Chapter 6
Control Charts For Attributes
We look at control charts for attributes, for the fraction of nonconforming units.
6.1
Introduction
We look at three types of sets of control charts for attributes:
• the p chart, fraction nonconforming or number nonconforming
for a collection of items (more than one item), using binomial model.
• the c chart, total number nonconformities per unit, using poisson model.
• the u chart, average number of nonconformities per unit, using poisson model.
We will use these control charts to determine the probability of a type I error (accidently saying process is out of control) and type II error (accidently saying the process
is in control).
6.2
Control Charts For Fraction Nonconforming
SAS program: att5-6-2-books-pchart
We look at the p chart in this section. The p chart is used when we are investigating
the number of defectives in a collection of items. The binomial distribution is used
as the underlying model; although either the normal or poisson approximation to the
binomial is often used in the analyses associated with the p chart.
Exercise 6.1 (Control Charts For Fraction Nonconforming)
1. Book binding defectives, Raw Data
The number and proportion of defective (nonconforming) book bindings, of
forty samples (lots) of n = 50 books each, is given below.
93
94
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
sample Di
1
7
2
9
3
7
4
8
5
7
6
12
7
16
8
6
9
8
10
15
11
7
12
5
13
16
14
8
15
11
16
15
17
10
18
16
19
8
20
15
p̂i
0.14
0.18
0.14
0.16
0.14
0.24
0.32
0.12
0.16
0.30
0.14
0.10
0.32
0.16
0.22
0.30
0.20
0.32
0.16
0.30
sample
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Di
12
5
11
5
11
9
10
11
10
9
13
8
13
9
11
10
11
11
10
10
p̂i
0.24
0.10
0.22
0.10
0.22
0.18
0.20
0.22
0.20
0.18
0.26
0.16
0.26
0.18
0.22
0.20
0.22
0.22
0.20
0.20
(a) Data binomial?
True / False The book binding data is best described by the binomial
distribution because each book is either defective or not, each lot size has
50 books and it is assumed the chance of choosing a defective book from
a lot of 50 remains constant from lot to lot. Furthermore, the binomial
distribution can be approximated by either a normal (if np > 5 and n(1 −
p) > 5) or poisson distribution (p small and n large).
(b) Normal probability plot
The SAS normal probability plot is more or less linear and so indicates the
number of defective book bindings (choose one) is / is not more or less
normally distributed.
(c) Notation
number of lots m = (choose one) 40 / 45 / 50
number in each lot n = (choose one) 40 / 45 / 50
number of defective for lot 7, D7 = (choose one) 15 / 16 / 17
proportion (of n = 50) defective for lot 7, p̂7 =
(choose one) 0.15 / 0.22 / 0.32
(d) p chart
From SAS, the p chart is (choose one) in / out of statistical control because
there are no out-of-control signals, runs, trends, or cycles.
Section 2. Control Charts For Fraction Nonconforming (ATTENDANCE 5)
95
(e) 3σ (LCL, UCL) for p chart, normal approximation1
FromPSAS, the average of all nonconforming proportions is,
Di
p̂i
p̄ = mn
=m
= (choose one) 0.15 / 0.20 / 0.25
The upper control limit and the lower control limit for p chart are,
r
r
p̄(1 − p̄)
0.20(1 − 0.20)
p̄ ± 3
= 0.20 ± 3
=
n
50
(choose one)
i. (0.03, 0.33)
ii. (0.01, 0.37)
iii. (0.03, 0.37)
(f) New (ten) lots in process control?, by graph
Ten new lots from the book binding process are plotted on the previously
determined control charts above.
sample Di
41
17
42
19
43
17
44
18
45
17
p̂i
0.34
0.38
0.34
0.36
0.34
sample Di
46
22
47
25
48
21
49
15
50
21
p̂i
0.44
0.50
0.42
0.30
0.42
From SAS, it appears the book binding process has gone out of control
and, more specifically, has shifted (choose one) downwards / upwards.
