The following are quality control data for a manufacturing process at Kensport Chemical Company. The data show the temperature in degrees centigrade at five points in time during a manufacturing cycle.
Sample |
x |
R |
---|---|---|
1 | 95.72 | 1.0 |
2 | 95.24 | 0.9 |
3 | 95.18 | 0.7 |
4 | 95.46 | 0.4 |
5 | 95.46 | 0.5 |
6 | 95.32 | 1.1 |
7 | 95.40 | 1.0 |
8 | 95.44 | 0.3 |
9 | 95.08 | 0.2 |
10 | 95.50 | 0.6 |
11 | 95.80 | 0.6 |
12 | 95.22 | 0.2 |
13 | 95.60 | 1.3 |
14 | 95.22 | 0.5 |
15 | 95.04 | 0.8 |
16 | 95.72 | 1.1 |
17 | 94.82 | 0.6 |
18 | 95.46 | 0.5 |
19 | 95.60 | 0.4 |
20 | 95.74 | 0.6 |
The company is interested in using control charts to monitor the temperature of its manufacturing process. Compute the upper and lower control limits for the R chart. (Round your answers to three decimal places.)
UCL=
LCL=
Sample | x | R |
1 | 95.72 | 1 |
2 | 95.24 | 0.9 |
3 | 95.18 | 0.7 |
4 | 95.46 | 0.4 |
5 | 95.46 | 0.5 |
6 | 95.32 | 1.1 |
7 | 95.4 | 1 |
8 | 95.44 | 0.3 |
9 | 95.08 | 0.2 |
10 | 95.5 | 0.6 |
11 | 95.8 | 0.6 |
12 | 95.22 | 0.2 |
13 | 95.6 | 1.3 |
14 | 95.22 | 0.5 |
15 | 95.04 | 0.8 |
16 | 95.72 | 1.1 |
17 | 94.82 | 0.6 |
18 | 95.46 | 0.5 |
19 | 95.6 | 0.4 |
20 | 95.74 | 0.6 |
average | 95.401 | 0.665 |
x̅̅ = 95.40
R̅ = 0.67
D4= 2.114
D3= 0
UCLr = D4*R̅ = 1.406
LCLr = D3*R̅ =
0.000
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