chapter 4 solutions doe
TRANSCRIPT
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4.3
Analysis of the data:
Factor Type Levels ValuesChemical random 4 1; 2; 3; 4
Bolt random 5 1; 2; 3; 4; 5
Analysis of Variance for Tensile Strengths, using Adjusted SS for
Tests
Source DF Seq SS Adj SS Adj MS F P
Chemical 3 12.950 12.950 4.317 2.38 0.121
Bolt 4 157.000 157.000 39.250 21.61 0.000
Error 12 21.800 21.800 1.817
Total 19 191.750Minitab output for problem 4.3
Conclusion:
The null hypothesis will be that all the treatment means
are equal. The alternative hypothesis is that they are not
all equal. From the Minitab output above, we can conclude
that there is no difference between the chemical types
because F= 3.49, which is greater that Fo= 2.38. Thus, do
not reject the null hypothesis. Also, this conclusion can
be obtained from the p-value for treatment effect
(Chemical), which is 0.1211 > 0.05. The p-value for the
block effect (Bolt), however, is very small (= 0.000), so
the block effect dominates the variations in the
experiment.
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4.8
a)
Analysis of the data:
Factor Type Levels ValuesDesign random 3 1; 2; 3
Region random 4 NE; NW; SE; SW
Analysis of Variance for Response, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Design 2 90755 90755 45378 50.15 0.000
Region 3 49036 49036 16345 18.06 0.002
Error 6 5429 5429 905
Total 11 145220
ANOVA Table from Minitab for Problem 4.8
From the Minitab output above, the designs do differ from
each other significantly. However, at this point we cannot
decide which design is best for our study.
b)
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev -+---------+---------+---------+--------
1 4 298.50 75.67 (--------*--------)
2 4 473.75 93.75 (-------*--------)
3 4 281.25 60.33 (--------*--------)
-+---------+---------+---------+--------
200 300 400 500
Pooled StDev = 77.79
Grouping Information Using Fisher Method
Design N Mean Grouping
2 4 473.75 A
1 4 298.50 B
3 4 281.25 B
Means that do not share a letter are significantly different.
Minitab output for Fisher LSD method for comparison
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From the Fisher LSD results we can conclude that Design 2
is significantly different in mean than both Design 1 and
3. Hence, Design 2 is the most effective design.
c)
50250-25-50
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Residual
Percent
500400300200
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-20
-40
Fitted Value
Residual
40200-20-40
6.0
4.5
3.0
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0.0
Residual
Frequency
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-20
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Observation Order
Residual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Response
Analysis of the residuals for Problem 4.8
There is violation of the normality assumption. Thus, a
transformation of the data is recommended for further
analysis. Therefore, all conclusions obtained in the
previous analysis are not valid now.
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4.13
a)
Analysis of the voltage data:
Factor Type Levels ValuesAlgorithms random 4 1; 2; 3; 4
Time Period random 6 1; 2; 3; 4; 5; 6
Analysis of Variance for Voltage, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Algorithms 3 0.002746 0.002746 0.000915 0.19 0.901
Time Period 5 0.017437 0.017437 0.003487 0.72 0.615
Error 15 0.072179 0.072179 0.004812
Total 23 0.092363
The ratio control algorithm does not affect the average
cell voltage at 5% level.
b) and c)
0.020.010.00-0.01-0.02
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Residual
Percent
0.150.100.050.00
0.02
0.01
0.00
-0.01
-0.02
Fitted Value
Residual
0.015
0.010
0.005
0.000
-0
.005
-0
.010
-0
.015
-0
.020
6.0
4.5
3.0
1.5
0.0
Residual
Frequency
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0.02
0.01
0.00
-0.01
-0.02
Observation Order
Residu
al
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Pot Noise
4.03.53.02.52.01.51.0
0.02
0.01
0.00
-0.01
-0.02
Algorithms
Residual
Residuals Versus Algorithms
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There are some concerns about the normality as well as the
variance of the residuals. It seems that bigger values of
algorithms give wider spread of residuals, while it should
be structure-less pattern. Hence, a log transformation can
be applied.
