# Full Factorial Design with 2 Factors and 5 Levels

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- This topic has 18 replies, 6 voices, and was last updated 2 years, 4 months ago by Robert Butler.

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- January 31, 2011 at 1:08 pm #53708

MaierParticipant@ralle2011**Include @ralle2011 in your post and this person will**

be notified via email.Hi Folks,

can someone provide help?

We have designed a Full Factorial Design with 2 Factors and 5 Levels and no replicates –> 25 trials, everything was fine

But when it comes to analyzing minitab tells me that there is no degree of freedom left to calculate ther error-term. I understand that, but are there any ways to analyze the DoE with main effects and interactions?

Some tricks?

Kind Regards

Ralf

0January 31, 2011 at 2:24 pm #191181

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.Since it’s already a done deal I won’t comment on the 5 level issue other than to say it is unlikely that you needed that many levels and that you probably spent more time and money on the problem than was necessary.

To your question – 2 factors at five levels with no replicates means you can look at main effects, the linear x linear interaction, quadratic effects, quadratic x linear interactions, quadratic x quadratic interactions, cubic effects, cubic x linear interactions , cubic x quadratic interactions, cubic x cubic interactions, quartic main effects, quartic x linear interactions quartic x quadratic interactions, quartic x cubic interactions and quartic x quartic interactions.

If you choose the normalized levels of -1 -.5, 0 , .5, 1 for the two variables and set up a check of condition indices you will find that even with this you have some really wild VIF’s and condition indices.

With no replicates the easiest thing to do would be to allow all terms in the model except the quartic x quartic interaction and run stepwise and see what you get then turn it around an run backwards and, again see what you get. If you wind up with a parsimonious model – one with very few terms – well and good and if the two approaches give you approximately the same model then you should be in good shape.

However, there is a good chance the machine will grind on and on and settle on a model with a large number of terms. If this is the case then you will have to go back to the program output and watch the behavior of the RMSE, R2, and Mallow’s Cp and look for for the iteration where one or more of them flatten out – that is do not change much when the next “significant” term is added. Truncate the model at that point and then run the usual regression analysis (residual checks, lof, etc.) to see if there is any appreciable difference between the truncated model and the model you got by just letting the machine grind on. If there isn’t -go with the simpler model.

Regardless of the final model form – if you should get cubic and quartic terms/interactions you will probably find it very difficult to give physical meaning to the terms and you will want to look at models where you limited the choices to mains, linear 2 way and quadratic and possibly cubic main effects. Again you will want to look at model behavior as above.

0January 31, 2011 at 11:02 pm #191185

MBBinWIParticipant@MBBinWI**Include @MBBinWI in your post and this person will**

be notified via email.I, however, will comment on your design. Are either of these factors continuous? If not, I’m convinced that you could have formed them in a fashion that they could have been. In that case, you could have used a 2 level design with center points and streamlined your activity, and learned as much or even more at less complication and lower costs. If they were, then you really need help on setting up your design approach.

0February 2, 2011 at 9:48 am #191196

MaierParticipant@ralle2011**Include @ralle2011 in your post and this person will**

be notified via email.Thank you both,

A few words on why the design is like it is. I dont know it. I wasnt involved. I just was asked if I could help to analyze the data. Obviously I cant, but I am interested in any solution which helps getting as much information out of this data a possible. I want to learn.

Minitab is telling me the following after I have choosen

Stat>DOE>Factorial>Analyze Factorial Design

———————————————————-

General Linear Model: Erg1 versus Faktor A; Faktor BFactor Type Levels Values

Faktor A fixed 5 1,5; 2,0; 2,5; 3,0; 3,5

Faktor B fixed 5 0,00; 0,25; 0,50; 1,00; 1,25Analysis of Variance for Erg1, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

Faktor A 4 11,45840 11,45840 2,86460 **

Faktor B 4 0,08640 0,08640 0,02160 **

Faktor A*Faktor B 16 0,07760 0,07760 0,00485 **

Error 0 * * *

Total 24 11,62240** Denominator of F-test is zero or undefined.

S = *

* NOTE * Could not graph the specified residual type because MSE = 0 or the

degrees of freedom for error = 0.———————————————————————-

And here stops my way. I understand that there are no DOF left. But there is also a response from Robert that I could analyze quadratic, qubic etc effects.

Thanks for your help .

Ralf

0February 2, 2011 at 11:54 am #191197

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.I don’t know Minitab so perhaps MBBinWI can offer something here. I’m puzzled by the word “Faktor” in the Minitab output – I’ve never seen that anywhere.

Your output is confusing. The first part says “General Linear Model” whereas the second part says “Analysis of Variance”.

ANOVA will treat A and B as class variables and the levels as type. Under these circumstances, with no replicates, all you can do is investigate the effect of A and B.

However, as written your post gives the impression that A and B are not class but continuous and therefore the levels are not type but are interval. If this is the case then you should be using regression methods such as general linear models.

So, question #1 is that already posed by MBBinWI – are A and B continuous variables?

