Assignment 4 In this assignment you are provided information on an experiment and you are required to
Question:
Assignment 4
In this assignment you are provided information on an experiment and you are required to investigate and interpret the output which is provided below.
Problem: Consider the three?variable central composite design shown in the table.
RunOrder | PtType | Blocks | Temperature | Time | Catalyst | Conversion (%) y1 | Viscosity y2 |
1 | 1 | 1 | -1 | -1 | -1 | 74 | 53.2 |
2 | 1 | 1 | 1 | -1 | -1 | 51 | 62.9 |
3 | 1 | 1 | -1 | 1 | -1 | 88 | 53.4 |
4 | 1 | 1 | 1 | 1 | -1 | 70 | 62.6 |
5 | 1 | 1 | -1 | -1 | 1 | 71 | 57.3 |
6 | 1 | 1 | 1 | -1 | 1 | 90 | 67.9 |
7 | 1 | 1 | -1 | 1 | 1 | 66 | 59.8 |
8 | 1 | 1 | 1 | 1 | 1 | 97 | 67.8 |
9 | -1 | 1 | -1.68179 | 0 | 0 | 76 | 59.1 |
10 | -1 | 1 | 1.681793 | 0 | 0 | 79 | 65.9 |
11 | -1 | 1 | 0 | -1.68179 | 0 | 85 | 60 |
12 | -1 | 1 | 0 | 1.681793 | 0 | 97 | 60.7 |
13 | -1 | 1 | 0 | 0 | -1.68179 | 55 | 57.4 |
14 | -1 | 1 | 0 | 0 | 1.681793 | 81 | 63.2 |
15 | 0 | 1 | 0 | 0 | 0 | 81 | 59.2 |
16 | 0 | 1 | 0 | 0 | 0 | 75 | 60.4 |
17 | 0 | 1 | 0 | 0 | 0 | 76 | 59.1 |
18 | 0 | 1 | 0 | 0 | 0 | 83 | 60.6 |
19 | 0 | 1 | 0 | 0 | 0 | 80 | 60.8 |
20 | 0 | 1 | 0 | 0 | 0 | 91 | 58.9 |
Modified fromDesign and Analysis of Experiments (9th Edition) [Texidium version]. (2017). Retrieved from http://texidium.com
Part 1- Conversion percent
A partial Minitab output for the full model is shown below.
Model Summary
S | R-sq | R-sq(adj) | R-sq(pred) |
4.71669 | 91.99% | 84.79% | 75.66% |
Fits and Diagnostics for Unusual Observations
Obs | Conversion % (y1) | Fit | Resid | Std Resid | |
20 | 91.00 | 81.09 | 9.91 | 2.30 | R |
R Large residual
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Model | 9 | 2555.73 | 283.97 | 12.76 | 0.000 |
Linear | 3 | 763.05 | 254.35 | 11.43 | 0.001 |
Temperature | 1 | 14.44 | 14.44 | 0.65 | 0.439 |
Time | 1 | 222.96 | 222.96 | 10.02 | 0.010 |
Catalyst | 1 | 525.64 | 525.64 | 23.63 | 0.001 |
Square | 3 | 601.30 | 200.43 | 9.01 | 0.003 |
Temperature*Temperature | 1 | 48.47 | 48.47 | 2.18 | 0.171 |
Time*Time | 1 | 124.48 | 124.48 | 5.60 | 0.040 |
Catalyst*Catalyst | 1 | 388.59 | 388.59 | 17.47 | 0.002 |
2-Way Interaction | 3 | 1191.37 | 397.12 | 17.85 | 0.000 |
Temperature*Time | 1 | 36.12 | 36.12 | 1.62 | 0.231 |
Temperature*Catalyst | 1 | 1035.12 | 1035.12 | 46.53 | 0.000 |
Time*Catalyst | 1 | 120.13 | 120.13 | 5.40 | 0.043 |
Error | 10 | 222.47 | 22.25 | ||
Lack-of-Fit | 5 | 56.47 | 11.29 | 0.34 | 0.869 |
Pure Error | 5 | 166.00 | 33.20 | ||
Total | 19 | 2778.20 |
Interpret the above output. As well as discussing factor significance you are expected to discuss the unusual residual and how it may (or may not) affect the overall analysis. Is there a specific recommendation that you would make?
The partial output for the reduced model is provided.
