# MSA, Distinct Categories and 5.15

Six Sigma – iSixSigma Forums Old Forums General MSA, Distinct Categories and 5.15

Viewing 27 posts - 1 through 27 (of 27 total)
• Author
Posts
• #45151

Cynic
Participant

I have never found anyone who could explain or provide the source / logic for using 5.15 in the calculation of distinct categories.  People usually say “see Wheeler” or something like that.  Or they cite some formula that still doesn’t explain where it came from.
I think it was made up and I therefore proclaim that the new value to use in the calculation of distinct categories will be 7.36
Unless someone can finally explain 5.15, my proclaimation of 7.36 will become law on Monday morning.

0
#146528

Haugen
Participant

5.15 standard deviations capture 99% of the area under the normal curve. Look it up in your table.

0
#146529

An idiot named Joe BB
Participant

5.15 is not used in the calculation of distinct categories. Get your question right and you may get an answer.

0
#146530

Idiot Joe
Participant

From Minitab –
Minitab calculates the number in this statement by dividing the standard deviation for Parts by the standard deviation for Gage, then multiplying by 1.41 and truncating this value. This number represents the number of non-overlapping confidence intervals that will span the range of product variation. You can also think of it as the number of groups within your process data that your measurement system can discern.
Imagine you measured 10 different parts, and Minitab reported that your measurement system could discern 4 distinct categories. This means that some of those 10 parts are not different enough to be discerned as being different by your measurement system. If you want to distinguish a higher number of distinct categories, you need a more precise gage.
The Automobile Industry Action Group (AIAG) [1] suggests that when the number of categories is less than 2, the measurement system is of no value for controlling the process, since one part cannot be distinguished from another. When the number of categories is 2, the data can be divided into two groups, say high and low. When the number of categories is 3, the data can be divided into 3 groups, say low, middle and high. A value of 5 or more denotes an acceptable measurement system.

0
#146551

Cynic
Participant

Ok, I did mistype.  I meant 1.41 but took a mental detour.
So now that you have responded with a fair amount of attitude, let’s see if you can answer the question.  And try not to be so pompous, I doubt it is warranted.

0
#146552

Cynic
Participant

Joe,
Brilliant!  I see that you can copy text from websites.  So what is the origin of 1.41?  I don’t see that in your answer, and this is in fact my question.
Want to try again?  I’m not trying to be rude, but for such arrogance in your previous post, I was expecting more.

0
#146553

Cynic
Participant

I don’t know what the Minitab text says but the calculation is actually
1.41 * part sigma / measurement system sigma
The origin of the 1.41 has been elusive.  Texts are referenced, other derivations mentioned but I haven’t found anyone who can offer the basis for 1.41 itself.

0
#146593

Happy and smart wanker
Participant

It sad to see the way some people is spreading the knowledge of the Six Sigma toolbox. Questions is a good thing, is that so hard to understand. I guess that what will follow Six Sigma is good manners and how to educate. No matter how good your method you have, if the educators is shit, you will end up with even more shit. And the educators on this page is… sometimes not so good. That is the backside of a open forum….
I wonder if they didn’t learn how to wank it off when they were out of sex, right?

0
#146609

Mikel
Member

Hey genius – 1.41 is the square root of two. go read the derivation of Wheeler’s discrimination ratio and you will understand.

0
#146859

Cynic
Participant

Stan,
I know that 1.41 is the square root of 2.  You are just giving out more of the same – please explain why 1.41 is used if you can.  As usual, “see Wheeler’s derivations” doesn’t explain.
If you don’t have the answer just say so.

0
#146863

Mikel
Member

Go do some work for yourself and stop the whinning.

0
#146864

EMP
Participant

Try the first edition of Wheeler’s Book Evaluating the Measurement Process where he uses the term Classification Ratio. The AIAG MSA third edition uses the letters ndc and in their footnote on page 117 says it is the same as Wheeler’s.
Suggest you read the material and if  that doesn’t please you, then email Wheeler and ask for more info. Before there was a lot of material on destructive testing he did studies tha tprovide vauable info on GAGE R&R and destructive testing. I received replies when I wanted to know more on the subject.
Just do it!

0
#146866

Ken Feldman
Participant

0
#146868

Cynic
Participant

Stan,
I’m not whining (and learn to spell).  I just asked a question because I thought it was interesting – and not really answered in any of the text.
You obviously don’t have the answer.  Instead of saying “look it up” as though you actually do have this knowledge but don’t want to provide it, just say you don’t know.
You are a joke “Stan”

0
#146869

TJ
Member

5.15 is the constant used in the P/T ratio formula.  5.15 is an industry standard for 99% of MS variation.  Minitab suses 6 sigma as default.  Six standard deviations account for 99.73% of MS variation.

0
#146870

Cynic
Participant

Darth,
thanks for the input.  I have reviewed this as well – it doesn’t really explain why the square root or 2, or 1.41 is used however.  My question really comes down to one of curiosity.  I believe the formula is informative, I just can’t find anywhere that Wheeler actually explains why 1.41 is used.
I was reaching a class a while back with a concultant who is a PhD and the question came up in class – neither one of us really had an answer other than “here is the formula, cited by Wheeler, etc.”

