Simple Question…?
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 This topic has 83 replies, 13 voices, and was last updated 17 years, 2 months ago by Peppe.

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July 23, 2004 at 9:24 am #36260
I’ve measured a quantity of 30 shafts for diameter. The specified diameter is 6.600 +0.1 / 0.0. The Largest is 6.686, the smallest is 6.665. The Mean is 6.676 and the Population Std deviation is 0.0044.What is the Sigma Value associated with the diameter turning process?The reason I ask this simple question is that when I attempt to calculate sigma using ztables, the calculation falls over as the process capability seems greatly better than the ztables will allow me to calculate. Have a go and you’ll see what I mean?Any help would be appreciated – I want to be able to describe to a nontechie audience how many PPM Defective are likely to occur.
0July 23, 2004 at 12:07 pm #104131
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.You don’t need a ztable at all.
average – closer specification limit / standard deviation = 6.6766.7/0.0044=5.45.
This means that the closer specification limit is 5.45 sigmas from the average, so the sigma level is 5.45.
The question is, is the process in control? If not, any capability calculation (as the sigma value) would be meaningless, as the process would not have a consistent behavior.
Also, probably this 30 shafts were produced cose in time one from the other, so this would be the “short term” sigma level. Probably, your process will not perform as good in the long run as it did when making this 30 shafts. Depending on the degree of process control achived, the long term capability can be only a little bit worse than the short term capability or much much worse.0July 23, 2004 at 12:19 pm #104132
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Alan H:
Your spec for shaft diameter is 6.600. The upper specification limit (USL) is 6.700 and the lower specification limit (LSL) is 6.600. No deviations below the spec are allowed in your operation. The largest diameter of 6.686 is outside the USL. The smallest diameter 6.665 is between the USL and the LSL. The mean 6.676 is inside the USL but far from LSL. So your shaft manufacturing process needs attention. I calculated the following Zscores for the largest and the smallest diameters and the mean and standard deviations you gave.
Mean = 6.676 Standard Deviation 0.004
Largest X = 6.686. For this Z = (X mean)/Std. Dev = 2.273
Smallest X = 6.665. For this Z = – 2.5
I don’t know all the 30 raw values of X. The Zscore is simply a different way of looking at the “spread” in the values of raw X scores. May be this “spread” follows what is mathematically referred to as a normal distribution. This is simply an idealization but will help your nontechnical audience grasp the meaning of what is going on in the real world with what the good stat books teach us. Did you prepare a stem plot or a frequency diagram using the 30 values? Can you post the 30 values? That would help others follow what you have initiated.
Second, I used Chebyshev’s theorem. This theorem makes no assumptions regarding the nature of the distribution. It does not have to be the famous normal, Gaussian, or the bell curve distribution, which your nontechnical audience must know for sure – everyone knows the Bell Curve!. By choosing a value of k, we calculate the fraction of the data that must fall within plus or minus k times sigma, where k is the number (whole or fraction) that we choose. This is the minimum fraction, or percent, that must fall within plus or minus k times sigma. Chevyshev’s theorem says, at least.
From Chebyshev’s formula, if k = 3, at least 88.89% of the diameter values must fall within the k times sigma, i.e, 3 sigma values. In your case the three sigma limits are 6.689 and 6.663. Is there an agreement with Chebyshev’s theorem. It would be very very surprising if the data reveals otherwise. If you have an “ideal” Gaussian distribution, then 99.74% must fall within the three sigm limits of 6.689 and 6.663. Your nontechnical audience should follow this.
Please check my calculations. Since I am getting somewhat curious now, do post the 30 values, if you can. Kind regards.
0July 23, 2004 at 12:25 pm #104133
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Alan H:
I did use 0.0044 for the standard deviation, as given in your post, to find the Zscores, not 0.004. In the meanwhile, I see Gabriel’s post. Very nice conclusion about the sigma level. Thanks Gabriel.0July 23, 2004 at 12:32 pm #104135I agree with Gabriel ‘s post .
Your manufacturing process has a mean and std dev which is neat and tight to reflect a Z of 5.45. But I would ensure the normality of the data ( of the 30 shaft diameters) as a check before I accepted this Z score.
Uma0July 23, 2004 at 12:40 pm #104137Hi Charmed,
6.686 is less than 6.700! ;> (spec is 6.6 +0.1 / 0.0)
I’ve typed in the 30 data items:
6.675
6.672
6.676
6.671
6.677
6.672
6.684
6.681
6.674
6.672
6.679
6.676
6.677
6.681
6.676
6.675
6.686
6.672
6.675
6.674
6.678
6.665
6.681
6.681
6.672
6.680
6.674
6.672
6.677
6.677
I’m not familiar with Chebyshev’s theorem – a mere novice at this game!
A0July 23, 2004 at 12:45 pm #104138
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Sorry, I forgot to address your last question:
About the PPM, it is impossible to tell from 30 parts. You could say “If the process was normally distributed, what we don’t know and probably won’t, and if the process was perfectly stable, which will not, and if this sample of 30 parts were perfectly representative of that process, what is not true, then the defectives rate would be 0.02PPM, all out of the upper specification limit”. In Excel: =(1NORMDIST(6.7,6.676,0.0044,TRUE))*1000000. Forget about trying to see if the distribution is normal or to fit any other distribution to the data. WIth only 30 samples you don’t have data in the tail where nonconformities will happen to see if any given distribution fits or not in that zone.
Someone might tell you about the 1.5 sigma shift: “Due to the 1.5 sigma shift, the long term capability would be that as if the average would have shifted 1.5 sigmas” With that in mind, you can continue with the speculation and make the same calculation with the shifted average that will give you 38PPM. In Excel: =(1NORMDIST(6.7,6.676+0.0044*1.5,0.0044,TRUE))*1000000. But as said in the other post, the difference between the long term and the short term capability can be much smaller or much larger than an equivalent 1.5 sigma shif of the average, depending on the degree of process control achived.
Instead of all that, I would just say something like “According to this sample, the closer specification limit is more than 5 sigmas from the average. It is not possible to make a statistacally valid estimation of the defectives rate based only in this small sample. But if we can make the process perform consistently close to what it did for this samples, the defectives rate will be very small, like PPM in two digits or better”0July 23, 2004 at 12:51 pm #104139
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Thanks Alan H for the 30 values. 6.686 is less than 6.700! Yes, I made a mistake there when I wrote about 6.686 and thinking mentally about 6.670 when I typed that. I juxtaposed “7” and “8” in my mind. The calculation of Z scores is OK.
0July 23, 2004 at 12:52 pm #104140Many thanks to all for your help – as Mr I Newton once said…
“If I have been able to see farther than others, it was because I stood on the shoulders of giants.”0July 23, 2004 at 1:08 pm #104141
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Alan,
6.686 is not outside the USL 6.7. Being the standard deviation 0.0044 it is, in fact, 3.2 sigmas inside.
I would not mess with Chebyshev. Chebyshev’s theorem address the “worst case” which, being very interesting from a theorwtical point of view, it is never the real case as far as I can tell.
For example, from Chebyshev’s theorem you have at least 99% of the population within the average ± 10 standard deviations (P(Xµ>10sigmas)<1/10^2=0.01). Can you imagine a reallife process with Cp=Cpk=3.33 and with 10,000 PPM?
As another example, you can have up to 100% of the population within ±1 sigma. The only distribution that meets that is one with 50% of its individuals at exactly one value and the other 50% at exactly another value. Can you imagine this shaft diamter process producing half of its parts eaxctly 6.6 and the other half exactly at 6.7 with no values in between?0July 23, 2004 at 1:14 pm #104142Gabriel,
Why did you not recommend he calculate a Zusl and Zlsl and then combine the ppm defective and come up with an overall Z?0July 23, 2004 at 1:15 pm #104143
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.z is, by definition, (Xµ)/sigma, whether the distribution is normal or not. Now, if the distribution is not normal then you cannot tell the fraction under the curve using a normal distribution table (also called z table).
About checking for normality, it will not be of much help. Even if you find that this 30 values match nearly perfect with the normal distribution, the datapoint that is closer form a specification limit is still more than 3 sigmas away from it. So the data will be of no help to tell where the distribution is normal or not in the zone where defects will hapen.0July 23, 2004 at 1:16 pm #104144
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.I agree with the need to check the normality of the data. Since Alan H was able to provide the 30 values measured, I prepared a histogram. The distribution is far from being a nice normal distribution. Since Gabriel is here, I would defer any further discussion of this point. I am thousands of steps, may be millions, behind him.
