Transcript Slide 1

Solutions
1. The tensile strength of concrete produced by
4 mixer levels is being studied with 4
replications. The data are:
Mixer 1
Mixer 2
Mixer 3
Mixer 4
Rep 1
20
30
45
25
Rep 2
20
35
45
25
Rep 3
25
25
50
20
Rep 4
15
30
40
30
Compute the MS due to mixers and plot the
means. For the greatest tensile strength,
which mixer would you choose?
Solution
1. To get the MS due to mixers, we need both the
grand mean and the mixer means. The means are:
Mixer 1 mean = 20
Mixer 3 mean = 30
Mixer 3 mean = 45
Mixer 4 mean = 25
Grand mean = 30
Mix er Means
50
Mean strength
45
40
35
30
25
20
15
1
2
3
4
Mix er
Mixer 3 provides the greatest tensile strength
4[(20- 30)2  (30- 30)2  (45- 30)2  (25- 30)2 ] 4 * 350
MS(mixers)

 466.67
4 -1
3
2. For the data in problem 1, compute
MS(error) and do the ANOVA table.
Solution
SS(error)  (20- 20)2  (20- 20)2  (25- 20)2  (15- 20)2 
(30- 30)2  (35- 30)2  (25- 30)2  (30- 30)2 
(45- 45)2  (45- 45)2  (50- 45)2  (40- 45)2 
(25- 25)2  (25- 25)2  (20- 25)2  (30- 25)2  200
SS e 200
MSe 

 16.67
dfe
12
Solution (continued)
2. The F ratio = 466.67/16.67 = 28
or = 4*350 * 12 = 28
3
200
The ANOVA table is
Source
SS
Mixers 1400
Error
200
Total
1600
df
3
12
15
MS
466.67
16.67
p
< 0.00001
3. Here are the cell and marginal means from
an experiment.
B low
B high
A marg
A low
55
15
35
A high
25
35
30
B marg
40
25
32.5
According to the ANOVA table, the A and B
main effects and the AB interaction are all
significant. If your objective is to minimize
the response, which values of A and B
would you select. Explain.
Solution
3. Since the AB interaction is significant, only
the combination means shown in the
interaction plot are interpretable.
AB Interaction
60
B = lo w
Mean Response
50
40
B =high
30
B = lo w
20
B =high
10
0
-1
0
1
Lev el of A
The minimum response is for combination
A low, B high.
4. A textile mill has a large number of looms to
weave its cloth. At random, 3 looms are
chosen to see if the looms are meeting the
standard cloth output. The data are
Loom 1
Loom 2
Loom 3
Rep 1
12
12
11
Rep 2
14
13
12
Rep 3
16
14
13
What type of ANOVA is this? Do the ANOVA
to see if there is any real variability among the
looms.
Solution
4. This is a random effects 1-way ANOVA.
To get the between-loom SS, we need the mean for each loom
and the grand mean. The means are:
Loom 1 = 14, Loom 2 = 13, Loom 3 = 12
Grand mean = 13
SS(between)=3[(14-13)2+(13-13)2+(12-13)2 ]=6
MS(between)= SS(between) / df(between) = 6/(3-1) = 3
SS(within) = (12-14)2+(14-14)2+(16-14)2 +
(12-13)2+(13-13)2+(14-13)2 +
(11-12)2+(12-12)2+(13-12)2 = 12
MS(within) = SS(within) / df(within) = 12/6 = 2
F = 3 / 2, which is not significant. The loom-to-loom variability
has not been shown to be > 0.
5. The E(MS) for the A effect is
E( MSA )   2  n 2  Kn 2
(a) What type of factor is this?
(b) How many factors are in the model?
(c) How would you set up the F-test?
Solution
5. (a) this is a random effects factor
because all of its components are
variances.
(b) There are 2 factors in the model
because there is only one interaction
term in the E(MS).
(c) The F-test is MSA / MSAB because
to test for the A effect, you must get
rid of the error and AB terms.
6. Because of limited resources, an
experiment with 4 factors at 2 levels
each is placed in an 8-run design.
(a) What kind of design is it?
(b) What is its generator?
(c) What is the design resolution?
(d) List all confounded effects.
(e) What is the defining relation of the
complementary fraction?
Solution
6. (a) This is a 24-1 fractional factorial
(b) The generator is D=ABC
(c) The design is resolution IV
(d) The confounded effects are
A + BCD AB + CD
B + ACD AC + BD
C + ABD AD + BC
D + ABC
(e) The defining relation for the
complemenary fraction is I = -ABCD.
