Quality Control Exam2 J.Sherrod (1)
pdf
keyboard_arrow_up
School
Florida Agricultural and Mechanical University *
*We aren’t endorsed by this school
Course
4234
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
Pages
6
Uploaded by The_Educated_King
Name:__________________
ESI 4234: Quality Control and Reliability Engineering Exam 2 –
Fall 2021 Instructions -
Closed-book and closed-notes. You may use one letter-sized formula sheet and your calculator. -
Be sure to NEATLY show your answers and write legibly. I cannot grade it if I cannot read it. -
SHOW AND EXPLAIN ALL YOUR WORK. You cannot get partial credit for wrong answers if you do not show any of your work. -
You have 75 minutes to complete the exam. -
Exam is 6 pages. Make sure you have all pages. -
Use spaces provided for your answers. Write on back of that page if additional space is needed Problem 1 (20 points) For a key joint width, the lower specification limit is 0.8100 mm and the upper specification limit is 0.8600 mm. The precision machining process used to make the joint has a standard deviation of 0.0060 mm and a mean of 0.8500 mm. Round all answers to 4 decimal places. a.
Find the capability indices ࠵?
࠵?
and ࠵?
࠵?࠵?
for the process. b.
Find the fallout of the process in ppm c.
The company wants to reduce its defect probability to operate at six-sigma quality. How can the engineers accomplish this by adjusting the centering and variability of the process? Jade
Sherrod
(a)
LSL
-0.8100mm
USL
-_
0.8600mm
cp=
USL
LSL
=
0.86-0.81
F-
0.0060mm
noooo
,
=
1.
3888921.389mm
6T
µ
-
-0.8500mm
Cpk
__
µ
Sl
-
M
,
M
LSL
)=min(
0.86-0.85
,
0.85-0.81
3T
3T
310.006
)
310.006
,
=min(
0<5556,2-2222
)
Cpk
-0.5556
MM
(b)
Fallout
__I(
'
SFM
)+
,
☒
(
Usyk
)
☒
(-6-6667)+1-01=(1-6667)
0+1-0.95204
=
0.047-96.106=47460
defective
ppm
(C)
By
adjusting
the
centering
you
need
thecpto
be
greater
than
the
Cpk
With
variability
it
must
be
reduced
to
where
Cp=2
Name:__________________
2 Problem 2(15 Points)
.
The following experiment was performed in which two factors A and B are varied and a material property is measured.
Replicate A B I II - - 45 49 + - 30 24 - + 80 74 + + 41 48 a.
Using the graph scales shown below, sketch the AB interaction plot. Be sure to clearly label the lines in your plot. b.
Based on the interaction plot, does the interaction appear to be highly significant? Be sure to explain your reasoning. c.
What levels should we set factors A and B to maximize the material property? 0
20
40
60
80
-1.5
-1
-0.5
0
0.5
1
1.5
Material Property
Factor A
Jade
Sherrod
YA
-
B
-
Y
#
B-
YA
-
Bt
Yat
Bt
•
YA
-
Bt
(7-7)
YA•-
B-
(
47
)
•
YA
'T
Bt
144.5
)
•
Yat
B-
127
)
The
interaction
is
insignificant
because
they
do
not
intersection
each
other
making
the
line
are
independent
The
levels
are
Factor
A
need
to
be
below
the
level
of
factor
B
Name:__________________
3 Problem 3 (15 points)
A two factor experiment was conducted in which factor A is varied over 4 levels, factor B is varied over 3 levels and the experiment is replicated 3 times. We obtained the ANOVA table for the response data as follows: a.
Find the values for the entries shown blank in the ANOVA table. b.
Controlling the probability of type I error at ࠵? = 0.05
level, state which effects are significant. Source
DF
SS
MS
F
A
3
37.26
12.42
B
2
82.91
41.46
AB
6
65.75
2.64
Error
24
99.73
4.16
Total
35
285.65
Jack
Sherrod
2.
