Question 10.P.5: Run tests. The number of defective items per sample for 11 s......
Run tests. The number of defective items per sample for 11 samples is shown below. Determine if nonrandom patterns are present in the sequence.
| SAMPLE | |||||||||||
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | |
| Number of defectives | 22 | 17 | 19 | 25 | 18 | 20 | 21 | 17 | 23 | 23 | 24 |
Since the median isn’t given, it must be estimated from the sample data. To do this, array the data from low to high; the median is the middle value. (In this case, there is an odd number of values. For an even number of values, average the middle two to obtain the median.) Thus,
The median is 21.
Next, code the observations using A/B and U/D:
Note that each test has tied values. How these are resolved can affect the number of observed runs. Suppose that you adhere to this rule: Assign letter (A or B, U or D) so that the resulting difference between the observed and expected number of runs is as large as possible. To accomplish this, it is necessary to initially ignore ties and count the runs to see whether there are too many or too few. Then return to the ties and make the assignments. The rationale for this rule is that it is a conservative method for retaining data; if you conclude that the data are random using this approach, you can be reasonably confident that the method has not “created” randomness. With this in mind, assign a B to sample 7 since the expected number of runs is
E(r)_{\text{med}}\, =\, N/2 + 1 = 11/2 + 1 = 6.5and the difference between the resulting number of runs, 5, and 6.5 is greater than between 6.5 and 7 (which occurs if A is used instead of B). Similarly, in the up/down test, a U for sample 10 produces six runs, whereas a D produces eight runs. Since the expected number of runs is
E(r)_{\text{u/d}} = (2N – 1) \div 3 = (22 – 1) \div 3 = 7it makes no difference which one is used: both yield a difference of 1. For the sake of illustration, a D is assigned.
The computations for the two tests are summarized below. Each test has a z – value that is within the range of ± 2.00. Because neither test reveals nonrandomness, you may conclude that the data are random.
| Sample | A/B | Number of Defectives | U/D |
| 1 | |A | 22 | — |
| 2 | |B | 17 | |D |
| 3 | |B | 19 | |U |
| 4 | |A | 25 | |U |
| 5 | |B | 18 | |D |
| 6 | |B | 20 | |U |
| 7 | tie | 21 | |U |
| 8 | |B | 17 | |D |
| 9 | |A | 23 | |U |
| 10 | |A | 23 | tie |
| 11 | |A | 24 | |U |
| Runs Observed | Expected | {\sigma_{r}} | z | Conclude | |
| Median | 5 | 6.5 | 1.58 | -.95 | Random |
| Up/down | 8 | 7.0 | 1.28 | .78 | Random |