Question 10.P.2: Control charts for means and ranges. Processing times for ne......
Control charts for means and ranges. Processing times for new accounts at a bank are shown in the following table. Five samples of four observations each have been taken. Use the sample data in conjunction with Table 10.3 to construct upper and lower control limits for both a mean chart and a range chart. Do the results suggest that the process is in control?
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample 5 | |
| 10.2 | 10.3 | 9.7 | 9.9 | 9.8 | |
| 9.9 | 9.8 | 9.9 | 10.3 | 10.2 | |
| 9.8 | 9.9 | 9.9 | 10.1 | 10.3 | |
| 10.1 | 10.4 | 10.1 | 10.5 | 9.7 | |
| Totals | 40.0 | 40.4 | 39.6 | 40.8 | 40.0 |
| TABLE 10.3 Factors for three-sigma control limits for \overline{X} and R charts | |||
| Number of Observations in Sample, n |
Factor for x Chart, A_{2} |
FACTORS FOR R CHARTS | |
| Lower Control Limit, D_{3} |
Upper Control Limit, D_{4} |
||
| 2 | 1.88 | 0 | 3.27 |
| 3 | 1.02 | 0 | 2.57 |
| 4 | 0.73 | 0 | 2.28 |
| 5 | 0.58 | 0 | 2.11 |
| 6 | 0.48 | 0 | 2.00 |
| 7 | 0.42 | 0.08 | 1.92 |
| 8 | 0.37 | 0.14 | 1.86 |
| 9 | 0.34 | 0.18 | 1.82 |
| 10 | 0.31 | 0.22 | 1.78 |
| 11 | 0.29 | 0.26 | 1.74 |
| 12 | 0.27 | 0.28 | 1.72 |
| 13 | 0.25 | 0.31 | 1.69 |
| 14 | 0.24 | 0.33 | 1.67 |
| 15 | 0.22 | 0.35 | 1.65 |
| 16 | 0.21 | 0.36 | 1.64 |
| 17 | 0.20 | 0.38 | 1.62 |
| 18 | 0.19 | 0.39 | 1.61 |
| 19 | 0.19 | 0.40 | 1.60 |
| 20 | 0.18 | 0.41 | 1.59 |
Source: Adapted from Eugene Grant and Richard Leavenworth, Statistical Quality Control, 5th ed. Copyright © 1980 McGraw-Hill Companies, Inc. Used with permission.
a. Determine the mean and range of each sample.
{\overline{{x}}}\;=\;{\frac{\Sigma\,x}{n}},{\mathrm{Range}}\;=\;\operatorname{Largest}\;-\;\mathrm{Smallest}b. Compute the average mean and average range:
\overset{=}{x}\,=\,{\frac{10.0+10.1+9.9+10.2+10.0}{5}}\,=\,{\frac{50.2}{5}}\,=\,10.04\\ \overline{{{R}}}\,=\,\frac{.4\,+\,.6\,+\,.4\,+\,.6\,+\,.6}{5}\,=\,\frac{2\,.6}{5}\,=\,.52c. Obtain factors A_{2} ,\ D_{4} ,\ \text{and}\ D_{3} from Table 10.3 for n = 4: A_{2} = .73,\ D_{4} = 2.28,\ D_{3} = 0.
d. Compute upper and lower limits:
{\mathrm{UCL}_{\overline{x}}}\,=\,{\overset{=}{x}}\,+\,A_{2}{\overline{{R}}}\,=\,10.04\,+\,.73(.52)\,=\,10.42\\ {\mathrm{LCL}_{\overline{x}}}\,=\,{\overset{=}{x}}\,-\,A_{2}{\overline{{R}}}\,=\,10.04\,-\,.73(.52)\,=\,9.66\\ \mathrm{UCL}_{R}~=~D_{4}\overline{{{R}}}~=~2.28(.52)~=~1.19\\ \mathrm{LCL}_{R}\ =\ D_{3}\overline{{{R}}}\ =\ 0(.52)\ =\ 0e. Verify that points are within limits. (If they were not, the process would be investigated to correct assignable causes of variation.)
The smallest sample mean is 9.9, and the largest is 10.2. Both are well within the control limits. Similarly, the largest sample range is .6, which is also within the control limits. Hence, the results suggest that the process is in control. Note, however, that for illustrative purposes, the number of samples is deliberately small; 20 or more samples would give a clearer indication of control limits and whether the process is in control.
| Sample | Mean | Range |
| 1 | 40.0/4=10.0 | 10.2 9.8=.4 |
| 2 | 40.4/4=10.1 | 10.4-9.8=.6 |
| 3 | 39.6/4=9.9 | 10.1-9.7=.4 |
| 4 | 40.8/4=10.2 | 10.5-9.9=.6 |
| 5 | 40.0/4=10.0 | 10.3-9.7=.6 |