A Simple Problem for the Experts

Tim Folkerts:Very nice work; however, you forgot to factor in the
envelope. The envelope must be a part of the total RSS
as well. It too has a tolerance and will vary in accordance
to the same standard deviation (S=.001). Dont forget to
compute the nominal gap: 4.976 – (1.240 * 4) = .016. So,
Z.gap is given as (0 – .016) / RSS.Great WorkReigle Stewart

Mostly I was interested in clarifying the question.  You stated that the engineering specs were 1.240 and 4.976; you implied (but didn’t specifically state) that the measured production specs were the same.  I just wanted to point out that I was using those numbers as the measured values and therefore as the basis of my calculations  (We all know that what the engineer asks for and what gets produced are often two different things!). 
 
Beyond the simple mathematical need to know the measured values, there are of course other challenges as you point out – the physical challenge of keeping the process performing as desired and the statistical challenge to get a good estimate of this performance.  I don’t pretend to know how well the postulated 1.5 sigma shift emulates either of these difficulties in real life. 
 
As you state “Perhaps one of the design goals should be to devise a set of tolerances that are “robust” to shift and drift.”  It seems that you can always make the process more robust in several ways:
1) you can create a design that is tolerant of variation.  2) you can use better methods of manufacture to reduce variation.3) you can measure a lot to catch changes and use feedback to correct the problems.
All of these cost money, so you have to balance the costs vs the benefits to decide how to most effectively “robustify” a process.
 
Tim
 

Reigle,
 
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 anti-DfSS.  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 Z-value that cuts of 1.7*10^-6 in each tail

Establish 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 = 0.001538

Establish 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 off-set 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.
 
Statman

Reigle,
 
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 Reject-Support significance tests and Accept-Support 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 under-estimate 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 reject-support 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 Accept-Support situation).  I do understand how easy it is for the novice to sometimes get confused and retains the conventions applicable to RS testing.
 
Statman
 

Tim,
I am not sure I understood you.
You said:
– “of all the ways to calculate a st dev of 1.0 from a sample of 30 peices, here are the odds that it came distribution with a specific ‘true’ standard deviation.”  In this particular case, there turns out to be approximately a 0.5% chance that the calculated st dev of 1.0 could have come from a distribution with a ‘true’ st dev greater than 1.5.
In fact, there was allways ONE “true” stdev, and it was allways 1.0. All the 300,000 individuals belong to a population with stdev 1.0, which is the “true” one. Then you took several samples from this population, and you found different “sample’s standard deviations”, which is used as an estimator of the “true” “population’s standard deviation”.
So, in fact, what you are saying is that if you take a sample of size 30 there is a 0.5% of chance to get a sample’s standard deviation of 1.5 or grater when the true standard deviation  is 1.0.
You must understand the difference between that statement and and a statement like “if I take a sample of size 30 and find a sample’s standard deviation of 1.0, there is an X% of chance that the actual standard deviation was in fact 1.5 or greater”.
To use a different example: There is a coin and I don’t know if it is fair (heads and tails are 50/50) or if it delivers more heads than tails.
I do not need a motecarlo simulation to know that, if the coin was fair, the rate of heads would be 50% (it’s a binomial distribution). But let’s do it anyway: I run a 10 columns by 10,000 rows where each cell has either a 0 or a 1 with equal probability (50/50), and count the number of ones (which will be taken as “heads”. I get the following result:

Heads
Cases
%
Theory

0
16
0.16%
0.10%

1
105
1.05%
0.98%

2
470
4.70%
4.39%

3
1166
11.66%
11.72%

4
2035
20.35%
20.51%

5
2472
24.72%
24.61%

6
2102
21.02%
20.51%

7
1102
11.02%
11.72%

8
425
4.25%
4.39%

9
102
1.02%
0.98%

10
5
0.05%
0.10%

Total
10000
100%
100%
The columns “Cases” and “%” correspond to the simulation run in Excel. The column “Theory” is according to the normal diustribution.
Now, that clearly means that, GIVEN THAT THE RATE OF HEADS OF THE COIN IS 50%, the chances to get 9 or 10 heads out of 10 trials is 1.07 or 1.08%.
BUT IT DOES NOT TELL YOU ANYTHING ABOUT THE CHANCES OF THE ACTUAL RATE OF HEADS OF THE COIN, GIVEN THAT YOU GOT ANY GIVEN RESULT OUT OF 10 TRIALS.
That is what Statman was trying to explain Reigle.
In other word, who cares about the chances to get a bad standard deviation in the sample given that the actual standard deviation is good. The real risk is that you get a good standard deviation in the sample, and assume that the actual one is also good, given that it is not.

Ok, I had missunderstood you. Thanks for the clarification.