control charts for non-normal process

8/3/2019 Control Charts for Non-normal Process

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CONTROL CHARTS FOR

NON-NORMALPROCESSES

P-CHARTS

np-CHARTS

C-CHARTS

U-CHARTS


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Stephen A. DeLurgio and MGraw-Hill, 2004 all rights reserved.

WHAT GENERATES ND OUTPUT?

IF AN EVENT IS THE RESULT OF A RELATIVELY LARGENUMBER OFSMALL, CHANCE, INDEPENDENTINFLUENCES, THEN ITS OUTPUT WILL BE ND.

MANY PROCESSES ARE ND BECAUSE:

WE HAVE WORKED HARD TO ELIMINATE THE VERY LARGEINFLUENCES, THUS ONLY A RELATIVELY LARGE NUMBEROF SMALL, INDEPENDENT INFLUENCES REMAIN.

WHAT IF A PROCESS IS NOT NORMALLY DISTRIBUTED?

THAT IS OUR FOCUS HERE!


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Statistical Sampling--Data

Attribute (Discrete, Go no-go information)

Defectives--refers to the acceptability of product

across a range of characteristics.

Defects--refers to the number of defects per unit--

may be higher than the number of defectives.

Variable (Continuous)

Usually measured by the mean and the standard

deviation.


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THE BINOMIAL DISTRIBUTION

Attribute (Go no-go information)

We monitor the number of defectives over time.

The relevant population parameter beingcontrolled is the population proportion Pie (P)

We want to assure that the population

proportion remains in control.

We want to make sure the populationproportion does not become defective.

The critical assumption is that Pie remains

constant over time.


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DISTRIBUTION OF SAMPLE PROPORTIONS

POP IS NOT ND

= .98

SAMPLE LOOKS

LIKE POP,

P= .99

DIST. OF SAMPLE

PS IS ND


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P-CHARTS

Require large samples nu30.

When population proportion is known:

P=T s Z T(1 - T)/n

When population proportion is unknown:

_ _ _P=P s Z P(1 - P)/n

Where P-Baris an estimate ofT


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Constructing a p-Chart

No. of defective Monitors

Sample n Defectives

1 100 4

2 100 2

3 100 5

4 100 35 100 6

6 100 4

7 100 3

8 100 8

9 100 1

10 100 2

11 100 3

12 100 2

13 100 2

14 100 8

15 100 3


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p =Total Number of Defectives

Total Number ofObservations

= p (1- p)

npS

UCL = p + Z

LCL = p - Z

p

p

s

s

P-CHART FORMULAS


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1. Calculate the sample proportion,

p, for each sample.

Defectives p

4 0.04

2 0.02

5 0.05

3 0.03

6 0.06

4 0.04

3 0.03

7 0.07

1 0.01

2 0.023 0.03

2 0.02

2 0.02

8 0.08

3 0.03

Stephen A. DeLurgio and The McGraw-Hill Companies, Inc., 1998Irwin/McGraw-Hill


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2. Calculate the average of the sample proportions.

0.037=1500

55=p

3. Calculate the standard deviation of the

sample proportion

.0188=

100

.037)-.037(1=

n

)p-(1p

=

sp

The McGraw-Hill Companies, Inc.,Irwin/McGraw-Hill


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4. Calculate the control limits.

3(.0188).037 sUCL = 0.093

LCL = -0.0197 (or 0)

p

p

sZ-p=LCL

sZ+p=UCL



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p-Chart (Continued)

5. Plot the individual sample proportions, the average

of the proportions, and the control limits

You will be asked to duplicate these results using SPSS.

What do you infer from the following control chart?


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Control Chart: PCHART

Sigma level: 3

15.00

14.00

13.00

12.00

11.00

10.00

9.00

8.00

7.00

6.00

5.00

4.00

3.00

2.00

1.00

Pro

portionNonconforming

.10

.08

.06

.04

.02

0.00

PCHART

UCL = .0942

Center = .0373

LCL = .0000


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Np-CHART

Sometimes we express binomial occurrences in units.

