Question

In: Statistics and Probability

A 2 ^3 full factorial screening experimental design was performed to identify the important factors affecting...

A 2 ^3 full factorial screening experimental design was performed to identify the important factors affecting the percent conversion in a chemical process. Two factors, temperature in C (x1) and reactant concentration (x2) were identified as the key process input variables and the method of steepest ascent was applied to identify the maximum percent conversion area. Create and analyze a central composite design using the data in the following table (in standard run order). Test the residuals assumption and comment on the adequacy of the model. Identify the temperature and reactant concentration setting that maximizes percent conversion.

Std. Run Temp Reactant % Conversion
1 200 15 53
2 250 15 88
3 200 25 79
4 250 25 83
5 189.65 20 58
6 260.35 20 88
7 225 12.93 75
8 225 27.07 84
9 225 20 86
10 225 20 89
11 225 20 83
12 225 20 81

Solutions

Expert Solution

Sol:

Regression Equation can be calculated from anova table and Correlation Coeficient table which is calculated through Excel functions:

ANOVA
df SS MS F Significance F
Regression 2 970.9791 485.4895 8.183364 0.009438
Residual 9 533.9376 59.3264
Total 11 1504.917
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept -29.5602 26.9144 -1.0983 0.300584 -90.4448 31.3244 -90.4448 31.3244
X Variable 1 0.407161 0.108936 3.737618 0.004643 0.160731 0.653592 0.160731 0.653592
X Variable 2 0.843277 0.54468 1.548206 0.15598 -0.38888 2.07543 -0.38888 2.07543

Regression Equation can be written as:

Conversion = 29.5 +0.40 Temp +0.8432 Reactant

For maximum prediction there are two sets these are (260.35,20) and (250,25)

Residuals can be calculated as follows:

Std. Run Temp Reactant% Conversion Prediction Residual
1 200 15 53 64.5078 11.5078
2 250 15 88 84.8628 -3.1372
3 200 25 79 72.9398 -6.0602
4 250 25 83 93.2948 10.2948
5 189.65 20 58 64.51032 6.510315
6 260.35 20 88 93.29229 5.292285
7 225 12.93 75 72.93988 -2.06012
8 225 27.07 84 84.86272 0.862724
9 225 20 86 78.9013 -7.0987
10 225 20 89 78.9013 -10.0987
11 225 20 83 78.9013 -4.0987
12 225 20 81 78.9013 -2.0987

The assumption of regression equation is That the regression residuals must be normal distributed

But here mean = -0.01 and Median = 2

So mean and Median are not equal so we can say that residuals are not Normally Distributed


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