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298 Part IV: Quality Assurance
Continued
Table 19.10 Data showing the lengths of tie rods for cars.
Sample number X – R Sample number X – R
1 274 6 14 268 8
2 265 8 15 271 6
3 269 6 16 275 5
4 273 5 17 274 7
5 270 8 18 272 4
6 275 7 19 270 6
7 271 5 20 274 7
8 275 4 21 273 5
9 272 6 22 270 4
10 273 8 23 274 6
11 269 6 24 273 5
12 273 5 25 273 6
13 274 7
Part IV.B.5
EXAMPLE 19.13
Suppose that the process in example 19.12 had some setback and as a result of that the
process had an upward shift. Furthermore, suppose that after the process experienced
this shift we took another set of 25 random samples of size n = 5 and these samples produced
X – – –
= 276 and R = 6. Clearly, in this example the value of X changed from 272
to 276 while R – remained the same.
Solution:
From the given values of X – – and R , we obtain
mˆ = 276 andsˆ = 2.58
Since 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 the
process will be
Continued