Gage RR
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 This topic has 11 replies, 6 voices, and was last updated 18 years, 5 months ago by Mikel.

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July 31, 2003 at 12:07 pm #32939
Why is 5.15*Std Dev used in the calculations for gage R&R (as opposed to 6*Std Dev)?
What is the significance of 5.15?0July 31, 2003 at 12:18 pm #88514The idea is to cover 99% of the area (leaving 0.5%) on either side.
If you see 99.5% of the area under the curve for oneside limit corresponds to 2.575 std. deviations from the mean.
Therefore for both sides 2.575+2.575 = 5.15 std. deviations cover 99% of the curve (leaving .5% on either side, therefore totally leaving 1% on both sides.)
Hope this helps.
Bee0April 26, 2004 at 1:11 pm #99092Bee
1. Why leave 0.5 % on either side ?
2. Why not cover 99.73 % which corresponds to 3 + 3 = 6 std. deviations ?
Awaiting further clarification.
prasad
0April 26, 2004 at 1:28 pm #99096When awaiting a reply to a 9 month old post – be patient.
Why do you want 99.73%? It is just a convention and the real trick is understanding how the rules for interpreting GR&R relate to whatever your +/ sigma decision is. Do you understand the interpretation rules and what they actually mean?0April 26, 2004 at 1:30 pm #99097
DevashishParticipant@Devashish Include @Devashish in your post and this person will
be notified via email.Dear all,
In fact as per AIAG RevIII it has been changed to 6*SD instead of earlier 5.15. to include 99.73%. It also makes calculation easier(multiply by 6 rather than 5.15)
Devashish0April 26, 2004 at 2:06 pm #991011. Thanks DD. It now seems more logical to have GRR = ( 6*sd ) / Tolerance. Awaiting yr other clarification as discussed this evening.
2. From Stan’s post i understand that 99 % is just a convention thanks for the same
3. Stan – as for a 9 month old post I thought the answer to the question raised by the previous person was not complete which u have now done. Besides I chkd the string for some info on the same subject.Further abt the significance of the GRR = ( 5.15 * SD ) / tolerance expressed as a % is ok by my understanding.
Prasad0April 26, 2004 at 2:31 pm #99107Did they change the acceptance criteria as well? If not, they just made it easier to have an acceptable gauge compared to total variation. That doesn’t make sense.
0April 26, 2004 at 3:09 pm #99114
Mario PerezWilsonParticipant@MarioPerezWilson Include @MarioPerezWilson in your post and this person will
be notified via email.To answer the question:
Why is 5.15*Std Dev used in the calculations for gage R&R (as opposed to 6*Std Dev)?
Measurement System Variability (Gauge Variability)
The reason you want to use 6 instead of 5.15.
When you compute 2 x Z (where Z is the standard zscore and can be equal to 2, 2.575, 3, etc.) times the standard deviation of the measurement error, óR, you are predicting or modeling the risk (uncertainty) with the normal distribution.
By computing 99.73% of the distribution, that is, multiplying +/ 3 (Z=3, 2Z=6) by the standard deviation of the measurement error, óR. This value, 6óR, is then a prediction of the precision of the measurement system or gauge, which is commonly referred to as the repeatability.
Why do we choose +/ 3óR (99.73%) instead of any other prediction?
Because, this prediction will later be compared against other predictions, such as, the product tolerance, process stability, process capability, process potential, all of which are computed using +/ 3 sigma limits.
The value 6óR, is then compared against the product tolerance, to compute the percent of
variation consumed by the measurement system (gauge) uncertainty.
This explanation has been modified from the book: Gauge R&R Studies. I hope the symbol for sigma shows as a Greek letter, as it does in my WordPad.
Mario PerezWilson (mpcps.com)0April 26, 2004 at 3:13 pm #99115
Mario PerezWilsonParticipant@MarioPerezWilson Include @MarioPerezWilson in your post and this person will
be notified via email.That wierd symbol ó is supposed to be sigma.
Mario PerezWilson (mpcps.com)0April 26, 2004 at 3:53 pm #99118Mario,
The choice of 5.15 or 6 is not relevant. The rules associated with that are. 5.15 gives 0.86 of the value 6 gives. If your rule is <10% P/T using 5.15, the same rule using 6 would be <11.6%.
If the rules don’t change going from 5.15 to 6, we just tightened the criteria for P/T by a little more than 16%. P/TV doesn’t change.
It turns out that the rules are quite conservative as it is, why make them more conservative?0April 26, 2004 at 5:53 pm #99136
Mario PerezWilsonParticipant@MarioPerezWilson Include @MarioPerezWilson in your post and this person will
be notified via email.Stan,
My post is not a response to your question about a change in acceptance criteria. I just replied to the original post, while reading the latest post in the thread, which happens to be yours.
Nevertheless, the acceptance criteria is arbitrary.
Mario PerezWilson (mpcps.com)
0April 26, 2004 at 6:14 pm #99138Mario,
Whether or not you were responding to my post, I responded to yours. You said we use 6 to facilitate comparisons – comparisons to what?
The acceptance criteria has its basis in the old 10:1 rule that test engineers have always used.
The choice of 5.15 or 6 is arbitrary. The systems question to be answered is how much uncertainty do we want at the edges of our acceptance criteria and what do we do about it. The secondary six sigma question is whether the measurement is one of the big bars on our pareto if we are trying to reduce variation. If it is, we need to address. If it is not, who cares?0 
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