University of Texas at El Paso IE 3352 Design of Experiments Sample Exam 2 J. Sa

Question

University of Texas at El Paso IE 3352 Design of Experiments Sample Exam 2 J. Sanchez, Ph.D. Name Date 1. An industrial engineer is investigating the effect of four assembly methods (1 2 on the assembly time for a color component. Four operators are selected for television the study. To account for the operator source of variability, the engineer uses the Randomized complete Block Design shown below. Analyze the data from this experiment (a 0.05) draw appropriate conclusions. Assembly Operator Method A B C D i 5 10 7 8 7 10 12 8 3 12 18 14 14 4 10 11 11 9 a. Do the assembly methods have the same mean time? Use a 0.05. b. Could the experiment be run completely random? c. Use Fisher's LSD to compare all possible pair of means and do the grouping. d. Suppose that assembly method 3 is currently used. Use Dunnett's procedure to compare the means. Which assembly method would you recommend? e. Analyze the residual for the model in part a) d comment on the adequacy of the model. What would you recommend? Based in your answer in part (d), conduct the appropriate transformation for the f. assembly time and make conclusions. g. Conduct a nonparametric analysis for this experiment and draw conclusions.

Explanation / Answer

2. Using Minitab

One-way ANOVA: Feed rate 0.2, Feed rate 0.25, Feed rate 0.30

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
Factor 3 Feed rate 0.2, Feed rate 0.25, Feed rate 0.30

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 2 1345 672.6 5.06 0.034
Error 9 1196 132.9
Total 11 2541

Model Summary

S R-sq R-sq(adj) R-sq(pred)
11.5265 52.94% 42.48% 16.34%

Means

Factor N Mean StDev 95% CI
Feed rate 0.2 4 79.25 15.82 (66.21, 92.29)
Feed rate 0.25 4 97.00 10.00 (83.96, 110.04)
Feed rate 0.30 4 104.50 6.95 (91.46, 117.54)

Pooled StDev = 11.5265

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor N Mean Grouping
Feed rate 0.30 4 104.50 A
Feed rate 0.25 4 97.00 A B
Feed rate 0.2 4 79.25 B

Means that do not share a letter are significantly different.

c)

Kruskal-Wallis Test: Response versus Feed rate

Kruskal-Wallis Test on Response

Ave
Feed rate N Median Rank Z
0.20 4 78.00 3.3 -2.21
0.25 4 96.00 6.5 0.00
0.30 4 104.50 9.8 2.21
Overall 12 6.5

H = 6.50 DF = 2 P = 0.039
H = 6.55 DF = 2 P = 0.038 (adjusted for ties

d)

Regression Analysis: Response versus Feed rate

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Regression 1 1275.13 1275.13 10.07 0.010
Feed rate 1 1275.13 1275.13 10.07 0.010
Error 10 1265.79 126.58
Lack-of-Fit 1 70.04 70.04 0.53 0.486
Pure Error 9 1195.75 132.86
Total 11 2540.92

Model Summary

S R-sq R-sq(adj) R-sq(pred)
11.2507 50.18% 45.20% 26.05%

Coefficients

Term Coef SE Coef T-Value P-Value VIF
Constant 30.5 20.2 1.51 0.162
Feed rate 252.5 79.6 3.17 0.010 1.00

Regression Equation

Response = 30.5 + 252.5 Feed rate

Hope this will helps you. Thanks and God Bless You.

For remaining part post question again saperately.

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