The formula is simply:

Mean = nP-bar (e.g., 100*.037 = 3.7

Were n = number in sample

P = best estimate of the population proportion

Sigma(np) = Sqrt(nP-bar(1-P-bar)


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DISTRIBUTION OF SAMPLE DEFECTS - nP

POP IS NOT ND

n = 100*.98 = 98

SAMPLE LOOKS

LIKE POP,

P= 100*.99 = 99

DIST. OF SAMPLE

nPS IS ND


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nP TheorynP-CHARTS

Require large samples nu30:

When population proportion is known:

nP = nT s Z nT(1 - T)When population proportion is unknown:

_ _ _

nP = nP s Z nP(1 - P)Where P-Baris an estimate ofT


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Constructing an np-Chart Step 1,

Sample n Defectives = np

1 100 4

2 100 2

3 100 5

4 100 35 100 6

6 100 4

7 100 3

8 100 8

9 100 1

10 100 211 100 3

12 100 2

13 100 2

14 100 8

15 100 3


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Calculate nP-bar

nsObservatioSample

DefectivesofNumberTotal=p

)p1(pn=Snp

np

np

sZ-pn=LCL

sZ+pn=UCL


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2. Calculate the average of the sample proportions.

3.7=150055*100=pn

3. Calculate the standard deviation of the

sample proportion

1.89=.037)-3.7(1

=

)p-(1np=snp



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4. Calculate the control limits.

3(1.896)3.73sUCL = 9.42

LCL =0

np

np

sZ-pn=LCL

sZ+pn=UCL



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p-Chart (Continued)

5. Plot the individual sample proportions, the average

of the proportions, and the control limits

You will be asked to duplicate these results using SPSS.

What do you infer from the following np-chart?


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Control Chart: PCHART

Sigma level: 3

15.

00

14.00

13.

00

12.00

11.

00

10.00

9.

00

8.00

7.

00

6.00

5.

00

4.00

3.

00

2.00

1.

00

NumberNonconfo

rming

10

8

6

4

2

0

PCHART

UCL = 9.4207

Center = 3.7333

LCL = .0000


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DEFECTS PER UNIT OF TIME

OR SPACE C Charts

Frequently, there are processes that yield

distributions that follow a Poisson Distribution.

The Poisson Distribution is a discrete distribution

which takes on the values X = 0, 1, 2, 3,...

It models the events per unit of time or space.

Determined by its mean,

___

C

___

C


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C-CHARTS POISSON DIST.

In Quality Control the Poisson Distributionmeasures:

Number of blemishes or defects per unit oftime or space such as:

Blemishes per sqft. on painted panel.

Defective ICs perWafer.

Number of defects per sqft. on furnituresurface.

Number of typos per page in your paper.


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Consider Defective ICs on a Wafer

How about 1 Giga Bit Chips


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http://www.austin.cc.tx.us/HongXiao/overview/history-semi/sld015.htm


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http://www.austin.cc.tx.us/HongXiao/

overview/history-semi/sld016.htm


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POISSON

GENERATING PROCESS

Probability of an occurrence is very very low.

No. of possible points is very very high.

Prob. remains constant.

Defined completely by its mean.

Variance = Mean

Standard Deviation = Sqrt(Mean) =sqrt( )

A Skewed distribution to the right.

High probability of low number, very low, but afinite probability of high number.

___

C

___

C


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POISSON AND EXPONENTIAL

If the time or space between events follows

an exponential distribution, then the rate of

occurrence of the event will likely follow a

Poisson.

That is, the Poisson and Exponential

Distribution are inverses of each other.


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Source:http://info.bio.cmu.edu/Course

s/03438/PBC97Poisson/PoissonPage.

html#distribution


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C-Chart

Mean =

UCL = + 3 sqrt( )

LCL = - 3 sqrt( )

Consider an example, a company measures

the number of defects per square foot of

expensive floor tile.

Service example, mistakes made per hour per

call center worker.