0July 23, 2004 at 1:18 pm #104145Gabriel,
Disregard the last post…after seeing the data points further down and graphically displaying, it was obvious that one side had zero ppm defective.0July 23, 2004 at 1:22 pm #104146
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.The LSL is 17 sigmas from the average (i.e. Zlsl=17). Please do tell me the PPM beyond 17 sigmas.
Do you now understand why I didn’t (and I don’t) recomend to combine both tails?0July 23, 2004 at 1:22 pm #104147
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.Charmed,
I always am glad to see what Gabriel has to say but I wonder why you say the distribution is “far from being a nice normal distribution”? It does pass the Anderson Darling normality test and looks graphically as “normally distributed” as I’ve worked with in other past circumstances.0July 23, 2004 at 1:24 pm #104148Gabriel very informative response. I learn a lot from the way you would present teh data. Thanks a lot!
0July 23, 2004 at 1:34 pm #104149
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Normal or Gaussian distribution is also a theoretical point of view, as is Chebyshev. Can you please tell us more why not to mess with Chebyshev in this shaft turning problem? The 30 values do not reveal a normal distribution either. Or, would you conclude otherwise. I am unable to agree with the counterpoints made regarding Chebyshev. I can make similar arguments about normal and Gaussian that we take as a holy cow. I think it (normal) is used only because it then gives us chance to make neat mathematical calculations (which may or may not have anything to do with reality!) This is meant as a sincere question, so that we can learn. I look forward to your kind reply. My warmest regards.
0July 23, 2004 at 1:37 pm #104150Gabriel, I’m LOL. Hope you saw that you wrote your question back at the same time as I wrote my clarification. I assume we are as straight as usual?
0July 23, 2004 at 1:57 pm #104151
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear CS:
I did not do any advanced tests for normality like the test that you have mentioned. I did the following. May be you can help me understand this too. I took the 30 values given by Alan H and rewrote them as whole numbers that fall between 0 to 100 by substracting 6.600 from each value and multiplying by 1000. So, 6.665 becomes 1000(6.665 – 6.600) = 65 and so on. The 30 values therefore fall between 65 and 86. The frequency of each value is as follows:
65 1, 71 1, 72 6, 74 3, 75 3, 76 3, 77 4, 78 1, 79 1, 80 1,
81 4, 84 1, and 86 1.
Then I simply plotted a frequency polygon. It does not look anywhere like a normal distribution to me. I would be happy to see what other experts here feel.
Also, if I do the same with the Upper and the Lower Spec Limits given by Alan H, the limits become 0 to 100 (6.600 + 0.1 /0.0). But, for Alan H, the 30 values fall between 65 and 86. So, I think the process should be centered around 50, i.e., 6.6500. This will be beneficial in the long run and will reduce the number of defective parts (which can drift towards the higher side of the diameter since now the highest frequency values are (72, 6) and (77, 4) and (81, 4). Look forward to your clarifications. Thanks for asking the question. With my warmest regards.0July 23, 2004 at 2:14 pm #104152Gabriel,
Even though the data is within the spec. limits (6.60 to 6.70) the shaft diameter average is around 6.67 and all data fall around the USL.
Given that the tolerance on spec. on the diameter is 6.60 +0.10 / 0.00, the designer wants the shaft diameter to be closer to 6.60. (I presume). Do you think it is fair to ask:
Are these 30 shafts performing without any problems?
What are the tolerances on the mating diameters?
If the shafts are performing as designed, should the tolerance and specs. be changed to 6.65 +/ 0.05?
Should the process be worked on to center the average?
Just a thought.
PB0July 23, 2004 at 2:18 pm #104153PB, I agree with you. Try to look at data using Taguchi Loss function.
Rgs,
Peppe0July 23, 2004 at 2:55 pm #104158You need to be careful with words like “far from normal”. This data set is not, in fact it is a pretty common looking histogram for 30 samples taken from a perfectly normal distribution.
I read your comments back to C Seider and I wonder why you went through so much trouble to make a histogram. Go to excel, use the data as is and plot the histogram.
If you don’t have Minitab or Jump or something like that to do a normality test, you can always go create a random data set of 30 several times and do the histograms in Excel before making claims of “far from normal”. Your statement is incorrect and you would know that if you had gone to see for yourself what samples of 30 from a normal distribution looked like.
Statistics from Minitab on the data’s normality:
pvalue = 0.288
do not reject the assumption of normality.0July 23, 2004 at 2:57 pm #104159
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.Charmed,
Your technique of doing plotting the values is appropriate. It was extra effort to “do the transform of subtracting 6.6 and multiplying by a 1000” but no harm done. If you plot even the original numbers, assuming the binning is equivalent you would get the same look of the histogram (plot of frequency vs diameter).
You say it doesn’t look anywhere like a normal distribution. One thing to be aware of, the more samples you take (in the original case it was 30) the more of a gaussian curve it should like like–assuming it is a normal population. Samples are the practical way of getting an idea of the mean and standard deviation for a population (I say an idea because there is uncertainty of the population mean, s.d. with a sample). In the same way, you begin to get an idea of the general shape of a population with a sample size of X but it won’t follow a typical Gaussian curve–the perfect normal distribution. That’s why some tools with some very advanced routines but easy usage were developed to test for normality (e.g. the Anderson Darling I mentioned in an earlier post).
Your point that the process should be centered around 6.65 may not take into account the business case. In the case in this string of posts, they had excellent control of the variability of the process so it was much more capable than many processes that are centered. The ppm defective of the sample in the process was VERY low even though not centered. There might be a business case for where their process is centered–lower cost due to lower waste, lower cycle time, or machine capability is not linearly perfect. Most customers would love to have a Ppk of 1.7+ since the process is very narrow relative to their specifications.
0July 23, 2004 at 3:20 pm #104162
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Charmed,
When you plot histograms you don’t plot frequencies of individual values. In theory, the probability that you have two individuals with exactly the same value is zero (I said “in theory” because the lack of infinite resolution will make some values to be truncated to the same number, like 0.9998 and 1.0002 will both be taken as 1.000 if only 3 digits are shwn recoded). So such a histogram will look as a lot of frquencies at zero with some isolated values with a frequency of 1. For example, the following table shows 30 values actually taken at random from a normal distribtuion with average 6.676 and standard deviation 0.0044:6.645
6.644
6.630
6.645
6.641
6.6376.640
6.637
6.644
6.634
6.635
6.6356.642
6.636
6.642
6.636
6.636
6.6356.635
6.633
6.633
6.636
6.641
6.6316.635
6.639
6.641
6.642
6.627
6.634
Plot them as you did, and you will not find anything close to a normal distribution either (but it is).
Now we know how histograms are NOT done. How should they be done, then?
Using “classes”. A class is a range of values. Each calss has a lower limit, an upper limit and a center that is ususlly taken as the value representative of that class. How many classes should one use? There are some “exact” formulas to calculate the correct number of classes, but I don’t like how they work. Accept for a minute that 5 clasees would be Ok for 30 values. Let’s make these classes as follows:>=
<
center
Freq6.665
6.670
6.6675
16.670
6.675
6.6725
106.675
6.680
6.6775
126.680
6.685
6.6825
66.685
6.690
6.6875
1
Now plot this histogram (or frequency polygon), and you will see it reminds you the Gauss bell somehow.
Another thing you can do is plot the 30 values on a normal probability paper and see if they form something like a stright line. In this case, they do.
But maybe most important: Before this post, I hadn’t even botheresd to check if the data was normal or not. Why? Because I would have changed nothing of what I said. Let’s say that the data was not normal: Ok, in this case one cannot use the Z table to estimate the PPM out of tolerance. Now let’s say that the data was normal: You cannot use the Z table anyway because the data you have is so far from the zone where the PPM will happen that it would be irresponsible to use the data you have to extrapolate the shape of the distribution that far.
By the way, I agree with you about centering. The process is off center. If it can be better centerd then there will be fewer defectives when the process is stable and also a more robust behavior against shifts and drifts.0July 23, 2004 at 3:28 pm #104164
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Read again all my posts in this thread. I am not saying that the normal distribution is a holy cow. Just a holy sheep (just a joke). In fact, I am saying exactly the contrary: We cannot assume normality to estimate the defectives even if the data looked perfectly normal. (for me the data doe not reveal a notnormal distribution, but this is irrelevant for what I have just said).
