A central composite design is run for analysis of a chemical process, resulting

Question

A central composite design is run for analysis of a chemical process, resulting in the experimental data shown in the MINITAB Analysis Packet. The objective of the analysis was to identify settings that would maximize process discharge. Refer to the packet and answer the following questions.
Is the regression significant? Explain. (Can you comment about any assumptions?) Compute the R2 and compare to the result provided by Minitab. Comment on the quality of the regression based on the magnitude of the R2 value. Based on your review of the Minitab output write out the regression model for this experiment. What is the approximate best setting to maximize the process discharge? SET UP the calculations for the stationary point. (Formulas must have all numbers provided but equations do not need to be solved.)

Analysis Results for Process Discharge
Term Coef SE Coef T P
Constant 79.9400 0.11896 671.997 0.000
A 0.9950 0.09405 10.580 0.000
B 0.5152 0.09405 5.478 0.001
A*A -1.3762 0.10085 -13.646 0.000
B*B -1.0013 0.10085 -9.928 0.000
A*B 0.2500 0.13300 1.880 0.102

S = 0.2660 R-Sq = 98.3% R-Sq(adj) = 97.0%
PRESS = 2.34577 R-Sq(pred) = 91.84%

Source DF Seq SS Adj SS Adj MS F P
Regression 5 28.2478 28.2478 5.64956 79.85 0.000
Linear 2 10.0430 10.0430 5.02148 70.97 0.000
Square 2 17.9548 17.9548 8.97741 126.88 0.000
Interaction 1 0.2500 0.2500 0.25000 3.53 0.102
Residual Error 7 0.4953 0.4953 0.07076
Lack-of-Fit 3 0.2833 0.2833 0.09443 1.78 0.290
Pure Error 4 0.2120 0.2120 0.05300

Total 12 28.7431

Explanation / Answer

a) H0: The regression is not significant
H1: The regression is significant
Let the los be alpha = 5%
From the output, P-value of regression is 0.000 < alpha 0.05, so we reject H0
Thus, we conclude that the regression is significant
b)
R-square = 98.3% percentage of variation in the dependent variable is explained by the independent variables
c)
The multiple regression of Process Discharge is
Process Discharge is = 79.94 + 0.9950A + 0.5152B - 1.3762A2 - 1.0013B2 + 0.25AB

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