99
2
9.97
10.95
24
99.73
f-
of
Source
A
=
MS
A
=
12.42=29855822.99
MS
error
4.16
DF
of
source
D=
SS
B
82-91=1.9997622
MS
B
=
41.46
F
of
Source
B.
=
MS
B
=
41.46
=
9.96635=9.97
MS
Error
4.16
MS
of
source
AB
=
SEE
}
=/
0.95
DF
of
error
=
abcn
-
D=
4(
3)
(3-1)=24
SJ
of
error
=
285.65-37.26-82.91-65.75=99.73
A=
Foos
,
3,24=3.01
Factors
A
and
B
3.
88>3.01
are
significant
B=Fo.os
,
2,24
=3
-40
at
4=0.05
but
4.06>3.4
Factor
AB
is
AB=Fo.os
,
6,24=2.51
1.9<2.51
Not
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Name:__________________
4 Problem 4 (15 points).
A 2
5
experiment was conducted to improve the projectile throw distance (
y
, measured in feet) by testing 5 factors based on the Low and High levels (measured in feet) shown in the below table. The statistical analysis of the experiment produced the predictive model: ࠵?
̂ = 215.5 − 8.70࠵?
࠵?
+ 19.82࠵?
࠵?
where ࠵?
࠵?
and ࠵?
࠵?
are the coded variables. Factor name Label Low (ft) High (ft) Height of Pivot ࠵?
10 18 Length of Weight ࠵?
2 5 Length of Sling ࠵?
0.5 2.5 Length of Short Arm ࠵?
1 4 Length of Long Arm ࠵?
2 8 We want to keep the Height of Pivot at 12 ft, Length of Weight at 2.5 ft, Length of Sling at 1.5 ft, Short Arm at 2.5 ft and Long Arm at 7.5ft. What is the predicted throw distance of the trebuchet? Jade
Sherrod
✗
B
=
Sj
.
}
-0.8571
j-zis.se
-8.706.857171-19.824.2
)
✗
E-
=
852=1.2
j
=
231.827
Name:__________________
5 Problem 5 (15 points)
A production process is monitored with the c-chart shown below. Total number of defects from 15 inspection units equaled to 68. Assuming that assignable causes of out-of-control points are found, recalculate the UCL, LCL and CL of the chart. After recalculating the control limits, clearly state if there are any units that fall outside the control limits. Use 3-sigma control limits.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Number of defects
Unit
UCL
CL
Jude
Sherrod
D=
6,83=4.533
recalculate
I
=
4,32
=3
-2301
UCE
E
-13
E
=
3.2301+3
3.2301=8.6218
CL
=D
=3
.
2301
LCL
=
E-
3fF=
3.231
-
3
3.231
=
-2.16170
there
are
no
units
outside
of
control
limits
so
they
are
in
control
.
Name:__________________
6 Problem 6 (20 Points).
D
elivery times (in days) of a supplier is measured as shown in below table (3 subgroups are measured and each subgroup contains 4 observations). The supplier promises a delivery time no longer than 20 days. Assume the process is normal and is in control. Delivery Time (days) Subgroup I II III IV 1 16.8 8.8 10.4 12.6 2 12.4 14.5 11.5 10.1 3 13.6 16.0 14.2 15.2 a.
Find process capability index ࠵?
࠵?
. Use the range method to estimate the standard deviation. b.
Find the process fallout in ppm. Jade
Sherrod
n
-20
Avs
012=3.735
12.15
12.125
}
/
3.0083
14.75
Avg
:
14.2667
13.1
12.03
12.63
Find
avg
range
→
avg
subgroup
→
dZatn=zo
is
3.73
Md
'
Cp
-
Usl
-
LSL
=
20-13.01
6T
=
0.3348
60
Fallout
-_
☒
(
"
㱺
+
1-
☒
lusty
)
0=(3-01,7)+1
-0=(2%-1)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help