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

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  • 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 x vector after mean centering.

0.007

-0.024

0.013

-0.016

-0.022

-0.018

The estimated score is -0.0373, so the residual is

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.

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