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Chapter 19: B. Statistical Process Control 297
EXAMPLE 19.12
Table 19.10 gives the summary statistics on X – and R for 25 samples of size n = 5 collected
from a process producing tie rods for certain types of cars. The measurement data are
the lengths of tie rods and the measurement scale is in mm.
The target value and the specification limits for the length of the rods are 272 and
272±8, respectively. Calculate the value of C p assuming that the tie rod lengths are
normally distributed, and find the percentage of nonconforming tie rods produced by
the process.
Solution:
In order to find the value of Ĉ p and the percentage of nonconforming tie rods produced
by the process, we first need to estimate the process mean m and the process standard
deviation s. These estimates may be found by using X – – and R /d2 , respectively. Thus,
we get
m
1
m ˆ = X = X i
m
i=
1
1
= ( 274+ 265+ 269+ ... + 273)=
272
25
m
1
s ˆ = R/ d = /
m R d i
2 2
i=
1
1
= ( + + ... + + )/ .
25 6 8 5 6 2 326
= 6/ 2. 326 = 2.
58
where the value of d 2 for different sample sizes is found from Appendix E. Substituting
the values of USL, LSL, and sˆ in equation (19.62), we get
ˆ 280
284
C p
= = 103 .
6258 .
( )
which indicates that the process is capable. To find the percentage of nonconforming
tie rods produced by the process we proceed as follows:
( )
Percentage of nonconforming = 1P(
264 X 280)
100%
= ( 1P( 3. 1 Z 3. 1)
) 100%
= 1
0.
9980 100% = 0. 2%
( )
Part IV.B.5
Thus, in this example, the percentage of nonconforming tie rods is as expected when
we consider the value of Ĉ p . But, as noted earlier, this may not always be the case since
C p does not take into consideration where the process mean is located. To better explain
this we use Example 19.13.
Continued