = S.st^2 + S.shift^2 and
S.shift = sqrt(1.25)*S.st = 1.118*S.st
Therefore, the standard deviation of the transient random shifts is greater than the short term (subgroup standard deviation). Since the control chart has subgroup of size 4, the standard deviation of the shifts is 2.24 times larger than the standard error of the control chart.
The transient shifts are random normal with mean 0 and standard deviation of S.shift in order for the control chart to not detect special causes and to apply normal theory. Therefore, with the condition of the ratio of S.lt to S.st of 1.5, (1NORMDIST(1.5,0,1.118,TRUE))*2 = 0.1797 or about 18% of the shifts will be greater than the 3 sigma control limits and a shift will be detected in at least 5 subgroups.
To test this I set up a random normal series of 200 data points with mean=100 and S.st = 10. I then added to that a random series of subgroup shifts with mean = 0 and S.shift = 11.18. Control charted with subgroups of 4 the data with the random transient shifts. The result are below. The shift was detected at data point 4. I have also performed a component of variation analysis on the data. You can see the S.st (within), the S.shift (Between) and the S.lt (total). C = 14.522/9.728 = 1.49.
Variance Components
Source Var Comp. % of Total StDev
Between 116.240 55.12 10.781
Within 94.640 44.88 9.728
Total 210.880 14.522
Xbar/S for C3
Test Results for Xbar Chart
TEST 1. One point more than 3.00 sigmas from center line.
Test Failed at points: 4 6 13 14 15 20 22 32 41 49 50
TEST 5. 2 out of 3 points more than 2 sigmas from center line
(on one side of CL).
Test Failed at points: 4 6 8 14 21 22 33 45
TEST 6. 4 out of 5 points more than 1 sigma from center line
(on one side of CL).
Test Failed at points: 4 6 31
Now I am sure you will come back an tell me that Dr. Harry already said this, or that He said it would only work when the planets were aligned a certain way or some silly thing like that. The point Mr. Stewart is the value of the 1.5 shift. Stay focused on that.
Focus….
Statman0January 22, 2004 at 2:55 am #94471Thank you Dave.
Statman0January 21, 2004 at 7:38 pm #94461John,
I hope that your comments about statistical thinking are in jest. Equating statistical thinking to muddled thinking is the reason the “popular press” can get away with the gross misstatements about our world. Many things can go wrong if we can not think statistically. Just as a writer can arrange words into convincing arguments or incoherent nonsense, data can be manipulated to be compelling, misleading, or simply irrelevant. The inability to think statistically can lead to overestimating the likelihood of rare events and underestimating common events, overestimation of the strength of evidence from small samples, and responding too strongly to how data is presented rather than to the strength of the evidence it contains.
Statistical thinking has often been confused with statistical methodology however, knowing the mechanics of applying a statistical method is not the same as thinking statistically. In fact, the confusion between statistical thinking and statistical methods may be at the root of why most people avoid the topic in the first place. Most introductory statistics courses and Six Sigma training focus heavily on the formulas and methods and avoid a detailed discussion of statistical thinking leaving the students with bewilderment on when and where they would ever apply what they have learned. I think that the common belief has been that if you teach the methods, the statistical thinking will follow. However, I believe that the opposite is true. The more one thinks statistically, the more likely one will seek out the methods for interpretation and analysis of data. In addition, just because someone has used a particular statistical method, it doesn’t follow that they have thought about the problem from a statistical point of view.
There have been several attempts to define statistical thinking. The American Society of Quality defines statistical thinking as a philosophy of learning and action based on three fundamental principles; all work occurs in a system of interconnected processes, variation exists in all processes, understanding and reducing variation are the keys to success. This definition is fine from a process improvement view, but it does not take into account variation and quantitative data in a scientific context. One of the best definitions I have read is “Statistical thinking concerns the relation of quantitative data to a realworld problem, often in the presence of variability and uncertainty. It attempts to make precise and explicit what the data has to say about the problem.” There is a very close tie between statistical thinking and scientific method. Rarely is scientific research presented in the form of absolutes. One of my favorite quotes relating to statistical thinking is “if mathematics is the language of science than statistical thinking is the language of scientific method”.
