5^2 DOE (Design of Experiments) with 3 Replications, 5 Levels 2 Factors

E

EricB503

#1
Hello, I am trying to set up a 5^2 DOE with 3 replications using minitab. What im measuring is interior weld strengths on car doors. Each door has 58 different interior welds. I will be measuring about 20 welds from each door. The 5 levels are the depths of the welds, ( 5 being the lowest and 1 being the highest). And the 2 factors are the types of doors tested, ( vinly or leather ). I will be testing 6 doors total, ( 3 vinly, 3 leather ). I have my expirment set up in minitab but I cant seem to figure out how to incorporate the different welds into my experimental design. As of right now I have 75 data yield points, ( which makes sense (5^2)*3 ) but I am still missing something. Any suggestions?

Thanks for your help,
 
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B

Barbara B

#3
I'm not quite sure if I understand your design: You've got 2 factors, 1 numeric factor with 5 levels (weldth depth) and 1 text factor with 2 levels (types of doors / vinyl vs. leather). Is that correct?

But how did you get 5^2=25 design points (with 1 replication)? Imho you must have 2 factors with 5 levels EACH to get 25 different combinations. With 5 levels for factor 1 (depth of weldth) and 2 levels for factor 2 (type of doort) you'll only have 5*2=10 different combinations of weldth depth and type of door :confused:

Usually you would start with a 2-level factorial design with center points. Is there a specific reason for using a design with 5 levels for depth of weldth? Otherwise I would recommend a 2-level factorial 2^2-design (=4 runs for the basic design, 4*3=12 runs with 3 replications) with 4 additional centerpoints (=8 runs due to the categorical factor "type of door"). In total there would be 12 + 8 = 20 runs.

If you get a significant curvature within your region of interest you could augment the design into a central-composite design (response surface design) with 5 levels for depth of weldth. But this would only be necessary, if the factorial model doesn't provide a fair enough explanation of the results.

Regards,

Barbara
 
S

supreecha

#4
Do you know for a fact that all 5 factors have a significant effect on the process?
If not, start with a fractional factorial of the 5 factors at 2 levels. Keep the resolution of the design at IV or greater to identify main effects and 2-way interactions. This is called a screening design. It is used to reduce the number of factors to those that are significant.

Once the number of factors has been reduced, the next step is to run the modeling and optimization experiment. The best approach is to run a response surface design on the remaining factors.

Another possibility depending on your process is EVOP (Evolutionary Optimization of Process). This involves making small incremental changes and tweaking the process along the path of steepest ascent.
Once the number of factors has been reduced, the next step is to run the modeling and optimization experiment. The best approach is to run a response surface design on the remaining factors.

General Linear Model: Flow Coat Th versus Spin Level, Spray Level, ...

Factor Type Levels Values
Spin Level fixed 3 4, 9, 14
Spray Level fixed 3 3, 6, 10
Dwell Level fixed 3 10, 17, 24


Analysis of Variance for Flow Coat Thickness, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P
Spin Level 2 88.653 88.653 44.327 19.52 0.000
Spray Level 2 0.038 0.038 0.019 0.01 0.992
Dwell Level 2 13.887 13.887 6.944 3.06 0.075
Spin Level*Spray Level 4 20.147 20.147 5.037 2.22 0.113
Error 16 36.330 36.330 2.271
Total 26 159.055


S = 1.50686 R-Sq = 77.16% R-Sq(adj) = 62.88%


Unusual Observations for Flow Coat Thickness

Flow Coat
Obs Thickness Fit SE Fit Residual St Resid
1 4.7800 8.3307 0.9618 -3.5507 -3.06 R
24 12.1000 9.7274 0.9618 2.3726 2.05 R

R denotes an observation with a large standardized residual.


Response Surface Regression: Flow Coat Th versus Spin Level, Spray Level, ...

The analysis was done using uncoded units.

Estimated Regression Coefficients for Flow Coat Thickness

Term Coef SE Coef T P
Constant 20.9833 3.09852 6.772 0.000
Spin Level -2.2437 0.51575 -4.350 0.000
Spray Level -0.4193 0.24278 -1.727 0.100
Dwell Level -0.2919 0.12180 -2.397 0.026
Spin Level*Spin Level 0.0674 0.02440 2.760 0.012
Spin Level*Spray Level 0.0473 0.02457 1.925 0.069
Spin Level*Dwell Level 0.0196 0.01232 1.588 0.128


S = 1.49431 PRESS = 79.3696
R-Sq = 71.92% R-Sq(pred) = 50.10% R-Sq(adj) = 63.50%


Analysis of Variance for Flow Coat Thickness

Source DF Seq SS Adj SS Adj MS F P
Regression 6 114.396 114.396 19.0660 8.54 0.000
Linear 3 83.475 44.818 14.9394 6.69 0.003
Spin Level 1 71.640 42.259 42.2591 18.93 0.000
Spray Level 1 0.009 6.662 6.6617 2.98 0.100
Dwell Level 1 11.826 12.828 12.8277 5.74 0.026
Square 1 17.013 17.013 17.0129 7.62 0.012
Spin Level*Spin Level 1 17.013 17.013 17.0129 7.62 0.012
Interaction 2 13.908 13.908 6.9539 3.11 0.066
Spin Level*Spray Level 1 8.277 8.277 8.2770 3.71 0.069
Spin Level*Dwell Level 1 5.631 5.631 5.6307 2.52 0.128
Residual Error 20 44.659 44.659 2.2330
Total 26 159.055


Unusual Observations for Flow Coat Thickness

Flow Coat
Obs StdOrder Thickness Fit SE Fit Residual St Resid
1 1 4.780 7.268 0.915 -2.488 -2.11 R
24 24 12.100 8.764 0.726 3.336 2.55 R

R denotes an observation with a large standardized residual.
 
Last edited by a moderator:
E

EricB503

#5
I might have designed it wrong. I have my two factors being vinyl and leather doors and my levels being the depths of the weld. But I came up with the conlclusion that it would take to much time and money to run the experiment becuase to complete the doe I would have to find 5 different doors with all 5 levels of weld depths present for each specific weld. With 3 replications I would have to gather well over 100 doors. Please correct me if i am wrong but thats my understannding of doe. I am fairly new to minitab and do not have alot of experience running doe's. Im thinking of doing a balanced anova to compare doors in a more generic sense. If you guys have any suggestions I would really appreciate it!
 

Bev D

Heretical Statistician
Leader
Super Moderator
#6
Your two factors are door type and weld depth.
the levels for the door type are: leather and vinyl
the levels for the weld depth are 1-5.

I would consider using only two levels of weld as a first pass. Even if you expect curvature 3 levels for weld depth (1, 3 and 5) still yield a very manageable experiment that will give you a LOT of useful infromation.

We almost always get more information from a series of smaller experiments than we do from a honkin big one....
 
E

EricB503

#7
Thanks for your help, When I get my info from the quality guy I'll let you know how it goes. Thanks again!
 
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