Hi everyone,
Does any of you used minitab for mixture design before.
I used it for a 4-component miture design. I wonder how they calculated the coefficient value for each component and 2-way and 3-way interactions?Also I don't quite understand only interactions have P-value, but single factors don't have.
I check Minitab guidebook, but I didn't find any answer for that.
A typical result sheet look like this:
The main effects are highly correlated with each other.
You only have "2-way" interactions and higher p-values because you can't estimate the coefficients for main effects.
If you were to drop a component from the mixture design, then you could estimate main effects p-values.
Minitab calculates the coefficients in the mixture model (no constant term in model) using least squares regression. The t-statistics and p-values are for the Null Hypothesis that each coefficient is equal to zero. Such a null hypothesis is appropriate for the cross-product terms which estimate curvilinear blending. The "single factor terms" are the predicted values of the response at each of the pure component blends (xi=1, all other Xjs = 0). Only in very rare cases is the null hypothesis bi=0 appropriate for such terms so Minitab does not compute a t-statistic and a p-value; an appropriate procedure in my view. See attached file for more discussion.
Thank you for the answer above and the file. According to the file, I assum the coefficient of single factors are taken from the response number, β1=y1, β2=y2, β3=y3. But refer to my data in Minitab, didn't find the coefficient it gave to me has a relationship with the response at pure component belends (xi=1) I got. Also got one example with negative coefficient (and my response is solubility, which can't be negative). So does minitab has additional processing for the coefficient for single factors in the special cubic model I used?
Can't effectively answer your question without seeing the raw data. I suspect that the design does not cover the full range of the all components 0-1. In such a case the single terms will be an extrapolation beyond the range of the data and negative single factor coefficients can result. One way around this is to base your interpretation on the contour plots.
Hi Ron, I used simplex lattice with four components input. Degree of lattice of 4 and augmenteddesign with axial points. I put the file in the attachment hopefully that could be helpful to explain my question. As you can see the coefficient of single factor for example MG (0.0289) didn't equal the response (solubility=0005) when MG=1.
Solubility is better analyzed on the log scale. The predicted values will not be negative when when expressed in solubility units.The coefficients, on log scale (base 10) and solubility scale are MG (-2.12, 0.008), BS (-2.51, ).003), PL (-1.68, 0.021) and QA (-1.04, 0.091). Good agreement with the observed values i believe.
The coefficient will not equal the observed values exactly because of the smoothing done when fitting by least squares
A residual analysis also supports the log transformation. The increasing relation between the observed and predicted values is less using the log scale.
The adjusted RSQ in creases from 52% to 79%. If the full cubic model is fit the adjusted RSQ increases to 87%
Question:. I note that there is a number of duplicate observations in the file. Do these data represent duplicate measurements for each experimental run or the results of single measurements on duplicate experimental runs? This info affects how you interpret the lack of fit test.
Thanks Ron. That's single measurements on duplicate experimental runs, just a check for experimental errors.
I did a log scale for solubility, so it is MG -2.38, BS-2.44, PL -1.78, OA -1.04, which is very close to Log Coefficient MG-2.12, BS -2.51, PL -1.68, OA -1.04 (but I'm not sure what if your number0.008, 0.003, 0.021 and 0.091 for?).
If I use the original solubility, it is MG 0.0042, BS 0.0036, PL 0.0165, OA 0.1002, and coefficient as MG 0.0289, BS -0.0262, PL 0.0486, OA 0.1119, which are more remotely related, is that because of the smoothing done as you said?
The numbers 0.008, 0.003, 0.021 and 0.0.091 are the antilogs of the coefficients for the log scale and are in solubility units. The antilogs compare well with the experimental results.
What order were the experiments run in?
How long did it take you to complete the experiment?
What is the background on this experiment? What is the purpose? What are you trying to accomplish?
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