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Gauge R&R - Reproducibility formula

processanalytic

Community Trekker

Joined:

Jul 22, 2014

Hi!!

Does anyone please let me know the exact formula used for repeatability & reproducibility in GRR studies in JMP. I am of the view that wrong formula is used for calculating reproducibility in case variability chart Gauge R&R (% of Tolerance). I have tried to compare it with MSA-4th edition book and I found the results from JMP are coming different. I am considering variation (6* std dev).

Thanks in advance...

1 ACCEPTED SOLUTION

Accepted Solutions
Solution

In this particular 2 factor crossed example, the large difference between estimates for EV and AV between methods is due largely to the fact that there is a significant Appraiser-Part interaction. The Average and Range method does not pick this up but the ANOVA method used by JMP does. One way (among several) to see the interaction is to run Measurement Systems Analysis > Parallelism Plots.

This situation is spelled out on p103 of the MSA manual:

The Average and Range method ( X & R) is an approach which will provide

an estimate of both repeatability and reproducibility for a measurement

system.

However, variation due to the interaction between the appraiser and the part/gage is not accounted for in the analysis

The ANOVA method can be used to determine the interaction between the gage and appraisers, if such exists.

Michael

8 REPLIES
mpb

Super User

Joined:

Jun 23, 2011

It can be quite confusing to compare 2 different implementations of R&R analyis for at least 2 reasons.

1. There are multiple ways to do the analysis which result in slightly different results. JMP can do the analysis using EMS, REML, or a Baysian method.

2. There are multiple ways to report results.

In any case, JMP is not "wrong". You just have to do an apples to apples comparison of results.

I imported the example from the MSA 4th edition on page 118 into jmp so it looks like this:

ID    Appraiser    Trial    Part    Y

1    A    1    1    0.29

1    A    1    2    -0.56

1    A    1    3    1.34

1    A    1    4    0.47

1    A    1    5    -0.8

1    A    1    6    0.02

1    A    1    7    0.59

1    A    1    8    -0.31

1    A    1    9    2.26

1    A    1    10    -1.36

2    A    2    1    0.41

etc.

Then I ran (interactively first) this script:

Variability Chart(

    Y( :Y ),

    X( :Appraiser, :Part ),

    Model( "Crossed" ),

    Max Iter( 100 ),

    Conv Limit( 0.00000001 ),

    Number Integration Abscissas( 128 ),

    Number Function Evals( 65536 ),

    Analysis Type( Use REML analysis ),

    Historical Sigma( 0 ),

    Std Dev Chart( 1 ),

    Gauge RR( 6, 0, 0, 0 ),

    Gauge RR Report( 1 )

);

Then I compared the R&R section of the report to the MSA report on page 129 and I found agreement with the "Anova" method results.

Note that I forced the REML method though EMS would, for this case, yielded the same results.

Or you can use this script to do the same thing:

EMP Measurement Systems Analysis(

    Y( :Y ),

    X( :Appraiser ),

    Part( :Part ),

    Model( "Crossed" ),

    Dispersion Chart Type( Range ),

    EMP Results( 1 ),

    EMP Gauge RR Results( 1 ),

    Average Chart( 1 ),

    Dispersion Chart( 1 ),

    Max Iter( 100 ),

    Conv Limit( 0.00000001 ),

    SendToReport(

        Dispatch(

            {"Average Chart"},

            "1",

            ScaleBox,

            {Rotated Labels( "Automatic" )}

        ),

        Dispatch( {"Range Chart"}, "1", ScaleBox, {Rotated Labels( "Automatic" )} )

    )

);

Hope this helps.

Michael

processanalytic

Community Trekker

Joined:

Jul 22, 2014

Dear mpb,

Thanks a lot for your reply, but sorry to say in this case even I have tried doing one such comparison with 2 factor crossed example from JMP with spec limit of 0.5 to 1.1

What I found is totally different result in case of % of tolerance. Following are my findings:

WITH JMP:

Repeatability : 64.33%

Reproducibility: 76.13%

Gauge R&R: 99.68%

WITH MSA MANUAL 4TH EDITION:

Repeatability : 52.18%

Reproducibility: 28.067%

Gauge R&R: 59.250%

If you let me know the email ID I can send you the file and my calculation. I feel my way of doing is correct.