Why is it so rare to see statistical thinking employed in critical decision making and choice of actions? We have all heard the desire that we “make data based decisions” yet we can all think of critical decisions that have lacked the basic rigor of statistical thinking with data. I mentioned earlier that one of the problems is how statistics is taught. But I think the larger issue is how science is taught in general as well as our view of knowledge. Knowledge is often confused with fact. However, knowledge has two often over looked components that are not considered part of fact. They are the supporting raw data and a measure of uncertainty that the raw data supports the knowledge. Knowledge will often change when new data is surfaced that raise the uncertainty. Statistical thinking is about measuring the uncertainty between the raw data and the knowledge. In an ideal world we would use our knowledge to make decisions. In reality, there are many other influences that affect our decisionmaking. What often happens is information –not knowledge is filtered through a sea of preconceptions, misconceptions, inattention, shortterm memory, politics, pressures, fears, and lack of time to reach a decision. The only basis that minimizes these influences is when knowledge, supported by data, is presented with a measure of uncertainty.
That’s statistical thinking.
Cheers,
Statman0January 21, 2004 at 6:52 pm #94454You said:
“Your indepth knowledge of operations will definitely compensate for your lack of experience or expertise with statistics. “ (Emphases mine)
So I would guess that you feel statistical thinking plays no role in understanding the dynamics of business and operations provided you can compensate with other “indepth knowledge”.
Why should he waist his time reading this book or for that matter become a BB. In your opinion, it sounds like the lack of the statistical thinking dimension doesn’t matter.
Maybe he should find out what the role and responsibilities of a BB are before he asks idiotic questions0January 21, 2004 at 4:14 pm #94445I’m sorry, but which MBA courses provide “indepth knowledge of operations”?
How does one achieve this “indepth knowledge of operations” without an ability to think statistically about operations?0January 21, 2004 at 4:06 pm #94444You have confused statisical methods with statistical thinking.
Sad that some have only grasped the nuts and bolts about statistics and have not made the connection between statistical thinking and a “comprehensive understanding of the business dynamics”
After all the teachings of Deming, Juran, Crosby etc.0January 21, 2004 at 4:56 am #94433I’m glad to hear that the expenses will be taken care of. I don’t think that my parents would have paid for the trip. I just hope that they let me skip school on those days if it is before June.
Statman0January 21, 2004 at 3:54 am #94429Yes,
I’m done. The topic is now dead….like the shift
Cheers,
Statman
PS what should we discuss?0January 21, 2004 at 3:16 am #94423Reigle,
“You talkthe talk, but will you be able to walkthewalk?”
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
0January 21, 2004 at 3:01 am #94421Got 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……0January 21, 2004 at 2:59 am #94420Reigle,
Any thoughts on how we will now handle the issue that a 6 sigma process short term will produce 32 PPM long term when for years we have been advertising as on 3.4?
That change of almost an order of magnitude in the number of defects in a 6 sigma process may be hard for some to swallow.
Just curious,
Statman0January 21, 2004 at 2:54 am #94419Yes, the MBA will be a big limiting factor. But with the right instructors, you should still survive.
Maybe0January 21, 2004 at 2:42 am #94418Reigle,
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 (1K) and K = 11/c, it is quite easy to show that your equation is equal to my equation:
(1k) = 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.0January 20, 2004 at 4:28 pm #94394Oops, cut off the proof:
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) = 1NORMSDIST(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.st1.5 = 51.5 = 3.5 and
Pr(Z>3.5) = 1NORMSDIST(3.5) (in excel) = 0.000233
0.000233*1,000,000 = 233 PPM
Proof:
Z.st = (USLT)/S.st
5.0 = (USL – 100)/10 USL = 150
C = S.lt/S.st
1.5 = S.lt/10 S.lt = 15
Z.lt = (USL – T)/S.lt = (150100)/15 = 50/15 = 3.3333 not 3.5
Statman0January 20, 2004 at 4:25 pm #94393Reigle,
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) = 1NORMSDIST(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.st1.5 = 51.5 = 3.5 and
Pr(Z>3.5) = 1NORMSDIST(3.5) (in excel) = 0.000233
0.000233*1,000,000 = 233 PPM
Proof:
Z.st = (USLT)/S.st
5.0 = (USL – 100)/10 USL = 150
C = S.lt/S.st0January 20, 2004 at 2:52 pm #94390No Reigle,
You are wrong about the use of “if” in a debate about the logic of an argument. The use of “if” is to declare a premise to the argument that shows that the argument is either invalid or limited. Your task is to now show why the premise is not valid.