0.500.250.00-0.25-0.50
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Residual
Percent
-2.0-2.5-3.0-3.5-4.0
0.2
0.0
-0.2
-0.4
Fitted Value
Residual
0.30.20.10.0-0.1-0.2-0.3-0.4
6.0
4.5
3.0
1.5
0.0
Residual
Frequency
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0.2
0.0
-0.2
-0.4
Observation Order
Residual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for ln(Pot Noise)
Analysis of the residuals after applying natural log transformation for Pot Noise
The variance of the residuals now looks better but the
normality plot shows small deviation!
Factor Type Levels Values
Algorithms random 4 1; 2; 3; 4
Time Period random 6 1; 2; 3; 4; 5; 6
Analysis of Variance for ln(Pot Noise), using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F PAlgorithms 3 6.16605 6.16605 2.05535 33.26 0.000
Time Period 5 0.94460 0.94460 0.18892 3.06 0.042
Error 15 0.92704 0.92704 0.06180
Total 23 8.03769
From the ANOVA table above, we can conclude that the ratio
control algorithm affects the pot noise.
4.03.53.02.52.01.51.0
0.3
0.2
0.1
0.0
-0.1
-0.2
-0.3
-0.4
Algorithms
Residual
Residuals Versus Algorithms
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d)
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev -----+---------+---------+---------+----
1 6 -3.0877 0.2422 (----*----)
2 6 -3.5086 0.3667 (----*----)3 6 -2.1998 0.2337 (----*----)
4 6 -3.3559 0.3558 (----*----)
-----+---------+---------+---------+----
-3.50 -3.00 -2.50 -2.00
Pooled StDev = 0.3059
Grouping Information Using Fisher Method
Algorithms N Mean Grouping
3 6 -2.1998 A
1 6 -3.0877 B
4 6 -3.3559 B C
2 6 -3.5086 C
Algorithms verses pot noise analysis
The ratio control algorithm 2 or 4 (no significant
difference between 2 and 4 according to Fisher LCD) will be
selected to minimize the pot noise, since algorithms have
no effect on average cell voltage.
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4.21
Factor Type Levels Values
Distance random 4 4; 6; 8; 10
Subject random 5 1; 2; 3; 4; 5
Analysis of Variance for Focus Time, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Distance 3 32.950 32.950 10.983 8.61 0.003
Subject 4 36.300 36.300 9.075 7.12 0.004
Error 12 15.300 15.300 1.275
Total 19 84.550
ANOVA Table for Problem 4.21
210-1-2
99
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Residual
Percent
108642
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0
-1
-2
Fitted Value
Residual
1.51.00.50.0-0.5-1.0-1.5-2.0
6.0
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3.0
1.5
0.0
Residual
Frequency
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-1
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Observation Order
Residual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Focus Time
Analysis of the residuals for problem 4.21
There seems to be an outlier observation, which is
observation 14. Maybe it is better to rerun this
observation. However, from the ANOVA table obtained, there
exists a significant difference in the distance. Blocking
is also significant.
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4.22
Analysis of Variance for Reaction Time, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Batch 4 15.440 15.440 3.860 1.23 0.348Day 4 12.240 12.240 3.060 0.98 0.455
Ingredients 4 141.440 141.440 35.360 11.31 0.000
Error 12 37.520 37.520 3.127
Total 24 206.640
ANOVA table for problem 4.22
In this experiment, batch and days are the block factors.
The experimenter is interested in the effect of the
Ingredients on the reaction time. Thus, an appropriate
hypothesis would be:
Ho: There is no significant difference in the Ingredients.
H1: There is significant difference in the Ingredients.
Since the p-value of the ingredients 0.000 is less than
0.05, we can conclude that there is significant difference
between the Ingredients. However, there is no strong
evidence of a difference on batches or days. It seems that
in this experiment we were unnecessarily concerned about
this source of variability.
