As 6 Control Charts for Attributes

1Zhisheng(James) Zhang, School of Mechanical Engineering, Southeast University, [email protected]

Ch6 Control Charts for Attributes

• p and np chart• c and u chart• Variable sample size• OC curve and ARL

属性、特征


Terminology• Nonconforming: Defective • Conforming: Nondefective• Attributes: Quality characteristics of conforming

or non conforming• The fraction nonconforming: the ratio of the

number of nonconforming items in a population to the total number of items in that population.

#ˆ#

ii

D of nonconformingpn Total of samples

= =


Review of Binomial Distribution

Let x = # defects in a sample of size n where defects follow a Bernoulli Process (two outcomes, p-constant, x - independent)

f (x) =nx

⎛ ⎝ ⎜ ⎞

⎠ px (1 − p)n − x

E(x) = np

V(x) = np(1 − p)


Sample Estimation of p

2

ˆL e t

( )ˆ( )

1 (1 )ˆ( ) ( )

xpn

E x n pE p pn n

p pV p V xn n

=

= = =

−= =


Review of the Basic Model of a Control Chart

Let w be a sample statistic that measures some quality characteristic of interest, andsuppose that the mean of w is μw and the standard deviation of w is σw. Then the centerline, the upper control limit, and the lower control limit become

UCL = μw + k σw

Center line = μw

LCL = μw - k σw

where k is the "distance" of the control limits from the center line, expressed in standarddeviation units


Fraction Nonconforming Control Chart (p-Chart)

(1 )ˆF o r n p la rg e ~ , p pp N pn−⎛ ⎞⇒ ⎜ ⎟

⎝ ⎠

Binomial (n>10, p close to 0.5) ⇒ normal

If LCLp <0, set LCLp=0

Remarks: When data points are plotted below LCL, they generally do not represent a real improvement. Actually, they are often caused by errors in the inspection rather than a process improvement

ˆ

ˆ

(1 ) 3

Centerline

(1 ) 3

p

p

p pUCL pn

p

p pLCL pn

−= +

=

−= −


How to Establish a p-Chart?

• Is there an assignable cause for out-of-control points or a nonrandom pattern? If so, find the root causes and delete these points, and then updatecontrol limits.

If p unknown, conduct a test and trial control limits with

p)p(Em

p

mn

Dp

m

1ii

m

1ii

=

==∑∑

==

pp =

m=20-25samples for constructing trial control limits


A P-Chart Example

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 10 20 30 40 50 60 70 80 90 100

Revised control chart after making adjustments

Out-of-control point


Example The following data give the number of nonconforming ROM chips in samplesof size 200. Construct a p chart for these data. Assume that any values beyond the controllimits have an assignable cause and revise the control limits as appropriate.

Sample Nonconforming Sample Nonconforming 1 19 12 182 7 13 173 11 14 214 29 15 165 24 16 166 24 17 237 15 18 148 25 19 49 11 20 21

10 10 21 2411 37 22 10


Example (Textbook problem 6-18) A fraction nonconforming control chart with center line0.10, UCLp = 0.19, and LCLp = 0.01 is used to control a process.

a. If 3-sigma limits are used, find the sample size for the control chart.

b. Use the Poisson approximation to the binomial to find the probability of type I error.

c. Use the Poisson approximation to the binomial to find the probability of type II error if theprocess fraction defective is actually p = 0.20.


Design of the Fraction Nonconforming Control Chart (p Chart)

• Three key parameters:– the sample size; the frequency of sampling; the width

of the control limits• General Guidelines:

– select n so that the probability of finding at least one nonconforming unit per sample is at least r Pr{D≥1} ≥r

– p small ---> n large

– Duncan approach: 50% chance of detecting a process shift, i.e. p1=p0+δ (δ>0). P{x>UCL| p1}=0.5

− δ small⇒ n large

– a positive lower control limit

)p1(pLn2

−⎟⎠⎞

⎜⎝⎛

δ=

round n to integer2Lp

p1n −>


Variable Sample Size• Variable width of control limits

corresponding to each sample size– not appropriate for nonrandom pattern check

• Constant width of control limits using average sample size– future sample size should not differ greatly

