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In an experimental program to investigate the properties of a thin film plastic coating on ceramic-based resistors, six factors were identified. X1 - Supplier of a basic component of the plastic...

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In an experimental program to investigate the properties of a thin film plastic coating on ceramic-based resistors, six factors were identified.
X1 - Supplier of a basic component of the plastic (there were 2) X2 - Viscosity of the coating at time of application X3 - Thickness of coating X4 - Temperature of the first bake (dry) cycle X5 - Temperature of the second bake (cure) cycle X6 - Speed of the conveyer
(a) List the experiments and a random list of run orders for a full factorial design with these six factors using symbolic - and + signs for the levels. Show the complete confounding table.
(b) List the experiments and random run orders if we are to run a ih fractional design .
(c) Describe how you would analyse the data in each of the situations defined in (a) and (b) above.
Question 2
You are trying to help an ice cream vendor to be able to predict how many ice cream cones he will sell in a given afternoon (so that he knows how much ice cream to buy). He decides, with your help, to run a half fraction of a 25 to study the influence of several factors on sales. The data are given on the next page.
From the data given, is a prediction equation justified? If so, what is the equation? Justify your answers.
Question 3
Consider the 25-2 fractional factorial design with generators: 4 = 12 and 5 = 123. Specify the factor level combinations for this design and determine the complete confounding pattern.
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Macquarie  University   Faculty  of  Science   Department  of  Chemistry  and  Biomolecular  Sciences   Analytical  Measurement  Uncertainty  and  Method  Validation   CBMS760  /  CBMS860     2013   2010     Tutorial  Set  4  /  Assignment  4     Hand  in  complete  solutions  /  working  for  all  three  questions  below  by  6  pm  on   Wednesday,  May  8,  2013.     Question  1     In  an  experimental  program  to  investigate  the  properties  of  a  thin  film  plastic  coating  on   ceramic–based  resistors,  six  factors  were  identified.     X  –  Supplier  of  a  basic  component  of  the  plastic  (there  were  2)   1 X  –  Viscosity  of  the  coating  at  time  of  application   2 X  –  Thickness  of  coating   3 X  –  Temperature  of  the  first  bake  (dry)  cycle   4 X  –  Temperature  of  the  second  bake  (cure)  cycle   5 X  –  Speed  of  the  conveyer   6   (a)   List  the  experiments  and  a  random  list  of  run  orders  for  a  full  factorial  design  with   these  six  factors  using  symbolic  –  and  +  signs  for  the  levels.    Show  the  complete   confounding  table.     (b)    List  the  experiments  and  random  run  orders  if  we  are  to  run  a  ½  fractional  design  .     (c)    Describe  how  you  would  analyse  the  data  in  each  of  the  situations  defined  in  (a)   and  (b)  above.     Question  2     You  are  trying  to  help  an  ice  cream  vendor  to  be  able  to  predict  how  many  ice  cream   cones  he  will  sell  in  a ...

Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
117 Votes
1
Q1.
a) Table of the experiments and random list of run orders for level 2
RUN
X1
X2
X3
X4
X5
X6
1
-
-
-
-
-
-
2
+
-
-
-
-
-
