Is this a real process with issues? And if so, at which point are you at the moment with the process analysis and optimization? (just being curious) 0.0641), but the process mean for pushout force=15 is significantly nearer to the target 0.4. The dimple depth from the process has the same variation as the dimple depth from the special study with pushout force=15 (0.0735 vs.
![two way anova minitab two way anova minitab](https://i.ytimg.com/vi/YaXIBlun4SI/maxresdefault.jpg)
![two way anova minitab two way anova minitab](https://s3.studylib.net/store/data/005848501_1-c9bc544f57e7c5c028b28051a3e04b48-768x994.png)
Minitab R16 has a nice new feature called "Assistant" and you could easily get some information about a capability before (=DimpleDepth) and after (=DimpleDepth-SpecialStudy) a process modificatation (see attachment). Or you could compare the out-of-tolerance-rate and capability of "DimpleDepth-SpecialStudy" from the tests with different workstations and pushout force=15 (12.15%, Pp=0.52, Ppk=0.48) with the rate based on column B "DimpleDepth" in your xls-file (36.05%, Pp=0.45, Ppk=0.13). If this is a currently running process maybe a Gage R&R was done and you could get some more details on the measurement uncertainty. Set "randomize all runs" (behind the button "Options") and double press OK.Īt this point without additional information you can only assume that the pushout force most likely has a positive linear effect on dimple depth, but you have not enough evidence to rank the workstations regarding their variance. Stat > Quality Tools > Gage Study (even if you're doing a different kind of experiment) > Create Gage R&R Study WorksheetĪnd fill in "part numbers" (PushoutForce 10, 15 and 20), "operator numbers" (Workstation 1, 2 and 3) and "replicates"=10 (10 DimpleDepths for each combination of PushoutForce-Level and Workstation). One way to get a worksheet with random order in Minitab is If the experiments / tests were made in consecutive order (as in the data provided), the smaller variation could also be occured due to an instable measurement system and not to a single workstation.Īs far as I understand the test situation the measurements are not-destructive, so I would take the pieces and start again with the measurements, but now in a random order so later on it could be detected if the variation effects occur due to a better workstation with less variation. 13-82) for Workstation 1, 2 and 3 show also less variation than observations at the beginning and at the end. Workstation 2 has the lowest variation if PushoutForce is considered, but on the other hand measurements in the middle of the data (obs no. But the probabilty for DimpleDepths within tolerance is small for the data provided (88% within / 12% below 0.3 or above 0.5).Īnd there are some effects in the variation, which could not be explained with the given data. IMHO at this point it is obvious that DimpleDepth increases for growing PushoutForce and for a tolerance of 0.4+/-0.1 PushoutForce=15 is best. Attached you'll find an analysis of DimpleDepths. To understand the structure of data it is often helpful to plot it previous to tests, as Bev and Allatar mentioned earlier. Subset the worksheet to remove workstation 2 and then you can obtain a result looking at differences by means for pushoutforce.īut that will still not change the fact that workstation 2 is doing something different. But it really is only workstation 2 that has a different variation. If you want a p-value for means, you can try splitting the workstations apart and analysing seperately. This is a significant effect and should be investigated. What can you do next? found out why workstation two has a different variation is something that I would want to know. Well it says to me workstation two has less variation.ĭespite P-values that will be inaccurate for the means, we can still see that push out force has a difference for means. Plotting by push out force will hide differences in variation by workstation 2 becuase all workstations are lumped together. Plotting by workstation, will mask differences in variation by workstation due to differences between the means of push out force.
![two way anova minitab two way anova minitab](https://i0.wp.com/www.statistikian.com/wp-content/uploads/2012/11/Two-Way-Anova-Dalam-Excel.jpg)
Plot the boxplots on one factor at a time to see this. Testing the variances of the factors on their own will not show a significant difference in variation. You can also see the effect of unequal variances on the residuals vs fits. The pooled will be too high for station 2 and too low for station 1 and 3. Now the problem is that the p-value here will not be accurate because its based on a pooled variation. Interactions and workstation are not significant differences. In a general linear model or two way Anova, we can see that their is differences between means of push out strength. The means do not look that different by workstation. My interpretation is that workstation 2 has less variation. A boxplot of Pushoutforce first and workstation second shows an interesting picture.