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)
Please rate my answer and comment for doubt