The data shown in Task 1 are the deviations from nominal diameter for holes dril

Question

The data shown in Task 1 are the deviations from nominal diameter for holes drilled in a carbon-fiber composite material used in aerospace manufacturing. The values reported are deviations from nominal in ten-thousandths of an inch. (a) Set up X bar and R charts in 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 C_p File Home InsertPage Layout Formulas DataReview View Aspen Properties Aspen Simulation W PROTECTED VIEW Be careful -files from the internet can contain vinuses. Unless you need to edit it's safer to stay in Protected View J6 Sample Number Observation 1 Observation 2 Observation 3 Observation 4 Observation 5 20 Task 1 Task 2 : Factor Table

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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