# 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, 4 months ago.

- AuthorPosts
- July 16, 2006 at 11:46 pm #44034
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,

Bill0July 17, 2006 at 1:01 am #140449Hi 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

0July 17, 2006 at 2:54 am #140450Bill:

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.

Paul0July 17, 2006 at 3:10 am #140451Paul: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

0July 17, 2006 at 3:13 am #140452Bill: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

Deep0July 17, 2006 at 11:05 am #140464Hello 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 31

–

–

74

78

732

+

–

65

64

693

–

+

74

76

784

+

+

85

88

91

A is temperature with range from 30 to 90

B is residence time with range from 15 to 45

Thanks.0July 17, 2006 at 12:47 pm #140469Bill: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 its 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

0July 17, 2006 at 1:21 pm #140475

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.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 Xs 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 for0July 17, 2006 at 1:38 pm #140478Robert: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

0July 17, 2006 at 3:22 pm #140480

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.Xcenterd-scaled = (Xvalue – A1)/A2

Where

A1 = (Xmax +Xmin)/2

and

A2 = (Xmax – Xmin)/2

0July 17, 2006 at 3:42 pm #140483Thanks Robert:I think both are the same. How would you modify this equation if you have three levels?? Just curious..anyway thanks once againDeep

0July 17, 2006 at 3:59 pm #140485

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.No modification necessary.

0July 17, 2006 at 10:09 pm #140511Thanks 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?

0July 19, 2006 at 3:35 pm #140645

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.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 1.0 - AuthorPosts

The forum ‘General’ is closed to new topics and replies.