The declaration to “agree to disagree” is often incorrectly thought to be a declaration that the debate is now at a stalemate. However, this is actually a declaration that you have lost the debate because you do not have a counter argument.
It has been, once again, enjoyable to engage in a mental joist with you.
Hope you can return once again when you have retooled your arguments.
Statman
0January 20, 2004 at 2:33 pm #94388Gabriel,
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 crackpots of the flat earth society are doing calculations based on a flat earth.
Cheers,
Statman0January 20, 2004 at 1:58 pm #94386thanks Gabriel,
Good to be back and good to hear from you
Statman0January 20, 2004 at 1:54 pm #94385Your first example is bogus. The capabilities of the two machines could be equivelent from an economic impact if the range of acceptability is wide relative to the process spread of the two machines. Remember Reigle, a difference is a difference only when it makes a difference.
Your second example is silly. You can keep all the databases you want but they only have meaning in the ability of the process to meet customer requirements.
You said: “After all, I do believe this is a key principle in DFSS … “establish specifications based on the existing capability, not the other way around”
I don’t think there is a practicioner in DFSS that will agree with you on that one. The principles of DFSS is to meet or exceed the customer expectations and produce the produce at a high level of capability.
Since you are struggling with my example, let me give you some additional information:
the Z.st = 5.0
So mean, Std dev short term are 100 and 10
PPM defective must be less than 400.
C = 1.5
Do we launch?
Statman0January 19, 2004 at 10:57 pm #94361Reigle,
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 offset 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 offset in the mean is not equivalent to an inflation of the standard deviation due to the nonequivalency 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
Statman0January 19, 2004 at 6:25 pm #94347Reigle,
This is the greatest amount of BS that you have ever posted:
“So if the points of unity are exactly aligned, then the sampling distribution and population distribution are probabilistically equivalent at the points of unity only. Again, remember that the points of unity define the limits of process capability (by quality convention, not my convention … like in CP we assume unity +/ 3s)”
Maybe you should recheck the formula for CP. You will notice that the formula for CP is the ratio of the specification range to the 6*sigma range. The CP STATISTIC DOES INCLUDE THE SPEC RANGE!!!!
You are adding nothing but confusion by trying to justify this unjustifiable shift.
Statman0January 19, 2004 at 4:47 pm #94341Reigle,
Congratulations!
You have successfully provided a simulation that proves that the upper (1alpha)*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 13: 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*(1510) = 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*(c1) where c= S/ s and since c for n=30 and alpha = 0.005 is approximately 1.5, shift* = 1.5.
Step 78: 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=(145100)/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=(130100)/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=(160100)/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.
Statman0December 23, 2003 at 5:57 pm #93805Reigle,
First of all, the topic of chaos theory and Tolerancing has been quite extensively written about. Just do a Google search on the topic. What Harry has written is not particularly unique or novel.
Secondly, Don Wheeler wrote a paper in 1995 describing process behavior and the application of chaos theory. What Wheeler shows is that Shewhart control chart theory is as robust for process modeling as the application of Chaos theory even when acknowledging the existence of chaotic behavior. A couple nice things about this paper, Don is a much better writer than Dr. Harry so the concept of chaos and process variation are described in a very straightforward manner and, the paper is a free download.
http://www.spcpress.com/ink_pdfs/Wh%20Chaos%20Theory.pdf
Third, one can’t help but to notice the irony that in the same book that Harry claims that the design engineer will only accept a “rule of thumb” linear shift of the mean for taking into account the presence of sampling error, he also expects the computational power and acceptance of solving sets of differential equations for Tolerancing.
Statman0December 23, 2003 at 5:25 pm #93804You asked:
“7) Do you know the short term standard deviation (if so, is it a population value or sample estimate)? 8) Do you know the longterm standard deviation (if so, is it a population value or sample estimate)?”
Describe a situation in which the short or longterm standard deviations are “a population value” when derived from process analysis.
Statman0December 22, 2003 at 8:38 pm #93774Thanks Reigle.
