The yield of a chemical process is being studied. The two most important variabl
ID: 3183545 • Letter: T
Question
The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data follow:
Temperature
Pressure
200
215
230
150
90.1
90.5
89.9
90.3
90.6
90.1
160
90.5
90.8
90.4
90.7
90.9
90.1
170
90.4
90.7
90.2
90.2
90.6
90.4
a. Analyze the data and draw conclusions. Use = 0.05.
b. Prepare appropriate residual plots and comment on the model’s adequacy.
c. Under what conditions would you operate this process?
please use minitab and show graphs if possible..Thanks!
Temperature
Pressure
200
215
230
150
90.1
90.5
89.9
90.3
90.6
90.1
160
90.5
90.8
90.4
90.7
90.9
90.1
170
90.4
90.7
90.2
90.2
90.6
90.4
Explanation / Answer
————— 01/04/2017 9:52:42 PM ————————————————————
Ho(null hypothesis):- No differ significance.
vs H1:- differ significance.
MTB > Name C4 "FITS1" C5 "RESI1" C6 "SRES1".
MTB > GLM;
SUBC> Response 'response';
SUBC> Nodefault;
SUBC> Categorical 'temparature' - 'pressure';
SUBC> Terms temparature pressure;
SUBC> TMethod;
SUBC> TAnova;
SUBC> TSummary;
SUBC> TCoefficients;
SUBC> TEquation;
SUBC> TFactor;
SUBC> TDiagnostics 0;
SUBC> GFOURPACK;
SUBC> Fits 'FITS1';
SUBC> Residuals 'RESI1';
SUBC> SResiduals 'SRES1'.
General Linear Model: response versus temparature, pressure
Method
Factor coding (-1, 0, +1)
Factor Information
Factor Type Levels Values
temparature Fixed 3 150, 160, 170
pressure Fixed 3 200, 215, 230
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
temparature 2 0.30111 0.15056 8.55 0.004
pressure 2 0.76778 0.38389 21.80 0.000
Error 13 0.22889 0.01761
Lack-of-Fit 4 0.06889 0.01722 0.97 0.470
Pure Error 9 0.16000 0.01778
Total 17 1.29778
Comment :- The p value less than 0.05 so that reject null hypothesis.
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.132691 82.36% 76.94% 66.19%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 90.4111 0.0313 2890.80 0.000
temparature
150 -0.1611 0.0442 -3.64 0.003 1.33
160 0.1556 0.0442 3.52 0.004 1.33
pressure
200 -0.0444 0.0442 -1.00 0.333 1.33
215 0.2722 0.0442 6.15 0.000 1.33
Regression Equation # this is the model
response = 90.4111 - 0.1611 temparature_150 + 0.1556 temparature_160 + 0.0056 temparature_170
- 0.0444 pressure_200 + 0.2722 pressure_215 - 0.2278 pressure_230
Fits and Diagnostics for Unusual Observations
Obs response Fit Resid Std Resid
12 90.1000 90.3389 -0.2389 -2.12 R
R Large residual
Residual Plots for response
from the we can say that ,data follow normally distribution and variance is constant in the versus fit.
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