The upper control limit and the lower control limit for p chart (based on
these ten lots) are,
r
r
p̄(1 − p̄)
0.38(1 − 0.38)
p̄ ± 3
= 0.38 ± 3
=
n
50
(choose one)
i. (0.01, 0.59)
ii. (0.03, 0.59)
iii. (0.18, 0.59)
(g) New (ten) lots in process control?, by test, normal approximation
Test if the proportion defective has changed from the old group of forty
lots to the new group of ten lots at α = 0.05.
1
The p control chart is based on a normal approximation to the underlying binomial distribution.
As we find out, some questions about the p chart involve using the binomial directly, some use
the normal approximation to the binomial and others still use the poisson approximation to the
binomial.
96
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
Statement
H0 : p1 = p2 versus
Ha : p1 6= p2
Test
Since p̂1 = 0.20, p̂2 = 0.38,
2 p̂2
and p̂ = n1np̂11 +n
= (40)(0.20)+(10)(0.38)
= 0.236
+n2
40+10
and so
Z=r
p̂1 − p̂2
³
p̂(1 − p̂) n11 +
1
n2
0.20 − 0.38
´=q
¡1
+
0.236(1 − 0.236) 40
1
10
¢ = −1.199
and so the p–value is 2 × P (Z < −1.199) =
(choose one) 0.02 / 0.13 / 0.23
Conclusions
accept null; the process remains in control, p1 = p2 ,
(which seems to contradict the p chart above)
(h) Probability of detecting a shift in proportion?, exact binomial
The probability2 a shift in (average) proportion to 0.38 will be detected
on the first subsequent sample (lot), after the initial forty (40) lots, where,
recall each lot is of size n = 50, is
=
=
=
=
=
=
=
P {detected 1st sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}
1 − P {LCL ≤ p̂ ≤ U CL|p = 0.38}
1 − [P {p̂ ≤ U CL|p = 0.38} − P {p̂ < LCL|p = 0.38}]
1 − P {np̂ ≤ nU CL|p = 0.38} + P {np̂ < nLCL|p = 0.38}
1 − P {np̂ ≤ 50(0.37)|p = 0.38} + P {np̂ < 50(0.03)|p = 0.38}
1 − P {np̂ ≤ 18.5|p = 0.38} + P {np̂ < 1.5|p = 0.38}
1 − 0.447 + 0 =
(choose one) 0.234 / 0.355 / 0.553
(i) Probability of detecting a shift in proportion?, poisson approximation
The probability3 a shift in (average) proportion to 0.38 will be detected
2
We will use the (exact) binomial distribution in this case; notice, in particular, that
P {np̂ ≤ 18.5|p = 0.38} = P {np̂ ≤ 18|0.38} = 0.447 using 2nd DISTR binomcdf(50,0.38,18)
and P {np̂ < 1.5|p = 0.38} = P {np̂ ≤ 1|p = 0.38} ≈ 0 using 2nd DISTR binomcdf(50,0.38,1).
3
We will use a poisson approximation to the binomial distribution in this case; notice, in particular, that P {np̂ ≤ 18.5|λ = 19} = P {np̂ ≤ 18|λ = 19} = 0.469 using 2nd DISTR poissoncdf(19,18)
and P {np̂ < 1.5|λ = 19} = P {np̂ ≤ 1|λ = 19} ≈ 0 using 2nd DISTR poissoncdf(19,1).
Section 2. Control Charts For Fraction Nonconforming (ATTENDANCE 5)
97
on the first subsequent sample (lot), after the initial forty (40) lots, where,
recall each lot is of size n = 50, is
=
=
=
=
=
=
=
P {detected 1st sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}
1 − P {LCL ≤ p̂ ≤ U CL|p = 0.38}
1 − [P {p̂ ≤ U CL|p = 0.38} − P {p̂ < LCL|p = 0.38}]
1 − P {np̂ ≤ nU CL|λ = np = (50)0.38} + P {np̂ < nLCL|λ = np = (50)0.38}
1 − P {np̂ ≤ 50(0.37)|λ = (50)0.38} + P {np̂ < 50(0.03)|λ = (50)0.38}
1 − P {np̂ ≤ 18.5|λ = 19} + P {np̂ < 1.5|λ = 19}
1 − 0.469 + 0 =
(choose one) 0.234 / 0.355 / 0.531
There is no need, really, to use the poisson approximation to the binomial
because it is as easy to work out the answer using the binomial directly, as it
is to work out the answer using the poisson approximation to the binomial.