0December 22, 2003 at 8:35 pm #93773Hi Rick,
The 1.5 comment was my feeble attempt at sarcasm.
I have not read the Montgomery’s discussion of the issue. I am only assuming based on my experience why he would have an issue with how Pp can be misused.
I understand where you are coming from on the issue of “normality” with Gabriel’s distribution B. However, I think that there needs to be clarity in that you are talking about the shape of the histogram of all the data points, not that the data is normal. If you subtract Gabriel’s process A data from his process B data you will find that Gabriel has simply taken a random normal data series (Process A) and added constants to the subgroups. So the underlying distribution in each process is normal but process B is not stable due to shifting the subgroups in various directions to various degrees. Therefore, Process A is a random normal and stable process where process B is a process that is inherently normal but not stable. In other words, process B has subgroups with normally distributed data but the averages of the subgroups do not differ due to only random normal variation.
There is a risk in using the histogram of all the data and subjecting it to a normality test. A process that is not stable can have a normal shape in the histogram of all the data. A process that is stable can have a nonnormal shape in the histogram. This is because you can have any of the four situations: Stable/Normal, Stable/Nonnormal, Not stable/normal, Not stable/nonnormal.
You are right that there is a risk of prediction future capability of a process based solely on the Cpk as the Cpk assumes a stable process. However, both Cpk and Pp assume normality when used to predict the nonconformance rate relative to the specs. There is no advantage (from a normality perspective) in one metric over the other.
As I showed in my last post, my preference is to use a COV approach to establish the percent contribution to the process due to process management and to get an estimate of the entitlement level of capability. If, as in the case of Gabriel’s data, the process can be assumed inherently normal, then the Cp will indicate the best possible performance of the process and the predicted nonconformance rate at that level.
Statman0December 22, 2003 at 7:45 pm #93770Hi Reigle,
Let me know how to contact you.
Statman0December 22, 2003 at 6:17 pm #93763Hi Reigle,
I very much welcome the opportunity to have direct dialogue with Dr. Harry about these “Mysteries of Six Sigma” in a public forum. The only condition is that the objective is to develop a consciences on practical application for practicioners. I think this is your objective as well.
As you know from our previous discussions, my concern is in the misapplication of statisical concepts. I have acknowledge in almost every post that under limited and appropriate conditions the 1.5 shift or PPM vs DPMO, have a value add application. The concern is in universal applications without concideration of assumptions or consequences.
Just let me know the details and I will make time to attend.
Regards,
Statman0December 22, 2003 at 6:04 pm #93761Yes,
No K does not equal 11/c, K = (11/c)*Zst
You are making this more complicated than it needs to be. If we both agree that there exists a c such that C = S.lt/S.st and S.lt = C*Slt, Then it is just as convienent and mathematically correct to find the estimate of Zlt by dividing Zst by 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.
By the way, my proof showed up garbelly goop when I posted it even though it looked fine in the editor. Was it clear on your end or should I repost it. It looks like the forum was having trouble translating the fonts.
Cheers,
Statman0December 22, 2003 at 3:26 pm #93755Reigle,
Yes, my responses to your posts last evening were a bit flippant and I apologize. It was late last evening when I had arrived at my destination and checked my email. I was not in the mood for a long response.
I don¡¦t think, however, that there is anything of value to debate or discuss. I certainly do not want to discuss or debate what Dr. Harry said or didn¡¦t say in one of his books as it appears you do. I don¡¦t want to debate whether I base the shift on the specs (as you said in one of your posts), because I don¡¦t believe that there is a linear Z shift.
What has been requested by several on this form is why is there a 1.5 shift, where did it come from, when should it be used. We have now reached a point of agreement on these issues. And that is that the shift has extremely limited application. Or as you put it: ¡§the convention of adding and subtracting the 1.5 is merely a convenient way to APPROXIMATE the stateofaffairs; hence, it is a general RULEOFTHUMB¡Kthe shift factor (1.5S) is a “quick and dirty” method¡K¡¨. Therefore, I see no value to anyone to further discuss the use of the 1.5S shift.
