99
3.96875
90
80
70
Res iduals
Norm al % probability
95
STATISTICS 402B
50
Spring 2016
0.96875
Homework Set#5
30
20
2
10
Solutions from Montgomery, D. C. (2012) Design-2.03125
and Analysis of Experiments, Wiley, NY
5
1. Problem 5.21 (Montgomery)
1
5.21. The yield of a chemical process is being studied. -5.03125
The two factors of interest are temperature and
pressure. Three levels of each factor are selected; however, only 9 runs can be made in one day. The
-5.03125
0.96875
3.96875
6.96875
71.91
97.47
experimenter
runs -2.03125
a complete
replicate
of the
design on each day.
The78.30
data are84.69
shown91.08
in the following
table.
Res idual
Predicted
Day 1
Day 2
Pressure
(e) Two three-factor interactions, ABC and
ABD, apparently have largePressure
effects. Draw a cube plot in the
Temperature
250
260
250 Repeat
260using the
270
factors A, B, and C with the average yields shown at270
each corner.
factors A, B, and D.
Low
84.0 Where
85.8would
86.1
85.2
87.3the process be run
Do these two plots
aid in data86.3
interpretation?
you recommend
that
88.5
87.3
89.0
89.4
89.9
90.3
with respect to Medium
the four variables?
High
89.1
90.2
91.3
91.7
93.2
93.7
B: B
B: B
Cube Graph
Cube Graph
Design Expert Output
yield
(a)
This
experiment
is
a
2-way
factorial
in
a
RCBD
(as
discussed
in
class
andyield
in Section 5.6 of the text)
Response: Yield
where
day
to
day
variation
is
to
be
controlled
by
blocking.
Identify
the
two
factors
under
study (factors
ANOVA for Selected Factorial
Model
86.53
76.34
86.00
83.50
Analysis of
variance
table [Partial
squares] factor in this experiment. How many treatment combinations are
that
are crossed)
andsum
the ofblocking
Sum of
Mean
F
there?
Source
Squares
DF
Square
Value
Prob > F
(b) Write
linear model for1 this experiment
Block down the13.01
13.01as in Example 5.6.
Model
109.81 84.22
8
13.73
25.84
< 0.0001
77.06
84.03
(c) B+
Use
JMP to analyze
the above data
(needed to
answerB+the 84.56
questions
that follow)
and attachsignificant
the JMP
A
5.51
2
2.75
5.18
0.0360
output.
B
99.85
2
49.93
93.98
< 0.0001
AB
4.45
4
1.11
2.10
(d) Construct a complete ANOVA table for the analysis of this data (as in Table0.1733
5.22). Extract numbers
Residual
4.25
8
0.53
from the JMP output for filling this out,
including
p-values.
C+
D+
Cor Total
127.07
17
85.41
77.47
94.75
74.75
(e) Test the hypothesis of no interaction using the above ANOVA table (use α = .05). State the F-statistic
The Modelwith
F-value
25.84 implies
the model
significant.
There is only
theofassociated
degrees
of isfreedom,
C: C the p-value, and your decision. How would you proceed
D: D based
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
on your decision?
C- model
B- intervals
D74.97
77.69 underscoring
93.28
83.94
Values
"Prob
> F" less
than 0.0500
indicate
termsusing
are significant.
(f)ofBCompare
Pressure
means
(or
effects)
confidence
or the LSD
method.
AA+
AA+
In this case(Extract
A,
B are significant
model
terms.
A: A from the JMP output to answer this question).
A: A
results
(g) Compare Temperature means (or effects) using confidence intervals or the LSD underscoring method.
(Extract results from the JMP output to answer this question).
Both main effects, temperature and pressure, are significant.
a concluding
Run(h)
the Write
process
at A low B statement
low, C lowsummarizing
and D high. the results from this experiment
5.22.
Consider
the data in Problem 5.7. Analyze the data, assuming that replicates are blocks.
2. Problem
6.8 (Montgomery)
Design
Output
6.8. Expert
A bacteriologist
is interested in the effects of two different culture media and two different times on
Response:
Warping
the growth of a particular
virus. She performs six replicates of a 22 design, making the runs in random
ANOVA for Selected Factorial Model
order.
Analysis of variance table [Partial sum of squares]
Mean
F
Data table on the Sum
nextofpage.