About Chebyshev, as I said it is far too pessimistic compared to any reallife case I have ever seen, normally distributed or not. It is just a worst (and vary unlikely) case. Have you checked the two examples I gave about this?0July 23, 2004 at 3:30 pm #104165
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.;)
0July 23, 2004 at 3:39 pm #104166
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.There will be one value that minimizes the “loss to society”. It would be great to find it and minimize the variation arround that target (the Cpm and Taguchi’s loss function concepts). Of course I am clueless about what that value is.
0July 23, 2004 at 4:06 pm #104169
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear All:
Thank you for answering my humble questions. I read all the posts where I found replies. It has been very helpful. I am glad to know, that although I did the histogram calculations differently, I did not make any blunders. I did the histograms in this way since the upper and lower specs are then between 0 to 100. May be just a coincidence, but we get nice neat whole numbers to work with. Of course, as Gabriel has pointed out, “binning” the data differently might give a different look to the histogram or frequency polygon. I did not do that yet (sorry Gabriel, too much to digest here!). If you all say it is a “normal” distribution for 30 pieces of data on shaft diameters, then “normal” it is.
I am glad that Gabriel was in agreement with my idea of the centering of the process. I just felt moving the mean to 6.650 would make for a more robust process since the upper and lower specs are between 0 to 100. A mean of 50 (which mean 6.650 in actual diameter values) makes more sense, or just plain common sense.
Finally, Gabriel, you were wondering about some kind of a minimum loss to society. I guess you mean developing a shaft turning process that would produce the “minimum” number of rejects and therefore the “maximum” profits. You are now turning into a philosopher. That’s a great idea, as long as the customer does not change the specification limits. If that changes, we have to start all over again. We must go back and ask Alan H what he does to be able to meet the USL and LSL so admirably. With my warmest regards to all the experts here. This has been a most exciting exchange. To repharse Alan H, I feel like a little boy standing on the shoulders of giants, looking far into the horizon and being able to see afar from normality (just kidding Stan).
0July 23, 2004 at 4:28 pm #104171Let me disagree. You are changing specs before it was “smaller is best” now you have created “nominal is best” and it is not allowable, because Alan’s data are drifted in upper side of specs, instead of lower as “smaller is best” require.
Furthermore, is still open the last question from Alan.
“I want to be able to describe to a nontechie audience how many PPM Defective are likely to occur”
What baout ?
Rgs,
Peppe
0July 23, 2004 at 4:31 pm #104173
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.“Finally, Gabriel, you were wondering about some kind of a minimum loss to society. […] You are now turning into a philosopher.”
Hey! I wish I had developed that concept, but it was Taguchi, not me. I’m just a “user”.
The idea is that, if the tolerance is 6.6 to 6.7, a part at 6.699 will be not different than a part at 6.701 for any practical purpose. But a part at 6.65 will certainly be different from them. There is no such a thing as black and white, or “up to here it’s perfect and beyond it’s trash”. There is an optimum value, and growing problems as you drift from that optimum.
Problems may be scrap, but not only that. For example, in this example maybe it is a grinding process and the grinding time would double if you wanted to reach 6.65. So if at an average of 6.676 all the parts work perfectly, why to go any further? The quality will not improve and the costs will go up.
Another reallife example: We buy tubes than then we turn. The diameters and thickness of the tube was defined long ago so as wou eill allways have enough material to remove in the turning operationj, taking into account the variations of the tubes manufacturing process according to the technology that was available at that time.
Today the variations in the tubes is much smaller than then. There is a worldwide standard that steel is sold by weight. The supplier now delivers all tubes perfectly in tolerance, but near the limit of the max material. So when we by X maters of tubes to turn a given number of pieces, we are buying more kilograms that what we used to. More cost per part, more material to remove, more chips to dispose, more turning time, more energy consuption, more oil used for the turning, more contamination…… Surely this suplier is not thinking in reducing the cost to society. Just to make more $$$. But it is a short range sight. We are now looking for a supplier that is willing to sell by meter. In this way, the supplier can look for the minimum matrial (instead of maximum) and sell the same meters of tubes with less steel (less cost for himself), and we would make more rings per Kg with less chips, less turning time, less oil, less energy…..0July 23, 2004 at 4:38 pm #104175
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.“6.6 +0.1/0” is not “smallest is best”. “6.7 max” would have been.
In this case, I don’t know what is best, but it is clear that they want the parts between 6.6 and 6.7 and that they are willng to call “defective” anything out of that range.
If you only told me to give you parts in 6.6 +0.1/0 with no other aditional information, I would do my best to make them at 6.65 (unless there is some “process efficiency” reason to make them somewhere else).0July 23, 2004 at 5:43 pm #104180
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
Have you read the book by Schumacher (I hope I have spelled it correctly) with the title Small is Beautiful? It made a great impact on me when I first read it, and also on many other millions. Also, to cite what I remember reading a book on the history of statistics – this not an exact quote, in am extrapolating a bit. The least squares or linear regression (bestfit) line always passes through the average value of X and Y in a data set. This paraphrased may be called the “search for mediocrity”. This “medicority” phrase appears in the discussion of the work of Galton (this was before Pearson and the modern interpretation of the meaning of Rsquared in linear regression). Galton studied biological traits like the heights of children and parents and how they were correlated. Then after finding what was a “crude” regression line for the heights, he made bold leaps of generalizations for all other biological traits. Thus, from mediocrity, we can go to small is beautiful, and then search for excellence – or what Gabriel says in a different post – the minimum loss to society. More later but thanks for the dissenting opinion, expressed frankly. All judicial rulings include dissenting opinions.0July 23, 2004 at 5:45 pm #104181
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Garbriel:
Again thanks for contributing to this discussion so nicely.0July 23, 2004 at 5:56 pm #104182
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
Let’s finish with the important point you raised which you see pasted below,
Furthermore, is still open the last question from Alan.
“I want to be able to describe to a nontechie audience how many PPM Defective are likely to occur”
What baout ?
If Alan H can put together many of the ideas discussed in this thread, go in stepbystep fashion explaining what he did (produced shafts, took 30 of them, measured their diameters) and then explain the meaning of average, SD, normal distribution etc., every one will love. it. I think he will become a hero and his nontechie audience, assuming the CEO is one of them, will surely grant his a big raise and a promotion for developing a shaft turning process that produces these shafts with such a tight control between the USL and the LSL. I am sure, we can agree on this point, at least. Are you the CEO, by any chance? Have a great weekend.0July 23, 2004 at 7:12 pm #104184
AnonymousParticipant@Anonymous Include @Anonymous in your post and this person will
be notified via email.The data has indication of special cause variation. I obviously don’t have any information regarding how or when the samples were taken or if you have a good measurement system. Control charting however indicates that something is having an impact on the process that produced these diameters.
It would be risky to make any statements about the capability of your process at this time.0July 23, 2004 at 10:16 pm #104196
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Anonymous,If you are still out there, I want to tell you that I agree with Stan not in the package of his posts (now deleted together with your other posts because of the rude language) but yes in their content. I have plotted an IXmR chart and found not a single individual value or moving range out of control limits or with any evidence of an outofcontrol pattern. (I am now sending to iSixSigma the control chart to attach to this message)And all that assuming that the data was time ordered, what was never mentioned in first place. If the data is not time ordered (for example, if Alan just took any 30 parts and recorded the values as they were measured, not in the order they were manufactured) then it is absolutely impossible to assess process stability.I stand with Stan (no pun intended). If you find that we are wrong, please do show it. Tell us how can you tell that the process is not in control: what SPC technique you used, what poin(s) are out of control and wat rule(s) it/they broke. I will be the first one to recognize the mistake and will be happy of having learnt something.PS: To the forum moderator – Thanks for reopenning this thread.
0July 23, 2004 at 11:06 pm #104198You are incorrect – the data does not have special cause variation
0July 24, 2004 at 12:30 am #104200Gabriel,Thank you for the support and also for taking the time to give good advice.Interesting thread, claims of nonnormality where they were wrong and claims of out of control where they were wrong. What are people being taught in Six Sigma? (the sky is falling)
0July 24, 2004 at 6:44 am #104208
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Stan and Gabriel:
The sky is not falling and Six Sigma is just fine!
