Download mt_1_s01_305_soln - University of Windsor

Name (print, please) _____________________________________________ ID ___________________________
Statistical Quality Control & Design 73-305 Summer 2001
Faculty of Business Administration
University of Windsor
Midterm Exam I Solution
Monday, June 18, 6:00 – 8:00 pm
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a one-sided formula sheet.
Time available: 2 hours
Instructions:
 This exam has 13 pages including this cover page and 5 pages of tables.
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/18
2
/14
3
/6
4
/8
5
/9
Total:
/55
Name:______________________________________________
ID:_________________________
Question 1: (18 points)
1.1 _______________ data are measured on a continuous scale with individual measurements rounded to
some desired number of decimal places.
a. Variables
b. Attributes
1.2 The glass cover of a pressure gauge may crack. A manufacturer wants to use a control chart to monitor
the performance of a new process that is suggested to reduce the number of finished items with one or
more cracks. The manufacturer may use a
a. variable control chart
b. attribute control chart
1.3 A _______________ is a failure of the part or article to conform to some one specification.
a. defect
b. defective
1.4 Nonconforming means
a. defect
b. defective
1.5 Cost of inspection is an example of
a. appraisal cost
b. internal failure cost
c. external failure cost
d. intangible cost
1.6 A process is described in control when only _______________ of variation seem to be operating.
a. common causes
b. assignable causes
1.7 A process is out of control if the variation is likely to be associated with some
a. common causes
b. assignable causes
1.8 Use of X and s charts instead of X and R charts is recommended if the subgroups are
a. large
b. small
1.9 If the process mean changes, which of the following is less likely to change?
a. standard deviation
b. range
1.10 The cost of collecting data for which of the following charts is likely to be less?
a. Variables chart
b. Attributes chart
1.11 If many quality characteristics are measured, it may be impracticable and uneconomical to
use
2
Name:______________________________________________
ID:_________________________
a. Variables control chart
b. attributes control chart
1.12
If defectives are monitored and the subgroup size is expected to vary, use
a. p chart
b. np chart
c. c chart
d. u chart
1.13
If defects are monitored and the subgroup size is expected to vary, use
a. p chart
b. np chart
c. c chart
d. u chart
1.14 Daily inspection records are maintained on the production of a special-design electronic
device. 100 items have been inspected each day for the past 31 days. A total of 651 items failed
during a particularly severe heat stress test. A p chart is constructed and proportion rejected of
all the 31 days are plotted. Only one point is outside the control limits. All the other points are
within the upper and lower control limits. The point that is outside the control limits corresponds to
the 15th day of production. On the 15th day, 51 items failed the test. Recommend an aimed-at
value of p 0 for continuous use of the p chart.
a. 0.15
651  51
b. 0.20
p0 
 0.20
(31  1)(100)
c. 0.21
d. 0.31
e. 0.651
f. 6.51
1.15
Specification limits are
a. obtained by statistical analysis on the process data
b. set by the buyer
1.16
If measurement of an item is outside the specification limit, then
a. the process is out of control
b. the item is defective
c. the process is incapable of meeting specification
d. none of the above conclusion is appropriate
1.17
If a subgroup mean is outside the control limit, then
a. the process is out of control
b. the item is defective
c. the process is incapable of meeting specification
d. none of the above conclusion is appropriate
If the process mean is within 3 of a specification limit, then
a. the process is out of control
b. the item is defective
1.18
3
Name:______________________________________________
ID:_________________________
c. the process is incapable of meeting specification
d. none of the above conclusion is appropriate
Question 2: (14 points)
Control charts for X and R are kept on the manufacture of a drive shaft. After 25 subgroups have
been drawn and inspected, and the diameters recorded,  X  518.4 mm and  R  12.2 mm. The
subgroup size is 4.
a. (4 points) Assuming the process is in a state of control, compute the X and R chart central line
and control limits for the next production period.
Answer:
 X  518.4  20.736 mm ( X chart central line)
X
k
25
R
  12.2  0.488 mm ( R chart central line)
R
k
25
For n  4, we get A2  0.73, D3  0, D4  2.28 from Table D
UCLX  X  A2 R  20.736  (0.73)(0.488)  21.0922 mm
LCLX  X  A2 R  20.736  (0.73)(0.488)  20.3798 mm
UCLR  D4 R  ( 2.28)( 0.488)  1.1126 mm
LCLR  D3 R  (0)(0.488)  0 mm
b. (2 point) Estimate the mean  and standard deviation  of the process assuming that it is in
statistical control.
Answer:
  X  20.736 mm
For n  4, we get d 2  2.059 from Table C

R 0.488

 0.2370 mm
d 2 2.059
4
Name:______________________________________________
ID:_________________________
c. (4 points) If the specified nominal dimension and tolerances were 20.800.50 mm, what
proportion of items are too large? What proportion of items are too small?
Proportion of items which are too large
 P( X  USLX )  P( X  21.30 |   20.736,  0.237)
21.30  20.736 

 P z 
  P( z  2.38)  Area 1
0.237


 1  P( z  2.38)  1  0.9913  0.0087
Probability
Answer:
USLX  20.80  0.50  21.30 mm
LSLX  20.80  0.50  20.30 mm
Area=0.0329
 0.237
Area=0.0087
(2)
(1)
X=20.30 =20.736 X=21.30
Diameter
Proportion of items which are too small
 P( X  USLX )  P( X  20.30 |   20.736,  0.23)
20.30  20.736 