Before calculators, though, the binomial calculation was more difficult than
the poisson calculation and so this is why the poisson approximation is still
used today in this text.
(j) Probability of detecting a shift in proportion?, normal approximation
The probability4 a shift in (average) proportion to 0.38 will be detected
on the first subsequent sample (lot), after the initial forty (40) lots, is
=
=
=
=
=
=
=
P {detected 1st sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}
1 − P {LCL ≤ p̂ ≤ U CL|p = 0.38}
1 − [P {p̂ ≤ U CL|p = 0.38} − P {p̂ < LCL|p = 0.38}]
1 − P {p̂ ≤ U CL|p = 0.38} + P {p̂ < LCL|p = 0.38}
U CL − p
LCL − p
1 − P {Z ≤ p
|p = 0.38} + P {Z < p
|p = 0.38}
p(1 − p)/n
p(1 − p)/n
0.37 − 0.38
0.03 − 0.38
1 − P {Z ≤ p
} + P {Z < p
}
0.38(1 − 0.38)/50
0.38(1 − 0.38)/50
1 − 0.442 + 0 =
(choose one) 0.234 / 0.355 / 0.558
The normal approximation is used if the number of defectives is thought
to follow a normal shape (a few number of small defective count per unit,
4
We will use a normal approximation to the binomial distribution in this case; notice, in par} = P {Z ≤ −0.146} ≈ 0.442 using the TI–83 function,
ticular, that P {Z ≤ √ 0.37−0.38
0.38(1−0.38)/50
normalcdf(−E99, −0.146).
98
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
a large number of middle defective count per unit and few number of large
defective count per unit), whereas the poisson approximation is used if the
number of defectives is thought to follow a poisson shape (a large number
of small defective count per lot, which tails off to a small number of large
defective count per lot).
(k) More probability of detecting a shift in proportion, poisson approximation
The probability a shift in (average) proportion to 0.38 will be detected on
(exactly) the fifth subsequent sample (lot), after the initial forty (40) lots,
is
=
=
=
=
=
P {detected 1st,2nd,3rd,4th,5th sample|p = 0.38}
1 − P {not detected 1st,2nd,3rd,4th,5th sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}
× · · · × P {not detected 5th sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}
× · · · × P {not detected 1st sample|p = 0.38}
1 − P {not detected 1st sample|p = 0.38}5
1 − 0.5315 =
(choose one) 0.234 / 0.655 / 0.958
where we have assumed the chances of not been detected on the 1st, 2nd,
3rd, 4th and 5th samples are all the same.
(l) Probability of type I error, poisson approximation
The probability of a type I error, of accidently saying the process (based
on the first 40 lots) is out of control, is
=
=
=
=
=
=
P {accidently saying process is out of control|p = 0.20}
1 − P {process is in control|p = 0.20}
1 − P {LCL ≤ p̂ ≤ U CL|p = 0.20}
1 − [P {p̂ ≤ U CL|p = 0.20} − P {p̂ < LCL|p = 0.20}]
1 − P {np̂ ≤ nU CL|λ = np = (50)0.20} + P {np̂ < nLCL|λ = np = (50)0.20}
1 − P {np̂ ≤ 50(0.37)|λ = (50)0.20} + P {np̂ < 50(0.03)|λ = (50)0.20}
1 − P {np̂ ≤ 18.5|λ = 10} + P {np̂ < 1.5|λ = 10} =
(choose one) 0.007 / 0.014 / 0.031
Notice the probability of a type I error is similar to the probability of
detecting a shift in the (average) proportion, only the proportions are
different in the two questions: p = 0.20 and p = 0.38, respectively.
