A 23, two-replicate, full factorial design was conducted to investigate the effe
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Question
A 23, two-replicate, full factorial design was conducted to investigate the effects of three manufacturing factors on the spring constant of leaf springs. The design factors were Xi furnace temperature (1840F, 1880F), X2 heating time (23 sec, 25 sec), and X3 transfer time (10 sec, 12 sec). The response is the measured spring constant k. The data are shown in the following table. Create the standard order factorial design in Minitab. Analyze the factorial design and, following the principle of hierarchical order, remove all interaction terms then main effects that are not statistically significant at ?-0.05 Analyze the reduced factorial design and output a normal probability plot of the residuals, residuals versus fitted values plot, and plots of the residuals versus each factor. State the final model and comment on its adequacy based on its R-square values and residuals analyses Test ConditioinXi Tem X: Heat Time | X3 Trans Time 65.3 81.3 53.3 69.9 61.8 77.4 73.9 89.9 63.2 83.0 55.4 68.1 59.7 79.6 74.5 88.5 4 13 15Explanation / Answer
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Full Factorial Design
Factors: 3 Base Design: 3, 8
Runs: 16 Replicates: 2
Blocks: 1 Center pts (total): 0
All terms are free from aliasing.
Factorial Regression: response versus A, B, C
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Model 3 1336.93 445.64 9.23 0.002
Linear 3 1336.93 445.64 9.23 0.002
A 1 1066.02 1066.02 22.08 0.001
B 1 0.30 0.30 0.01 0.938
C 1 270.60 270.60 5.61 0.036
Error 12 579.29 48.27
Lack-of-Fit 4 566.03 141.51 85.37 0.000
Pure Error 8 13.26 1.66
Total 15 1916.22
Model Summary
S R-sq R-sq(adj) R-sq(pred)
6.94798 69.77% 62.21% 46.26%
Coded Coefficients
Term Effect Coef SE Coef T-Value P-Value VIF
Constant 71.55 1.74 41.19 0.000
A 16.33 8.16 1.74 4.70 0.001 1.00
B 0.28 0.14 1.74 0.08 0.938 1.00
C 8.23 4.11 1.74 2.37 0.036 1.00
Regression Equation in Uncoded Units
response = 71.55 + 8.16 A + 0.14 B + 4.11 C
Alias Structure
Factor Name
A A
B B
C C
Aliases
I
A
B
C
Result: Factor B (heating time) is significant factor which has 0.938 p-value is greater than 0.05 level of significance. Here R squared value is 69.77%. Hence 69.77% variation in data.
Effects Plot for response
Interpretation: this graph shows factor A(furnace temperature) and factor C(transfer time) are not significant and factor B(heating time)is significant.
Residual Plots for response
Interpretation: In Normal probability plot, the observations are not along the straight line. Hence data is not normal. In residual vs fitted value plot the points are not spread well along the line. Hence variance is not constant and in histogram, the graph is not symmetrical.
Reduced Model: response (measured spring constant) = 0.14 B (heating time)
Analysis of reduced model:
One-way ANOVA: response versus B
Method
Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level ? = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
B 2 -1, 1
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
B 1 0.30 0.303 0.00 0.963
Error 14 1915.92 136.851
Total 15 1916.22
Model Summary
S R-sq R-sq(adj) R-sq(pred)
11.6983 0.02% 0.00% 0.00%
Means
B N Mean StDev 95% CI
-1 8 71.41 9.78 (62.54, 80.28)
1 8 71.69 13.34 (62.82, 80.56)
Pooled StDev = 11.6983
Result:P-value for factor B is 0.938 which is greater than 0.05 and R squared value is almost zero hence there is no variation in data.
Residual Plots for response
Interpretation: In normal probability plot the observations are along the straight line. Hence data is normal. In residual vs fitted value plot points are not that much spread hence variance is not constant. Histogram is not symmetrical.
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