Source
Squares
DF
Square
Value
Prob > F
Block
11.28
1
11.28
(a) Calculate (by hand computation) the main effects of Time and Culture Medium, and the interaction
Model
968.22
15
64.55
9.96
< 0.0001 significant
effect
by Culture Medium.
A of Time698.34
3
232.78
35.92
< 0.0001
B
156.09
3
52.03
8.03
0.0020
AB
113.78
9
12.64
1.95
0.1214
Residual
97.22
15
6.48
1
Cor Total
1076.72
31
The Model F-value of 9.96 implies the model is significant. There
is only
6-12
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Culture Medium
Time
12 hr
18 hr
1
2
21
22
25
26
23
28
24
25
20
26
29
27
37
39
31
34
38
38
29
33
35
36
30
35
Expert Output
(b)Design
Construct
a graphic as in Fig. 6.1 of the text (reproduced on p. 12 of the notes set #9) for this data.
Response:
Virus growth
ANOVA
for Selected Factorial
(c) Draw
an interaction
plot forModel
the Time × Culture Medium interaction with Time on the x-axis, showing
Analysis of variance table [Partial sum of squares]
all
data points
(you
may
use
Graph
Builder for
this).
Sum of
Mean
F
Source that the
Squares
DFof squares
Squareis 793.63,
Value
Prob
F
(d) Given
total sum
construct
a >complete
anova table for the analysis of this
Model
691.46
3
230.49
45.12
< 0.0001
significant
experiment(p-values
not
needed).
A
9.38
1
9.38
1.84
0.1906
B
590.04
1
590.04
115.51
< 0.0001
AB
92.04
0.0004
3. A 23 factorial
experiment
with1 factors 92.04
A, B, and C18.02
was carried
out in a completely randomized design with
Residual
102.17
20
5.11
2 replications
per
treatment
combination
and
the
response
variable
yield was measured . The following
Lack of Fit
0.000
0
resultsPure
have
Errorbeen obtained:
102.17
20
5.11
Cor Total
793.63
23
Treatment Yield Total
(1) There is 5only 3
The Model F-value of 45.12 implies the model is significant.
a 0.01% chance that a "Model F-Value" this large could occur
a due to noise.
6 5
b are significant.
8 7
Values of "Prob > F" less than 0.0500 indicate model terms
In this case B, AB are significant model terms.
ab
10 12
c
6 9
ac
5 4
Normal plot of residuals
Residuals vs. Predicted
bc
12
13
4.66667
abc
9 14
99
Res iduals
Norm al % probability
(a) Run a95
JMP analysis of the data (available in 2.66667
homework5-3.jmp) and extract values to fill out the
90
following
table:
80
70
Source of Variation d.f. SS
2
50
Treatment 0.666667
30
Error
20
2
10
Total
-1.33333
5
(b) Calculate
(by hand) estimates of all effects using contrasts of observed treatment totals.
1
(c) Use the values you calculated above to breakdown
-3.33333 the treatment SS in the above anova table into
single degree of freedom corresponding to each of the effects in part (b).
-1.33333
0.666667
2.66667
4.66667
-3.33333
23.33
26.79
30.25
33.71
37.17
(d) Perform F-tests by completing the anova table to include columns for MS, F, and p-value. (You may
idual output from JMP.) What are the effects
Predicted
check these with the Res
values
that are significant at α = .05?
(e) Interpret each significant interaction effect using a suitable 2×2 table of means and a figure representing
simplerate
effects.
Growth
is affected by factor B (Time) and the AB interaction (Culture medium and Time). There is
some
very
slight
indication
of inequality
of variance
shown
by the
smallappropriate
decreasing funnel
shape in intervals.
the plot
(f) The significance
of these
effects can
also be
tested
using
confidence
Use the
of residuals versus predicted.
JMP output to extract the relevant 95% confidence intervals.
Note: Need to present written answers to each part, with calculation shown for parts that you are required to
do hand computation. Use the JMP output to obtain numbers for answering other parts. Attach edited JMP
output when you use the JMP output to extract numbers as part of the analysis. The JMP data files 5-21.jmp
and homework5-3.jmp are available to download.
Due Wednesday, March 30th, 2016 (turn-in at the beginning of class)
6-13
2