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DOE Question

Six Sigma – iSixSigma Forums Old Forums General DOE Question

This topic contains 13 replies, has 4 voices, and was last updated by  Robert Butler 13 years ago.

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  • #44034

    lin
    Participant

    Hello All,
    I have completed a simple 2 factor full factorial design with three replications.  The ANOVA table based on the coded factors (-1 to 1 for the two levels) shows factor B and the interaction AB to be signficant.  Both p values are significantly less than 0.05.  The p value for factor A is .35.
    Now, if I run a regression using the actual values for A, B and AB, the results indicate that all three are significant with a p value less than 0.05.
    Why does this occur?  Am I missing something?
    Thanks,
    Bill

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    #140449

    Deep
    Participant

    Hi BillThat is interesting. Did you try running regression with coded values? Try that and see what you get. I cannot think about anything else, or wait for other experts. In the mean time try to run regression using coded values (factors coded value instead of the actual values). ThanksDeep

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    #140450

    Sorour
    Participant

    Bill:
    P value alone does not determine strength of the correlation. What is the r-sq. value, or the correlation coefficient? If that is small, it implies there is no correlation with 95% confidence.
    Paul

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    #140451

    Deep
    Participant

    Paul:I am not sure what you are mentioning here. See he got a P value of 0.35 when he did ANOVA. So i am expecting similar kind of results with regression. It cannot be less than 0.05, right? (thats is what the poster got) Besides, both should give almost the same r-square values, right?Since its a DOE, the X’s are independent and all that… What else can go wrong? ThanksDeep

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    #140452

    Deep
    Participant

    Bill:Just curious, When you said “””I run a regression using the actual values for A, B and AB, the results indicate that all three are significant with a p value less than 0.05.”””””How did you calculate the AB interaction column for the actual values? Did you multiply A and B ? (assuming u are using Mintab)Thanks
    Deep

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    #140464

    lin
    Participant

    Hello Deep,
    Yes, I multiplied A and B.  Also, I ran the regression on the coded factors and got the same results as for the DOE ANOVA.  Below is the data:
     

    Standard Run Number
    A
    B
    Result 1
    Result 2
    Result 3

    1


    74
    78
    73

    2
    +

    65
    64
    69

    3

    +
    74
    76
    78

    4
    +
    +
    85
    88
    91
    A is temperature with range from 30 to 90
    B is residence time with range from 15 to 45
    Thanks.

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    #140469

    Deep
    Participant

    Bill:I did not get a chance to look at your data. Since you got the same result your problem is solved, I guess. In many situations it is necessary to use coded levels. In this case it is easy since it is a DOE and all coded levels are already there and moreover it’s a two level design. Other cases you have to code the levels, just like transforms that use to get standard normal z units or from sample variance to Chi – square (usually subtract the center from the level and divide by the distance from mean to that level). If you need additional help on this, you can email me (microsoft_sucks0; I will send you a pdf regarding this. Or you can wait, I am sure some one will give you the details.ThanksDeep

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    #140475

    Robert Butler
    Participant

    I copied your data and ran it as an ANOVA and also as a regression (both stepwise and just an OLS with all factors present).  If you use the coded values for all of the analysis (ANOVA and OLS/Stepwise) you get the same results – the interaction and rtime were significant whereas temp was not.  In addition all three types of SS (Type I, Type II, and Type III) have the same values. 
      With uncoded data all three SS types disagree.  In the second instance ANOVA type I and type III have the same values and give the same results as the coded responses listed below.  For the regression Type I SS is in agreement with Type I and Type III of the ANOVA and the results below, however Type III now declares all three variables to be significant as does the Type II SS of the stepwise/backward elimination.  A discussion of the issues surrounding the various types of  SS is far beyond the scope of a simple post to this forum.  From your standpoint what this means is that you have a nice example of the pitfalls of not centering and scaling your X variables. 
      As you can see from the summary statistics below – X’s centered and scaled to the -1, 1 range give consistent results regardless of SS type and at the end of the day both your ANOVA and your regression are telling you the same story.
    ANOVA results were:
    Source                     DF    Type III SS    Mean Square   F Value   Pr > F
    temp                         1          6.750             6.750                    1.00    0.3466rtime                         1      396.750       396.7500                   58.78   <.0001temp*rtime              1      330.750       330.7500                   49.00   0.0001
    OLS results were:
                                                          StandardParameter          Estimate                   Error    t Value    Pr > |t|
    Intercept         76.250              0.750               101.67      <.0001temp                  0.750              0.750                   1.00      0.3466rtime                  5.750              0.750                   7.67      <.0001temp*rtime       5.250               0.750                  7.00      0.0001
    Stepwise gave the same significance levels – obviously the coefficient estimates were slightly different.
    In all three cases the various SS types give the same results for

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    #140478

    Deep
    Participant

    Robert:When you said “”pitfalls of not centering and scaling your X variables. “””how do you scale and center the data? This is what i do…say coded value = X’Center = X0high Level = X1and difference from X0 to high = Delta.

    X’ = X1-X0 / DeltaAre there any other way to do this?Just curiousThanksDeep

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    #140480

    Robert Butler
    Participant

    Xcenterd-scaled = (Xvalue – A1)/A2
    Where
    A1 = (Xmax +Xmin)/2
    and
    A2 = (Xmax – Xmin)/2
     

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    #140483

    Deep
    Participant

    Thanks Robert:I think both are the same. How would you modify this equation if you have three levels?? Just curious..anyway thanks once againDeep

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    #140485

    Robert Butler
    Participant

    No modification necessary.

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    #140511

    lin
    Participant

    Thanks for all your effort Robert.  Your results confirm what I got.  I am not clear on what you mean  about not centering and scaling the X variables.  The coding (using the equation you gave as I did) does that, doesn’t it?

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    #140645

    Robert Butler
    Participant

      In your initial post you indicated that you had run ANOVA using the coded values of your X’s (-1 and 1) whereas for the regression you had used the actual values.  My point was that if you used the coded values (that is, centered and scaled) for the regression then your ANOVA and your regression answers are in agreement. In this case, the coded values and the centered and scaled values are identical.
      It’s my understanding that some packages do an automatic scaling and run everything on the resultant -1,1 matrix but I also understand there are others that do not.  The fact that you got different answers when you ran ANOVA on scaled and regression on actual would suggest your package doesn’t do automatic centering and scaling.
      Personally, I prefer a package where you have to do this manually for the simple reason that there are situations where -1 to 1 scaling is wrong and will lead to incorrect results.  The most common is the area of mixture design and analysis.  For these you need to scale things from 0 to 1.

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