IET 421/621 (Spring 2016)
Homework 7
Total Points: 10
Due date: 04/28/2016, 10:00 PM EST
Shane Holbrook
1.
(5 points) The percentage of hardwood concentration in raw pulp, the vat pressure, and
the cooking time of the pulp are being investigated for their effects on the strength of paper. Three
levels of hardwood concentration, three levels of pressure, and two cooking times are selected. A
factorial experiment with two replicates is conducted, and the following data are obtained:
Percentage of
Hardwood
Concentration
Factor
time
pressure
%
Type
fixed
fixed
fixed
Cooking Time 3.0 Hours
Cooking Time 4.0 Hours
Pressure
Pressure
400
500
650
400
500
650
196.6
196
197.7
196
199.8
199.4
198.4
198.6
199.6
200.4
200.6
200.9
198.5
197.2
196
196.9
198.4
197.6
197.5
198.1
198.7
198
199.6
199
197.5
196.6
195.6
196.2
197.4
198.1
197.6
198.4
197
197.8
198.5
199.8
Levels
2
3
3
Values
3, 4
400, 500, 650
2, 4, 8
Analysis of Variance for Strength
Source
time
pressure
%
time*%
pressure*%
time*pressure
time*pressure*%
Error
Total
S = 0.604612
DF
1
2
2
2
4
2
4
18
35
SS
20.2500
19.3739
7.7639
2.0817
6.0911
2.1950
1.9733
6.5800
66.3089
R-Sq = 90.08%
MS
20.2500
9.6869
3.8819
1.0408
1.5228
1.0975
0.4933
0.3656
F
55.40
26.50
10.62
2.85
4.17
3.00
1.35
P
0.000
0.000
0.001
0.084
0.015
0.075
0.290
R-Sq(adj) = 80.70%
(a) Analyze the data and draw conclusions. Use = 0.05. Judging by the data above, we can clearly see
that the p-value is below the significance level of 0.05. This pertains to all of the main effects of:
concentration, time, and pressure. It can be seen that the p-value however for time vs pressure vs
concentration is high therefore not significant. As well as Time vs concentration. The rest are significant.
(b) Prepare appropriate residual plots and comment on the models adequacy.
Residual Plots for Strength
Versus Fits
1.0
90
0.5
Residual
Percent
Normal Probability Plot
99
50
10
1
-1.0
-0.5
0.0
0.5
0.0
-0.5
-1.0
1.0
196
198
Residual
Histogram
Versus Order
1.0
0.5
Residual
Frequency
12
6
3
0
200
Fitted Value
-0.8
-0.4
0.0
Residual
0.4
0.8
0.0
-0.5
-1.0
10
15
20
25
30
35
Observation Order
Above we can see that everything is normal and nothing is out of place.
Residuals Versus time
(response is Strength)
1.0
Residual
0.5
0.0
-0.5
-1.0
3.0
3.2
3.4
3.6
3.8
4.0
time
Looking at the residual versus time, we can see that everything is symmetric, so the test is fine.
Residuals Versus pressure
(response is Strength)
1.0
Residual
0.5
0.0
-0.5
-1.0
400
450
500
550
600
650
pressure
4
The same can be seen with the pressure residuals.
Residuals Versus %
(response is Strength)
1.0
Residual
0.5
0.0
-0.5
-1.0
2
And finally we have a normal pattern with the concentration %.
(c) Under what set of conditions would you run the process? Why?
To do this we need to analyze some more graphs:
Interaction Plot for Strength
Data Means
200
pressure
400
500
650
M ean
199
198
197
196
3
time
Here we can see the interaction plot of cooking time versus pressure
Interaction Plot for Strength
Data Means
200.0
%
2
4
8
199.5
M ean
199.0
198.5
198.0
197.5
197.0
3
time
6
Here is the interaction plot for cooking time versus concentration
Judging from the above mentioned graphs and data, I would say that you should use the 2%
concentration, the pressure around 650, and the time at 4 hours. The reason is clearly viewed as
stronger by any other means of the tests shown above.