(m) Probability of type II error, poisson approximation
The probability of a type II error, of accidently saying the process (based
Section 2. Control Charts For Fraction Nonconforming (ATTENDANCE 5)
99
on the first 40 lots) is in control (p = 0.20), when, in fact, p = 0.42, is
β =
=
=
=
=
=
P {accidently saying process in control|p = 0.42}
P {LCL ≤ p̂ ≤ U CL|p = 0.42}
P {nLCL ≤ np̂ ≤ nU CL|λ = np}
P {np̂ ≤ nU CL|λ} − P {np̂ < nLCL|np}
P {np̂ ≤ 50(0.37)|λ = np = (50)0.42} − P {np̂ < 50(0.03)|λ = np = (50)0.42}
P {np̂ ≤ 18.5|λ = 21} − P {np̂ < 1.5|λ = 21} =
(choose one) 0.074 / 0.143 / 0.302
(n) Average run length, poisson approximation
The average run until the process goes out of control, since β = 0.302, is
ARL1 =
(choose one) 1 / 3 / 5
1
1
=
= 1.432 ≈
1−β
1 − 0.302
2. Another p chart example, summary data
A fraction nonconforming control chart with n = 350 has the following parameters.
UCL = 0.345
Center line = 0.1725
LCL = 0
(a) Probability of detecting a shift in proportion?, normal approximation
The probability5 a shift in (average) proportion to 0.45 will be detected
on the first subsequent sample (lot), after the initial 350 samples, is
=
=
=
=
=
=
=
P {detected 1st sample|p = 0.45}
1 − P {not detected 1st sample|p = 0.45}
1 − P {LCL ≤ p̂ ≤ U CL|p = 0.45}
1 − [P {p̂ ≤ U CL|p = 0.45} − P {p̂ < LCL|p = 0.45}]
1 − P {p̂ ≤ U CL|p = 0.45} + P {p̂ < LCL|p = 0.45}
LCL − p
U CL − p
|p = 0.45} + P {Z < p
|p = 0.45}
1 − P {Z ≤ p
p(1 − p)/n
p(1 − p)/n
0.345 − 0.45
0 − 0.45
1 − P {Z ≤ p
} + P {Z < p
}
0.45(1 − 0.45)/350
0.45(1 − 0.45)/350
1 − P {Z ≤ −3.949} + P {Z < −19.922} =
5
We will use a normal approximation to the binomial distribution in this case; notice, in par} = P {Z ≤ −0.146} ≈ 0.442 using the TI–83 function,
ticular, that P {Z ≤ √ 0.37−0.38
0.38(1−0.38)/50
normalcdf(−E99, −0.146).
100
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
(choose one) 0 / 0.567 / 1
(b) Critical value, L, normal approximation
Determine the critical value of the control limits, L, assuming the control
limits are based on the the normal approximation to underlying binomial
distribution.
Using the upper limit.
Since
UCL = p + L
r
p(1 − p)
n
then the standardized limit, L, is
UCL − p
0.345 − 0.1725
L= q
=q
≈
p(1−p)
n
0.1725(1−0.1725)
350
(choose one) 1.23 / 3.45 / 8.54
Using the lower limit.
Since
LCL = p − L
r
p(1 − p)
n
then the standardized limit, L, is
p − LCL
0.1725 − 0
L= q
=q
p(1−p)
n
0.1725(1−0.1725)
350
≈
(choose one) 1.23 / 3.45 / 8.54
Comparing two methods.
It (choose one) does / does not matter whether we use the upper or
lower control limits when determine the critical value, L.
What is L?
True / False The L is the distance between the center line and either
one of the limits in standard deviation units.
(c) Comparing p and np chart
Rather than use the 3σ p (proportion) chart,
r
p̄(1 − p̄
(LCL, UCL) = p̄ ± 3
= (0, 0.345)
n
it is often informative to also look at the related 3σ np (number, or count)
chart
(nLCL, nUCL) = (350(0), 350(0.345)) =
Section 3. Control Charts for Nonconformities (ATTENDANCE 5)
101
(choose one)
i. (0, 120.50)
ii. (0, 120.75)
iii. (0, 121)
6.3
Control Charts for Nonconformities
SAS program: att5-6-3-book-cchart,uchart
We look at the c chart and the u chart in this section. The c chart and u charts are
both used when we investigate the number of defects for one unit, as compared to
many units, as was the case for the p chart above. The poisson distribution is used
as the underlying model for both of the c chart and the u chart.
