5.
Process Improvement
5.4. Analysis of DOE data 5.4.7. Examples of DOE's


Data Source  
This example uses data from a NIST high performance ceramics experiment 
This data set was taken from an experiment that was performed a few
years ago at NIST by Said Jahanmir of the Ceramics Division in the
Material Science and Engineering Laboratory. The original analysis
was performed primarily by Lisa Gill of the Statistical Engineering
Division. The example shown here is an independent analysis of a
modified portion of the original data set.
The original data set was part of a high performance ceramics experiment with the goal of characterizing the effect of grinding parameters on sintered reactionbonded silicon nitride, reaction bonded silicone nitride, and sintered silicon nitride. Only modified data from the first of the three ceramic types (sintered reactionbonded silicon nitride) will be discussed in this illustrative example of a full factorial data analysis. The reader can download the data as a text file. 

Description of Experiment: Response and Factors  
Response and factor variables 
Purpose: To determine the effect of machining factors on ceramic strength
Response variable = mean (over 15 repetitions) of the ceramic strength Number of observations = 32 (a complete 2^{5} factorial design)
Factor 1 = Table Speed (2 levels: slow (.025 m/s) and fast (.125 m/s)) Factor 2 = Down Feed Rate (2 levels: slow (.05 mm) and fast (.125 mm)) Factor 3 = Wheel Grit (2 levels: 140/170 and 80/100) Factor 4 = Direction (2 levels: longitudinal and transverse) Factor 5 = Batch (2 levels: 1 and 2) 

The data 
The design matrix, with measured ceramic strength responses, appears
below. The actual randomized run order is given in the last column. (The
interested reader may download the data as a
text file.)
speed rate grit direction batch strength order 1 1 1 1 1 1 680.45 17 2 1 1 1 1 1 722.48 30 3 1 1 1 1 1 702.14 14 4 1 1 1 1 1 666.93 8 5 1 1 1 1 1 703.67 32 6 1 1 1 1 1 642.14 20 7 1 1 1 1 1 692.98 26 8 1 1 1 1 1 669.26 24 9 1 1 1 1 1 491.58 10 10 1 1 1 1 1 475.52 16 11 1 1 1 1 1 478.76 27 12 1 1 1 1 1 568.23 18 13 1 1 1 1 1 444.72 3 14 1 1 1 1 1 410.37 19 15 1 1 1 1 1 428.51 31 16 1 1 1 1 1 491.47 15 17 1 1 1 1 1 607.34 12 18 1 1 1 1 1 620.80 1 19 1 1 1 1 1 610.55 4 20 1 1 1 1 1 638.04 23 21 1 1 1 1 1 585.19 2 22 1 1 1 1 1 586.17 28 23 1 1 1 1 1 601.67 11 24 1 1 1 1 1 608.31 9 25 1 1 1 1 1 442.90 25 26 1 1 1 1 1 434.41 21 27 1 1 1 1 1 417.66 6 28 1 1 1 1 1 510.84 7 29 1 1 1 1 1 392.11 5 30 1 1 1 1 1 343.22 13 31 1 1 1 1 1 385.52 22 32 1 1 1 1 1 446.73 29 

Analysis of the Experiment  
Five basic steps  The experimental data will be analyzed following the previously described five basic steps. The analyses shown in this page can be generated using R code.  
Step 1: Look at the data  
Plot the response variable 
We start by plotting the response data several ways to see if any trends
or anomalies appear that would not be accounted for by the standard
linear response models.
First, we look at the distribution of the response variable regardless of factor levels by generating the following four plots.
Clearly there is "structure" that we hope to account for when we fit a response model. For example, the response variable is separated into two roughly equalsized clumps in the histogram. The first clump is centered approximately around the value 450 while the second clump is centered approximately around the value 650. As hoped for, the runorder plot does not indicate a significant time effect. 

Box plots of response by factor variables 
Next, we look at box plots of the response for each factor.
Several factors, most notably "Direction" followed by "Batch" and possibly "Wheel Grit", appear to change the average response level. 

