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Minitab GR&R (Gage R&R) - Minitab is giving different results for distinct categories

Geoff Cotton

Quite Involved in Discussions
Minitab is giving me different results for distinct catagories when analysing GR&R data through the XBARR method then by ANOVA.

Any ideas why?
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Atul Khandekar

Without looking at the data, it is difficult to say. Does your ANOVA table show a significant Part*Appraiser interaction?
ANOVA calculations would be more accurate since Range-Average method does not take interaction into consideration.

Geoff Cotton

Quite Involved in Discussions

Sorry for thr delay in my reply, we have just had our TS audit so I was took out of the frame for a few days. (We have five minors.)

Looks like I can't attach a Minitab File so I've included the data below. The distinct catagories says the study fails i.e DC's = 2 using the ANOVA method. I think I know why, but would like to hear your thoughts.

The total tolerance for the part is 0.2mm

Thanks in advance

measurment sample operator
4.954 1 A
4.967 2 A
4.992 3 A
4.923 4 A
4.972 5 A
4.954 6 A
4.980 7 A
4.959 8 A
4.970 9 A
4.988 10 A
4.955 1 A
4.972 2 A
4.994 3 A
4.922 4 A
4.957 5 A
4.971 6 A
4.974 7 A
4.971 8 A
4.968 9 A
4.968 10 A
4.957 1 A
4.974 2 A
4.998 3 A
4.923 4 A
4.972 5 A
4.956 6 A
4.974 7 A
4.959 8 A
4.973 9 A
4.986 10 A
4.971 1 B
4.973 2 B
5.007 3 B
4.921 4 B
4.965 5 B
4.971 6 B
4.974 7 B
4.971 8 B
4.969 9 B
4.982 10 B
4.971 1 B
4.971 2 B
5.004 3 B
4.928 4 B
4.968 5 B
4.972 6 B
4.978 7 B
4.971 8 B
4.960 9 B
4.989 10 B
4.960 1 B
4.972 2 B
5.005 3 B
4.926 4 B
4.969 5 B
4.979 6 B
4.979 7 B
4.974 8 B
4.970 9 B
4.988 10 B
4.943 1 C
4.954 2 C
4.932 3 C
4.925 4 C
4.956 5 C
4.972 6 C
4.976 7 C
4.964 8 C
4.969 9 C
4.973 10 C
4.956 1 C
4.969 2 C
4.988 3 C
4.929 4 C
4.967 5 C
4.976 6 C
4.976 7 C
4.967 8 C
4.972 9 C
4.964 10 C
4.970 1 C
4.963 2 C
4.994 3 C
4.922 4 C
4.962 5 C
4.956 6 C
4.979 7 C
4.968 8 C
4.964 9 C
4.975 10 C


Minitab R&R


Let me add my 5 cents, because I guess I can help you a little.

Number of Distinct Categories: NCD is calculated as

(ProcessSTDEV / MeasSTDEV)*1,41

By analyzing your data, is easy to notice that for ANOVA ProcessSTDEV = 0,0169 and MeasSTDEV = 0,00999. On the other hand, for Xbar/R, ProcessSTDEV = 0,0208 and MeasSTDEV = 0,00765.

For ANOVA, the NCD = 2,38 and rounded to nearest integer = 2.
For Xbar/R the NCD = 3,83 so = 4.

The difference between the exact NCD values is not too big (and leads you to the same decision, according to AIAG manual).
The additional problem here is related to the way that Minitab has rounded each one of the values (in opposite directions).

Differences in STDEVs are consequence of different methods of estimating sigma, but the differences in your example are not too big...

Hope this helps!

Atul Khandekar


I tried to analyse the data using my own software. Attached are two report files: 1. ANOVA calculations with interaction (charts included) and 2. Range-Average calculation.

Though you do have a choice of using either ANOVA or Range-Average, ANOVA is said to be a more accurate estimate. The Range Average method uses RBar/d2 method for estimation of variation and ANOVA uses Sum Squares (first principles, if you like).

Look at the readings of Appraiser C (esp. Part 3, Reading 1). This is very different from the other 8 readings for this part. Also observe the XBar and R Charts for the operators and the interaction plot. There is clearly an assignable cause for this error (may be improper method, or just carelessness!). There is a large variation in measurements by this operator. This probably is why ANOVA shows an interaction. (I still have an issue whether this interaction should be shown separately as significant or should it be pooled with EV - of course, either way it does not affect the DC much.) In Range Average method, this will inflate the repeatability error. If you were to change this one reading to, say. 4.992, you would get better value of nDC.

In Range Average method, since interaction is not separated, a part of it gets confounded with PV, hence a larger value of nDC (as compared to ANOVA).

Looking at the data, I also noticed that the total range of data is 0.086 which is much lower than the tolerance of 0.2. There can be two reasons for this: 1. Process Capability Cp is high and selected parts represent the entire process spread OR 2. Part selection is wrong - it does not cover the entire tolerance range. This may be one reason, GRR makes up a large percentage of PV.



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