Error in ANOVA table in the AIAG MSA manual? Table A4

[keithljelp]

Hello,

In the ANOVA tables in the AIAG MSA Manual's; 3rd and 4th the F-value for the data in table 24, is 79.41 for Appraiser and 492.29 for Parts. My question is where did they get these numbers? I was taught to take the MSE-sub appraiser/MSE-sub within to calculate the F-value. Well, 1.58363/0.04598 is 34.44171 by my trusty HP 42S and 34.440 with I use R.

Am I missing something here???

Thanks,

Keith

 

[Stijloor]

Re: Error in ANOVA table in the AIAG MSA manual? Table 24

Hello,

In the ANOVA tables in the AIAG MSA Manual's; 3rd and 4th the F-value for the data in table 24, is 79.41 for Appraiser and 492.29 for Parts. My question is where did they get these numbers? I was taught to take the MSE-sub appraiser/MSE-sub within to calculate the F-value. Well, 1.58363/0.04598 is 34.44171 by my trusty HP 42S and 34.440 with I use R.

Am I missing something here???

Thanks,

Keith

MSA Experts?

Thanks.

Stijloor.

 

[Miner]

From AIAG MSA 3rd Edition errata:

[FONT=Arial, Helvetica, sans-serif]Page 190: Table 19a, F values changed from 34.44 to 79.41, and 213.52 to 492.29 in order to maintain consistency with the random effects model (footnote 79, page 187). MSA3p190.pdf [/FONT]
​

 

[Rameshwar25]

using MSA 4th manual and after analysing worksheet examples from Elsmar.com, i have understood that:
F(Appraiser)=MS (Appraiser)/MS(App by part)= 1.58363/0.01994=79.41
F(Parts)=MS (Partsr)/MS(App b y part)= 9.81799/0.01994=492.29

Refer following text from MSA manual, page 105

" F-ratio column is calculate only.......... It is determined by the mean square value of interaction devided by mean square error."


I think it will do.
Rameshwar

 

[keithljelp]

Y'all,

Thanks for the info. I missed the errata.

Could someone point me to a detailed discussion of AIAG's reasoning. I am not sold just because they said so. I inherently distrust AIAG because of the way they calculate percent total variation. I agree with Dr. Wheeler the math is wrong there.

Thanks,

Keith

 

[Miner]

In this case, AIAG is following the correct approach.

In the standard ANOVA, all factors are considered FIXED. This means that the levels tested are the only levels of interest. In this situation, your calculation of the F - ratio is the correct approach.

However, in a Gage R&R study, the factors are RANDOM. The parts are selected at random to represent all parts. Usually, the operators are selected at random to represent all operators. In this situation, the General Linear Model (GLM) is used instead of ANOVA (yes, the table looks the same). The calculations shown by AIAG in the appendix are the correct calculations for GLM.

You can also have a MIXED model where the parts are random and the operators are fixed (because that all there are). In this situation GLM must also be used.

If you are a Minitab user, Minitab 15 and earlier assume a fully RANDOM model. Minitab 16 allows you to choose between a RANDOM and a MIXED model.

 

[keithljelp]

Miner, Thanks.

I will study up on GLM and random ANOVA.

Keith