Step 2: Create the theoretical model  
Theoretical model: assume all fourfactor and higher interaction terms are not significant  For a 2^{5} full factorial experiment we can fit a model containing a mean term, five main effect terms, ten twofactor interaction terms, ten threefactor interaction terms, five fourfactor interaction terms, and a fivefactor interaction term (32 parameters). However, we start by assuming all fourfactor and higher interaction terms are nonexistent. It's very rare for such highorder interactions to be significant, and they are very difficult to interpret from an engineering viewpoint. The assumption allows us to accumulate the sums of squares for these terms and use them to estimate an error term. We start with a theoretical model with 26 unknown constants, hoping the data will clarify which of these are the significant main effects and interactions we need for a final model.  
Step 3: Fit model to the data  
Results from fitting up to and including thirdorder interaction terms 
The ANOVA table for the 26parameter model (intercept not shown) follows.
Summary of Fit RSquare 0.995127 RSquare Adj 0.974821 Root Mean Square Error 17.81632 Mean of Response 546.8959 Observations 32 Sum Source DF of Squares F Ratio Prob>F X1: Table Speed 1 894.33 2.8175 0.1442 X2: Feed Rate 1 3497.20 11.0175 0.0160 X1: Table Speed* 1 4872.57 15.3505 0.0078 X2: Feed Rate X3: Wheel Grit 1 12663.96 39.8964 0.0007 X1: Table Speed* 1 1838.76 5.7928 0.0528 X3: Wheel Grit X2: Feed Rate* 1 307.46 0.9686 0.3630 X3: Wheel Grit X1:Table Speed* 1 357.05 1.1248 0.3297 X2: Feed Rate* X3: Wheel Grit X4: Direction 1 315132.65 992.7901 <.0001 X1: Table Speed* 1 1637.21 5.1578 0.0636 X4: Direction X2: Feed Rate* 1 1972.71 6.2148 0.0470 X4: Direction X1: Table Speed 1 5895.62 18.5735 0.0050 X2: Feed Rate* X4: Direction X3: Wheel Grit* 1 3158.34 9.9500 0.0197 X4: Direction X1: Table Speed* 1 2.12 0.0067 0.9376 X3: Wheel Grit* X4: Direction X2: Feed Rate* 1 44.49 0.1401 0.7210 X3: Wheel Grit* X4: Direction X5: Batch 1 33653.91 106.0229 <.0001 X1: Table Speed* 1 465.05 1.4651 0.2716 X5: Batch X2: Feed Rate* 1 199.15 0.6274 0.4585 X5: Batch X1: Table Speed* 1 144.71 0.4559 0.5247 X2: Feed Rate* X5: Batch X3: Wheel Grit* 1 29.36 0.0925 0.7713 X5: Batch X1: Table Speed* 1 30.36 0.0957 0.7676 X3: Wheel Grit* X5: Batch X2: Feed Rate* 1 25.58 0.0806 0.7860 X3: Wheel Grit* X5: Batch X4: Direction * 1 1328.83 4.1863 0.0867 X5: Batch X1: Table Speed* 1 544.58 1.7156 0.2382 X4: Directio* X5: Batch X2: Feed Rate* 1 167.31 0.5271 0.4952 X4: Direction* X5: Batch X3: Wheel Grit* 1 32.46 0.1023 0.7600 X4: Direction* X5: Batch 

This fit has a large R^{2} and adjusted R^{2}, but the high number of large (>0.10) pvalues (in the "Prob>F" column) makes it clear that the model has many unnecessary terms.  
Stepwise regression  Starting with the 26 terms, we use stepwise regression to eliminate unnecessary terms. By a combination of stepwise regression and the removal of remaining terms with a pvalue larger than 0.05, we quickly arrive at a model with an intercept and 12 significant effect terms.  
Results from fitting the 12term model 
Summary of Fit RSquare 0.989114 RSquare Adj 0.982239 Root Mean Square Error 14.96346 Mean of Response 546.8959 Observations (or Sum Wgts) 32 Sum Source DF of Squares F Ratio Prob>F X1: Table Speed 1 894.33 3.9942 0.0602 X2: Feed Rate 1 3497.20 15.6191 0.0009 X1: Table Speed* 1 4872.57 21.7618 0.0002 X2: Feed Rate X3: Wheel Grit 1 12663.96 56.5595 <.0001 X1: Table Speed* 1 1838.76 8.2122 0.0099 X3: Wheel Grit X4: Direction 1 315132.65 1407.4390 <.0001 X1: Table Speed* 1 1637.21 7.3121 0.0141 X4: Direction X2: Feed Rate* 1 1972.71 8.8105 0.0079 X4: Direction X1: Table Speed* 1 5895.62 26.3309 <.0001 X2: Feed Rate* X4:Direction X3: Wheel Grit* 1 3158.34 14.1057 0.0013 X4: Direction X5: Batch 1 33653.91 150.3044 <.0001 X4: Direction* 1 1328.83 5.9348 0.0249 X5: Batch 

Normal plot of the effects 
Nonsignificant effects should effectively follow an approximately
normal distribution with the same location and scale. Significant
effects will vary from this normal distribution. Therefore, another
method of determining significant effects is to generate a normal
probability plot of all 31 effects. The effects that deviate
substantially from the straight line fit to the data are
considered significant.
Although this is a somewhat subjective criteria, it tends to work
well in practice. It is helpful to use both the numerical output
from the fit and graphical techniques such as the normal probability
plot in deciding which terms to keep in the model.
A normal probability plot of the effects is shown below. (To reduce the scale of the yaxis, the largest two effects, X4: Direction and X5: Batch, are not shown on the plot. In addition, these two effects were not used to compute the normal reference line.) The effects we consider to be significant are labeled. In this case, we have arrived at the exact same 12 terms by looking at the normal probability plot as we did from the stepwise regression.
Most of the effects cluster close to the center (zero) line and follow the fitted normal model straight line. The effects that appear to be above or below the line by more than a small amount are the same effects identified using the stepwise routine, with the exception of X1. Some analysts prefer to include a main effect term when it has several significant interactions even if the main effect term itself does not appear to be significant. 

