In: Statistics and Probability
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 |
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