With all due respect, and humbly so, I would still like to offer a different opinion again on the question of normality of the data. Alan H has only provided us with 30 values of the shaft diameters. We don’t know much else about what is going on his shaft turning process. Surely, all of us – you, Gabriel, and me included, agree that there are different ways of binning the data. What might look “far from normal”, as I said, might look “very normal” to experts like yourself. Please allow me to persist, however. You are both, I can see, interested in carrying on an honest and sincere debate.
Let’s look at this problem of the 30 diameters differently. Imagine the process of taking a photograph of a beautiful scene, or a beautiful statue in a museum, or a beautiful model. In one view, the beautiful model could be very far away, walking on the beach near the seashore. We can see just her silhouette. In another close up photograph, we see her lovely image, from head to head. In yet another photograph we see only her smiling face. We can zoom further, and then the beautiful face disappears and we see something very very different. We might not like what we see.
And so it is with “binning” of data. This is no different from the manipulation of the image of the beautiful model by the photographer, by choosing different lenses, lighting, and magnifications.
The choice of the interval or class width is discussed in many statistics textbooks. This affects what is known as “graininess”, just like the “graininess” in the photograph of the beautiful model as we start zooming in. It all depends on the eye of the beholder.
Gabriel gave the following example of binning of data in his post earlier in this thread.
Using “classes”. A class is a range of values. Each calss has a lower limit, an upper limit and a center that is ususlly taken as the value representative of that class. How many classes should one use? There are some “exact” formulas to calculate the correct number of classes, but I don’t like how they work. Accept for a minute that 5 clasees would be Ok for 30 values. Let’s make these classes as follows:>=
<
center
Freq6.665
6.670
6.6675
16.670
6.675
6.6725
106.675
6.680
6.6775
126.680
6.685
6.6825
66.685
6.690
6.6875
1
Now plot this histogram (or frequency polygon), and you will see it reminds you the Gauss bell somehow.
Another thing you can do is plot the 30 values on a normal probability paper and see if they form something like a stright line. In this case, they do.
And therefore, I humbly remain unconvinced that the world is always “normal”. It seems to me nothing more than a convenient mathematical ploy that helps us determine the probability of the occurrence of defects in Six Sigma. The area under the normal curve from – X to + X is the “yield”. The area outside this, or what we call the “tails” of the normal distribution curve are the “rejects” or the “defects”. We can do various statistical tests, only if we know the areas. The normal, or Gaussian, curve becomes just a mathematical “crutch”. Statisticians have no choice but to use, it seems to me at least, to develop various “tests” and “confidence levels”.
Please, do not let me know if there is anything fundamentally wrong in taking such a view, albeit a minority view. After all, Six Sigma is now being extended and applied to many many industries far beyond the manufacturing sector where all of this began – first with Shewhart, then Deming, and finally the Motorola researchers. Shewhart based everything was based on actual observations. He used drawings from his “normal bowl” to explain how samples might look like a “normal” distribution. But, he also provides examples of how what looks like a “normal” distribution could also come from a rectangular or a triangular universe. Mathematics is just a tool. May be we should keep an open mind and be willing to see the world as it presents itself without these “tinted or colored” glasses that makes everything look like it must be normal. This is just my humble opinion and is certainly subject to revision and modification as I learn more.
0July 24, 2004 at 6:47 am #104210
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Stan and Gabriel:
By the way, I have no clue as to what Anonymous is talking about and the special cause variation. Looks like I missed something when I was gone. With my warmest regards to both of you.0July 24, 2004 at 7:39 am #104215
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Stan and Gabriel:
I failed to edit my post. There are two important corrections.
Please, do not (please delete not) let me know if there is anything fundamentally wrong in taking such a view, albeit a minority view. After all, Six Sigma is now being extended and applied to many many industries far beyond the manufacturing sector where all of this began – first with Shewhart, then Deming, and finally the Motorola researchers. Shewhart based everything was based on (please delete underlined) actual observations. He used drawings from his “normal bowl” to explain how small samples from a large population might look like a “normal” distribution. (I added the underlined here.)0July 24, 2004 at 7:49 am #104217
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel:
The article was obtained by doing a Google search by just typing Taguchi loss function.
http://www.dnh.mv.net/ipusers/rm/loss.htm
After you mentioned the Taguchi loss function, I found this article. I would like your opinion about the technical content of this article. Author is not known. But he says, one “normal” distribution curve has a larger area, or the tail regions, outside the specification limits and the customer is therefore not very happy, more often. Do you agree with this main message? I just want to think some more about the minimum loss to society that you mentioned. Thanks in advance.
0July 24, 2004 at 7:06 pm #104236
AnonymousParticipant@Anonymous Include @Anonymous in your post and this person will
be notified via email.Thank you. I am incorrect. After 6 weeks on the road, I should have known better than to do any analysis.
I took the sample data from Alan H’s post and copied it, then absent mindedly sorted it to eliminate the spaces. It ordered the data and obviously created special causes in the control chart.
A lesson to all; don’t operate heavy machinery or Minitab without sufficient sleep.
My apologies.
Regards,
RGD0July 25, 2004 at 4:39 am #104244
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.So you got a run up or down of 30 consecutive points. WOW! That WOULD HAVE been a BIG out of control signal ;)
In fact, it was. One of the possible special causes for an outofcontrol signal is “the data is wrong”.
Had you only explained what you found, the discussion would have been much shorter and less “violent”.
Anyway, I am happy that you didn’t escape and came back to face your mistake.0July 25, 2004 at 4:53 am #104247
Noname expertParticipant@Nonameexpert Include @Nonameexpert in your post and this person will
be notified via email.Are You Anna ?if yes,welcome back to the forum,regards.
0July 25, 2004 at 5:20 am #104251
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Charmed:
What you say is correct. As in the photos, there is an optimum distance or number of bins for each case.
However, making a histogram is not the best way to check for normality (or lack of it). Bins are needed, but also bins neglet information such as that different values within the same bin are different. I only proposed you to plot a histogram with the bins because you introduced the concept that the distirbution was nothing close to normal after plotting a frequency polygon in a way tha, even when it is technically right for a frequency polygon, it is useless to compare with a model distribution.
But I also said:
“Another thing you can do is plot the 30 values on a normal probability paper and see if they form something like a stright line. In this case, they do”
The normal probability paper is a paper with a grid, a not linear grid, where cummulative normal distributions for a stright line. Much like exponential functions look like stright lines when plotted on a logarithmic paper. When you plot datapoints sampled from a normal distribution in this paper, the points do not form a stright line but a cloud of “random” points arround the stright line that represents the distribution they belong to. Any nonnormal distribution will look cambered (like a U or an S) on the normal probability paper. If the sample datapoints belong to a clearly not normal distribution, they will form a cloud of random points arround some pattern that will clearly not be a stright line.
If you do this in this case, you will find that the cloud of the 30 point closely match the stright line that would correspond to a normal distribution with average 6.676 and standard deviation 0.0044.
There are also some mathematical tests to check how likely is to get such a set of points if they belonged to a normal distribution. These test get rid of the visual and subjective judgment of how nicely the histogram fits the bell shape or how close are the point to a stright line.
But, most importnat, no test, and I mean NO test, will ever tell you that a sample belongs to a normal distribution. The best you can say is “it is unlikely that such a sample could be obtained from a normal distribution” or, in the other hand, “there is no evidence to reject that this sample belongs to a normal distribution”. In fact, as the normal distribution does not exist in real life, no real life sample will ever belong to a normal distribution. I never meant to say that the data was normal. Just that there was no reason to say that it was not normal. A person can be the criminal even if you fail to prove it. And a distribution can be not normal even if you fail to prove that it is not normal.
“Mathematics is just a tool. May be we should keep an open mind and be willing to see the world as it presents itself without these “tinted or colored” glasses that makes everything look like it must be normal.”
Who wants to make everything look normal? Reread my posts. Everywhere I am advising not to rely on the normal distribution to predict the defectives rates.
Box Cox said “All models are wrong. Some models are useful”. A real life distribution is NEVER fully normal. Yet, when it is not too far, the normal distribution can be used as a useful model.
