In: Statistics and Probability
The following data represents sample data of 6 oz. bottles of after-shave. Develop the X-bar and R control charts (mean chart and range chart) for the following sample. A process that is considered to be in control measures an ingredient in ounces. Below are the samples taken from the process.
Sample |
1 |
2 |
3 |
4 |
5 |
6 |
Sum |
X-bar |
R |
1 |
6.00 |
5.90 |
6.00 |
5.90 |
6.00 |
5.90 |
|||
2 |
6.00 |
6.00 |
6.30 |
6.04 |
6.02 |
6.00 |
|||
3 |
5.90 |
6.00 |
6.06 |
5.90 |
6.00 |
5.90 |
|||
4 |
5.88 |
6.00 |
6.03 |
5.90 |
6.05 |
5.92 |
|||
5 |
6.10 |
6.00 |
5.96 |
6.00 |
6.10 |
6.02 |
|||
Average= |
a) What are the control limits for the mean chart? (4 points)
b) What are the control limits for the range chart? (4 points)
c) Is the process in control? Show your explanation. (2 points)
Solution:
# SAMPLE |
1 |
2 |
3 |
4 |
5 |
6 |
AVERAGE |
RANGE |
1 |
6 |
5.9 |
6 |
5.9 |
6 |
5.9 |
5.95 |
0.1 |
2 |
6 |
6 |
6.3 |
6.04 |
6.02 |
6 |
6.06 |
0.3 |
3 |
5.9 |
6 |
6.06 |
5.9 |
6 |
5.9 |
5.96 |
0.16 |
4 |
5.88 |
6 |
6.03 |
5.9 |
6.05 |
5.92 |
5.963 |
0.17 |
5 |
6.1 |
6 |
5.96 |
6 |
6.1 |
6.02 |
6.03 |
0.14 |
AVERAGE |
5.993 |
0.174 |
XBAR = 5.993
RBAR = 0.174
A2 VALUE CORRESPONDING TO N = 6 = 0.483
D3 & D4 VALUES CORRESPONDING TO N = 0, 2.003
CONTROL LIMITS FOR XBAR
UCL X = XBAR + (A2 * RBAR) = 5.993 + (0.483 * 0.174) = 6.077
LCL X = XBAR - (A2 * RBAR) = 5.993 - (0.483 * 0.174) = 5.909
CONTROL LIMITS FOR RBAR
UCL R = RBAR * D4 = 0.174 * 2.003 = 0.349
LCL R = RBAR * D3 = 0.174 * 0 = 0
2. YES, SINCE ALL THE POINTS ARE WITHIN THE CONTROL LIMITS, IT SHOWS THAT THE PROCESS SHOWS NATURAL VARIATION AND IS IN CONTROL.
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