• Standardized Control Chart– can be used to check a nonrandom pattern– no reference to the actual process fraction defective

ip

ip n

)p1(p3pLCL;n

)p1(p3pUCLii

−−=

−+=

nsobservatioof#totaldefectsof#total

n

Dp m

1ii

m

1ii

==

∑

∑

=

=

m

nn

m

1ii∑

==

0CL;3LCL;3UCL;pp;

n)p1(p

ppZ

i

ii =−===

−−

=

MEAN 0.223

LWL

UWL

LCL

UCL

0

0.1

0.2

0.3

0.4

0.5

Def

ects

0 10 20 30

Row Numbers

P Chart Using Cricketgraph III

Remarks: Use Approaches 1 and 3 together (Example, Table 6-4, Figs. 6-6, 6-7, 6-8; P281-283)


I n(i) D(i) pi=D(i)/n(i) sigma=sqrt(pbar*(1-pbar)/n(i)) LCL (ni) UCL(ni) LCL (n bar) UCL(n bar) Zi LCL(stand)UCL(stand)1 100 12 0.120 0.029 0.007 0.184 0.007 0.185 0.81 -3 32 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 33 80 6 0.075 0.033 0.000 0.194 0.007 0.185 -0.64 -3 34 100 9 0.090 0.029 0.007 0.184 0.007 0.185 -0.20 -3 35 110 10 0.091 0.028 0.011 0.180 0.007 0.185 -0.18 -3 36 110 12 0.109 0.028 0.011 0.180 0.007 0.185 0.47 -3 37 100 11 0.110 0.029 0.007 0.184 0.007 0.185 0.48 -3 38 100 16 0.160 0.029 0.007 0.184 0.007 0.185 2.17 -3 39 90 10 0.111 0.031 0.003 0.188 0.007 0.185 0.49 -3 310 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 311 110 20 0.182 0.028 0.011 0.180 0.007 0.185 3.06 -3 312 120 15 0.125 0.027 0.015 0.176 0.007 0.185 1.08 -3 313 120 9 0.075 0.027 0.015 0.176 0.007 0.185 -0.78 -3 314 120 8 0.067 0.027 0.015 0.176 0.007 0.185 -1.09 -3 315 110 6 0.055 0.028 0.011 0.180 0.007 0.185 -1.48 -3 316 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 317 80 10 0.125 0.033 0.000 0.194 0.007 0.185 0.88 -3 318 80 7 0.088 0.033 0.000 0.194 0.007 0.185 -0.26 -3 319 90 5 0.056 0.031 0.003 0.188 0.007 0.185 -1.30 -3 320 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 321 100 5 0.050 0.029 0.007 0.184 0.007 0.185 -1.56 -3 322 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 323 100 10 0.100 0.029 0.007 0.184 0.007 0.185 0.14 -3 324 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 325 90 9 0.100 0.031 0.003 0.188 0.007 0.185 0.13 -3 3

sum 2450 234average 98 pbar= 0.096

p bar=total defective/total samples

Fig 6-6 control chart w ith variable sample size

0.0

0.1

0.2

0.3

1 4 7 10 13 16 19 22 25

sample index

p

0.0

0.1

0.1

0.2

0.2

0.3

1 4 7 10 13 16 19 22 25

p

standardized control chart

-4.0-3.0-2.0-1.00.01.02.03.04.0

1 4 7 10 13 16 19 22 25 28

index

zi


np Control Chart (The number of nonconforming items)

Rather than plotting the fraction nonconforming, we plot the number ofnonconforming items with an “np Chart”:

UCLX = np + 3 np(1–p) Center line = npLCLX = np – 3 np(1–p)

• This chart, too, can be "standardized", adjusted for unequal ni

, etc.

nsobservatioof#totaldefectsof#total

n

DP m

1ii

m

1ii

==

∑

∑

=

=pp =

Example 6-2, P298


Example: The number of transmission cases that required deburring in a 16-day sample of 100 each was as follows:

Day Number Day Number1 5 9 42 4 10 63 3 11 154 2 12 45 6 13 56 3 14 77 9 15 38 6 16 6

Prepare an np chart with trial control limits. Assume that any points plotting out of control have assignable causes, and continue to refine the control limits until no points plot out of control.


