Excel spreadsheet containing Gage R&R with Anova Method wanted

C

Caroline54

Hello Everyone,

I need an Excel spreadsheet containg the GRR with Anova Method?
I have to submit the "GRR done with ANOVA method" to my customer because my ppap was rejected as I used the Basic GRR with Xbar and R.

Thank you kindly for your help.

Caroline.
 
Y

Yew Jin

Re: GR&R with ANOVA Method

May be you can post the raw data in excel spreadsheet so that we can help to use minitab to analysis the data for you
 
T

Ted Schmitt

Re: An Excel spreadsheet containing GR&R with Anova Method wanted

Hello Everyone,

I need an Excel spreadsheet containg the GRR with Anova Method?
I have to submit the "GRR done with ANOVA method" to my customer because my ppap was rejected as I used the Basic GRR with Xbar and R.

Thank you kindly for your help.

Caroline.

Caroline,

I got this example a while back... it´s in Portuguese, if you think this is what you need, I´ll transalate it and send it again... pls let me know

Hope this helps you ! :)
 

Attachments

  • Estudo R&R.xls
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C

Caroline54

Thank you everyone for your help.

I entered my data into the excel Anova Spreed sheet. But this method shows that "Gage needs improvement" I am attaching a copy of the results
How do we interpret the results? what needs to be improved?
Can somebody verify with minitab or other software if my results are correct?

Thank you so much for your help.

Caroline

GAGE REPEATABILITY AND REPRODUCIBILITY DATA SHEET
ANOVA METHOD

Part Number Gage Name Appraiser A Ana
NUMBER
Part Name Gage Number Appraiser B Eva
NAME xxx
Characteristic Specification Gage Type Appraiser C Sandy
-1.5 1.5
Characteristic Classification Trials Parts Appraisers Date Performed
3 10 3

APPRAISER/ PART AVERAGE
TRIAL # 1 2 3 4 5 6 7 8 9 10
1. A 1 -0.70 -0.92 -0.23 -1.02 -0.94 -0.52 -0.89 -0.25 -1.01 -1.01 -0.749
2. 2 -0.72 -0.90 -0.23 -1.03 -0.94 -0.52 -0.89 -0.25 -1.01 -1.01 -0.750
3. 3 -0.70 -0.90 -0.25 -1.00 -0.94 -0.52 -0.89 -0.25 -1.01 -1.01 -0.747
4. AVE -0.71 -0.91 -0.24 -1.02 -0.94 -0.52 -0.89 -0.25 -1.01 -1.01 xa= -0.749
5. R 0.02 0.02 0.02 0.03 0.00 0.00 0.00 0.00 0.00 0.00 ra= 0.009
6. B 1 -1.03 -1.02 -0.40 -1.10 -1.01 -0.60 -0.90 -0.35 -1.08 -1.15 -0.864
7. 2 -1.03 -1.02 -0.40 -1.08 -1.01 -0.60 -0.90 -0.35 -1.08 -1.15 -0.862
8. 3 -1.03 -1.03 -0.45 -1.10 -1.01 -0.60 -0.92 -0.35 -1.08 -1.15 -0.872
9. AVE -1.03 -1.02 -0.42 -1.09 -1.01 -0.60 -0.91 -0.35 -1.08 -1.15 xb= -0.866
10. R 0.00 0.01 0.05 0.02 0.00 0.00 0.02 0.00 0.00 0.00 rb= 0.010
11. C 1 -0.69 -0.90 -0.40 -0.14 -0.89 -0.84 -0.70 -0.99 -0.98 -0.96 -0.749
12. 2 -0.69 -0.90 -0.40 -0.14 -0.89 -0.84 -0.70 -0.99 -0.98 -0.96 -0.749
13. 3 -0.69 -0.90 -0.40 -0.14 -0.88 -0.85 -0.71 -0.99 -0.98 -0.96 -0.750
14. AVE -0.69 -0.90 -0.40 -0.14 -0.89 -0.84 -0.70 -0.99 -0.98 -0.96 xc= -0.749
15. R 0.00 0.00 0.00 0.00 0.01 0.01 0.01 0.00 0.00 0.00 rc= 0.003
16. PART X= -0.788
AVERAGE -0.81 -0.94 -0.35 -0.75 -0.95 -0.65 -0.83 -0.53 -1.02 -1.04 Rp= 0.689

Anova Table
Source DF SS MS F Sig
Appraiser 2 0.27 0.14 2200.07 *
Parts 9 4.02 0.45 7184.67 *
Appraiser-by-Part 18 3.03 0.17 2704.38 *
Equipment 60 0.00 0.00
Total (N-1) 89 7.33
* Significant at a = 0.05 level

Anova Report Standard Deviation (s) % Total Variation % Contribution

Repeatability (EV) 0.20 66.7% 44.4%
Reproducibility (AV) 0.06 19.3% 3.7%
Appraiser by Part (INT) 0.13 43.8% 19.2%
GRR 0.21 69.4% 48.2%
Part-to-Part (PV) 0.21 72.0% 51.8%
Gage
system needs improvement

Note:
Tolerance = 3.00 Total variation (TV) = 0.295802107
Number of distinct data categories (ndc) = 1 Gage discrimination low
 

Miner

Forum Moderator
Leader
Admin
I do not have access to my analysis software at this time to validate your analysis, but I can provide feedback on the results. Search this forum and you will find a validation data set from Ford that you can used to validate the spreadsheet.

As to your analysis:
The ANOVA table shows statistical significance between operators (that Reproducibility) and an interaction between the Operators and the parts. This means that different operators measured different parts differently. This should show up in a graph.

The MSA results show that your Repeatability (variation within the same operator) is the highest at 66.7% of the total variation. The interaction between operator and part was next at 43.8%, with the Reproducibility (variation from Operator A to Operator B) following at 19.3%

You have problems with all three and the gage is of no use for process control at a %GRR of 69.4% and an ndc=1

If you will provide information on the product, the characteristic measured and the measuring device, we can better help you. I could guess at some common issues, but I would rather get more specific for you.

Repeatability issues are commonly (but not always) caused by variation in form of the part itself with no established method of measurement. An example would be an oval shaft where one operator randomly measure the diameter. Throw in one operator that usually measures the max diameter and another that measures the max and min diameter then averages them and you have poor Reproducibility and a possible interaction.

Poor part fixturing can also be an issue.

Poor Reproducibility is typically caused by not having a clearly defined method of measurement.

Again, these are common issues, but may not be your issue.
 
A

Atul Khandekar

Caroline, can you please verify if data is correctly enterd for Part No. 4 & 8 for Operator C?
 
O

OceanDiver

I have been reading this discussion, and others, on ANOVA trying to understand it. I am trying to develop a MSA study at my facility and write up procedures and make a spreadsheet to perform the calculations.

I have a questions regarding the degrees of freedom, I notice that in tedschmitt spreadsheet above there is a column in the ANOVA calculations for degrees of freedom. The equipment row states that there are 60 degrees of freedom, the same as is listed in the MSA manual. However, I cannot find any way that this is actually calculated. Can someone please explain how the this particular degrees of freedom is found.

Thanks
 
O

OceanDiver

Well, I ended up finally finding the answer to my question. For those interested:

take the rows and make them correspond to
row 1 to i
row 2 to j
row 3 to k

row 4 degree of freedom = i * j * (k-1)

Thanks
 
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