The one point we don¡¦t agree on is that a sigma shift of K in the numerator of the Z equation does not equate to a Z with a K*sigma inflation. My proof of this is below:
Let D = SLT, sst = Sigma short term, and slt= Sigma long term and C = slt /sst. So slt = C*sst
Let K define the equivalent linear shift to equate Zst to Zlt
Zst ¡V K = C
D/sst ¡V K = D/slt
D/sst ¡V K = D/(C*sst)
D ¡V K* sst = D/C
K* sst = D ¡V D/C
K = D*ƒv(1 1/C)/sst = (11/C)* D/sst
Therefore, K = Zst *(11/C)
Since K is not equal to C and K depends on Zst, A sigma shift of K in the numerator of the Z equation does not equate to a Z with a K*sigma inflation.
Furthermore,
Zlt = Zst ¡V K = Zst – Zst *(11/C) = (1/C)*Zst
Zlt = Zst /C
As you can see, the mathematically correct form of compensating for long term is to divide Zst by C. This is just as ¡§convenient¡¨ as the mathematically incorrect form of subtracting C.
The important issue in all of this and a topic that is certainly worth discussing on this forum is the proper methods and considerations in estimating C. This is where engineers and SS practitioners can find some value. I am willing to have this discussion. But first you have to get beyond this shift thing.
Statman0December 22, 2003 at 3:15 am #93743“Dr. Harry’s book where he says the shift factor is a constant of 1.5. In fact, Dr. Harry says (repeatedly) it is a variable quantity, not a constant”
Still shrinking…….still shrinking…still shrinking0December 22, 2003 at 3:09 am #93742He DOES NOT say that the use of a 1.5 shift is UNIVERSAL. He does say that if no other data is available, 1.5 can be used to APPROXIMATE the short term Z.
Still shrinking…..still shrinking……..0December 21, 2003 at 5:51 pm #93732Thank you for your post. Unfortunately, this post does not offer any new information than what I have already summarized in my first post. I am glad, however, that you posted this. This is a sampling of the almost incomprehensible manner in which Dr. Harry’s book was written. Fortunately, I am versed in the language of “Phdeze”. So allow me to translate your post in a couple of bullets:
Ø Qualification of a process with the following requirements:
o 3* sDesign will not exceed ¾ of oneside of the design spec range (USLT), Therefore sDesign = ¼ *(USLT) and USL is at Z = 4.
o alpha risk due to sampling error does not exceed 0.005 for the estimate of sDesign with a 30 part qualification sample
o The Upper 99.5% bound for the sDesign will be: Upper 99.5%= sDesign *sqrt(29/c2.995,29) = 1.487*sDesign and can be rounded to 1.5*sDesign
Ø To define a qualification standard deviation (sQual ) such that it meets the above requirement is squal = 2/3*sDesign .
Ø Plugging this relationship back into the requirement of sDesign = ¼ *(USLT) reveals that squal = 1/6th of the design spec and USL is at Z=6 for squal .
Ø And T+1.5*squal +3*squal = T+3*sdesign
All of this I have acknowledged in my other posts. This is not the issue. The issue is the universal use of the 1.5 shift for conversion of short term to long term for any value of Zst. More specifically,
1. The shift only equals 1.5 in the unique case that Zst = 4.5
2. There is no constant shift of 1.5 (or any other number)
3. A six sigma process short term will be 4sigma long term (owing to the worse case sampling error) not a 4.5 sigma process long term
4. Because of point #3, a six sigma process will have a DPMO of 31.69 not 3.4 long term.
5. A sigma shift of K in the numerator of the Z equation does not equate to a Z with a K*sigma inflation.
So I have considered your design situation twice, once in Harry’s book and now with this post. It may explain how one came up with the 1.5 shift but does not explain why one would ever use the 1.5 shift since it only works for one specific case of Zst = 4.5.
The book may be better titled as “The Incredible Shrinking Shift” because the more Dr. Harry tries to rationalize it, the more limited becomes the application. We have seen this go from application to any process – to application to stable processes only – to application in design considerations only – to applications in design consideration with 30 part alpha = .005 qualification only – to only when short term Z = 4.5 . . . . . .