I Di UCL(trial) LCL(trial) UCL LCL1 5 12.34 0 11.32 0.002 4 12.34 0 11.32 0.003 3 12.34 0 11.32 0.004 2 12.34 0 11.32 0.005 6 12.34 0 11.32 0.006 3 12.34 0 11.32 0.007 9 12.34 0 11.32 0.008 6 12.34 0 11.32 0.009 4 12.34 0 11.32 0.00

10 6 12.34 0 11.32 0.0011 15 12.34 0 11.32 0.0012 4 12.34 0 11.32 0.0013 5 12.34 0 11.32 0.0014 7 12.34 0 11.32 0.0015 3 12.34 0 11.32 0.0016 6 12.34 0 11.32 0.00

sum 88 UCL(trial) LCL(trial)p bar=88/(100*16)= 0.055 12.34 -1.34 set to zeronp 5.500eliminate point 11pbar=(88-15)/(100*15) 0.048667 UCL LCLnp 4.866667 11.32 -1.59 set to zero

np chart with trial control limits

0.0

5.0

10.0

15.0

20.0

1 3 5 7 9 11 13 15

np chart after eliminate outliers

0.0

5.0

10.0

15.0

20.0

1 3 5 7 9 11 13 15

Di

UCL(trial)

LCL(trial)

UCL

LCL


np Chart Properties

• Advantage– np chart is a scaling of the vertical axis by the constant n,

provide the same information as p chart– np chart needs less calculation ( no need to calculate Di/ni)– often used when n is constant and p is small

• Limitation– not easy for interpretation when n is varied (UCL LCL and Ctr

line all vary)– only plot # of defects without considering sample size, hard to

take action


Example #8: (Textbook Exercise 6-15) A control chart is used to control the fraction nonconforming for a plasticpart manufactured in an injection molding process. Ten subgroups yield the following data:

Sample Number Sample Size No. Nonconforming1 100 102 100 153 100 314 100 185 100 266 100 127 100 258 100 159 100 810 100 8

a. Set up a control chart for the number nonconforming in samples of n = 100.b. For the chart established in part (a), what is the probability of detecting a shift in the process fractionnonconforming to 0.30 on the first sample after the shift has occurred?


OC Curve and ARL

• Type II error for the p chart (OC curve see Fig 6-11, P307)

• ARL0=ARLin-control=1/α• ARL1=ARLout-of-control=1/(1-β)

}|{}|{}|ˆ{}|ˆ{ 1111 pnLCLDPpnUCLDPpLCLpPpUCLpP pppp ≤−<=≤−<=β


Example (Textbook Exercise 6-20) Consider the control chart in Exercise 6-18. Find the average run length if the process fraction nonconforming shifts to 0.20.


• Why need it?– Control the total number of nonconformities in a sample or the

average number of nonconformities per unit• nonconformity/defect: a specification of the quality characteristic is not

satisfied result in a defect or nonconformity, e.g., – weld spots on a car– paint dent on a car body

• A unit may not be “nonconforming”, even though it has several nonconformities. So, nonconforming ≠defects or nonconformities

• Assumption: The occurrence of nonconformities in samples of constant size is well modeled by the Poisson distribution.

– The number of potential location for nonconformities is the infinitely large, and the probability of occurrence of a nonconformity at any location is small and constant

Control Charts for Nonconformities (Defects)- C and U Charts


Statistical Basis of Control Charts for Nonconformities (Defects)

- c and u ChartsLet X = total # of defects in a sample

Assume X ~ Poisson ( E(X) = C )For large C ⇒ X ~ N ( E(X) = C , Var(X) = C )

Further, suppose that a sample size of n inspection units(e.g. 100 yd2 , 144 microprocessors, 1100 employees)Note: n is not necessarily integer.Then, Y = average # of defects per unit in a sample = X / n

E ( Y ) = C / n = UVar ( Y ) = C / n2 = U / n

Note: C vs. U distinction is similar to X vs X–

common in "normal" analyses

• c chart: total number of defects in a sample • u chart: average number of defects per unit in a sample size of n inspection units

X=0,1,2,...!