7. Three different circuit designs are being
studied to find the least amount of noise
present. The data are
Circuit
Design
Noise present
A
2
3
1
B
5
6
7
C
4
3
5
(a) How many replications does this
experiment have?
(b) Do the ANOVA.
(c) Find the best circuit design.
Solution
7. (a) The design has 3 replications.
(b) The means are
Circuit A: 2
Circuit B: 6
Circuit C: 4
Grand Mean: 4
SS(error)= (2-2)2 + (3-2)2 + (1-2)2
+(5-6)2 + (6-6)2 + (7-6)2
+(4-4)2 + (3-4)2 + (5-4)2
=6
MS(error) = SS(error)/df(error) = 6/6 = 1
Solution (continued)
7. SS(circuits) = 3[(2-4)2 +(6-4)2 +(4-4)2]
= 24
MS(circuits)= SS(circuits)/df(circuits) = 24/2
= 12
Source
SS df MS
p
Circuits 24
2
12
<0.008
Error
6
6
1
Total
30
8
(c) There is evidence of a difference among
circuits. The best circuit design is A.
8. The effects of a 2 x 2 fixed effects factorial
design are:
A effect = 1
B effect = -9
AB effect = -29
Y = 30.5
(a) Write the fitted regression model for this
design.
(b) Plot the interaction effect.
Solution
8. The fitted regression model is
Yˆijk  30.5  0.5 X 1  4.5 X 2  14.5 X 1 X 2
To plot the interaction effect, we need the
means at all 4 corners of the design, that is
(-1,-1), (+1,-1), (-1,+1), and (+1,+1).
Substituting into the regression equation we
get
ˆ
Yijk  30.5  0.5(1)  4.5(1)  14.5(1)(1)  20
 30.5  0.5(1)  4.5(1)  14.5(1)(1)  50
 30.5  0.5(1)  4.5(1)  14.5(1)(1)  40
 30.5  0.5(1)  4.5(1)  14.5(1)(1)  12
Solution (continued)
8. The interaction plot is
AB Interaction
60
Response
50
B lo w
40
B high
30
20
B lo w
B high
10
0
-1
0
Lev el of A
1
9. A factory produces grain refiners in 3
different furnaces, each of which has its
own unique operating characteristics. Each
furnace can be run at 3 different stirring
rates.
The process engineer knows that stirring
rate affects the grain size of the product, so
he decides to run an experiment testing the
three stirring rates on his 3 furnaces.
(a) What type of design is this? Why?
(b) Set up the experiment.
Solution
9. (a) This is a randomized blocks design. Each
stirring rate is run on all three furnaces so that
furnace effects will not contaminate the
stirring rate effect.
(b) The experiment is
Furnace
Stirring rate
A
B
C
5 rpm
obs
obs
obs
10 rpm
obs
obs
obs
15 rpm
obs
obs
obs
10. The effect of 5 different ingredients
on reaction time is being studied. Each
batch of material is large enough for
only 5 runs. Moreover, only 5 runs can
be made in a day. Design the
experiment.
Solution
10. This is a Latin Square design, where A, B,
C, D, E are the five ingredients under
study.
Batches of material
Days
1
2
3
4
5
1
A
B
C
D
E
2
B
C
D
E
A
3
C
D
E
A
B
4
D
E
A
B
C
5
E
A
B
C
D
11. The yield of a chemical process is
being studied using 5 batches of raw
material and 5 acid concentrations,
which are nuisance factors.
Two factors are suspected to be
important: standing time (5 levels) and
type of catalyst (5 levels).
(a) What kind of experiment is this?
(b) Design the experiment.
Solution
11.This is a Graeco-Latin Square design.
Batches of raw material
Acid
concentration
1
2
3
4
5
1
Aα
Bγ
Cε
Dβ
Eδ
2
Bβ
Cδ
Dα
Eγ
Aε
3
Cγ
Dε
Eβ
Aδ
Bα
4
Dδ
Eα
Aγ
Bε
Cβ
5
Eε
Aβ
Bδ
Cα
Dγ
In this design, Roman letters are the five
standing times and Greek letters are the five
types of catalyst.
12. The mean results for a completely balanced
experiment are
Pressure
Temperature
Low
Medium
High
Low
7
6
8
Medium
6
5
13
High
11
10
15
With these cells means, compute orthogonal
contrasts and their SS for:
(a) High Pressure vs the average of Low and
Medium Pressure.