3
-
+
-
-
-
-
4
+
+
-
-
-
-
5
-
-
+
-
-
-
6
+
-
+
-
-
-
7
-
+
+
-
-
-
8
+
+
+
-
-
-
9
-
-
-
+
-
-
10
+
-
-
+
-
-
11
-
+
-
+
-
-
12
+
+
-
+
-
-
13
-
-
+
+
-
-
14
+
-
+
+
-
-
15
-
+
+
+
-
-
16
+
+
+
+
-
-
17
-
-
-
-
+
-
18
+
-
-
-
+
-
19
-
+
-
-
+
-
20
+
+
-
-
+
-
21
-
-
+
-
+
-
22
+
-
+
-
+
-
23
-
+
+
-
+
-
24
+
+
+
-
+
-
25
-
-
-
+
+
-
26
+
-
-
+
+
-
27
-
+
-
+
+
-
28
+
+
-
+
+
-
29
-
-
+
+
+
-
30
+
-
+
+
+
-
31
-
+
+
+
+
-
32
+
+
+
+
+
-
33
-
-
-
-
-
+
34
+
-
-
-
-
+
35
-
+
-
-
-
+
36
+
+
-
-
-
+
37
-
-
+
-
-
+
38
+
-
+
-
-
+
39
-
+
+
-
-
+
40
+
+
+
-
-
+
41
-
-
-
+
-
+
42
+
-
-
+
-
+
43
-
+
-
+
-
+
44
+
+
-
+
-
+
45
-
-
+
+
-
+
46
+
-
+
+
-
+
47
-
+
+
+
-
+
48
+
+
+
+
-
+
49
-
-
-
-
+
+
50
+
-
-
-
+
+
51
-
+
-
-
+
+
52
+
+
-
-
+
+
53
-
-
+
-
+
+
54
+
-
+
-
+
+
55
-
+
+
-
+
+
56
+
+
+
-
+
+
57
-
-
-
+
+
+
58
+
-
-
+
+
+
59
-
+
-
+
+
+
60
+
+
-
+
+
+
61
-
-
+
+
+
+
62
+
-
+
+
+
+
63
-
+
+
+
+
+
64
+
+
+
+
+
+
There is No Confounding rule present for level 2
b) For ½ fractional design
RUN
X1
X2
X3
X4
X5
X6
1
-
-
-
-
-
-
2
+
-
-
-
-
+
3
-
+
-
-
-
+
4
+
+
-
-
-
-
5
-
-
+
-
-
+
6
+
-
+
-
-
-
7
-
+
+
-
-
-
8
+
+
+
-
-
+
9
-
-
-
+
-
+
10
+
-
-
+
-
-
11
-
+
-
+
-
-
12
+
+
-
+
-
+
13
-
-
+
+
-
-
14
+
-
+
+
-
+
15
-
+
+
+
-
+
16
+
+
+
+
-
-
17
-
-
-
-
+
+
18
+
-
-
-
+
-
19
-
+
-
-
+
-
20
+
+
-
-
+
+
21
-
-
+
-
+
-
22
+
-
+
-
+
+
23
-
+
+
-
+
+
24
+
+
+
-
+
-
25
-
-
-
+
+
-
26
+
-
-
+
+
+
27
-
+
-
+
+
+
28
+
+
-
+
+
-
29
-
-
+
+
+
+
30
+
-
+
+
+
-
31
-
+
+
+
+
-
32
+
+
+
+
+
+
Confounding rule-
X6 = X1* X2*X3*X4*X5
c) We can analyze the data after getting the response variable in both the cases, however the complexity in analysis will be lesser in second case as number of runs are lesser as well as number of factor can be reduced. Also confounding rules are available in the case.
In first case data will be analyzed using Leaf Spring experiment. After looking on the result and applying effect hierarchy principle, we can reduced the number of runs by eliminating the 6, 5 and even 4 factor interaction based on values.
Further model can be frames based on response variable. The model in both the cases should not be significant different if we had eliminated less effect factor interaction.
Further the impact of each factor on the model can be determine by seeing the p value for each of the cases. The significance of overall model can be determine by looking the overall p and F value of model.
Q2 Below is the given data-
RUN
X1
X2
X3
X4
X5
Response
1
1
24
cloudy
5
egula
happy
105
2
1
25
cloudy
5
egula
surly
122
3
1
24
sunny
10
egula
happy
92
4
1
25
sunny
10
egula
surly
149
5
1
24
sunny
5
waffle
happy
106
6
1
25
sunny
5
waffle
surly
153
7
1
24
cloudy
10
waffle
happy
111
8
1
25
cloudy
10
waffle
surly
113
9
2
24
sunny
5
egula
surly
100
10
2
25
sunny
5
egula
happy
146
11
2
24
cloudy
10
egula
surly
105
12
2
25
cloudy
10
egula
happy
125
13
2
24
cloudy
5
waffle
surly
105
14
2
25
cloudy
5
waffle
happy
126
15
2
24
sunny
10
waffle
surly
94
16
2
25
sunny
10
waffle
happy
156
Main Effect Plot
Interaction plot
Numerical optimizer result
Results-
DESIGN DETAILS
Design type: Two-level
Design description: 1/2 Fraction
Number of factors: 5
Number of runs: 16
Resolution: 4
Number of blocks: 2
FACTORS
Factors and Levels:
__________________________________________________
Factor Label Low Center High
__________________________________________________
X1 temperature 24 24.5 25
X2 weather cloudy 0 sunny
...
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