 P z 
  P( z  1.84)  Area 2
0.237


 0.0329
d. (1 point) Estimate the standard deviation  X of the sample mean.
Answer:
x 

n

0.237
4
 0.1185 mm
e. (3 points) Suppose that a sudden change in the process occurs that increases  by 0.02 mm but
does not change  X . What is the probability that a point will fall outside the control limits on the
X chart after the change in the process mean.
Answer:
  20.736  0.02  20.756 mm
 P( X  21.0922 |   20.756, X  0.1185)
 P( X  20.3798 |   20.756, X  0.1185)
21.0922  20.756 
20.3798  20.756 


 P z 
  P z 

0.1185
0.1185




 P( z  2.84)  P( z  3.17 )  Area 1  Area 2
 [1  P( z  2.84)]  P( z  3.17 )
 [1  0.9977]  0.00076  0.00306
5
Probability
P( X  UCLX )  P( X  LCL X )
Area=0.00076
X 0.1185
Area=0.0023
(1)
(2)
X=20.3798 =20.756 X=21.092
Diameter
Name:______________________________________________
ID:_________________________
Question 3: (6 points)
An item is made in lots of 150 each. The lots are given 100% inspection. The record sheet for the
first 30 lots inspected showed that a total of 90 items did not conform to specifications.
a. (2 points) Determine the trial control limits for a p chart.
Answer:
p
p(1  p )
0.02(1  0.02)
90
3
 0.0343
 0.02 , 3
n
150
(30)(150)
UCLp  p  3
p(1  p )
 0.02  0.0343  0.0543
n
LCL p  p  3
p(1  p )
 0.02  0.0343  0 (negative)
n
b. (2 points) If the average fraction of nonconforming items remains unchanged, what is the
probability that the 31st lot will contain 9 or more nonconforming units?
Answer:
Expected number of nonconforming units in the 31 st lot,
90
 np 
 3 0r,  np  n p  (150)(0.02)  3
30
P(c  9 |  np  3)  1  P(c  8 |  np  3)  1  0.996  0.004
c. (2 points) If the average fraction of nonconforming items remains unchanged, what is the
probability that the 31st lot will contain 4, 5 or 6 nonconforming units?
Answer:
P(c  4 |  np  3) + P(c  5 |  np  3) + P(c  6 |  np  3)
= P(c  6 |  np  3) - P(c  3 |  np  3)
= 0.966 – 0.647 = 0.319
6
Name:______________________________________________
ID:_________________________
Question 4: (8 points)
A c chart is used to monitor surface imperfections on a certain electrical insulator. Past history
indicates a mean of 3.
a. (2 points) Calculate standard 3-sigma limits for this process.
Answer:
c  3 , 3 c  3 3  5.196
UCLc  c  3 c  3  5.196  8.196
UCLc  c  3 c  3  5.196  0 (negative)
b. (3 points) Use Table G to find the approximate Type I error probability under the assumption that
the process is operating at a mean of 3.
Answer:
c  3
P(Type I error)  P(c  UCLc )  P(c  LCLc )
 P(c  8 |  c  3)  P(c  0 |  c  3)
 [1  P(c  8 |  c  3)]  P(c  0 |  c  3)
 [1  0.996]  0
 0.004
c. (3 points) Assume that the process is actually operating at a mean of 6. What is the Type II error
probability?
Answer:
c  6
P(Type II error)  P( LCLc  c  UCLc )
 P(0  c  8 |  c  6)
 P(c  8 |  c  6)
 0.847
7
Name:______________________________________________
ID:_________________________
Question 5: (9 points)
A spherical plastic ball is to be produced against a specification of 1.600.01 cm (diameter).
Suppose that the process is in control with a   1.598 and   0.003 . Calculate the process
capability indexes
a. (4 points) C p and C pk .
Answer:
USLX  1.60  0.01  1.61 cm
LSLX  1.60  0.01  1.59 cm
USLX  LSL X 1.61  1.59
Cp 

 1.111
6
6(0.003)
  LSL X 1.598  1.59
LSL X   1.61  1.598
C pL 

 0.889 , C pU 

 1.333
3
3(0.003)
3
3(0.003)
C pk  min( C pL , C pU )  min( 0.889,1.333)  0.889
b. (4 points) What is your estimate of the percent nonconforming product?
Probability
Answer:
Proportion of nonconforming product
 P( X  USLX )  P( X  LSL X )
 P( X  1.61 |   1.598,  0.003)
 P( X  1.59 |   1.598,  0.003)
1.61  1.598 
1.59  1.598 


 P z 
  P z 

0.003 
0.003 


 P( z  4)  P( z  2.67)  Area 1  Area 2
Area=0.0338
 0.003
Area (1)0
(2)
X=1.59 =1.598
Diameter
 0  0.0038
 0.0038
X=1.61
c. (1 point) Choose the best recommendation from the following:
i. Center the mean
(Since C pL  1, mean is within 3 of the LSL )
ii. Reduce the variation
(Since C p  1 , variation is small)
iii. Change the specification
iv. Both (i) and (ii)
v. No action is necessary
USL  LSL
. So, the variation is small.
6
 1, so the mean is near the LSL and needs centering.
Since C pk  1, the process is incapable. Since C p  1,  
However, C pL
8