6- The data shown in Table 6E.4 are the deviations from nominal diameter for hol

Question

6- The data shown in Table 6E.4 are the deviations from nominal diameter for holes drilled din a carbon- fiber composite material used in aerospace manufacturing. The values reported are deviations from nominal in ten-thousandths of an inch. TABLE 6E.4 Hole Diameter Data for Escrcise 6.9 Sample Number r r, r. 1 30 +50 -20 10 +30 0 +50 -60 -20 +30 3 50 +10 +20 +30 +20 4 -10 10 +30 -20 +50 5 +20 -40 +50 +20 +10 0 0 +10 -40 +20 0 0 +2020-10 8 470 30 +30 -10 0 0 0 +20 -20 +10 10 +10 +20 +30 +10 +50 11 +40 0 +20 0 +20 12 +30 +20 +30 +10 +40 13 430 -30 0 +10 +10 14 +30-10 +5010 30 15 +10 -10 +50 +40 0 16 0 0+30 -10 17 +20 +20 +30 +30-20 18 410 -20 +50 +30 +10 19 +50-10 +40 +200 20 +50 0 0 +30 +10 (a) Set up x and R charts on the process. Is the process in statistical control? (b) Estimate the process standard deviation using the range method. (c) If specifications are at nominal 100, what can you say about the capability of this process? Calculate the PCR Cp

Explanation / Answer

the ranges are obtained for each of the 20 samples.

where range=max(x1,x2,x3,x4,x5)-min(x1,x2,x3,x4,x5)

the data obtained as

sample   x1   x2   x3   x4   x5   range
1 -30   50   -20   10   30   80
2 0   50   -60   -20   30   110
3 -50   10   20   30   20   80
4 -10   -10   30   -20   50   70
5 20   -40   50   20   10   90
6 0   0   40   -40   20   80
7 0   0   20   -20   -10   40
8 70   -30   30   -10   0   100
9 0   0   20   -20   10   40
10 10   20   30   10   50   40
11 40   0   20   0   20   40
12 30   20   30   10   40   30
13 30   -30   0   10   10   60
14 30   -10   50   -10   -30   80
15 10   -10   50   40   0   60
16 0   0   30   -10   0   40
17 20   20   30   30   -20   50
18 10   -20   50   30   10   70
19 50   -10   40   20   0   60
20 50   0   0   30   10   50

then Rbar=mean of the ranges obtained=63.5

here each sample contains n=5 observations.

estimate of the process standard deviation is s=Rbar/d2

from table of constants of control charts for n=5 d2=2.326

hence estimated process standard deviation=63.5/2.326=27.300 [answer]

we have upper specification limit=USL=nominal+100

lower specification limit=LSL=nominal-100

PCR Cp=(USL-LSL)/6s=(nominal+100-nominal+100)/6s=200/(6*27.3)=1.22>1

it means that if the process is in the control, the quality of the products will automatically conform to the specifications desired. this signifies that the process is well in statistical control

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