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