In: Statistics and Probability
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 | 0.9 |
4 | 95.46 | 0.4 |
5 | 95.46 | 0.5 |
6 | 95.32 | 1.1 |
7 | 95.40 | 0.9 |
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 | 95.56 | 1.3 |
14 | 95.22 | 0.6 |
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 =
Construct the R chart.
Compute the upper and lower control limits for the
x
chart. (Round your answers to three decimal places.)
UCL = LCL =
Construct the
x
chart.
What conclusions can be made about the quality of the process?
The R chart indicates that the process variability is ---Select--- in control out of control . ---Select--- No samples fall One sample falls Two samples fall More than two samples fall outside the R chart control limits. The
x
chart indicates that the process mean is ---Select--- in control out of control . ---Select--- No samples fall One sample falls Two samples fall More than two samples fall outside the
x
chart control limits.
Data:
Sample | x | R |
1 | 95.72 | 1 |
2 | 95.24 | 0.9 |
3 | 95.18 | 0.9 |
4 | 95.46 | 0.4 |
5 | 95.46 | 0.5 |
6 | 95.32 | 1.1 |
7 | 95.4 | 0.9 |
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 | 95.56 | 1.3 |
14 | 95.22 | 0.6 |
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 |
no. of samples = 20, size of each sample = 5.
Mean = | 95.40 | 0.68 |
= 95.40 and = 0.68
Now,
For R chart, | |
LCL = | D3Rbar |
UCL = | D4Rbar |
At n = 5, D3 = 0, D4 = 2.114
Then, LCL = 0 and UCL = 0.68*2.114 = 1.438 and CL = 0.68
R chart:
Process is under statistical control (R bar)
Now, x chart,
For X chart, | |
LCL = | xbar - A2Rbar |
UCL = | xbar + A2Rbar |
At, n=5, A2 = 0.577
then, LCL = 95.4 - 0.577 * 0.68 = 95
and UCL = 95.4 + 0.577*0.68 = 95.79
and CL = 95.4
then, x chart:
The process is statistically out of control (x bar)
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