Last comment: You wrote:
“When the quantity T + 3S.qual = T + 3S.design or when T – 3S.qual = T – 3S.design, the statistical probability of encroaching the innermost limits of DM is exactly the same in both cases (p = .00135); however, the probability of violating USL or LSL is not the same. This point often leads to great confusion among novice practitioners because of the natural propensity to focus only on the specification limits (and not the inner most limits of design margin). “
Check my math on this but if T + 3S.qual = T + 3S.design then S.qual = S.design. If they are equal, of course they will have the same probability for a given critical value. I will assume you mean that when we align the tail probabilities of the qualification and design distributions? But where does that alignment occur?
AT 4.5*S.qual.
Secondly, thank god for the novices that are “focused only on the specification limits” because the probability of the process producing product out of spec is why we are doing the qualification.
Statman0December 19, 2003 at 8:56 pm #93721Hi Gabriel,
Is it because process B has a 1.5 sigma mean shift? J
Just to add to your point. The variation calculated within the subgroups for Cpk is an estimate of the inherent, or entitlement level of variation. The variation calculated as the RMS of all the data is an estimate of the total variation including any temporal, cyclical, or transient special cause variation. There is risk in using the Ppk as a prediction of future process performance, particularly if the special causes are primarily of the random transient variety. Even though the Ppk will take in to account the drifts or shifts over the period of data collection, there is no way to predict that the pattern of special causes will exist in the future. I think this is why Montgomery gets frustrated over the use of Ppk.
One good use of these metrics is in components of variation context. If I perform a COV on your data:
Process A:
Variance Components
Source Var Comp. % of Total StDev
Between 0.014* 0.00 0.000
Within 1.256 100.00 1.121
Total 1.256 1.121
Process B :
Variance Components
% of
Source Var Comp. Total StDev
Between 1.274 50.37 1.129
Within 1.256 49.63 1.121
Total 2.530 1.591
You can see that Process A is performing at it’s level of entitlement. Where process B sees a 50.37% of it’s total variation contribution from process management (for lack of a better term) with the same potential performance level (entitlement) as process A.
The proper use of Ppk should be to assess the amount of improvement that can be made through DMAIC projects vs the need for process redesign. Makes more since than adding a 1.5 shift to process A.
Cheers,
Statman0December 19, 2003 at 6:58 pm #93718Hi Gabriel,
My post wouldn’t have had the impact if I had set the parameters so that the annual cost of the alpha and beta risk was $50 dollars a year J.
What you have done is emphasized my point. Which is that you should not accept some general “rule of thumb” about an inspection system without a cost/benefit analysis of the consequences.
Since it would be unacceptable to have such a high cost of unnecessary internal repair due to the inspections system alpha risk, we would have to compensate by adding additional gates. But OK, let’s go with that. The parts are inspected two more times before repair, each at the same alpha and beta level of the first (.05).
Therefore,
Test 2 will see 47 false positives and 64.65 actual defectives per day and have these results:
47*.05 = 2.35 false positives going on to inspection number 3
64.65*.05 = 3.23 false negative shipped to the customer
Test 3 will see 2.35 false positives and 61.32 actual defectives per day with these results:
2.35*.05 = .1175 false positives per day on to repair and
61.32*.05 = 3.066 false negatives shipped to the customer
Then there will be (3.35 + 3.23 + 3.066) = 9.646 false positives per day shipped to the customer and
0.1175 false negatives repaired
Then my cost of unnecessary repair or scrape = $2,000*0.1175 = $235 per day
And cost due to field repair or replacement = ($10,000 – $2,000)*9.646 = $77,168per day
Total annual cost of the “qualified” inspections system with 2 additional inspections = $77,403 * 365 = $28,252,095
So yes, we can reduce the cost of false positives with more internal inspection. However, we increase the number of defectives shipped to the customer as well as the cost of additional inspection, complexity, WIP, and process cycle time, and lower customer satisfaction.
To your second point, if there was a variables measurement with in which safety limits could be set up, you would go with that measurement as opposed to the pass/fail attribute gage.
Don’t get me started on the number of defects associated with my PC K.
Cheers,
Statman0December 19, 2003 at 6:03 pm #93715I’m sorry 3 sigma is 66800 PPM not 4 sigma
Oops.0December 19, 2003 at 5:48 pm #93714OK,
Let me see if I understand the experts correctly.
