Control Attributes Charts
Variables Control Charts for continuous measurements
Attribute Control Charts for discrete measurements
P Charts for Fraction Defective
Monitors fraction defective
Defective is unusable
fraction of broken glass plates
fraction of defective chips on a wafer
fraction of defective truffle shells
go-no-go gauge
2
P Chart Based on Binomial Distribution
sample size n
for sample i
Xi = number of defectives in sample of n
p = fraction defective in popln
= P(unit is defective)
r. v. X i Binomial(n, p)
n
P(X i = x ) = p x ( 1 p ) n x
x
E(X)=np V(X)=np(1-p)
3
x = 0,1, . . . , n
Sample Statistic is Fraction Defective
At each sample time plot
pi =
xi
n
Note:
1
E( x i ) = p
n
1
np ( 1 p ) p ( 1 p )
V (pi ) = 2 V ( xi ) =
=
n
n2
n
E( p i ) =
Control Limits
UCL = p + 3
p(1 p )
n
CL = p
LCL = p 3
p(1 p )
n
4
Some Notes on P Charts
If LCL is negative, make it zero
Test statistic is not normal - don't use zone rules
If pi = 0, process in-control
5
Estimating p From Initial Data
Data:
draw m (say 20) subsamples of size n
X1, X2, ... , Xm
Sample Statistics:
p1, p2, ... , pm
Estimate for CL = p
p=
x1 + . . . + x m
nm
6
Example for P Charts
A process that produces bearing housings is
investigated. 10 samples of size 100 are selected. Is
process in-control?
Sample #
# Noncon
1
5
2
2
3
3
4
8
5
4
6
1
7
2
8
6
9
3
10
4
Make initial control chart:
m
D
total # defectives i
p=
= i =1
= 0.038
total # sampled
mn
UCL = 0.038 + 3
0.038(1 0.038)
= 0.095
100
CL = 0.038
LCL = 0.038 + 3
0.038(1 0.038)
= 0.02 0
100
7
P Chart for C1
Proportion
0.10
3.0SL=0.09536
0.05
P=0.03800
0.00
- 3.0SL=0.000
0
1
2
3
4
5
6
7
8
9
10
Sampl e Number
No iterations are needed all points in the baseline sample
fall within the control limits.
Sample
#
#
Nonco
n
pi
1
2
3
4
5
6
7
8
9
10
5
2
3
8
4
1
2
6
3
4
.
.
.
.
.
.
.
.
. .04
05 02 03 08 04 01 02 06 03
8
P-Chart and Average Run Length
If the fraction defective shifts from its current value
0.038 to 0.060, find the probability of detecting this
on the first sample following the shift.
r. v. D = number of defectives in sample of 100
r.v. D Binomial(n = 100 , p = 0.06)
P ( sample point above UCL)
= P ( p > UCL / p = 0.06)
D
= P(
> UCL / p = 0.06) = P ( D > 100 * 0.095)
100
9.5 6
= P(Z >
) = P ( Z > 1.47) = 0.07
100 * .06 * .94
Find the expected number of samples until detection
ARL =
1
= 14.3
0.07
9
Selecting a Sample Size for P-charts
Criterion: Select a sample size such that the probability
of finding at least one defective in a sample
exceeds 95%
Example: The fraction defective is 1%. Recommend a
sample size using the above criterion.
r.v.D Binomial (n, p = 0.01)
P ( D 1) > 0.95 1 P ( D = 0) > 0.95
n 0
1 p (1 p ) n > 0.95 0.05 > 0.99 n
0
n > 300
10
C for Charts Number of Defects
Monitors Number of Defects (not defectives)
Examples:
# of defects (contaminants) per sample of recycled
plastic
# of defects (all types) per sample of cars
# of defects (scratches, bubbles, etc.) per sample
of a painted surface
An item may be acceptable with a few defects
11
The Defect Distribution is the Poisson
r. v. X = # of defects per sample
parameter: c = mean # of defects per sample
e ccx
P( X = x ) =
x!
E(X) = c
x = 0 , 1, 2, . . .
Var(X) = c
Control Limits for C Chart
# of Defects per Sample
UCL = c + 3 c
LCL = c - 3 c
CL = c
Estimating C from Initial Data
ci = number of defects per sample
initial data ci, i=1...m
estimate mean number of defects per sample with
c=
m
1
m
c
i
i=1
12
Example for C Charts
6.40 The number of nonconformities found on final
inspection of a cassette deck is shown below.
The initial sample consists of 18cassette
decks. Is the process in statistical control?
0
1
1
0
2
1
1
3
2
1
0
3
2
5
1
2
1
1
Create a c chart
c=
27
= 1.5 defects per sample
18
UCL = c + 3 c = 5.17
LCL = c - 3 c = 2.17 0
CL = c = 1.5
Yes process is in-control. All points within limits.
13
Suppose we implement a c chart to monitor the cassette
decks with a sample size of 4 unitsGive control
limits this c chart
If I use 4 cassettes, the average number of defects per
sample will be 4*1.5=6.0 and the c chart is
UCL = c + 3 c = 13.35
LCL = c - 3 c = 1.35 0
CL = c = 6.0
14
U Chart for Number of Defects per Unit
Convenient Y axis
sample size n
count # of defects ci
sample statistic - plot: ui=ci/n
From Baseline data c1cm
Estimate
m
u=
c
i =1
i
n*m
Control Limits
u
UCL = u + 3
n
u
CL = u LCL = u 3
n
15
Example for U Charts
sampl
e
1
2
3
4
5
6
7
8
9
10
defect
s
10
15
18
20
13
12
15
13
14
16
There are 10 samples in the baseline and each consists
of 2 disk drive assemblies. The number of defects
in the sample is gi ven above. Show a standard 3sigma u-chart that would be appropriate for
monitoring this data.
m
c
i
146
u=
=
= 7.3
n * m 20
i =1
16
u
= 9.86
n
CL = u = 7.3
UCL = u + 3
LCL = u 3
u
= 4.74
n
17
sampl
e
1
2
3
4
5
6
7
8
9
10
Defect
s in
sampl
e
10
15
18
20
13
12
15
13
14
16
Defect
s per
unit
5
7.
5
9
10
6.
5
6
7.
5
6.
5
7
8
UCL=9.86
CL=7.3
LCL=4.74
Sample 4 is outside UCL iterate and recompute limits
with only 9 samples.
m
u=
c
i =1
i
n*m
=
136
= 7.56
18
18
CL=7.56
UCL=10.17
LCL=4.95
19
What would be the control limits for this u-chart with a
false alarm rate of 1 percent?
UCL = u + 2.54
u
= 9.77
n
CL = u = 7.56
LCL = u 2.54
u
= 5.35
n
20
Summary of Attributes Control Charts
Chart
Sample
Statistic
CL
p
pi fraction
defective
p
c
c i # defects
per sample
ui # defects
per unit
c
u
u
UCL/LCL
p3
p (1 p )
n
c3 c
u3
21
u
n