Exercise 6.2 (Control Charts for Nonconformities)
1. c chart: letter misprints per book page, raw data
The number of letter misprints (nonconformities, defects) per page of a book,
where 24 pages have been taken at random from this book, is given below.
page
1
2
3
4
5
6
7
8
9
10
11
12
c i = Di
7
0
7
8
17
2
6
6
8
5
3
5
page
13
14
15
16
17
18
19
20
21
22
23
24
c i = Di
2
5
1
5
1
9
0
1
0
9
3
8
(a) Data poisson?
True / False The mistakes per page data is best described by the poisson6
6
This data could also be described by a binomial distribution with parameters p and n, but as n
becomes larger and larger, the calculations involved become increasingly difficult. Try it: use your
calculator to determine a binomial distribution where n = 100, then n = 1000, then n = 10000
where p = 0.01 and x = 3, say. We use the poisson distribution instead of the binomial in this
section because n is very large.
102
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
distribution because we assume the chance (p) of any letter on a page to
be mistyped to be small and also the number of letters (n) on any page to
be large7 .
(b) Notation
number of typos per page for page 6, c6 = D6 = (choose one) 1 / 2 / 3
average number of typos per page8 , c ≈ c̄ = (choose one) 1.8 / 4.9 / 6.3
number of pages, m = (choose one) 12 / 24 / 36
(c) c chart
From SAS, the c chart is (choose one) in / out of statistical control because
there are is an out-of-control signal9 at page (choose one) 4 / 5 / 6
(d) 3σ (LCL, UCL) for c chart, poisson approximation
FromPSAS, the average of number of letter misprints per page is,
c̄ = mci = (choose one) 1.5 / 2.0 / 4.9
The upper control limit and the lower control limit for c chart are,
√
√
c̄ ± 3 c̄ = 4.9 ± 3 4.9 = (−1.7, 11.6) =
(choose one)
i. (−1.7, 0)
ii. (0, 11.6)
iii. (0, 12.6)
(e) Probability of type I error, poisson approximation
The probability of a type I error, of accidently saying the process (based
on the 24 pages) is out of control (has more letter misprints than usual),
is
α =
=
=
=
=
P {accidently saying process out of control|c = λ = 4.9}
1 − P {LCL ≤ D ≤ U CL|c = λ = 4.9}
1 − [P {D ≤ U CL|λ} − P {D < LCL|c = λ = 4.9}]
1 − P {D ≤ 11.6|c = λ = 4.9} + P {D < 0|c = λ = 4.9}
1 − P {D ≤ 11|c = λ = 4.9} + P {D < 0|c = λ = 4.9} =
(choose one) 0.012 / 0.003 / 0.005
(f) Probability of type II error, poisson approximation
The probability of a type II error, of accidently saying the process (based
7
Notice that n is assumed to be so large, it is not even specified in the book data here, unlike
previously, for the book binding data, where it was specified.
8
Use SAS output.
9
Out of control means in this case the typical constant number of letter misprints is different for
the one page as compared to the other pages.
Section 3. Control Charts for Nonconformities (ATTENDANCE 5)
103
on the 24 pages) is in control (c̄ = 4.9), when, in fact, c = 2.3, is
β =
=
=
=
P {accidently saying process in control|c = λ = 2.3}
P {LCL ≤ D ≤ U CL|c = 2.3}
P {D ≤ U CL|c = 2.3} − P {D < LCL|c = 2.3}
P {D ≤ 11.6|c = λ = 2.3} − P {D < 0|c = λ = 2.3} =
(choose one) 0.774 / 0.897 / 0.999
(g) Average run length, poisson approximation
The average run until the process goes out of control, since β = 0.999, is
ARL1 =
1
1
=
1−β
1 − 0.897
(choose one) 1 / 9.7 / 10.5
2. u chart: letter misprints per five page samples of a book, raw data
The number of letter misprints (nonconformities, defects) per five (n = 5) pages
of a book, where 18 groups of five pages (5 × 18 = 90 total pages) have been
taken at random from this book, is given below.