I have a process that produces 1,000 parts per day. The average cost of scrap or repair internally is $2,000 per part. The average cost of repair or replacement externally is $10,000. The process has a current capability of 4 sigma (66800 PPM defective) and therefore produces 67 defective parts per day on average. My gage R&R on the inspection was acceptable at .90. For simplicity, lets say that the alpha and beta risk of the inspection were equal components of this at 0.05 each.
Then
Number of false positives per day = (1,00067)*.05 = 47
Number of false negatives = 67*.05 = 3.35
Then my cost of unnecessary repair or scrape = $2,000*47 = $94,000 per day
And cost due to field repair or replacement = ($10,000 – $2,000)*3.35 = $26,800 per day
Total annual cost of the “qualified” inspections system = $120,800 * 365 = $44,092,000
Well I guess I have a different definition of an “acceptable” inspection system!
Statman0December 18, 2003 at 6:39 pm #93689John,
You forgot one.
Write a book and get published.
Statman0December 18, 2003 at 6:17 pm #93687Yes, of course, Minitab statguide is always correct. They would never choose simplistic over accuracy in there definitions.
Tell me, if the statguide is correct, how can you logically have a transformation minimize the pooled Std. Dev. when converted back to the original metric? Wouldn’t a conversion back to the original metric produce the same pooled Std. Dev.?
One should understand the mechanics of a methodology before they pass judgement.
Statman0December 18, 2003 at 3:39 pm #93682Your answer is wrong.
The optimal value of lambda minimizes the differences in within group variation. It has nothing to do with the “pooled Standard Deviation”.
Or another way to put it, it “stabilizes” the variation by making the variation independent of the group averages. It is not a normalizing transformation. The assumption is the groups are normal with different levels of variation that are dependent on the group mean.
I guess anyone can be a MBB.
How does one know that the process is stable with a dependency between the group mean and variation or that the need for a “lambda” transformation is due to special causes?
Statman0December 17, 2003 at 4:41 pm #93656Hi Sqe,
I should have been clearer in this post and defined PPM as PPM defective. The reason for the clarity is that PPM is often used without the definition of what one is counting. I have seen PPM and DPMO used interchangeably by some authors with out explanation.
My definitions will probably not agree with others and I run the risk of creating a whole new string on defining PPM but here is the context:
DPU = (# of defects)/(# of units processed)
PPM (defective) = 1,000,000* (# of Defective units)/(# of units processed)
Where # of Defective units is the number of units that do not meet acceptance criteria
The reason DPU*1,000,000 does not necessarily equal PPM (defective) is that the distribution of defects across units may not be uniform. In other words, all defects may be on one or two units and the other units have no defects. If you had 20 defects and 20 total units, 12 defects on unit one and 8 on unit two. 1,000,000*DPU = 20/20 = 1,000,000 and PPM = 100,000.
Statman
0December 17, 2003 at 4:09 pm #93652Hi PB,
Thank you for the correction and the compliment.
I Should have used excel when I made the table rather than calculating it in tabular form.
Cheers,
Statman0December 17, 2003 at 2:46 am #93638Repost with the table set up better
Hi Gabriel,
You asked:
In that case, what is the added value of DPMO over DPU or PPM that pays for the added complexity and uncertainty of the measurement?
It really depends on the objective and the approach to the project. Lets say we have the following information on 5 subprocesses within an overall process:
Sub A Sub B Sub C Sub D Sub E Final
# of Defects: 500 25 300 50 25 200
OPU 500 150 180 100 200 1130
# of Units 300 250 200 150 125 125
Defectives 80 50 30 20 15 10
# of Opp 250,000 37,500 36,000 15,000 25,000 141,250
DPU 1.67 0.10 1.50 0.33 0.20 1.60
DPMO 2,000 666.67 8,333.33 3,333.33 1,000.00 1,416
PPM 266,666 200,000 150,000 133,333 120,000 80,000
Ø If the objective was to reduce defect rate, you would focus on Sub C and use DPMO as your primary metric. The relatively high number of defects and low opportunity count makes it a candidate for improvement
Ø If the objective of the project was to reduce complexity, you would focus on Sub A and use OPU as the primary metric
Ø If the objective was to improve yield, you would focus on Sub A or Sub B (depending on cost of scrap which will probably increase as you move down the process) and use PPM as your primary metric
Ø If the objective was to reduce repair cost at final, you would Pareto Defects at final to determine the area for improvement and use DPU at final as your primary metric.
