University of Windsor
Odette School of Business
Statistical Quality Control & Design 73-305 Fall 2003
Solution to Assignment 5
Problem 1 (16 points)
a. (4 points)
 n  15  15  15  18  18  18  20  20  20  23  23  23  228
 np  6  0  1  0  8  1  0  1  9  0  1  9  36
 n  228  19
n 
ave
12
m
Centerline, p 
 np  36  0.1579
 n 228




UCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2510  0.4089
nave
19
LCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2510  0.0931  0 (negative)
nave
19
Proportion nonconforming
b. (2 points)
P Chart Using nave
0.5
0.4
0.3
UCL
p
pbar
LCL
0.2
0.1
0
-0.1
0
1
2
3
4
5
6
7
8
9
10
11
12
Sample
c. (1 point) Check points: 1, 5, 9 and 12 (points near or outside UCLp)
d. ( 4 points)
Check point: 1
The point is in control when UCLp is computed using n ave
13
Since nind  15  19  nave , the individual sample size is less and the individual
control limits are wider. The point will be in control if UCLp is computed using
individual sample size. Hence, it is not necessary to compute UCLp using individual
sample size. The point is in control.
Check point: 5
The point is out of control when UCLp is computed using n ave
Since nind  18  19  nave , the individual sample size is less and the individual
control limits are wider. The point may be in control if UCLp is computed using
individual sample size. Hence, it is necessary to compute UCLp using individual
sample size.


p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.4157
n
18
Since p  0.4444  0.4157  UCL p , the point is out of control.
UCL p  p  3
Check point: 9
The point is out of control when UCLp is computed using n ave
Since nind  20  19  nave , the individual sample size is more and the individual
control limits are narrower. The point will be out of control if UCLp is computed
using individual sample size. Hence, it is not necessary to compute UCLp using
individual sample size. The point is out of control.
Check point: 12
The point is in control when UCLp is computed using n ave
Since nind  23  20  nave , the individual sample size is more and the individual
control limits are narrower. The point may be out of control if UCLp is computed
using individual sample size. Hence, it is necessary to compute UCLp using
individual sample size.


p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.3860
n
23
Since p  0.3913  0.3860  UCL p , the point is out of control.
UCL p  p  3
e. (5 points) Individual control limits are computed below:
Sample size: 15




UCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2825  0.4404
n
15
LCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2825  0.1246  0 (negative)
n
15
Sample size: 18




UCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2578  0.4157
n
18
LCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2578  0.0999  0 (negative)
n
18
Sample size: 20




UCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2446  0.4025
n
20
LCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2446  0.0867  0 (negative)
n
20
Sample size: 23




UCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2281  0.3860
n
25
LCL p  p  3
p 1 p
0.1579 1  0.1579 
 0.1579  3
 0.1579  0.2281  0.0702  0 (negative)
n
25
Problem 2 (3 points)
Centerline c  4
P Chart using Individual Limits
0.5
Proportion nonconforming
The same conclusion is
obtained from the p chart
computed using the individual
sample sizes and shown on the
right. Point 1 is in control, but
and Points 5, 9 and 10 are out
of control.
0.4
0.3
UCL
p
pbar
LCL
0.2
0.1
0
-0.1
0 1 2 3 4 5 6 7 8 9 10 11 12 13
UCL c  c  3 c  4  3 4  4  6  10
LCL c  c  3 c  4  3 4  4  6  2  0 (negative)
Problem 3 (3 points)
 n  10  25  250
 c  109  112  95  102  86  88  119  145  69  75  1000
 c  1000  4
Centerline, u 
 n 250
Sample
UCL u  u  3
u
4
 43
 4  1.2  5.2
n
25
LCL u  u  3
u
4
 43
 4  1.2  2.8
n
25
Problem 4 (14 points)
N  800, n  40, c  4
a. (2 points) Pa  Px  c |   np 
 Px  4 |   np  40  0.05  2.0  0.947 (Appendix4)
b. (1 point) Producer’s risk = 1- Pacceptance  1  Pa  1  0.947  0.053

c. (3 points) Pa  P x  c |   np,   np1  p 


Pa  P x  4 |   40  0.20  8,   40  0.201  0.20   2.5298
 P y  4.5 |   8,   2.5298 

4.5  8


 P z 
 1.3835 
2.5298


 Pz  1.3835 
= 0.0838 (Appendix 1)
d. (1 point) Consumer’s risk = Pacceptance  Pa  0.0838
e. (2 points) At p  0.05
P N  n p 0.947800  400.05
AOQ  a

 0.0450
N
800
At p  0.20
P N  n p 0.0838800  400.20
AOQ  a

 0.0159
N
800
f. (2 points) The OC curve is shown below:
Probability of acceptance,
Pa
Operating Characteristic Curve
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Proportion of defective, p
0.8
1
g. (3 points) AOQL = max (AOQ) = 0.0607 (see the AOQ values below)
AOQ Curve
0.07
0.06
AOQ
0.05
0.04
0.03
0.02
0.01
0
0
0.2
0.4
0.6
0.8
1
Proportion of defective, p
AOQL (defectives are replaced)
Sample size, n
Acceptable number of defectives, c
Lot size, N
Proportion
Defective
(Input)
(p)
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0.055
0.06
0.065
0.07
0.075
0.08
0.085
0.09
0.095
0.1
0.105
0.11
0.115
0.12
0.125
40
4
800
Probability
of c or less
Defects
(Pa)
=BINOMDIST(c,n,p,TRUE)
0.999998223
0.999950846
0.999677372
0.998824808
0.996899623
0.993329894
0.98753334
0.978977703
0.967227004
0.95197174
0.933043773
0.910418131
0.88420454
0.854631635
0.822026552
0.786792233
0.749384339
0.710289168
0.6700036
0.629017697
0.587800281
0.546787591
0.50637491
0.466910965
0.428694772
AOQL
=max(all AOQ)
0.060729724
AOQ
Pa*(N-n)*p/N
0.004749992
0.009499533
0.014245403
0.018977671
0.023676366
0.028309902
0.032835484
0.037201153
0.041348954
0.045218658
0.048751537
0.051893833
0.05459963
0.056833004
0.058569392
0.05979621
0.060512785
0.060729724
0.060467825
0.059756681
0.058633078
0.057139303
0.055321459
0.05322785
0.050907504
Problem 5 (4 points)
Pa , p 0.05  Px  1 | n  10, p  0.05   0.9139 (From Appendix 3) (1 point)
Pa , p 0.20  Px  1 | n  10, p  0.20   0.3758 (From Appendix 3) (1 point)
The OC curve is shown below. (1 point)
Probability of acceptance,
Pa
Operating Characteristic Curve
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Proportion of defective, p
The sampling plan with n  40, c  4 yields steeper OC curve. (1 point)