Gage RR

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• #29018

Rouge BB
Member

I am trying to compare to gage systems to prove that one of them is at least 75% better than the other. Any ideas on the best way to go about doing this.

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#73247

DrSeuss
Participant

This one is pretty simple…..Just do a simple paired t-test comparing the differences between the two gages and zero, the ideal difference if both gages & measuring processes were exactly the same.  A paired t-test is nothing but a 1 sample t-test using the differences between the two gage reading for each part rather than the individual part readings. I hope this helps.  Look at the confidence intervals to determine what differences you need to see for a 75% change.  Good luck!

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#73249

Pipkin
Participant

Hello all, I have been working on a gage R&R, but some of the parameters that I am measuring only have USL (no LSL, ex: Good above 5), therefore I can not calculate the P/T. I have been thinking on infering the LSL using the following formula :
LSL=USL – (USL-Target)*2Does anyone have any idea?
Is it statiscally correct? Best regards,

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#73253

Hemanth
Participant

What is this gage? does it measure attribute or variable characterisitic?
This is what I think can be done, I am not sure, but do an MSA for both the gages the reduction in %R&R value will give you the comparison for both gages. Select more than 20 samples if you are doing MSA on attribute characteristic.

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#73256

RougeBB
Member

Thanks for the adivce. My main issue is that I am not able to run the same part through the gage twice. This prevents me from the traditional GR&R route. I am quite sure that the new MS is far better than the old one. I was able to perform a variable GR&R on the old system. (31% total R&R) Can the paired T be used to compare the new system against a master standard?

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#73257

aa
Participant

Both DrSeuss (paried t-test and null hyp. as stated) and Jack (expression ) comments are extermely and valid useful.
I would like to add this comment with a real LSL and USL values.
Say,  USL = 9.8 and Target of = 7.8 and LSL can be determine, using Jack expression – I would rewrite it in this fashion :
LSL = USL   –   2 (USL – Target)
The way Jack expression is written may look like value 2 is to the power, if this valid then LSL = 7.5, I would disagree with that. But in fact it is multipling the content of the bracket with 2.  If this have cause some confusion!
Therefore,
LSL = 9.8 – 2(9.8 – 7.5) = 5.2
This is right, USL and LSL specification limit of our product 9.8 and 5.2, where we target, 7.8. But due to the process varation, sample mean (target is between 7 and 8).
Hope this worked example and comments, are useful. If you need any further help, post your email, will respond, significant experience in Gage RR and paried T-test vs specification development and customer and supplier specifcation reviews.

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#73268

Pipkin
Participant

Hello People,By the way, there were some parameters that doesn’t have any target, so that I was thinking on inferring the missing spec. using instead of the target value the grand mean from the data that I am analyzing.
Ex:
LSL= USL – 2(USL-Grand Mean)
Any commentis welcome? Later,

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#73269

vin
Member

RogueBB,
If you have a lot of parts that you can measure, could you try the following?
Take 60 parts from one, homogenous (as best as you can tell) population.
Randomly divide the 60 into 2 groups of 30.
Measure first group with old gage, then second group with new gage.
Calculate variance of each group.
Total Variance = Variance of parts + Variance of gage, so assuming parts came from same population, the difference between variance of group 1 and variance of group 2 is the improvement in your gage.  Take the square root of the difference of the two variances, and that should theoretically be the reduction in your sigma from the old gage to the new one.  Since you have Total R and R numbers from the old system, I assume you can get the sigma of the old gage.  Divide the sigma you calculated above (square root of difference of two variances) by the sigma of the old system, multiply by 100, and that should give you % improvement (actually percent reduction of measurement variability) from old process to new process.
The above sounds crazy even to me, but it may be valid.  I look forward to hearing your comments.

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#73285

Mike Carnell
Participant

Rouge,
There are a couple ways to do this. If your sample is destoyed or can’t be retested then you can use hypothesis testing to validate an assumption of homogeneity so you can use a batch for your samples. If it can be said to be homogenous the you don;t use the same sample over and over you can take another sample and assume it is the same.
If that will not work then you can run a nested design. Minitab has a R&R for destructive testing.
On the lower spec limit. I wouldn’t worry about it. You can use percent contribution or percent study varion. It will calculate either without a two sided spec.
Good luck.

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#73300

Hemanth
Participant

Yes you can do a paired T test for MSA but I am not sure how can measure twice on a same sample if your test is destructive (I believe this is what you meant when you said that you cannot run the test twice on the same gage.). Otherwise paired T test can be used, only constraint which I see is that be sure that your data is normal atleast to some (p=0.03 or greater) extent. I hope this helps.
hemanth

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#73369

Wing
Member

I am not sure, but I read from some materials on Measurement Capability and Correlation that when there is only one-side specification limit, the P/T can be calculated by the fomula below:
P/T=3*¦Ò/(USL-¦Ì) or P/T=3*¦Ò/(¦Ì-LSL)

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#73370

Mike Carnell
Participant

We used to use the paired t test before we found the GR&R stuff. The basic issue is as the name suggests (pair) that it allows you to compare 2 things. If you have more than two readings for repeatability , there is a problem. If you have more than 2 operators (reproducibility) there is also a problem. That is one of the advantages to ANOVA.

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#73380

RR Kunes
Member

Wrong!
A paired T will only tell you if their is a difference between the  two it does not quantify and therefore is relatively useless in this instance.

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