Question 10.6: Twenty sample means have been taken from a process. The mean......
Twenty sample means have been taken from a process. The means are shown in the following table. Use median and up/down run tests with z = 2 to determine if assignable causes of variation are present. Assume the median is 11.0.
The means are marked according to above/below the median and up/down. The solid lines represent the runs.
The expected number of runs for each test is
E(r)_{\mathrm{med}}~=~{\frac{N}{2}}~+~1\;=\;{\frac{20}{2}}~+~1\;=\;11\\ E(r)_{u/d}~=~\frac{2N\,-\,1}{3}\,=\,\frac{2(20)\,-\,1}{3}\:=\:13The standard deviations are
\sigma_{\mathrm{med}}~=~\sqrt{\frac{N-1}{4}}~=~\sqrt{\frac{20~-~1}{4}}~=~2.18\\ \sigma_{u/d}\;=\;\sqrt{\frac{16N-29}{90}}\;=\;\sqrt{\frac{16(20)-29}{90}}\;=\;1.80The z_{\text{test}} values are
z_{\mathrm{med}}~=~{\frac{10~-~11}{2.18}}~=~-.46\\ z_{u/d}\;=\;\frac{17\;-\;13}{1.80}\;=\;+2.22Although the median test does not reveal any pattern, because its z_{\text{test}} value is within the range ± 2, the up/down test does; its value exceeds + 2. Consequently, nonrandom variations are probably present in the data and, hence, the process is not in control.
If ties occur in either test (e.g., a value equals the median or two values in a row are the same), assign A/B or U/D in such a manner that that z_{\text{test}} is as large as possible. If z_{\text{test}} still does not exceed 2 ± ( ± 1.96, etc.), you can be reasonably confident that a conclusion of randomness is justified.
| Sample | A/B | Mean | U/D | Sample | A/B | Mean | U/D |
| 1 | |B | 10.0 | — | 11 | |B | 10.7 | |D |
| 2 | |B | 10.4 | |U | 12 | |A | 11.3 | |U |
| 3 | |B | 10.2 | |D | 13 | |B | 10.8 | |D |
| 4 | |A | 11.5 | |U | 14 | |A | 11.8 | |U |
| 5 | |B | 10.8 | |D | 15 | |A | 11.2 | |D |
| 6 | |A | 11.6 | |U | 16 | |A | 11.6 | |U |
| 7 | |A | 11.1 | |D | 17 | |A | 11.2 | |D |
| 8 | |A | 11.2 | |U | 18 | |B | 10.6 | |D |
| 9 | |B | 10.6 | |D | 19 | |B | 10.7 | |U |
| 10 | |B | 10.9 | |U | 20 | |A | 11.9 | |U |
A/B: 10 runs\qquadU/D: 17 runs