MY OPINION IS THAT THIS IS NOT ONE OF THESE CASES. Even when there is absolutely no reason to say that the distribution where this 30 points were taken from is not normal, it is very unlikely that a normal distribution based on the sample average and sample standard deviation can give you a good prediction of the defectives that will be found under the tail simply because in these 30 values there is no one single point in the tail where defects will happen (above the upper specification limit), and in fact the point that is closer to that tail is still more than 3 sigmas away well into the “in tolerance ” zone. Making predictions extrapolating a normal distribution 3 sigmas away of the last datapoint lacks of any scientific, logical, or rational foundation. If you want to play with numbers, there you have the lotery.0July 25, 2004 at 5:46 am #104252
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.I didn’t found your quote in that article. The most similar I found are these two:
“Curves A and B represent the frequencies of performance of two designs during a certain time period. B has a higher fraction of “bad” performance and therefore is less desirable than A.”
It refers to the traditional way, also called “goal post mentality”, wehre a product anywhere within tolerance is equally good and a product anywhere outside tolerance is equally bad. The figure referred in that paragraph shows two normal distributions, both of them centered in the middle of the tolerance, but B has more spread, and the loss function associated with the traditiona “goal post” mentality (zero lost within tolerance, a constant loss out of tolerance).
“If two products have the same average but different variance, then the product with the smaller variance has better quality figure 4. Product B performs near target less often than its competitor.”
This one is in the context of the Taguchi loss function. The associated graph shows the loss function alaTaguchi (zero lost only on target and an increasing loss as you ove away from the target) and two normal distributions with the same spread, one with the average close to the target and the other one with the average farther from the target. If you imagine a narrow zone arround the target (in the horizontal axis) you will see that the area under the curve A in that zone is much larger than the area of the curve B in that zone, meaning that you find individuals in that zone (i.e. near the target) for the case A more often than for the case B.
I don’t know if this ansers your question: I like the Taguchi way.0July 25, 2004 at 10:55 am #104260
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel:
A very charming reply. I pasted it here so we can study each point together.
What you say is correct. As in the photos, there is an optimum distance or number of bins for each case. (How nice to hear!)
However, making a histogram is not the best way to check for normality (or lack of it). Bins are needed, but also bins neglet information such as that different values within the same bin are different. I only proposed you to plot a histogram with the bins because you introduced the concept that the distirbution was nothing close to normal after plotting a frequency polygon in a way tha, even when it is technically right for a frequency polygon, it is useless to compare with a model distribution. (No problem, we agree.)
But I also said:
“Another thing you can do is plot the 30 values on a normal probability paper and see if they form something like a stright line. In this case, they do”
The normal probability paper is a paper with a grid, a not linear grid, where cummulative normal distributions for a stright line. Much like exponential functions look like stright lines when plotted on a logarithmic paper. When you plot datapoints sampled from a normal distribution in this paper, the points do not form a stright line but a cloud of “random” points arround the stright line that represents the distribution they belong to. Any nonnormal distribution will look cambered (like a U or an S) on the normal probability paper. If the sample datapoints belong to a clearly not normal distribution, they will form a cloud of random points arround some pattern that will clearly not be a stright line.
If you do this in this case, you will find that the cloud of the 30 point closely match the stright line that would correspond to a normal distribution with average 6.676 and standard deviation 0.0044. (I did not do this. This is still just like using a loglog plot. On a loglog plot you can make things look like straight lines. Does that mean the law is ln y = ln A + B ln x? Not necessarily, more tests are needed. We should agree, Gabriel.)
There are also some mathematical tests to check how likely is to get such a set of points if they belonged to a normal distribution. These test get rid of the visual and subjective judgment of how nicely the histogram fits the bell shape or how close are the point to a stright line. (You have done this, and I accept your conclusion, also CS conclusion.)
But, most importnat, no test, and I mean NO test, will ever tell you that a sample belongs to a normal distribution. (I fully agree with you.) The best you can say is “it is unlikely that such a sample could be obtained from a normal distribution” or, in the other hand, “there is no evidence to reject that this sample belongs to a normal distribution”. In fact, as the normal distribution does not exist in real life (this is my main point), no real life sample will ever belong to a normal distribution. I never meant to say that the data was normal. (I did not think so, either, no disagreement between us.) Just that there was no reason to say that it was not normal. A person can be the criminal even if you fail to prove it. (Only if we know that he is, how do we know that, if we cannot prove it?) And a distribution can be not normal even if you fail to prove that it is not normal.
“Mathematics is just a tool. May be we should keep an open mind and be willing to see the world as it presents itself without these “tinted or colored” glasses that makes everything look like it must be normal.”
Who wants to make everything look normal? Reread my posts. Everywhere I am advising not to rely (I have seen this and I agree, I come from different background) on the normal distribution to predict the defectives rates.
Box Cox said “All models are wrong. Some models are useful”. (How Correct, I fully agree with this from a great and honored statistician.) A real life distribution is NEVER fully normal. Yet, when it is not too far, the normal distribution can be used as a useful model. (Exactly my point also – it is just a useful model but does not mean we are stuck with it forever, are we?)
MY OPINION IS THAT THIS IS NOT ONE OF THESE CASES. (And I will accept that, since you and CS and Stan have performed some other tests to support this statement. BUt, now take the other side of the criminal story. Just because you convict somebody as being criminal, does not mean he/she committed the crime. We may also be missing something and coming to the wrong conclusion, so I persist, especially since Six Sigma is now being extended to many areas and will soon become a joke if we are not careful about the fundamental scientific and mathematical basis of what is a very useful tool to analyze business and management deficiencies.) Even when there is absolutely no reason to say that the distribution where this 30 points were taken from is not normal, it is very unlikely that a normal distribution based on the sample average and sample standard deviation can give you a good prediction of the defectives that will be found under the tail simply because in these 30 values there is no one single point in the tail where defects will happen (above the upper specification limit), and in fact the point that is closer to that tail is still more than 3 sigmas away well into the “in tolerance ” zone. Making predictions extrapolating a normal distribution 3 sigmas away of the last datapoint lacks of any scientific, logical, or rational foundation. If you want to play with numbers, there you have the lotery. (I don’t like to play lottery, but I win many times when I play. Just lucky! Nothing to do with statistics. If there is no luck you will never see people winning $200 million or $230 million. An old lady, who did house cleaning all her life won the highest lottery recently. Sorry, I did purchase tickets, but that was her luck on that day. God’s Will for Her, not Me.) So we agree in many ways. But there are also few points where we don’t. Never mind. Thanks a lot Gabriel.0July 25, 2004 at 11:02 am #104261
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel: Please allow me to be brief now. There are many articles on the Taguchi Loss Function. It says Loss = k (x – mean)^n, where x is the deviation from the mean value. Notice I say “n” but Taguchi says n = 2. Of course, we must also know how to fix k. It would be wonderful if there is a way to test if n = 2, even approximately, and how deviation from 2 affects k. The parabolic function used for loss is just a simple way to solve a much more complex problem. With my warmest regards.
0July 25, 2004 at 11:06 am #104262
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Gabriel:
Sorry: I made an incorrect staement here to discuss Taguchi Loss Function. It (x – mean) that is the deviation from the mean value and x is the value of interest.0July 25, 2004 at 2:25 pm #104264
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.“So we agree in many ways. But there are also few points where we don’t. Never mind. Thanks a lot Gabriel.”
I failed to find a single point where we don not agree. Maybe I was not clear in one point. In the paragrapha that I began with:
“MY OPINION IS THAT THIS IS NOT ONE OF THESE CASES”
I mean to say that my opinion is that in this case the normal model is NOT a useful model to predict defectives even when there is no evidence to support that the data is not normal.
It seems to me that you understood the opposite from your comment:
“And I will accept that, since you and CS and Stan have performed some other tests to support this statement.”
About the other side of the criminal story, you are right. One is found guilty when it has been proven “beyond any reasonable doubt”. Not “beyond any doubt”. So there is still an acceptable margin of error, or risk to condemn an inocent. In statistics this risk is called Alfa risk. Remember I said that one conclusion of the test could be “it is not likely to obtain this data IF the distribution was normal” Not likely is different from impossible. The gap between thenm is the alfa risk (which is the probability to get such a data froma normal distribution).0July 25, 2004 at 2:31 pm #104265
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Charmed:
If you ever want to discuss in some depth the “Taguchi’s loss function”, then find another one. I will be glad to read the exchange, but I am not the right one to provide help with the details. I am not uset to it, never used the formulas, and only know the basic concepts.
