Chapter 4 solutions DOE

8/9/2019 Chapter 4 solutions DOE

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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.


2/13

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


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


3/13

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)

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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.


4/13

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


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)

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Residual Plots for Pot Noise

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Algorithms

Residual

Residuals Versus Algorithms


5/13

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.

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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

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d)


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



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.


7/13

4.21


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


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

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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.


8/13

4.22

Analysis of Variance for Reaction Time, using Adjusted SS for Tests


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.


9/13

3.01.50.0-1.5-3.0

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Residual Plots for Reaction Time

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Residuals Versus Day

(response is Reaction Time)

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Residuals Versus Batch

(response is Reaction Time)

Residuals analysis for problem 4.22


10/13


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



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


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.


11/13

4.42


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


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.


12/13

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Residuals analysis for problem 4.42


13/13


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


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


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.

Chapter 4 solutions DOE

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