Model appears to account for most of the variability 
At this stage, the model appears to account for most of the variability
in the response, achieving an adjusted R^{2} of 0.982. All the
main effects are significant, as are six 2factor interactions and
one 3factor interaction. The only interaction that makes little
physical sense is the " X4: Direction*X5: Batch"
interaction  why would the response using one batch of material
react differently when the batch is cut in a different direction as
compared to another batch of the same formulation?
However, before accepting any model, residuals need to be examined. 

Step 4: Test the model assumptions using residual graphs (adjust and simplify as needed)  
Plot of residuals versus predicted responses 
First we look at the residuals plotted versus the predicted responses.
The residuals appear to spread out more with larger values of predicted strength, which should not happen when there is a common variance. 

Next we examine the distribution of the residuals with a normal quantile
plot, a box plot, a histogram, and a runorder plot.
None of these plots appear to show typical normal residuals and the boxplot indicates that there may be outliers. 

Step 4 continued: Transform the data and fit the model again  
BoxCox Transformation 
We next look at whether we can model a transformation of the response
variable and obtain residuals with the assumed properties. We
calculate an optimum BoxCox transformation by finding the value of
that maximizes the negative log likelihood.
The optimum is found at = 0.2. A new Y: Strength variable is calculated using:


Fit model to transformed data  When the 12effect model is fit to the transformed data, the "X4: Direction * X5: Batch" interaction term is no longer significant. The 11effect model fit is shown below, with parameter estimates and pvalues.  
The fitted model after applying BoxCox transformation 
The 11Effect Model Fit to Tranformed Response Data Response: Y:NewStrength Summary of Fit RSquare 0.99041 RSquare Adj 0.985135 Root Mean Square Error 13.81065 Mean of Response 1917.115 Observations (or Sum Wgts) 32 Parameter Effect Estimate pvalue Intercept 1917.115 <.0001 X1: Table Speed 5.777 0.0282 X2: Feed Rate 11.691 0.0001 X1: Table Speed* 14.467 <.0001 X2: Feed Rate X3: Wheel Grit 21.649 <.0001 X1: Table Speed* 7.339 0.007 X3: Wheel Grit X4: Direction 99.272 <.0001 X1: Table Speed* 7.188 0.0080 X4: Direction X2: Feed Rate* 9.160 0.0013 X4: Direction X1: Table Speed* 15.325 <.0001 X2: Feed Rate* X4:Direction X3: Wheel Grit* 12.965 <.0001 X4: Direction X5: Batch 31.871 <.0001 

Model has high R^{2}  This model has a very large R^{2} and adjusted R^{2}. The residual plots (shown below) are quite a bit better behaved than before  
Residual plots from model with transformed response 
The plot of the residuals versus the predicted values indicates that the
transformation has resolved the problem of increasing variace with increasing
strength.
The normal probability plot, box plot, and the histogram of the residuals do not indicate any serious violations of the model assumptions. The run sequence plot of the residuals does not indicate any time dependent patterns. 

Step 5. Answer the questions in your experimental objectives  
Important main effects and interaction effects  The magnitudes of the effect estimates show that "Direction" is by far the most important factor. "Batch" plays the next most critical role, followed by "Wheel Grit". Then, there are several important interactions followed by "Feed Rate". "Table Speed" plays a role in almost every significant interaction term, but is the least important main effect on its own. Note that large interactions can obscure main effects.  
Plots of the main effects and significant 2way interactions 
Plots of the main effects and the significant 2way interactions are
shown below.
Next, we plot 2way interaction plot showing means for all combinations of levels for the two factors. The labels located in the diagonal spaces of the plot grid have two purposes. First, the label indicates the factor associated with the xaxis for all plots in the same row. Second, the label indicates the factor defining the two lines for plots in the same column. For example, the plot labeled r*s contains averages for low and high levels of the rate variable (xaxis) for both levels of speed. The blue line represents the low level of speed and the pink line represents the high level of speed. The two lines in the r*s plot cross, indicating that there is interaction between rate and speed. Parallel lines indicate that there is no interaction between the two factors. 

Optimal Settings 
Based on the analyses, we can select factor settings that
maximize ceramic strength. Translating from "1" and "+1" back to
the actual factor settings, we have: Table speed at "1" or .125m/s;
Down Feed Rate at "1" or .125 mm; Wheel Grit at "1" or 140/170;
and Direction at "1" or longitudinal.
Unfortunately, "Batch" is also a very significant factor, with the first batch giving higher strengths than the second. Unless it is possible to learn what worked well with this batch, and how to repeat it, not much can be done about this factor. 

Comments  
Analyses with value of Direction fixed indicates complex model is needed only for transverse cut 


Half fraction design 


Natural log transformation 