I advice you to open a new thread on this subject. I’m sure you will get some proffesional help from the forum.0July 25, 2004 at 3:06 pm #104266Charmed,
Taguchi’s loss function uses the same principle that is used in regression – that losses are proportional to the square of the deviation. According to Taguchi, a Quality Loss is proportional to the deviation from target. (The constants are described on page 170 of ‘Six Sigma for Electronics Design and Manufacturing,’ by Sammy Shina Ph.D., which incidently is one of the most sensible books I’ve read on Six Sigma. According to the author the 1.5 sigma shift makes no sense whatsoever.)
If one expands a normal distribution using Taylor’s expansion we can derive a similar law to Taguchi’s, therefore, we might speculate that the Loss is proportional to the complementary normal distribution [1 – N(0,1)] of the mating component. (A Japanese company is now investigating this approach as it can be quite difficult to determine ‘real ‘ tolerances in some systems.)
Accordingly, we might further postulate that Quality is an indirect measure of ‘matching’ – between a component and a mating component, or in general between a supplier and a vendor. In other words, the function of Quality is to minimise losses due to mismatching. This is in use in common language and we often ask if an applicant a good match for a certain position.
In a recent post by Reigle, he cited the case of a power transformer in an attempt to demonstrate a univariate quality relationship – what he failed to recognise is that a tranformer matches the source and output impedances using the turns ratio of the transformer, n1/n2. The function of the transformer laminates are only to minimise eddy current flow, thereby reducing heat losses. If the laminates are too thin, then there is a risk of vibration and delamination.
Please corrent me if I’m wrong, but is the next step in your argument that since we can’t guarantee that a distribution is normal, then we should appy a shift? (You see I suspect that you are realy Reigle.)
My apologies to both of you if I am wrong.
Just my two cents …
Respectfully,
Andy U
0July 25, 2004 at 3:12 pm #104267
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel:
“MY OPINION IS THAT THIS IS NOT ONE OF THESE CASES”
I mean to say that my opinion is that in this case the normal model is NOT a useful model to predict defectives even when there is no evidence to support that the data is not normal.
It seems to me that you understood the opposite from your comment:
Yes, in this case, it does look like I understood the opposite of what you meant. And, according to you, there are many points where we disagree and a few where we agree. I try to be say something positive first. “Technically speaking” it may be incorrect, but that’s how we use the language and how most human interactions are (or should be) conducted. Again, thanks a lot.0July 25, 2004 at 3:34 pm #104268
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Andy U:
This is Charmed, and thanks for your participation. This is not Riegle.
Please correct me if I’m wrong, but is the next step in your argument that since we can’t guarantee that a distribution is normal, then we should appy a shift? (You see I suspect that you are realy Reigle.)
Please allow me to address only the main point pasted above – since we can’t guarantee that a distribution is normal, then we should appy a shift? NO. That’s not my intent. I have seen the recent threads where the 1.5 sigma shift was discussed. This is an “old” controversy. Never seems to die. In my humble opinion, there is a way out. But this means we accept that the world is not necessary “normal”.
First of all, I would like to recommend that we BANISH the use of the term “normal” since it leads to misinterpretations. Statisticians started using this term “normal” to described what should be called “Gaussian” distribution since it seemed to be confirmed so often – this was pure manipulation, as I discussed in my exchange with Gabriel (there was agreement on that point). Calling all distributions “normal”, oops Gaussian, is like manipulating the photograph of a beautiful model and insisting that we should only look at that photograph at a certain magnification, lighting, etc. – this is also called Lyihng with Statistics. Statisticians use such selfdeprecating terms but continue to do what they do and seem to also think that the conclusions justify the wisdome that is being offered. That’s really what we do when start binning data and look only at the “normal”, hereafter “Gaussian” distribution.
There is a way out. Let’s take one step at a time. Abandoning the concept of “normal”, or hereafter “Gaussian” distribution, entails paying a severe price. We can invoke Chebyshev’s theorem which involves no such assumptions. Can we practice Six Sigma and perform other statistical tests without using normal, hereafter to be only called Gaussian, distribution? I think we can. Coincidentally, I had just finished analyzing some Gage R & R data when I was offline. This may be found in the link given below.
Attribute Agreement Analysis White Paper
Author: Issac Newton, Minitab Technical Center
First Publication: February 2004http://www.minitab.com/ltd/Resources/VirtualStories/attributeagreement.pdf
I would be very happy to discuss this in more detail later. Let’s take on step at a time. I would be overjoyed if we do start asking questions about the “normal”, or Gaussian, distribution. Thanks, Andy U, for your probing questions. Please be assured this is not Riegle.
Just Charmed0July 25, 2004 at 3:53 pm #104269Andy,
Just my vote, but I am pretty sure Charmed is not Reigle. Reigle’s language is very structured and he could not have had this many exchanges without invoking the gospel according to Mikel.
I do think a good discussion of Taguchi’s Loss Function and how it should be applied in the Six Sgima community is in order. The implication of all the 1.5 shift discussion is we are trying to center our processes, but in practice most Six Sigma consultants never say the simpliest of things like – “center your processes”. This is a huge mistake, instead we measure DPMO against what is probably a poor understanding of our customer.
Just my opinion.0July 25, 2004 at 4:07 pm #104270
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Andy U:
Please permit me now to briefly touch upon the other part of your post. This is pasted below for reference.
Taguchi’s loss function uses the same principle that is used in regression – that losses are proportional to the square of the deviation. According to Taguchi, a Quality Loss is proportional to the deviation from target. (The constants are described on page 170 of ‘Six Sigma for Electronics Design and Manufacturing,’ by Sammy Shina Ph.D., which incidently is one of the most sensible books I’ve read on Six Sigma. According to the author the 1.5 sigma shift makes no sense whatsoever.)
If one expands a normal distribution using Taylor’s expansion we can derive a similar law to Taguchi’s, therefore, we might speculate that the Loss is proportional to the complementary normal distribution [1 – N(0,1)] of the mating component. (A Japanese company is now investigating this approach as it can be quite difficult to determine ‘real ‘ tolerances in some systems.)
The Taguchi loss function then:uses the same principle that is used in regression – that losses are proportional to the square of the deviation.
can be justified by expanding a normal distribution using Taylor’s expansion, therefore, etc.
This is a classic method used in many theoretical analyses. The first step is always to simply modify an existing theory – just tweak it a bit and see if it works. Often, this is fine. We are happy. That does not mean all is fine. Sometimes, we must seek a different solution or a wholly new approach since the “flaws” in the present approach are usually carefully hidden. And there are many adherents. This makes it difficult for an alternative viewpoint to even get across. I say this only because Six Sigma methodologies are now penetrating products/services far beyond the manufacturing sector, where Motorola researchers developed it. The 1.5 sigma shift is a part of that legacy. I am not in a position to offer any opinions on that matter. I see simply an “number” that is being added to the calculated value of the Process Sigma using the equation
Process Sigma = Normsinv[1 – (Defects/Opportunities)] + 1.5
If we agree that distributions other than normal, or oops Gaussian, can be observed and must taken seriously by all Six Sigma practitioners, this equation for Process Sigma must clearly be replaced by some other, more meaningful, equation. Can such an equation be developed? May I therefore humbly and respectfully suggest we get back to the drawing board – and start taking a hard look at “data” or “observations”, including even manufacturing activities. Certainly, we have to look at service, financial, healthcare, government, national security to name just a few areas where Six Sigma practitioner boldly want to go. To paraphrase another thread – One Small Step for one man but a giant leap for Six Sigma.
The 9/11 commission, which just finished its report on the terrorist attacks on the United States, feels we lacked “imagination” in our analysis of the “data” being gathered by intelligence agencies. In other words, we do not still know how to deal with what we call “outliers”. We seek “mediocrity”. The least squares, or the bestfit, or the linear regression line must always pass through the average values of X and Y. This has always bothered me. This enshrines “mediocrity”. We must search for EXCELLENCE, which lies in the outliers. How do we make sense of this? If we had understood this, we might have been able to prevent the 9/11 attacks. Unfortunately, in my humble opinion – just one no name person’s opinion – statistics failed us with its search for mediocrity.
Thanks again for asking these probing questions. It gives me a chance to spell these points out at least while the mathematical edifice is far from completed. We can at least think aloud while waiting for guidance. With my warmest regards.0July 25, 2004 at 4:42 pm #104272
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Stan:
In my post on Friday, regarding the “distribution” of the 30 diameters given by Alan H, I said,
So, I think the process should be centered around 50, i.e., 6.6500.
