# Solution 2.16 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

## Education Article

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

1. The design matrix, together with the computed values of the parameters, is given below.

2. There are various ways in which this can be done, the simplest being to convert the design matrix to raw values, and recalculate the model. The new coefficients become the following.

 b0 b1 b2 b3 b11 b22 b33 b12 b13 b23 -24.926 1.195 2.588 5.691 -0.013 -0.054 -0.465 -0.004 -0.004 0.034

3. The calculation is given below, the total error sum of squares equalling 4.222.

4. The sum of squares replicate error is 1.939.

5. The sum of squares lack-of-fit error is 2.283.

6. It can easily be seen that 1.939 + 2.283 = 4.222.

7. R = 5 degrees of freedom are used to determine the replicate error. N-P-R degrees of freedom are used to determine the lack-of-fit error. Hence each error should be divided by 5 to give mean square errors of 0.388 (replicate) and 0.457 (lack-of-fit). For this particular design both errors have the same number of degrees of freedom. The ratio between the mean square errors is 1.178. This implies that the lack-of-fit error is only just larger than the replicate error, so probably is not very significant, i.e. a reasonable model has been found.

8. Removing the b1 term results in a model containing 9 of the terms above, to give the following errors.

Removing the b13 term has the following effect.

Hence removing the first factor increases the residual sum of square error from 4.222 to 45.350 which is a huge increase, so this factor is very significant, whereas the interaction between the first and third factors increases this residual sum of squares to only 4.254, which is a very small increase, meaning it is not very significant. Alternative approaches such as the t-test could be employed to assess the significance of these parameters, but the approach is question 8 usually provides sufficient guidance. It is important to recognise that most traditional tests assume detailed models about sampling and noise that may not be applicable to all chemical experiments.

The new equation become

y = b0 + b1x1 + b2x2 + b3x3 + b11x12 + b22x22 + b33x32 + b12x1x2

The following equations are obtained.

From the third equation we see that

The second equation allows us to express x1 in terms of x2 so that

Hence substituting in the first equation gives a value for x2

The values at the optimum are x1 = -0.332, x2 = -0.086 and x3 = 0.841 in coded values.

The true values of the parameters at the optimum are

 Power (%) Time (s) Cycles 40.01 24.57 6.84

It is not, of course possible to use a non-integer number of cycles, so 7 cycles is chosen. However, the fact that the optimum calculated value is within the data range, suggests that the range of conditions studied is adequate.

The predicted efficiency is 50.54%. Note that this is similar to the observed value for the last experiment, but more than the average value for the experiments at the centre of experimentation. The root mean square replicate error is 0.622%. From question 3, the highest estimated response is 50.17%. This mechanism of response modelling allows experimental errors to be taken into account when computing an optimum.

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