)(xcexp

xc−

=


Control Charts for Nonconformities (Defects)- c Chart

• Control limits for the c chart with a known c

• If unknown c, c is estimated from preliminary samples of inspection units for constructing trial control limits

• The preliminary samples are examined by the control chart using the trial control limits for checking out-of-control points

Example 6-3, P310

c3cUCL

cCLc3cLCL

+=

=+=

If LCL<0, set LCL=0

samplesofnumbersamplesallindefectsof#total

m

ccc

m

1ii

===∑

=


Control Charts for Nonconformities Per Unit- u Chart

• c: total nonconformities in a sample of n inspection units (n is not necessary be integer)

• u: average # of nonconformities per inspection unit in a sample

• If unknown u, is estimated from preliminary samples of inspection units for constructing trial control limits

Example 6-4, P317

nu3uUCL

uCLnu3uLCL

+=

=

−=

m

uu;

ncu

m

1ii

i

ii

∑===

u


Variable Sample Size of Control Charts for Nonconformities

• If sample size varies, it is always to use a u chart rather than a c chart

• Approaches– Control limits varies with each sample size, but the center

line is constant

– Use a control limits based on an average sample size

– Use a standardized control chart (this is preferred option), with UCL=3, LCL=-3, Center line=0.

• This chart can be used for pattern recognition

uCL;nu3uUCL;

nu3uLCL

ii

=+=−=

m

nn

m

1ii∑

==

i

ii

nu

uuZ −=

Example 6-5, P319


OC Curve and ARL for c and u Charts

• Type II error for the c chart (OC curve see Fig 6-20, P291)

• ARL0=ARLin-control=1/α• ARL1=ARLout-of-control=1/(1-β)

}|{}|{}|{}|{

11

11

unLCLcPunUCLcPuLCLxPuUCLxP

uu

uu

≤−<=≤−<=β


Example (Textbook Exercise 6-45) Find the 3-sigma control limitsfor:a. A c chart with process average equal to four nonconformities.b. A u chart with c = 4 and n = 4.


Following are the number of nonconformities in 20 samples of 50 letter-quality printer cases.Develop the trial control limits for a c chart and if any values are out of control, assume that thecause is assignable. Modify the control limits accordingly.

Sample Nonconf. Sample Nonconf.1 19 11 372 21 12 163 14 13 44 23 14 285 13 15 176 21 16 297 15 17 258 24 18 159 20 19 1110 19 20 19


Day RollsProduced

Number ofImperfections

Day RollsProduced

Number ofImperfections

1 18 12 11 18 182 18 14 12 18 143 24 20 13 18 94 22 18 14 20 105 22 15 15 20 146 22 12 16 20 137 20 11 17 24 168 20 15 18 24 189 20 12 19 22 2010 20 10 20 21 17

Example (Textbook Problem 6-37 ) A paper mill uses a control chart to monitor the imperfection infinished rolls of paper. Production output is inspected for 20 days, and the resulting data are shown below.Use these data to set up a control chart for nonconformities per roll of paper. Does the process appear to bein statistical control? What center line and control limits would you recommend for controlling currentproduction?


Example (Textbook Problem 6-56) A control chart is to be established on a process producingrefrigerators. The inspection unit is one refrigerator, and a control chart for nonconformities is tobe used. As preliminary data, 16 nonconformities were counted in inspecting 30 refrigerators.

a. What are the 3-sigma control limits'?b. What is the α-risk for this control chart?c. What is the β-risk if the average number of defects is actually 2 (i.e., if c = 2.O)?d. Find the average run length if the average number of defects is actually 2.

As 6 Control Charts for Attributes

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