Solution 2.1 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant
Education Article
- Published: Jan 1, 2000
- Channels: Chemometrics & Informatics
1. The design matrix for the raw (uncoded) data is given by
|
x0 |
x1 |
x2 |
x1x2 |
|
1 |
0.1 |
2 |
0.2 |
|
1 |
0.1 |
4 |
0.4 |
|
1 |
0.2 |
2 |
0.4 |
|
1 |
0.2 |
4 |
0.8 |
2. The coefficients can be calculated by
b = D-1 .y
given by
|
b0 |
35 |
|
b1 |
20 |
|
b2 |
-4 |
|
b12 |
4 |
so that
y = 35 + 20x1 - 4x2 + 4 x1 x2
3. The design matrix now becomes
|
x0 |
x1 |
x2 |
x1x2 |
|
1 |
-1 |
-1 |
1 |
|
1 |
-1 |
1 |
-1 |
|
1 |
1 |
-1 |
-1 |
|
1 |
1 |
1 |
1 |
so that
y = 27.8 + 1.6x1 – 3.4x2 + 0.2 x1 x2
Note the latter equation could have been obtained by a more traditional, non matrix based method of adding and subtracting coefficients, although this is not asked in the problem.
Because the overall uncoded concentration range is 0.1 mM whereas the coded concentration range is 2 units (the difference between –1 and +1), the coefficients for x1 are much smaller in question 3 compared to question 2. Time covers the same range, so the coefficients are of approximately similar size.