Two Way ANOVA Test For Equal Variances

optomist1

A Sea of Statistics
Super Moderator
Good Day,

I am new to the site, the layout is great, nice to see a Minitab forum.

I am exploring a problem involving test for equal variances (TEV)...as part of a Two Way ANOVA; two factors both factors have three levels...all conducted with Minitab 15.,

My dilemma is this, when I run the TEV three ways, each factor alone & then together; each factor alone confirms equal variances, yet both factors together the Bartletts and Levene's test are p = 0.000 & 0.002 respectively or evidence of unequal variances. When a Two Way ANOVA is run, a statistically significant difference in means is revealed for factor 2.

Simply put is the result of the two factor TEV of concern? If, so what? Again my goal is to evaluate whether there is a statistically significant difference in means relative to my process output, diameter, via a Two Way ANOVA.

Thank you in advance for your assistance....
 

Bev D

Heretical Statistician
Leader
Super Moderator
statistics is not a math question - it's an essay question.

A good experiment only needs a statistical answer to confirm the obvious.
Your best analysis is to plot your data - the raw data not the averages.

If you could post the results of your experiment we can look at it and then we can understand the inconclusive statistical analysis
 

optomist1

A Sea of Statistics
Super Moderator
Good Day Bev,

First thank you to responding so quickly...I am a statitician in training so take it easy on me!!! By the way, great quote...

I have run x bar/r charts (subgroup 3) and of course have some factor setting(s) superior (pushout force = 15) to others, both charts in control.

My output is "Dimple Depth" spec = 0.4 +/-0.1, the two inputs are "workstation" and "pushout force"

I've attached the excel version of the data....my conclusions are that
1) workstation # is not a significant factor,
2) pushout force is significant...setting of approximately 15 is optimal.

As I stated earlier, the test for equal variance for both factors together yielded results or evidence of unequal variances, yet separately they did not.

I guess I am looking for some form of affirmation regarding my conclusions & techniques, realizing there are many ways to accomplish this.

Thank you for your time and insight...

Regards,
Marty
 

Attachments

  • Excel - Dimple Test.xls
    20 KB · Views: 128

optomist1

A Sea of Statistics
Super Moderator
Hi Bev,

In my haste to respond I did not attach a copy of some of my Minitab output....sorry for the delay. I hope I clearly, stated my dilemma or question.

Thank you for your time...

Regards,
Marty
 

Attachments

  • Minitab Output Two Way ANOVA.doc
    152 KB · Views: 122
A

Allattar

Well from the results.

A boxplot of Pushoutforce first and workstation second shows an interesting picture.

My interpretation is that workstation 2 has less variation.
The means do not look that different by workstation.

In a general linear model or two way Anova, we can see that their is differences between means of push out strength. Interactions and workstation are not significant differences. Now the problem is that the p-value here will not be accurate because its based on a pooled variation. The pooled will be too high for station 2 and too low for station 1 and 3.
You can also see the effect of unequal variances on the residuals vs fits.

Testing the variances of the factors on their own will not show a significant difference in variation. Plot the boxplots on one factor at a time to see this.
Plotting by workstation, will mask differences in variation by workstation due to differences between the means of push out force. Plotting by push out force will hide differences in variation by workstation 2 becuase all workstations are lumped together.

But what does this mean?
Well it says to me workstation two has less variation.
Despite P-values that will be inaccurate for the means, we can still see that push out force has a difference for means.

What can you do next? found out why workstation two has a different variation is something that I would want to know. This is a significant effect and should be investigated.

If you want a p-value for means, you can try splitting the workstations apart and analysing seperately. But it really is only workstation 2 that has a different variation. Subset the worksheet to remove workstation 2 and then you can obtain a result looking at differences by means for pushoutforce.
But that will still not change the fact that workstation 2 is doing something different.
 
B

Barbara B

optomist1,

to understand the structure of data it is often helpful to plot it previous to tests, as Bev and Allatar mentioned earlier. Attached you'll find an analysis of DimpleDepths.

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. But the probabilty for DimpleDepths within tolerance is small for the data provided (88% within / 12% below 0.3 or above 0.5).

And there are some effects in the variation, which could not be explained with the given data. Workstation 2 has the lowest variation if PushoutForce is considered, but on the other hand measurements in the middle of the data (obs no. 13-82) for Workstation 1, 2 and 3 show also less variation than observations at the beginning and at the end. 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.

As 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. One way to get a worksheet with random order in Minitab is
Stat > Quality Tools > Gage Study (even if you're doing a different kind of experiment) > Create Gage R&R Study Worksheet
and 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). Set "randomize all runs" (behind the button "Options") and double press OK.

Hope this helps,

Barbara
 

Attachments

  • DimpleDepths - SpecialStudy 2011 01 17.xls
    25 KB · Views: 95
  • Two way ANOVA and test for equal variances 2011 01 17.pdf
    762 KB · Views: 114

optomist1

A Sea of Statistics
Super Moderator
Good Day Barbara,

If you are familiar w/Mike Meyers from Saturday Night Live....I must say I am verklempt....thank you for the extensive response!

Much of your response confirms my minitab tests....the pushout force = 15, and separate factor analysis.

The simultaneous test (both factors) for equal variance threw me. Without the ability to do follow up test or to obtain additional data, I am left with the following question(s);

Does workstation #2 have a signifcantly higher variation than the other two stations

Does workstation #1 have significantly larger mean from workstations 2 & 3

or that there is no statistically significant difference between work stations.

Based solely on this info, it would appear that there is no statistically significant difference between work stations, am I off track?

Thank you again for the treatise and assistance..

Marty
 
B

Barbara B

Marty,

at 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.

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. 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).

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). 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. 0.0641), but the process mean for pushout force=15 is significantly nearer to the target 0.4.

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)

Best regards,

Barbara
 

Attachments

  • Process Capability DimpleDepth 2011 01 17.pdf
    209.5 KB · Views: 126

optomist1

A Sea of Statistics
Super Moderator
Hi Barb,

Thank you again for your valuable input, the more I learn the more interesting this becomes.

As it turns out my "hunch", informed by posts such as yours is that there is insufficient information....no evidence to suggest a difference between machines, and yes push or punch out force of 15 is best setting.

I currently use Minitab 15, what is your experience with 16? It seems to have some worthwhile features.

This may be statistical semantics, you refered to workstations as "text data", I assume that this synonomous with "nominal data"?

Many Thanks,
Marty

"Aerdynamics Is For Those Who Cannot Build Engines" - Enzo Ferrari
 
B

Barbara B

Hi Marty,

sometimes it's just semantics which make it difficult to understand what a method requires or how it is applied :bigwave:

Minitab's "text data" contains variables which are nominal (no direct ranking possible, like auto brands) or ordinal (ranking possible but distance between categories not necessarily equal, like results of a race where the distance in seconds between first and second is not known and could differ, but you know that the first is faster than the second). In some Minitab-menus you'll find "categorical" as a synonym for "text data".

My favourite feature in Minitab R16 is the prediction option (in GLM and Regression with categorical factors / General Regression), because the math behind is really tricky and it's much easier to get prediction and prediction intervals now. The new menu "tolerance interval" is highly appreciated (calculation of tolerance intervals out of sample data or summarized data / sample characteristics). For users who don't do statistics and maths on a daily basis, the "Assistant" menu is very helpful. (And no, I won't get money for this promotion on Minitab!)

Best regards,

Barbara
 
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