This therefore means I cannot be Riegle or most of the Six Sigma consultants that you mention here in your exchange with Andy.
Sorry, for interjecting. I usually try not. Exchanges such as this, even in a public forum, should be treated as “private”. I try to address only posts that are directly addressed to something I have said, or are directly addressed to me. Please permit me this one exception.
Finally, I agree with you also on the need to discuss the Taguchi Loss Function and to better understand it within the context of Six Sigma. Or, let’s put it differently. How do we minimize the loss (this is another way of saying the lowest cost for the product) while at the same time meeting the customers requirements? Not easy. But, then nothing is easy, even if we think it is, such as the birth of a beautiful …..0July 25, 2004 at 5:15 pm #104273Stan,
I agree with you on both counts …
I only read one of the posts – I should have read more of them – and it looked like the argument was heading in a certain direction.
Cheers,
Andy0July 25, 2004 at 5:26 pm #104274Stan,
Charmed has made a very interesting assertion in his public/private communication with you. He asserts that Taguchi’s loss function minimises the unit manufacturing costs. Do you think this might be the a common belief?
My understanding is that the Taguchi loss function minimises the loss to society, which includes manufacturer, supplier, and customer.
In other words, if I have a ‘tricky’ process and by customer depends upon my efforts, I should not ‘walk away’ just because I can’t achieve a z=4.5 sigma.
What is your take on this ..
Cheers,
Andy0July 25, 2004 at 5:36 pm #104276Dear All, let me to try to eplane better my point of view on this matter. Maybe my Fridays post was too brief.
From Alan data I see : well controlled process (maxmin of values is very low) ,but all the values are in the upper side of distribution, infact if you move the target to 6.65 the process is well centred. Now the point is: which is Alan target ? For me it is 6.6 not ,6.65 as stated in previous post, because the customer limits are 6.6 +0.1/+0.0. Now if the procees give results always in the upper side of distribution, this means it is out of control or better it have an offset of +0.05, so the process is stable but drifted to upper side, so because the prosses is very stable, as you have an out of control you’ll have no 0.1 available, but just 0.05. For me it not the best solution to move your target to 6.65 (you can obtain the same effect moving the machine set up to target of 6.55), but putting in place the right action to reduce the drifting from upper side. How to do it, is another story (maybe just calibrating the machine, maybe changing machine not able to achieve it, etc..). The reason why I said “smaller is best” is beacuse you can consider 6.6 as your zero and the limit of +0.1 as value that must tend to zero. The reason of I mentioned the Taguchi loss function is bacuse it take care of “cost” to set up limits, that at least is the most importat parameter. This is just my point of view, of course can be wrong.
P.S. Apologies if this post is just descriptive without details of data, but it is sunday and I have data in my office PC and now I’m writing from home (it is raining, unfortunately …).
Rgs, Peppe0July 25, 2004 at 6:00 pm #104277I agree completely.
By the way, most of0July 25, 2004 at 6:03 pm #104278I agree completely.
By the way most of these folks on here only know of z >/= 4.5 in theory anyway. Most do not center their processes nor do the utilize Stnadard Work, SMED, and other tools that are about consistency and centering.0July 25, 2004 at 6:07 pm #104279I agree with measuring loss with respect to 6.6 with the way Alan defined the problem, but you cannot target your process at 6.6 — you instead target your process offset above 6.6 to achieve minimum product below 6.6. This is hopefully <<6.65.
0July 25, 2004 at 7:20 pm #104281
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
Thanks for participating, even from your home, when it is raining (What, I don’t see any rain! You are not in the same place I am, at this time. I know that with absolute certainty. It has not rained all morning.)
The problem that you mention, especially about where to “center” the process can be solved mathematically. Just be a little more open and let’s enjoy this. I have some ideas that are churning in my head now. Of course, I was very intrigued by the fact that someone (Alan H) came to the forum with such a simple, but very important, question. He had 30 measurements of the shaft diameters. What next? So, I wanted to find out what the 30 values were. He obliged and this thread has continued, now to a discussion of the Taguchi Loss Function. Wonderful!
If Alan H returns and is willing to share more information and more historical data and what his (customer’s) requirements, I think, we can take a crack at solving the minimum loss problem as well. The first step in minimum loss calculation should be minimum loss to the company producing the shafts. The problem can be tackled without Alan H’s data. May be you can supply us with some instead.
Society is a boundaryless and abstract concept that really means nothing, at its very core. As the old Aesop fable goes, you cannot satisfy everyone. This also goes for satisfying “society”. There are only two people to satisfy and I am very clear about that: Boss and the Customer – in that order. If the Boss is not satisfied, I have no customer to satisfy. If the Boss wants me to satisfy customer, I do. He says smile, I do. He says, jump I do. (But not off a cliff, my wife will not like that!)
It’s that simple. Hope rain stops and you go back to office soon and reply. The 30 diameters are in Alan H’s posts. No more excuses. With my warmest regards or shall I say Cheers!0July 25, 2004 at 7:32 pm #104282
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear All:
There is an interesting article on this website about Six Sigma and the SocioTech Issue that we have ventured into. I am sure many have seen it. May be worth taking a second look.
https://www.isixsigma.com/library/content/c020902a.asp0July 25, 2004 at 11:03 pm #104286
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Charmed:
You said:
“And, according to you, there are many points where we disagree and a few where we agree.”
Why do you say that? This is from my previous post:
“I failed to find a single point where we do not agree”
If my English is Ok (what could be not the case) this means that, as far as I found, we agree on every point. I couldn’t find (or “I failed to find”) one point (“a single point”) where we disagree (“where we do not agree”).0July 25, 2004 at 11:37 pm #104288
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Peppe,
Sorry for this question: Are you sure you understand what “in control” and “stable” mean?
In control and stable are the same thing, and means consisten behaviour over the time.
Quote:
“Now if the procees give results always in the upper side of distribution, this means it is out of control or better it have an offset of +0.05, so the process is stable but drifted to upper side, so because the prosses is very stable, as you have an out of control …”
“…if the procees give results always in the upper side of distribution, this means it is out of control…” If the process gives consistent results (like “allways in the upper side”) then it is stbale or incontrol by definition. If the process gives garbage all the time then it is stable or in control. Further more, there is no way a process can allways give results in the upper side of the distribution, because the distribtuion is the result of the process. It would be like saying “the process gives results allways above the process average”. It is impossible, by definition of average.
“…so the process is stable but drifted to upper side…” If the process is stable it does not drift. “Drift” is one of the forms of unstability. What IS possible is that the process was stable in one possition, then become unstable and drifted, and then stablized again in a new possition. But we do not have any info that the process was ever in a different possition than in the 30 points infromed by Alan.
“…the prosses is very stable, as you have an out of control …” This is self contradictory. A process can not be very stable and out of control at the same time, because stable means in control, and out of control means unstable.
Another point:
“The reason why I said “smaller is best” is beacuse you can consider 6.6 as your zero and the limit of +0.1 as value that must tend to zero”
You can consider 6.6 as your zero if you want, but this not make this a “smaller is best case”. In this case it is perfectly possible to get a part at, for example, 6.55 (that would be 0.05 if you take 6.6 as zero). If it was a smaller is best case, the part at 6.55 would be better than a part at 6.6, what is not true because 6.6 is the lower specification limit and then anything below 6.6 is considered nonconforming.
What makes a case be “smaller is best” is when you want it as small as it is physically feasible. For example ovality. You want it as close to zero as possible, but you will never get an ovality below zero. So the smaller the better.0July 26, 2004 at 12:35 am #104292
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel, Gabriel, Gabriel:
Sweeter words were never uttered. “I failed to find a single point where we do not agree. “
Somehow, somehow, I misread this and “failed” to see how “failed” appears in this sentence. I must have read it as “I fail” etc. and missed the “not” later in the sentence. Why, because Gabriel, I wasn’t expecting this from you. This is called subconscious juxtaposition (I don’t know the correct psychobabble). What a truly joyous statement!
If my English is Ok (what could be not the case) this means that, as far as I found, we agree on every point. I couldn’t find (or “I failed to find”) one point (“a single point”) where we disagree (“where we do not agree”).