Model Summary
S | R-sq | R-sq(adj) | R-sq(pred) |
5.05856 | 88.95% | 82.50% | 69.18% |
Coded Coefficients
Term | Coef | SE Coef | T-Value | P-Value | VIF |
Constant | 79.59 | 1.75 | 45.45 | 0.000 | |
Temperature | 1.03 | 1.37 | 0.75 | 0.467 | 1.00 |
Time | 4.04 | 1.37 | 2.95 | 0.012 | 1.00 |
Catalyst | 6.20 | 1.37 | 4.53 | 0.001 | 1.00 |
Time*Time | 3.12 | 1.33 | 2.35 | 0.036 | 1.01 |
Catalyst*Catalyst | -5.01 | 1.33 | -3.78 | 0.003 | 1.01 |
Temperature*Catalyst | 11.37 | 1.79 | 6.36 | 0.000 | 1.00 |
Time*Catalyst | -3.88 | 1.79 | -2.17 | 0.051 | 1.00 |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Model | 7 | 2471.13 | 353.02 | 13.80 | 0.000 |
Linear | 3 | 763.05 | 254.35 | 9.94 | 0.001 |
Temperature | 1 | 14.44 | 14.44 | 0.56 | 0.467 |
Time | 1 | 222.96 | 222.96 | 8.71 | 0.012 |
Catalyst | 1 | 525.64 | 525.64 | 20.54 | 0.001 |
Square | 2 | 552.83 | 276.42 | 10.80 | 0.002 |
Time*Time | 1 | 141.78 | 141.78 | 5.54 | 0.036 |
Catalyst*Catalyst | 1 | 365.42 | 365.42 | 14.28 | 0.003 |
2-Way Interaction | 2 | 1155.25 | 577.62 | 22.57 | 0.000 |
Temperature*Catalyst | 1 | 1035.12 | 1035.12 | 40.45 | 0.000 |
Time*Catalyst | 1 | 120.13 | 120.13 | 4.69 | 0.051 |
Error | 12 | 307.07 | 25.59 | ||
Lack-of-Fit | 7 | 141.07 | 20.15 | 0.61 | 0.736 |
Pure Error | 5 | 166.00 | 33.20 | ||
Total | 19 | 2778.20 |
Image transcription text
Residual Plots for Conversion (%) y1 Normal Probability Plot Versus Fits 10 90 Percent 50 Residual 10
10 50 70 90 Residual Fitted Value Histogram Versus Order 10 4.5 Frequency 3.0 Residual 1.5 -S -5.0
-2.5 0.0 2.5 5.0 7.5 10.0 12.5 2 4 6 10 12 14 16 18 20 Residual Observation Order
State (with justification) your proposed model equation for Conversion percent.
Part 2
A reduced model for viscosity was developed in Minitab
Coded Coefficients
Term | Coef | SE Coef | T-Value | P-Value | VIF |
Constant | 59.948 | 0.417 | 143.65 | 0.000 | |
Temperature | 3.583 | 0.395 | 9.07 | 0.000 | 1.00 |
Catalyst | 2.230 | 0.395 | 5.65 | 0.000 | 1.00 |
Temperature*Temperature | 0.823 | 0.381 | 2.16 | 0.046 | 1.00 |
Model Summary
S | R-sq | R-sq(adj) | R-sq(pred) |
1.45933 | 88.14% | 85.91% | 63.02% |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Model | 3 | 253.204 | 84.401 | 39.63 | 0.000 |
Linear | 2 | 243.264 | 121.632 | 57.11 | 0.000 |
Temperature | 1 | 175.352 | 175.352 | 82.34 | 0.000 |
Catalyst | 1 | 67.912 | 67.912 | 31.89 | 0.000 |
Square | 1 | 9.939 | 9.939 | 4.67 | 0.046 |
Temperature*Temperature | 1 | 9.939 | 9.939 | 4.67 | 0.046 |
Error | 16 | 34.074 | 2.130 | ||
Lack-of-Fit | 11 | 30.421 | 2.766 | 3.78 | 0.077 |
Pure Error | 5 | 3.653 | 0.731 | ||
Total | 19 | 287.278 |
Fits and Diagnostics for Unusual Observations
Obs | Viscosity (y2) | Fit | Resid | Std Resid | |
9 | 59.100 | 56.250 | 2.850 | 2.98 | R |
10 | 65.900 | 68.302 | -2.402 | -2.51 | R |
R Large residual
Image transcription text
Residual Plots for Viscosity (y2) Normal Probability Plot Versus Fits 3.0 1.5 Percent 50 Residual 10 -1.5
-3.0 4 SE 70 Residual Fitted Value Histogram Versus Order 3.0 1.5 Frequency Residual A -1.5 -3.0 -2 -1 2
2 4 10 12 14 16 18 20 Residual Observation Order
Comment on this model. What further action might you take based on the residuals?
Part 3
We are required to optimize the two responses.
Contour plots are as shown
Image transcription text
Contour Plots of Conversion (%) y1 Time*Temperature Catalyst*Temperature Hold Values Temperature
0 Time 0 Catalyst 61 0 -1- $7 . 47 -1 0 1 Catalyst* Time -1 1
Image transcription text
Contour Plot of Viscosity y2 vs Catalyst, Temperature 1.5 62.5 65.9 69.3 59.1 1.0 0.5 67.6 Catalyst 0.0
$5.7 -0.5 -1.0 04 2 -1.5 54.0 57.4 60.8 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Temperature
On the assumption that we want to maximize conversion percent while maintaining viscosity between 55 and 60 where would you propose that the inputs be set? Clearly justify your response on the basis of the contour plots.
Your manager has decided to change the viscosity requirement to be a target of 65. Interpret the response optimizer output below
Image transcription text
Optimal Temperat Time Catalyst D: 1.000 High 1.6818 1.6818 1.6318 Cur [1.0984] [1.6818] [0.0552] Low
-1.6818 -1.6818 -1.6818 Composite Desirability D: 1.000 Viscosit Targ: 65.0 y = 65.0 d = 1.0000 Conversi
Maximum y = 97.0 d = 1.0000
Applied Regression Analysis and Other Multivariable Methods
ISBN: 978-1285051086
5th edition
Authors: David G. Kleinbaum, Lawrence L. Kupper, Azhar Nizam, Eli S. Rosenberg