# Solution 5.8 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

## Education Article

**Published:**Jan 1, 2000**Channels:**Chemometrics & Informatics

1. The mean centred data for samples 1 to 9 is as follows.

2. The relevant matrices are as follows.

First PLS component

Second PLS component

The samples can be exactly modelled by two components as can be seen by adding the values of ** t.q** for the first two components together plus the mean (of the first 9 samples). So for sample 1 this equals 4.667 - 3.368 - 0.299 = 1, and so on

3. The steps are as follows.

The new**vector after mean centering.**

*x*
0.007 |
-0.024 |
0.013 |
-0.016 |
-0.022 |
-0.018 |

0.012 |
-0.003 |
0.025 |
-0.008 |
0.004 |
-0.007 |

The contribution to the mean centred concentration is -0.215.

The value is -0.030.

The contributions to the *x* block modelling is identical to the residual in step (b). The contribution to the *c* block is -0.452.

4. Adding the mean of samples 1 to 9 onto the predicted contributions obtained using 1 and 2 PLS components gives 4.667 -0.215 -0.452 = 4, which is an exact prediction.

5. The reason for this is that the loadings are not obtained simply by using the ** x** data block unlike PCA. Another problem is that the loadings are not orthogonal. The aim of PLS is to maximise the covariance between the

*x*and

*c*data which is obtained via the vector

*h*.