Your English is very very OK Amigo, it’s my Spanish that needs brushing. I prefer Swedish, though. A Nord friend taught me some wise words in Swedish to practice on a fabulous Swedish blonde. Since then I have become very comfortable with my Swedish. I just cannot speak French, though. Sorry. With my warmest regards. Have a wonderful week.0July 26, 2004 at 12:39 am #104294Hey Charmed, you can always practice your Swedish on Stan’s Swedish flight attendant alter ego.
0July 26, 2004 at 12:43 am #104296
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Darth: You mean they still fall for that old trick I learned! In English, it means… well, I can’t tell you. Then I will lose my edge. Have a great week.
0July 26, 2004 at 1:00 am #104297
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Gabriel:
Very nice discussion with Peppe. I really loved the last paragraph, reproduced below. Sorry for interjecting.
What makes a case be “smaller is best” is when you want it as small as it is physically feasible. For example ovality. You want it as close to zero as possible, but you will never get an ovality below zero. So the smaller the better.
Now a simple question for you on ovality. Oval means out of round, but it looks like round – say an ellipse or an egg shape, or something similar. That’s my understanding of ovality. How do we measure ovality, precisely, that’s one question. It would be nice to know, Gabriel. Please tell me if you know about this. Second, if we take ovality to mean simply offround, when ovality goes to zero, you must get a circle. The circle is considered a “perfect” shape. The full moon is very nearly (I think) a perfect circle. The sun’s disk is very nearly (I think) a perfect circle. The earth is very nearly (I think) a perfect sphere. Now, let’s go further. Let the ‘oval’ become a circle. What happens next? The circle can get smaller and smaller. Ultimately, the circle vanishes and becomes what ….. a point! From a point grows a space called a circle or a sphere, and in this space we find an oval if the growth of the shape from the point is not uniform in all directions. Would you agree, Gabriel? If I am wrong, tell me. If not, have a wonderful week. (I go practice Swedish next week and after that it will be Spanish week! I always loved Sophia Loren and Evita.)0July 26, 2004 at 2:46 pm #104324Dear All, first of all apologies for my imperfect english, but it is easy to understand, english is not my native language. About data, Charmed, I don’t need excuse, I just mean I have data already analized in the office, not at home, anyway I did it again here.
To answer to Gabriels questions, yes I know what means stable and in control, but maybe I have explaned bad what I mean. The Alan process is quite stable (maxmin=0.021=20% of available tolerance), but it is not distribuited normally within the tolerance because is quite far from target (min=6.665=0.065 far from 6.6), than the top limit (max=6.686=0.014 near top limit). 29 of 30 valueas are within 6.56.7 range that is just 50% of tolerance, but the more far from target. Of course we need more data to have full picture of all and predict the trend, that is too (too to be real in my view) flat (the mean is 6.676 and the equation is : y = 1E05x + 6,6759 and R2 = 0,0007. Let me also disagree on target. What I understand is that you must achieve 6.6 and if ideally you achieve it, you are doing the best, not a nonconformity. The specs are 6.6 +0.01,0.0, so at least your customer is asking for 6.6, so, smaller is the best is the right approach for this, otherwise you have also a lower limit with tolerance. But this could be only explaned by Alan the customer requirements. Just to give you an examples,we have, sometimes, some requirement where tolerances are not admitted (i.e. optical wavelenght in multiwave systems), they must be just the exact value.
Rgs, Peppe0July 26, 2004 at 2:56 pm #104326
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
Nice to hear from you again. Now you say, …..but it is not distribuited normally within the tolerance ……. This is pasted from your message. So, you too believe the data is not obeying the famous “Gaussian” mathematical law describing the distribution of the numbers X (in this case diameters)? Please give a simple yes or no answer. You can even use your native language or shake your head (sorry, we don’t have video clips with this, may be soon!)0July 26, 2004 at 4:06 pm #104343Dear Charmed, it was very clear that data are normally distribuited around the mean (it is very clear also just simply plotting data), but what the problem is that the mean is too far from target and too near upper limits. I repeat, this is just my point of view and can be wrong.
Rgs, Peppe0July 26, 2004 at 5:19 pm #104359
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
I prepared the same plot that Gabriel posted which shows how the data falls between the upper and the lower specs and the three sigma values from the mean. I think, you are insisting on looking at “target” very differently from I am and may be some others are as well. Cannot do much about that. Each man (or woman) has his or her own “target”. With my warmest regards.0July 26, 2004 at 5:57 pm #104365
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.With “ovality” I meant “difference form a perfect circular shape”, also called “lack of roundness” or “circularity error”. “Ovality” was not a happy choice.
About the “shape growing from a point” etc…, I have no comments.
0July 26, 2004 at 6:19 pm #104370
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Peppe,
Now it’s clear: You don’t understand what stability means (not at least in the field of statistical quality control and statistical process control).
“The Alan process is quite stable (maxmin=0.021=20% of available tolerance)”
Alan’s process can be 0.0002%, 20% or 20000% of the available tolerance and that by itself will not make it more satble or more unstable. Meeting the specifications has absolutely nothing to do with stability ot process control. A process can be stable even if a tolerance and/or a target do not exist at all. Stability has to do with the evoution of the process behavior over the time. Read the post from Anonymous (near the bottom of this thread) where he initially said that the process was not stable, then I answered that in fact was stable and finally he recognized it was stable and explained what his mistake was. You will see that in all that discussion the tolerance is not even mentiond. You will also see a control chart, that is the way to tell if a process is stable or not.
About “smaller is best”. Tell me one thing: Which is best? 6.2 or 6.6? Which is samller?
And finally: “we have, sometimes, some requirement where tolerances are not admitted (i.e. optical wavelenght in multiwave systems), they must be just the exact value” I can’t argue that because I don’t know. I just wonder what kind of measurement system you use to measure a wavelength perfectly free of any error or uncertainty.0July 27, 2004 at 7:18 am #104414I agree with you. In fact the problem is just that. From your point of view the data fall within rules of 3 sigma and for you are OK. For me no, or better if I had that situation in my process I wasn’t happy of that, until I drop down the values nearest as possible to target. Of course it is just a way to approach the problem, but from your perspective it could be not a problem, from my side yes. Rgs, Peppe
0July 27, 2004 at 9:56 am #104420
CharmedParticipant@Charmed Include @Charmed in your post and this person will
be notified via email.Dear Peppe:
You say, From your point of view the data fall within rules of 3 sigma and for you are OK.
I never said that. Please read my earlier post carefully. You jumped after I talked about where the process should be centered. So, I said, you have your view and I have mine. But, my view was not what you have stated here. Again, with my warmest regards. If have to sign off now. These threads are going nowhere and have become very counterproductive (I don’t mean you Peppe, in general what is going on in this forum now.)0July 27, 2004 at 10:17 am #104421Dear Charmed, I believe, from my point of view, here often is loose the real objective of discussion. To check caracteristic of some data is very easy for everyone using excel or other tool. Different thing is to think about a “question” and its effects of product, processes and money. Of course this simple thread was just an exercise and not a real situation, but it give indication about different approach to problems by different people. Thanks a lot for your respect to my view, also if they could be wrong. It is the right way to discuss, from my point of view.
Rgs,
Peppe0July 27, 2004 at 12:05 pm #104428
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Peppe,
The fact that the control chart shows no signal only means that the process is not unstable. Not that it is Ok.
Stable or incontrol only means that the behavior is consistent over the time. Imagine a process tha all the time produce parts in the range 14.0 to 16.0, with more parts near 15.0 than near the extremes of the range, and that when plotting a control chart no point goes beyond the control limits and no nomrandom patterns are identified. This process is stable. Is the process Ok to meet the specification? How to know? The specification has not been stated yet.
If the specification was 15±5 then yes, the process is very good. If the specification was 12 +4/0 then it is marginal, but it could be much improved if it was possible lower the the average. If the specification was 10±2 then the process is awful, with 100% out of tolerance. Yet, if the process location is controlable and can be adjusted to match something close to 10 it can become very good. Now if the pecification is 15±0.5, even when the process is perfectly centerd it has too much variation and produces many parts out of tolerance.
Nothing of this changes the fact that the process is stable.
By the way, in the same way that stability is not necessarily something good, unstability is not necessarily something bad. For example, improvement is a particular case of change of behavior, and change of behavior is unstability by definition. A process that remains stable does not improve.0July 27, 2004 at 12:40 pm #104435Gabriel, I fully agree with you. This is the exact point : improvement.
Many thanks for your reply. Rgs, Peppe0 
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