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298 Part IV: Quality AssuranceContinuedTable 19.10 Data showing the lengths of tie rods for cars.Sample number X – R Sample number X – R1 274 6 14 268 82 265 8 15 271 63 269 6 16 275 54 273 5 17 274 75 270 8 18 272 46 275 7 19 270 67 271 5 20 274 78 275 4 21 273 59 272 6 22 270 410 273 8 23 274 611 269 6 24 273 512 273 5 25 273 613 274 7Part IV.B.5EXAMPLE 19.13Suppose that the process in example 19.12 had some setback and as a result of that theprocess had an upward shift. Furthermore, suppose that after the process experiencedthis shift we took another set of 25 random samples of size n = 5 and these samples producedX – – –= 276 and R = 6. Clearly, in this example the value of X changed from 272to 276 while R – remained the same.Solution:From the given values of X – – and R , we obtainmˆ = 276 andsˆ = 2.58Since the process standard deviation did not change, the value of Ĉ p remained the same,that is, Ĉ p = 1.03. However, the percentage of nonconforming tie rods produced by theprocess will beContinued

Chapter 19: B. Statistical Process Control 299( )Percentage of nonconforming = 1P(264 X 280)100% 264276 X 276 280276= 1P 100258 . 258 . 2.58 %( ( ))= 1P 4. 65 Z 1. 55 100%= 606 . %.ContinuedThus, even though the value of C p did not change after the process mean experienced ashift, the process is producing nonconforming units at a rate 30 times more than in theprevious example. This implies that C p did not measure the effect that the upward ordownward shift had on the ability of the process to produce products within the specificationlimits.This major drawback of C p makes it less reliable than many other processcapability indices available in the literature. We will study some of them here.However, before we study other PCIs let us see another alternative but equivalentinterpretation of C p that is given by finding the percentage of specification bandused, that is1Percentage of specification band used = × 100.CpA smaller percentage of specification band used indicates a better process.Again, for reasons discussed above, this interpretation sometimes can also bemisleading.The other two process capability indices, first used by the Japanese, are C pland C pu . These indices are related to the lower specification limit and upper specificationlimit, respectively, and are defined as follows:Cpl=m LSL3 s(19.64)Part IV.B.5Cpu =The estimates of C pl and C pu are given byUSL m. (19.65)3sCˆCˆpl=pu =X LSL3s ˆUSL X.3s ˆ(19.66)(19.67)To illustrate the computation of Ĉ pl and Ĉ pu we use the information in Examples19.12 and 19.13.

Chapter 19: B. Statistical Process Control 299

( )

Percentage of nonconforming = 1P(

264 X 280)

100%

264

276 X 276 280

276

= 1

P

100

258 . 258 . 2.

58

%

( ( ))

= 1P 4. 65 Z 1. 55 100%

= 606 . %.

Continued

Thus, even though the value of C p did not change after the process mean experienced a

shift, the process is producing nonconforming units at a rate 30 times more than in the

previous example. This implies that C p did not measure the effect that the upward or

downward shift had on the ability of the process to produce products within the specification

limits.

This major drawback of C p makes it less reliable than many other process

capability indices available in the literature. We will study some of them here.

However, before we study other PCIs let us see another alternative but equivalent

interpretation of C p that is given by finding the percentage of specification band

used, that is

1

Percentage of specification band used = × 100.

Cp

A smaller percentage of specification band used indicates a better process.

Again, for reasons discussed above, this interpretation sometimes can also be

misleading.

The other two process capability indices, first used by the Japanese, are C pl

and C pu . These indices are related to the lower specification limit and upper specification

limit, respectively, and are defined as follows:

C

pl

=

m LSL

3 s

(19.64)

Part IV.B.5

C

pu =

The estimates of C pl and C pu are given by

USL m

. (19.65)

3s

pl

=

pu =

X LSL

3s ˆ

USL X

.

3s ˆ

(19.66)

(19.67)

To illustrate the computation of Ĉ pl and Ĉ pu we use the information in Examples

19.12 and 19.13.

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