Question 40.1: x and R Charts Eight samples ( m = 8) of size 4 ( n = 4) hav...
\bar{x} and R Charts
Eight samples ( m = 8) of size 4 ( n = 4) have been collected from a manufacturing process that is in statistical control, and the dimension of interest has been measured for each part. It is desired to determine the values of the center, LCL , and UCL for \bar{x} and R charts. The calculated x values (units are given in cm) for the eight samples are 2.008, 1.998, 1.993, 2.002, 2.001, 1.995, 2.004, and 1.999. The calculated R values (cm) are, respectively, 0.027, 0.011, 0.017, 0.009, 0.014, 0.020, 0.024, and 0.018.
The calculation of \bar{x} and R values above is step 1 in the procedure.
In step 2, the grand mean of the sample averages is computed:
\bar{\bar{x}}=(2.008+1.998+\cdot\cdot \cdot +1.999)/8= 2.000
In step 3, the mean value of R is computed:
\bar{R}=(0.027+0.011+\cdot\cdot \cdot +0.018)/8= 0.0175
In step 4, the values of LCL and UCL are determined based on factors in Table 40.2 .
| Sample Size n | \bar{x} Chart A_2 | R Chart | |
| D_3 | D_4 | ||
| 3 | 1.023 | 0 | 2.574 |
| 4 | 0.729 | 0 | 2.282 |
| 5 | 0.577 | 0 | 2.114 |
| 6 | 0.483 | 0 | 2.004 |
| 7 | 0.419 | 0.076 | 1.924 |
| 8 | 0.373 | 0.136 | 1.864 |
| 9 | 0.337 | 0.184 | 1.816 |
| 10 | 0.308 | 0.223 | 1.777 |
First, using Equation (40.3) for the \bar{x} chart,
LCL =\bar{\bar{x}}-A_2\bar{R} and UCL =\bar{\bar{x}}+A_2\bar{R} (40.3)
LCL = 2.000 – 0.729 (0.0175) = 1.9872
UCL = 2.000 + 0.729 (0.0175) = 2.0128
and for the R chart using Equation (40.4) ,
LCL =D_3\bar{R} and UCL = D_4\bar{R} (40.4)
LCL = 0 (0.0175) = 0
UCL = 2.282 (0.0175) = 0.0399
The two control charts are constructed in Figure 40.2 with the sample data plotted in the charts.