(b) Low Pressure vs Medium Pressure
(c) Medium Pressure vs High Pressure
Solution
12. To use orthogonal contrasts, we need Main
effect means. These means for pressure are
Low Level:
(7+6+11)/3 = 8
Medium Level: (6+5+10)/3 = 7
High Level:
(8+13+15)/3 =12
(a) The appropriate orthogonal contrast is
n[(-1)(Low) +(-1)(Med) +(2)High]=
n[(-1)(8) +(-1)(7) +(2)(12)] = 9n
Its SS = (nContrast)  (n9)  81n
n(6)
6
n c
2
J
j 1
2
j
2
Solution (continued)
12.(b) The contrast is
n[(1)(Low) +(-1)(Med) +(0)High]=
n[(1)( 8 ) + (-1)(7) +(0)(12)]= 1n
Its SS =
(nContrast)2
J
n c 2j
(n1)2 n


n(2) 2
j 1
(c) This contrast is unavailable, given the
other two, because Pressure has 3 levels
and thus only 2 df. Each contrast uses up
1 df so after two contrasts, no more are
available using the orthogonal contrast
method.
13. A semiconductor engineer is studying the effect of
lamination temperature (55˚C and 75˚C), lamination
time (10 seconds and 25 seconds), lamination
pressure (5 tn and 10 tn), and firing temperature
(1580˚C and 1620˚C) on the curvature of the
substrates produced. He must finish his study in one
day so he can do only 8 runs. He is not very good at
experimental design and doesn’t know how to do it.
You are his statistical consultant and he asks you to
design an experiment for him. Design the experiment
and explain what problems he will have after he gets
his results.
Solution
13. The statistical consultant suggests a
24-1 fractional factorial design.
Lamination
Lamination
Lamination
Firing
Temperature
Time
Pressure
Temperature
(tn)
(˚C)
( ˚C )
(seconds)
55
10
5
1580
75
10
5
1620
55
25
5
1620
75
25
5
1580
55
10
10
1620
75
10
10
1580
55
25
10
1580
75
25
10
1620
Solution (continued)
13. The statistician explains that when the engineer gets
his results, he won’t be sure of exactly what they
really are because the experiment is confounded by
the following aliases:
A+BCD C+ABD AB+CD AD+BC
B+ACD D+ABC AC+BD
But the statistical consultant says that some of the
interactions might make no sense in terms of the
process and the engineer might be able to rule them
out, so he recommends that he do the design.
14.What do you mean by a completely
randomized design?
Solution
14. A completely randomized design is
one in which all sources of bias are
removed by (1) randomly assigning
the experimental units to the various
treatment levels and (2) doing the
experimental runs in random order.
15. What is the difference between a
fixed effects model, a random effects
model, and a mixed model?
Solution
15. A fixed effects model is one in which
we are testing the effects of
treatments in terms of their means,
and the conclusions apply only to
those treatments under test.
Solution (continued)
15. A random effects model is one in which all
factors are random factors, that is, there are
many levels of each factor, but only J of them
are randomly selected for use in the
experiment. In a random effects model, we
are interested in estimating the variance
components, that is the portion of the total
variance due to each of the factors.
A mixed model is one is which at least one of
the factors is random.
16. What is the purpose of blocking in a
design?
Solution
16. Blocking is used to remove the
effects of a nuisance factor, which is
known and controllable, by having all
levels of all other factors tested in each
block. Removing the effect of a
nuisance factor reduces the residual
error and makes the design more
powerful.
17. What are two ways of dealing with
nuisance variables in the design we
have studied? Why do they work?
Solution
17. We can deal with nuisance variables
by blocking, by a Latin Square, or by a
Graeco-Latin Square. They all work
because all levels of the factors of
interest are tested at each
combination of levels of the nuisance
factors.
18. What is the logic behind
decomposition of the total SS in
ANOVA?
Solution
18. The logic is that the total SS includes
variability due to the factor(s) and
variability due to error. We can
separate these and look at the effects
of the factors under test.
19.
Why is a single-replicate design
undesirable? How does Cuthbert
Daniel’s idea help in this situation?
Solution
19. A single replicate design does not
allow for a pure error term to test
effects of factors. Instead we must
use higher-order interactions as error
under the assumption that they are not
significant.
Solution (continued)
19. Daniel’s approach allows us to look
at all effects by plotting them on normal
probability paper. If the effects fall on a
line, they behave like error and are not
significant. This gives us a way of
looking at all effects and choosing
those that fall off the line to include in
the model. All effects along the line are
used as residual.
20. How do you check model adequacy?
Solution
20. Checking model adequacy involves
residual plots. Normality is checked
by plotting residuals on normal
probability paper and making sure they
all fall along a line. Constant error
variance is checked by plotting
residuals against fitted values and
making sure they all have about the
same spread with no outliers.