Analysis of the residuals below shows no evidence of any
violation of any assumption in this experiment.
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3.01.50.0-1.5-3.0
99
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Residual
Percent
108642
2
1
0
-1
-2
Fitted Value
Residu
al
210-1-2
6.0
4.5
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1.5
0.0
Residual
Fr
equency
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-1
-2
Observation Order
R
esidual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Reaction Time
54321
3
2
1
0
-1
-2
Day
Residual
Residuals Versus Day
(response is Reaction Time)
54321
3
2
1
0
-1
-2
Batch
Residual
Residuals Versus Batch
(response is Reaction Time)
Residuals analysis for problem 4.22
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Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ----+---------+---------+---------+-----
A 5 8.400 1.140 (------*-----)
B 5 5.600 2.074 (-----*------)
C 5 8.800 1.643 (------*------)D 5 3.400 2.074 (------*-----)
E 5 3.200 1.924 (------*------)
----+---------+---------+---------+-----
2.5 5.0 7.5 10.0
Pooled StDev = 1.806
Grouping Information Using Fisher Method
Ingredients N Mean Grouping
C 5 8.800 A
A 5 8.400 A
B 5 5.600 B
D 5 3.400 B CE 5 3.200 C
Means that do not share a letter are significantly different.
Comparing means of Ingredients for problem 4.22
From the above comparison figure, we may conclude that if
the objective of this study is to minimize the reaction
time then either ingredient D or E should be selected.
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4.42
Factor Type Levels Values
Concentration fixed 7 2; 4; 6; 8; 10; 12; 14
Days random 7 1; 2; 3; 4; 5; 6; 7
Analysis of Variance for Strength, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Concentration 6 2037.62 1317.43 219.57 10.42 0.002
Days 6 394.10 394.10 65.68 3.12 0.070
Error 8 168.57 168.57 21.07
Total 20 2600.29
ANOVA Table for problem 4.42
This is a BIBD experiment. The objective is to test if
there is any difference among concentrations.
H0: There is no significant difference in concentrations.
H1: There is significant difference in concentrations.
From the ANOVA table we can conclude we can reject Hoand
say there is a significant difference in concentrations.
There is no indication of any violation of assumptions from
the residual graphs on the next page.
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840-4-8
99
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Residual
Percent
150140130120110
5.0
2.5
0.0
-2.5
-5.0
Fitted Value
Residual
6420-2-4-6
8
6
4
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Residual
Frequency
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5.0
2.5
0.0
-2.5
-5.0
Observation Order
Residua
l
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Strength
1412108642
7.5
5.0
2.5
0.0
-2.5
-5.0
Concentration
Resid
ual
Residuals Versus Concentration
(response is Strength)
7654321
7.5
5.0
2.5
0.0
-2.5
-5.0
Days
Resid
ual
Residuals Versus Days
(response is Strength)
Residuals analysis for problem 4.42
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Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ---------+---------+---------+---------+2 3 117.00 3.00 (------*-----)
4 3 121.67 3.79 (-----*------)
6 3 129.33 10.79 (------*-----)
8 3 139.67 10.07 (-----*------)
10 3 146.00 3.61 (------*-----)
12 3 120.33 2.52 (-----*------)
14 3 131.00 4.58 (-----*------)
---------+---------+---------+---------+
120 132 144 156
Pooled StDev = 6.34
Grouping Information Using Fisher Method
Concentration N Mean Grouping
10 3 146.000 A
8 3 139.667 A B
14 3 131.000 B C
6 3 129.333 B C
4 3 121.667 C D
12 3 120.333 C D
2 3 117.000 D
Means that do not share a letter are significantly different.
Mean analysis for concentrations.
From the mean comparison graph, we can conclude that if the
objective is to have a maximum strength for the paper
product, then either 10% or 8% hardwood concentration must
be considered.