Ø Etc.
I could go on and on. There are many other primary metrics that could be used for various projects ( e.g. cycle time, WIP, Cpk).
There are two important points from this example. First, there needs to be congruency between the problem statement, objective, and primary metric for a BB project. Secondly, these metrics are of course related but not one to one. An improvement in one may not correlate to an improvement in another.
Hope this helps.
Statman
0December 17, 2003 at 2:43 am #93637Hi Gabriel,
You asked:
In that case, what is the added value of DPMO over DPU or PPM that pays for the added complexity and uncertainty of the measurement?
It really depends on the objective and the approach to the project. Lets say we have the following information on 5 subprocesses within an overall process:
Sub A Sub B Sub C Sub D Sub E Final
# of Defects: 500 25 300 50 25 200
OPU 500 150 180 100 200 1130
# of Units 300 250 200 150 125 125
Defectives 80 50 30 20 15 10
# of Opp 250,000 37,500 36,000 15,000 25,000 141,250
DPU 1.67 0.10 1.50 0.33 0.20 1.60
DPMO 2,000 666.67 8,333.33 3,333.33 1,000.00 1,416
PPM 266,666 200,000 150,000 133,333 120,000 80,000
Ø If the objective was to reduce defect rate, you would focus on Sub C and use DPMO as your primary metric. The relatively high number of defects and low opportunity count makes it a candidate for improvement
Ø If the objective of the project was to reduce complexity, you would focus on Sub A and use OPU as the primary metric
Ø If the objective was to improve yield, you would focus on Sub A or Sub B (depending on cost of scrap which will probably increase as you move down the process) and use PPM as your primary metric
Ø If the objective was to reduce repair cost at final, you would Pareto Defects at final to determine the area for improvement and use DPU at final as your primary metric.
Ø Etc.
I could go on and on. There are many other primary metrics that could be used for various projects ( e.g. cycle time, WIP, Cpk).
There are two important points from this example. First, there needs to be congruency between the problem statement, objective, and primary metric for a BB project. Secondly, these metrics are of course related but not one to one. An improvement in one may not correlate to an improvement in another.
Hope this helps.
Statman0December 16, 2003 at 9:06 pm #93630Hi Praveen,
That’s what I’m talkin’ about!
There is this perception that if you don’t accept the premises of the Sigma/DPMO metrics then you are anti Six Sigma and are blind to the benefits of application of Six Sigma (statistical thinking) in strategic business planning.
In my case anyway, this could be no further from the truth. Metrics drive action and the application of statistical thinking and a Balanced Scorecard/Six Sigma Business Scorecard approach along with a Y =f(x) cascading to projects level metrics will provide the proper use of metrics to drive the right actions.
Cheers,
Statman0December 16, 2003 at 8:38 pm #93626Hi Mike,
Great response! A very well thought out response as well.
I think the key to this is as you say: “For some reason we don’t seem to be getting the idea that there is no single measurement system that gets you through this”.
The fact is, no business leader can make strategic decisions based on a single metric as there is no such thing as a single metric that can be used to compare performance across different products and processes. My goal (or bias if you will) is not to discredit the DPMO metric, as it has its place and value, but to point out the fallacy that the DPMO and the resulting Sigma metric is the “be all end all” metric for process performance. Strategic decisions like resource planning, customer value and satisfaction, make vs. buy, product line complexity, growth and revenue improvement, profitability, etc. are too important of challenges to be cheapened by the broad brush Sigma level generalities. Fortunately, our brains are advanced enough to handle more than one metric when making strategic decisions.
The critical question for Six Sigma practitioners is what is the proper application of DPMO?
Does it measure complexity? No, as you stated, it is normalized (actually I prefer standardized as it does not conotate a normalizing transformation) across processes. Process 1 having 2.5 times the DPMO of process 3 does not necessarily have more complexity. There may be far fewer opportunities but higher occurance of defects.
Does it measure opportunity for improvement ($’s to the bottom line)? Not necessarily. Process 3 may have achieved the level of 4 sigma by actually increasing cost rather than reducing cost. For example, Process 3 could have had two independent