Significance of Center Points
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 This topic has 8 replies, 9 voices, and was last updated 18 years, 2 months ago by Statman.

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March 27, 2003 at 9:33 pm #31817
Can someone explain the significance of center point analysis in DOE? I am currently in training and my instructor is not explaining it very well…
0March 27, 2003 at 10:01 pm #84260Centerpoints are a great way to test for curvature between or within factors, without fully characterizing the curvature. It will signal curvature is present, leaving the choice to you as to how to pursue characterizing the curvature. This is expecially useful if you’re following response surface methodology (method of steepest ascent/descent).
It can tell you that your range is not wide enough, although hopefully you don’t ever have this happen, because it’s an expensive way to find out. A hint: don’t design centerpoints into a screening experiment. You’re trying to find big signals, so a centerpoint won’t add much to your objective.
Centerpoints are great when you want to do one round of experimenting, analyze the results, and then want to do another one, adding the 2nd round of results to the first. If you test 2 factors at two levels with three centerpoints, and you later want to test the axial points to flesh out a better model, you can run the axial points with three more centerpoints, and analyze both experiments’ results simultaneously because the centerpoints can account for differences between experiments.
This isn’t everything, but others can better inform you than me.0March 27, 2003 at 10:23 pm #84263Brian,
Center points have these additional benefits:They are a way to establish an estimate of variability within the design space vs. running replicates and/or repeats for every single run of the DOE. Therefore an economical balance vs. using MSe as a way of creating a surrogate estimate of variability.
They can be used as a guard against background/noise factors from influencing or ‘biasing’ the reponses from a DOE.
For point two, if you can’t run all of the experiments within a day/shift/location in one fell swoop, you may look to center points as a diagnostic to see if there are shifts in the average experimental response across the days/shifts/locations. Of course, the other option would be to run the DOE, under multiple days/shifts/locations as blocks.
As has been mentioned earlier, center points are a way of testing one of the fundamental assumptions of factorial experimentation at two levels (i.e. linear response within the design space). Hang in there!!!
Regards,
Erik0April 14, 2003 at 8:37 am #84842
Anne on a mouseParticipant@Anneonamouse Include @Anneonamouse in your post and this person will
be notified via email.As I understand, DOE assumes linearity of response in the design space. The center points allow you to confirm this linearity which is particularly important if you intend to do some predictive work.
Your tutor may use this site too.0April 14, 2003 at 9:43 am #84846I ve gone into the statements made about this topic, can any one help me understand – what is centre point in an organisation,
In a very simple way.0April 14, 2003 at 1:10 pm #84850I assume from the question that you are discussing two level factorial designs. Since each factor is set at only two levels, one must assume that the relationship between the response and each factor is a straight line. The highest order model that can be fit to two points is obviously a straight line. Center points are additional experimental treatments where all of the continuous factors are simulataneously set to a mid level — the midrange for each factor. It is a good idea to augment a two level design with center points for several reasons: (1) It allows for a test of curvature between the factors and the response; (2) Center points are an ideal location in the design space at which to replicate (one typically only replicates at the center points); and (3) It can be shown that the center points provide more precision in model predictions made within the design space. Unfortunately, center points cannot be used to estimate higher order model terms (such as squared terms) and if significant curvature is present, then more runs, such as axial points (from a Central Composite Design) must be added in order to estimate a model that accommodates the curvature between the response and the factors.
Hope this is useful.0April 14, 2003 at 5:31 pm #84860While conducting experiements we assume the regression model to be a first order model… but we should be aware of the possibility of a second order model ( one having quadratic terms..i.e x square terms). center points 1) provid info abt the curvature( absence and presence of quadratic terms), 2) donot affect the effect estimates, 3) provide an independent estimate of error in unreplicated design. without increasing the size of the experiement too much.But you can employ center points only when the factors are quantitative not with categorical/ qualitative factors.
0October 30, 2003 at 7:04 pm #91827When using center points, the Sum of squares calculation found in D. Montgomery 5th ed, p272 multiplies the the number of center points by the number of corner points by the squarred distance between the mean of the corner points and the center points. But why do does the formula divide this sum of squares by the total number of data points used ?
When we calculate the sum of squares for main effects, technically we take the sum of squared distances between the factor level means and the grand mean and multiply by the number of samples per level (in a balanced design). We don’t divide by the degrees of freedom until we calculate the mean square. So why do we divide by the total number of data points in the calculation of the center point Sum of Squares ?0October 30, 2003 at 8:42 pm #91836Hi Nick,
This is because you can not assess the significance of the center point with the same MSE that you are using for the main effects and interactions also, the center point test is a one degree of freedom test so the SS = MS. This formula makes the curvature effect equivalent to the main effects and interaction effects calculations in the ANOVA table.
This is easier to explain using the standard error. Recall that the standard error for any ttest is:
S*sqrt(1/n1 + 1/n2) where S is the pooled standard deviation. and if n1 = n2 the this formula becomes 2S/sqrt(N) where N = n1 +n2. When you are testing the main effects and interactions, N is the total number of runs in the balance factorial since half the runs are at the high level and half the runs are at the low level.
When testing for curvature however, you are testing the difference between the average of the runs at the center point to the average of the corner points. So the formula for the standard error is:
S*sqrt(1/N +1/nc)
If you work through the math, you will see that this explains the calculations in Montgomery.
One important aspect of this, since center points do not have the hidden replication that the factor effects have, they require more replicates to achieve an equivalent power and usually are a waist of experimental resources.
Statman
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