Also in JMP the formula used specially for GRR (% of tolerance calculation is incorrect)

These are huge difference between JMP and MSA ways. What to believe?

Kindly help

mpb

Super User

Joined:

Jun 23, 2011

Please run either or both of the scripts embedded in the larger attached JMP data table. Scroll to the bottom to see the GRR report and compare with the results reported on page 127 and 129 of the MSA manual. You will see the JMP report contained in the second attached JMP data table.

You will see that the results agree with the results shown for the example on page 127 and 129 of the MSA manual.

In particular (p127):

JMP GRR = 0.302373 (agrees with MSA std dev (GRR) )

JMP Repeatability = 0.039973 (agrees with MSA variance estimate (EV)))

JMP Reproducibility = 0.051455 (agrees with MSA variance estimate (AV)))

Michael

processanalytic

Community Trekker

Joined:

Jul 22, 2014

Dear Michael,

Thanks again for your reply and I totally agree with what you have shown in your reply above but my worry is not about the standard deviation or % of variance components. My real worry is regarding % of Tolerance components. During these calculations the values are not matching. Kindly have a look at the attached files. I have tried to do same comparison with example from JMP database - "2 factor crossed".

laural_jmp

Staff

Joined:

Apr 7, 2014

Hello,

After looking at the pdf of the 2 factor Crossed example with the AIAG results in Excel, it appears that the method used here is the same as the method used in the manual on page 119 using ranges.  JMP does not use this method to compute variance components and Gauge results, so that is why JMP's results do match this example.  For example, the Excel file computes the Equipment Variation as 0.3131 whereas JMP computes it as 0.3860052.  Since the Percent of Tolerance is calculated using these values, they will be different. 

% of Tolerance = 100*Variation/Tolerance

JMP % of Tolerance = 100 * 0.3860052 / 0.6 = 64.33%

Excel % of Tolerance = 100 * 0.3131 / 0.6 = 52.183%

Neither method is "incorrect", just different ways to estimate statistics.  However, most would consider it better to use the ANOVA method (also in AIAG) and other methods in JMP because of their ability to give better estimates.

I hope this helps.

Solution

In this particular 2 factor crossed example, the large difference between estimates for EV and AV between methods is due largely to the fact that there is a significant Appraiser-Part interaction. The Average and Range method does not pick this up but the ANOVA method used by JMP does. One way (among several) to see the interaction is to run Measurement Systems Analysis > Parallelism Plots.

This situation is spelled out on p103 of the MSA manual:

The Average and Range method ( X & R) is an approach which will provide

an estimate of both repeatability and reproducibility for a measurement

system.

However, variation due to the interaction between the appraiser and the part/gage is not accounted for in the analysis

The ANOVA method can be used to determine the interaction between the gage and appraisers, if such exists.

Michael

processanalytic

Community Trekker

Joined:

Jul 22, 2014

Dear Laural/Michael,

Thanks for your reply. This is what I want to prove that there is gap in the way calculation is done in MSA and JMP methods. See even the difference in Reproducibility:

JMP % of Tolerance = 76.13%

Excel % of Tolerance = 28.06%

is huge. Now which one to consider correct? This is really confusing. Suppose we consider JMP as best and starts reducing variation from 76.13% to 40.0%. In this case still it is above the satisfactory level of 20% to 30%. But in reality as per MSA may be it is below 20%. BIG CONFUSION

Could you please tell me the exact formula used for both repeatability and reproducibility in case of JMP?

For MSA following Formula are used:

Repeatability : (Rbarbar) x (K1) x (6 Sigma)

Reproducibility :

(6 Sigma)  x SQRT{(Xbar diff x K 2)^2 - (EV/[(# parts) x (# trials)])}

Thanks

Ashok

Steven_Moore

Super User

Joined:

Jun 4, 2014

I suggest you read Dr. Donald Wheeler's white paper (It is in the JMP site under "White Papers") on the subject.  His explanations and examples are very informative.

Steve