2.
(5 points) A study investigated the effects of cyclic loading and environmental conditions
on fatigue crack growth at a constant 22 MPa stress for a particular material. The data from this
experiment are shown below (the response is crack growth rate).
Frequency
Air
2.29
2.47
2.48
2.12
10
0.1
Environment
H2O
2.06
2.05
2.23
2.03
Salt H2O
1.90
1.93
1.75
2.06
2.65
2.68
2.06
2.38
3.20
3.18
3.96
3.64
3.10
3.24
3.98
3.24
2.24
2.71
2.81
2.08
11.00
11.00
9.06
11.30
9.96
10.01
9.36
10.40
ANOVA: Growth Rate versus Freq, Environment
Factor
Freq
Environment
Type
fixed
fixed
Levels
3
3
Values
0.1, 1.0, 10.0
1, 2, 3
Analysis of Variance for Growth Rate
Source
Freq
Environment
DF
2
2
SS
209.893
64.252
MS
104.946
32.126
F
522.40
159.92
P
0.000
0.000
Freq*Environment
Error
Total
S = 0.448211
4
27
35
101.966
5.424
381.535
R-Sq = 98.58%
25.491
0.201
126.89
0.000
R-Sq(adj) = 98.16%
(a) Analyze the data from this experiment (use = 0.05).
As we can see from the results above, the p-value is less than that of the level of significance, so
this shows there is a significant difference in the effects.
(b) Analyze the residuals.
Residual Plots for Growth Rate
Normal Probability Plot
Versus Fits
99
0.5
Residual
Percent
90
50
10
-0.5
-1.0
-1.5
-1
Fitted Value
Histogram
Versus Order
0.5
0.0
4
2
0
Residual
Residual
Frequency
0.0
10
-0.5
-1.0
-1.5
-1.5
-1.0
-0.5
Residual
0.0
0.5
10
15
20
25
30
35
Observation Order
Residuals Versus Environment
(response is Growth Rate)
1.0
Residual
0.5
0.0
-0.5
-1.0
-1.5
1.0
1.5
2.0
2.5
3.0
Environment
Residuals Versus Freq
(response is Growth Rate)
1.0
Residual
0.5
0.0
-0.5
-1.0
-1.5
0
10
Freq
Looking at the graphs above, we can see that all of the patterns are normal so there are no errors.
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(c) Repeat the analyses from parts (a) and (b) using ln(y) as the response. Comment on the results.
(d)
(e)
ANOVA: (y1) versus Freq, Environment
(f) Factor
Type
Levels Values
(g) Freq
fixed
3
0.1, 1.0, 10.0
(h) Environment fixed
3 1, 2, 3
(i)
(j)
(k) Analysis of Variance for (y1)
(l)
(m) Source
DF
SS
MS
F
P
(n) Freq
2
7.5702 3.7851 404.09 0.000
(o) Environment
2
2.3576 1.1788 125.85 0.000
(p) Freq*Environment
4
3.5284 0.8821
94.17 0.000
(q) Error
27
0.2529 0.0094
(r) Total
35 13.7092
(s)
(t)
(u) S = 0.0967827
R-Sq = 98.16%
R-Sq(adj) = 97.61%
Residual Plots for (y1)
Versus Fits
0.2
90
0.1
Residual
Percent
Normal Probability Plot
99
50
10
1
-0.2
-0.1
0.0
0.1
0.0
-0.1
-0.2
0.2
0.5
1.0
1.5
Residual
Histogram
Versus Order
0.1
Residual
Frequency
2.5
0.2
4
2
0
2.0
Fitted Value
-0.15 -0.10 -0.05
0.00
0.05
Residual
0.10
0.15
0.0
-0.1
-0.2
10
15
20
25
30
35
Observation Order
10
Here we can clearly see by the data above, specifically the p-value that all the effects are
significant, due in part because they are well below the significance level of 0.05. When looking
at the graphed residual plots, we can see that everything is well in pattern and normally linear.
Although there is a slight variation, there is nothing really to suggest that the test is out of
balance.
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