5–page sample
1
2
3
4
5
6
7
8
9
n xi
5 7
5 0
5 7
5 8
5 7
5 2
5 6
5 6
5 8
ui
1.4
0
1.4
1.6
1.4
0.4
1.2
1.2
1.6
5–page sample
10
11
12
13
14
15
16
17
18
n xi
5 2
5 5
5 1
5 5
5 1
5 9
5 0
5 1
5 0
ui
0.4
1.0
0.2
1.0
0.2
1.8
0
0.2
0
(a) C chart or u chart?
The c chart is used when we know and use the number of defects per one
unit in our analysis. The u chart is used when do not know the number
of defects for one unit, but do know the total number of defects for more
than one, n, say, units. In this second case, we approximate the number of
defects per unit by the average number of defects, u, which is calculated
by dividing the total number of defects, x, for the n units by n.
True / False The u chart is most appropriate in this case because we are
given the total number of letter misprints per five page samples (and not
one page) from a book.
104
Chapter 6. Control Charts For Attributes (ATTENDANCE 5)
(b) Notation
number of pages in each of 18 samples, n = (choose one) 4 / 5 / 6
number of typos in 5–page sample 6, x6 = (choose one) 1 / 2 / 3
average number of typos in 5–page sample 6,
u6 = 25 = (choose one) 0.4 / 0.8 / 1.4
(c) u chart
From SAS, the u chart is (choose one) in / out of statistical control
because there are no out-of-control signals, runs, trends, or cycles.
(d) 3σ (LCL, UCL) for u chart, poisson approximation
From SAS, the average number of misprints per page is,
ū = (choose one) 0.83 / 1.4 / 4.9
The upper control limit and the lower control limit for u chart are,
r
r
ū
0.83
ū ± 3
= 0.83 ± 3
= (−0.39, 2.06) =
n
5
(choose one)
i. (−1.7, 0)
ii. (0, 2.06)
iii. (0, 12.6)
(e) Probability of type I error, poisson approximation
The probability of a type I error, of accidently saying the process (based
on the 18 groups of 5-page samples) is out of control (has more letter
misprints than usual), is
α =
=
=
=
=
P {accidently saying process out of control|u = λ = 0.83}
1 − P {LCL ≤ D ≤ U CL|u = λ = 0.83}
1 − [P {D ≤ U CL|λ} − P {D < LCL|u = λ = 0.83}]
1 − P {D ≤ 2.06|u = 0.83} + P {D < 0|u = λ = 0.83}
1 − P {D ≤ 2|u = 0.83} + P {D < 0|u = λ = 0.83} =
(choose one) 0.047 / 0.052 / 0.488
(f) Probability of type II error, poisson approximation
The probability of a type II error, of accidently saying the process (based
on the 18 groups of 5-page samples) is in control (ū = 0.83), when, in fact,
ū = 2.3, is
β =
=
=
=
P {accidently saying process in control|u = λ = 2.3}
P {LCL ≤ D ≤ U CL|u = 2.3}
P {D ≤ U CL|u = 2.3} − P {D < LCL|u = 2.3}
P {D ≤ 2.06|u = 2.3} − P {D < 0|u = 2.3} =
(choose one) 0.496 / 0.743 / 0.999
Section 4. Choice between Attributes and Variables Control Charts (ATTENDANCE 5)105
(g) Average run length, poisson approximation
The average run until the process goes out of control, since β = 0.496, is
ARL1 =
1
1
=
≈
1−β
1 − 0.496
(choose one) 2 / 3 / 5
6.4
Choice between Attributes and Variables Control Charts
Attribute charts are used if attributes (qualitative data) is involved; variable charts
are used if variable (quantitative data) is involved.
6.5
Guidelines for Implementing Control Charts
This section discusses how to implement control charts; in particular, which processes
to control, where the charts should be implemented and the type of control charts to
use.