0
#146875

sqe
Member

Are you serious?  You can’t see from the post from Darth where the 1.41 comes from?   If you go through the explanation, it’s there.

0
#146877

Cynic
Participant

I can see where it is in the formula.  Nowhere does the formula or the derivation say where it comes from (meaning why the square root of 2 is used).
Are you serious?

0
#146883

Mikel
Member

You thought 1.41 was so interesting that you called it 5.15.
Cynic honey (for Stevo), I think your mommy is calling you.

0
#146885

Mikel
Member

My god, you are an idiot.
With Darth spoonfeeding you, you still are crying like a baby.
Grow up and find something useful to ask.

0
#146887

Cynic
Participant

Want to discuss this in person?  Perhaps I could learn better that way?
What do you say?

0
#146890

Ronald
Participant

Cynic-
The alegbra is correct.
DR=SQRT[(Variance Total+ Variance Part-to-Part)/(Variance Measurement System)]
where Variance Measure=Variance Total-Variance P-t-P
thus
DR=SQRT[((Variance P-t-P)+(Variance Measurement)+Variance P-t-P))/(Variance Measurement)]
THIS IS WHERE THE 2 COMES IN (2*Variance P-t-P) RESULTS.
DR=SQRT[[2*(Variance Part-to-Part)/(Variance Measurement)]+1]
Minitab states that it drops the +1 to be conservation.
DC=SQRT(2)*SQRT[(Variance Part-to-Part)/(Variance Measurement)]
Your question may be where does Wheeler come up with the DR formula.
Trying to be more helpful than “Look it up” or “Can’t you get it” but it is straightforward from the original formula of:Total Variation = Part-To-Part + Total Gage R&R
Good Luck
Lee
A quick search got me this:
________________________________________
http://healthcare.isixsigma.com/forum/showmessage.asp?messageID=788
1.41 in Discrimination Index Calculation

Posted by: SheriPosted on: Monday, 2nd August 2004, 1:54 PM.
This is from the Minitab on-line help…
The total variation is the sum of the variation due to the difference between parts (Part-To-Part) and the variation due to the gage and operators (Total Gage R&R), which can be stated as the equation: 1. Total Variation = Part-To-Part + Total Gage R&R You can find estimates for these three sources of variation in the “VarComp” column in the MINITAB Session window output for a Gage R&R analysis
A relative measure of how much of the total variability is due to the Part-To-Part variability is: 2. r = Part-To-Part / Total Variation Note: r (rho) can vary from 0 to 1. Wheeler defines the discrimination ratio as: 3. D = sqrt((1+r)/(1-r)) The square root is in the formula to return the value to the original units of the data (standard deviation units). Note: D can vary from 1 to infinity. Using algebra and the above three formulas, you can derive the result: D = sqrt(((2*Part-To-Part)/Total Gage R&R) + 1) MINITAB’s formula is: Number of Distinct Categories = sqrt((2*VarComp for Part-To-Part)/VarComp for Total Gage R&R), rounded to the nearest integer MINITAB’s estimate for the number of distinct categories is conservative because the “+ 1” is dropped from the formula. It is mathematically possible to get 0 for the number of distinct categories because the variance components are estimated. Finally, the formula contains the square root of 2. MINITAB approximates the square root of 2 with 1.41

0
#146891

A thugster, run Stan!!
Participant

Yikes.   Not only a cynic, ..but a punk.    How wonderful.

Not often does someone like you just wander into a technical forum, pitch a fit and then REALLY try to pick a fight.

I envision a hunched over, skinny kid, pimply face, nasty brown English teeth (sorry Andy), rolling and clinking a stack of coins around in his left hand, and in his right hand a dog-eared, much used but little understood edition of Statistics for Dummies.

Am I right?    Even close?    Bet I am.    What say we meet and Ill size you up and let the forum know how close Ive pegged you?

0
#146898

Cynic
Participant

You are in fact very close in the discription!
You are describing your mama exactly.  You forgot the hairy butt though – nastiest thing I’ve ever seen…..

0
#146899

Cynic
Participant

Lee,
THANK YOU!!!  I good response indeed.  I appreciate what you had to say completely (not hostile like Stan or Sam).
I do actually understand the algebra, and you are correct when you said that I may be looking for where Wheeler actually come up with the DR calculation.  I find it interesting that the 1 is dropped, resulting in radical 2 (or 1.41) and am just questioning how Wheeler originally determined the initial calculation from which 1.41 is ultimately derived.
It isn’t a question that has any relevance in most problems as the calculation certainly does provide information about distinct categories for a MSA.  Just a curiosity that I haven’t found a complete answer to.
Thank you.

0
#146900

Mikel
Member

Sure, I’ll babysit you. Do you prefer Huggies or Pampers?

0
#146903

Cynic
Participant

I prefer huggies.  Don’t try to powder my bottom though.  I strictly like women!
I don’t have anything against homosexuality though – what you do in your own time is fine with me Stan.  I just don’t go that way.

0
Viewing 27 posts - 1 through 27 (of 27 total)

The forum ‘General’ is closed to new topics and replies.