0February 21, 2011 at 1:28 pm #191269

MaierParticipant@ralle2011**Include @ralle2011 in your post and this person will**

be notified via email.They are continous variables…

0February 21, 2011 at 1:49 pm #191270

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.If they are both continuous then it looks like there must be something you have to do to let Minitab know this. If you don’t then it appears as though Minitab treats A and B as class variables and, as I noted, given what has been done, the Minitab output will be as you have described.

0February 21, 2011 at 2:07 pm #191271

MaierParticipant@ralle2011**Include @ralle2011 in your post and this person will**

be notified via email.The question now is what &how to tell minitab ?

0February 21, 2011 at 2:24 pm #191272

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.I’m afraid I can’t offer much help on that end since I don’t know Minitab. I understand their site is a good one as is their support. You might want to run a search on the problem and/or call them. The generic issue is the analysis of a design without replicates.

0June 1, 2011 at 3:04 am #191509With no replicate and I assume, no center point. No way for the DOE to calculate the error.

0June 1, 2011 at 3:20 am #191510

MBBinWIParticipant@MBBinWI**Include @MBBinWI in your post and this person will**

be notified via email.You can force Minitab to evaluate the DOE by eliminating higher order interactions. This frees up DOF to be used for error evaluation.

0December 30, 2012 at 10:37 am #194515

elenaGuest@elenamurray**Include @elenamurray in your post and this person will**

be notified via email.I am usng two lwvwl five factor factorial design and while analysing data thru minitab getting degree freedom for error as zero and therefore the F and P values rae missing for the output someone pld help

0December 30, 2012 at 10:54 am #194516

elenaGuest@elenamurray**Include @elenamurray in your post and this person will**

be notified via email.I am using a two level five factor factorial design with 32 runs and no replicates. while analysing data thru minitab the software is taking the error degree of freedom as zero anf therefore F anf P value coulmn are empty someone pls help.

0December 30, 2012 at 6:32 pm #194517

Eric MaassParticipant@poetengineer**Include @poetengineer in your post and this person will**

be notified via email.Elena,

I suggest you click on “terms” and get rid of at least one of the highest order interactions, like “ABC” .

When you reduce the terms by taking a term out of the analysis, the degrees of freedom available for the error terms will increase and you should be able to see p values.

Also, you might want to click on the graphs button and select the Pareto diagram of effects

0March 2, 2013 at 9:48 am #194793Hi, Eric is right. I was also running an experiment with large factors and boggled mind why F and P values are not displayed. Then thought any how higher order interactions are not necessary let me remove them. Surprisingly got F and P values. Just to remove higher order interactions…

0March 5, 2013 at 6:39 am #194810Late on this, as I was looking around as Minitab seem to have removed this error message. MSE=0. What it is saying is that it can’t calculated MSE or Std Dev as it has no replicated for corner points. By looking at the first Pareto you can see which Main effect has the least effect on the model. TO analyse the design, go to terms and remove this main effect (e.g C) and any related interactions (e.g BC, AC, CD, CE, ACE (i.e any with the main effect in it). THis is to kid MT into thinking there are corner points and dulicates (i.e it will treat each data point for these as though the factor removed were not there, giving you duplicates.

AS I mentioned MT for some reason have removed the MSE = O error and replaced it with “S= * Press = *”. THe effect is the same, that the ANOVA Section shows the F and P with *. The Residual error is 0, and no SS are caculated.

0March 6, 2013 at 5:05 pm #194816

MBBinWIParticipant@MBBinWI**Include @MBBinWI in your post and this person will**

be notified via email.It is best practice to start with removing highest order interactions that are not statistically significant, one level at a time. In this case, with only 2 factors, you only have 2nd order interactions available, so remove those that are not statistically significant, then re-run the analysis.

More generally, if you had, for example 5 factors, you would first run the analysis and see if the 5th order interaction was significant. If not, remove. Re-run the analysis and look at the 4th order interactions. Those that are not significant, remove. Then the third, then the 2nd. You’ll end up with the best transfer function.

It’s possible to remove all the non-signficant terms at once, but occaisionally will miss a significant interaction that was being overshadowed by noise from another interaction or main effects, so better to do it iteratively.0May 17, 2018 at 7:18 pm #202559

Mike ScullyParticipant@mscully26**Include @mscully26 in your post and this person will**

be notified via email.I have a 3 factor (2 with 3 levels and 1 with 2 levels) and I am having issues. I am receiving the ‘can’t graph due to MSE=0’ error. It has been a while since I’ve messed with minitab so please go easy on me. Can someone guide me in the right direction please? I can share my DoE if needed. Thanks for any help I can get.

0May 18, 2018 at 2:43 pm #202561

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.Apparently the Isixsigma admin’s decided your other post with this same topic was the redundant one so – as I said in my response to the same post – you will have to tell us more about what it is that you are trying to do. Simple data plotting is just a matter of putting data points on an x-y plot and has nothing to do with MSE.

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