___

C

___

C

___

C

___

C

___

C


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SAMPLE DEFECTS/SQFT

1 5

2 4

3 7

4 65 8

6 5

7 6

8 5

9 16

10 10

11 912 7

13 8

14 11

15 9

16 5

17 7

18 619 10

20 8

21 9

22 9

23 7

24 5

25 7


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SPSS DESCRIPTIVE STATISTICS

Statistics

.

.

.

a

.

.

.

a

ss ng

N

ean

e an

o e

.

ev a on

ar ance

um

u p e mo es ex s . e sma es va ue s s own.


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POISSON

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. . .

. .

.

.

.

.

.

.

.

.

.

o a

Valid

Cumulative


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POISSON

16.014.012.010.08.06.04.0

10

8

6

4

2

0

Std. Dev = 2.57

Mean = 7.6

N = 25.00


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C-Chart What Do You Infer?

Control Chart: POISSON

Sigma level: 3

25.00

23.00

21.00

19.00

17.00

15.00

13.00

11.00

9.00

7.00

5.00

3.00

1.00

Nonconform

ities

20

10

0

POISSON

UCL = 15.

8086

Center = 7.5600

LCL = .0000


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U-CHART Nonconformities

Per Unit

Used in same situation as c-Chart but when

sample sizes vary.

This means area of time or space varies.

Consider situation with different size panels

or furniture.

Service: Mistakes made for employees in a

call center, but actual phone time varies.

We want to control defects per square foot.


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SAMPLE SQMETERS NONCONF

1 200 5

2 300 14

3 250 8

4 150 8

5 250 12

6 100 6

7 200 20

8 150 10

9 150 6

10 250 10

11 300 9

12 250 16

13 200 12

14 250 10

15 100 6

16 200 8

17 200 5

18 100 5

19 300 14

20 200 8

MEANS 205 9.6


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Whats the problem here?

NONCONFO

20.017.515.012.510.07.55.0

7

6

5

4

3

2

1

0

Std.

Dev = 4.

06Mean = 9.6

N = 20.00


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NONCONFORMITY/100 SQM

Looks more like a Poisson

NONP100

20.018.016.014.012.010.08.06.0

6

5

4

3

2

1

0

Std. Dev = 3.56

Mean = 9.9

N = 20.00


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u-Chart for Carpet Data

Mitra page 354

Control Chart: NONCONFO

Sigma level: 3

20.00

19.00

18.00

17.00

16.00

15.00

14.00

13.00

12.00

11.00

10.00

9.00

8.00

7.00

6.00

5.00

4.00

3.00

2.00

1.00

Fractionof

Nonconformities

.12

.10

.08

.06

.04

.02

0.00

NONCONFO

UCL

Center = .0468

LCL


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So:http://www.mathsrevision.net/alev

el/statistics/normal_distribution2.php

Poisson Approximation

The normal distribution can also be used to

approximate the Poisson distribution for largevalues of C (the mean of the Poisson

distribution).

If X ~ Po(C) then for large values of C, X ~

N(C, C) approximately. This last statementdenotes that X is ND with mean of C and

variance of C.


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ND APPROXIMATIONS FOR

BINOMIAL

For Binomial, the higher the value of N andthe closer p is to .5, the better the NDapproximates the Binomial. (See link on next

slide.) When n*p nd n*(1-p) are muchgreater than 5 then the ND approximates theBinomial Dist.

For large n (say n > 20) and p not too near 0

or 1 (say 0.05 < p < 0.95) the distributionapproximately follows the Normal distribution.


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SIMULATING APPROXIMATIONS

Fun with statistics:

Normal Approximation of Binomial

http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html


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So:http://www.mathsrevision.net/alev

el/statistics/normal_distribution2.php

Binomial Approximation

The normal distribution can be used as an

approximation to the binomial distribution, under

certain circumstances, namely:

If X ~ B(n, p) and if n is large and/or p is close to ,

then X is approximately N(np, npq)

(where q = 1 - p).

In some cases, working out a problem using the

Normal distribution may be easier than using a

Binomial.

control charts for non-normal process

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