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Centering polynomials calculation
I'm trying to understand how JMP centers polynomials. Here’s an example with two independent variables, x1 and x2, and a response variable y:
x1 | x2 | y |
2 | 12 | 80 |
4 | 14 | 85 |
6 | 16 | 90 |
8 | 18 | 95 |
10 | 20 | 100 |
12 | 22 | 105 |
When using the Analyze->Fit Model option in JMP with x1, x2, x1*x1, and x2*x2 (keeping the "Center Polynomials" checkbox checked), I get the following parameter estimates:
- Intercept: 75.375
- x1: 3.0714286
- x2: 0
- (x1-7)*(x1-7): -0.089286
- (x2-17)*(x2-17): 0
I wanted to test what happens if I center manually by subtracting the mean from each value of x1 and x2.
The mean of x1 is 7, and the mean of x2 is 17. So, for the x1 column, I subtracted 7 from each value (2, 4, 6, etc.). Here’s the resulting table:
x1 | x2 | y |
-5 | -5 | 80 |
-3 | -3 | 85 |
-1 | -1 | 90 |
1 | 1 | 95 |
3 | 3 | 100 |
5 | 5 | 105 |
Using these new centered values for x1 and x2 in the Fit Model option with "Center Polynomials" unchecked, I get the following parameter estimates:
- Intercept: 96.875
- x1: 3.0714286
- x2: 0
- (x1-7)*(x1-7): -0.089286
- (x2-17)*(x2-17): 0
As you can see, all parameter estimates are correct except for the intercept. Why is the intercept different? Could someone please explain how it is calculated?
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Re: Centering polynomials calculation
X2 is linearly dependent on X1 (i.e., X1 = X2 + 10). You should be seeing a Singularity Details outline box as well as zeroed estimates for everything besides the Intercept and X1. The estimates your showing don't match what you should have gotten given the data above.
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Re: Centering polynomials calculation
I admit it was not a good example. Here is a better one:
x1 | x2 | y |
2 | 1 | 80 |
4 | 5 | 85 |
6 | 3 | 95 |
8 | 2 | 100 |
10 | 8 | 105 |
12 | 3 | 110 |
With "Center polynomials" checked:
Intercept: 76.337952
x1: 3.1705139
x2: -0.497888
(x1-7)*(x1-7): -0.120141
(x2-3.66667)*(x2-3.66667): 0.1013011
Using the x1/x2 centered values with unchecked "Center polynomials" option
x1 | x2 | y |
-5 | -2.667 | 80 |
-3 | 1.333 | 85 |
-1 | -0.667 | 95 |
1 | -1.667 | 100 |
3 | 4.333 | 105 |
5 | -0.667 | 110 |
leads to:
Intercept: 96.705792
x1: 3.1705139
x2: -0.497888
x1*x1: -0.120141
x2*x2: 0.1013011
What additional calculation is needed to get 76.337952 ?
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Re: Centering polynomials calculation
Hi @MathAvgPorpoise : This kind of thing was discussed recently here.
https://community.jmp.com/t5/Discussions/Intercept-of-a-parabola/m-p/805020
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Re: Centering polynomials calculation
Hi @MathAvgPorpoise : And only the quadratic terms are centered...the main effects are not. So, you have to be careful.
i.e., Using the x1/x2 centered values with unchecked "Center polynomials" option, you will need to enter the squared centered values, but the linear terms should be the uncentered values. It will all match then.
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Re: Centering polynomials calculation
Centering the predictors is a transformation. It translates the coordinates. This shifts the origin of the space to the mean.