In order to summarize and review what was shown on this module, we gonna go through two final examples that cover control charts and process capability assessment.

# Final Example 1

The first final example is the Cylinder Liner Boring Case Study, which go through of how to use the $ \overline{X} $ and $R$ chart in real life. This case study is very important because those two control charts are the most used ones. It is available at http://www.win.tue.nl/~adibucch/2WS10/lect18Sutherland.pdf.

For a complete comprehension of this case study, you must know what is a Pareto Chart and a Cause and Effect Diagram. You can find additional information about those on this link: http://www.academyofaerospacequality.org/node/1330.

# Final Example 2

A decision needs to be made regarding the possible purchase of some new casting equipment. To aid in this process, a capability study is being performed to assess the performance of the current equipment. An important part that will consume much of the run time of the equipment is to be used for the study. The width of the cast part at a crucial spot should be 3.000 cm ± 0.025 cm, and rational sampling has produced the data below.

Perform the capability study on this process using the samples below and comment. The samples, of size n=5, were collected every half hour over two continuous shifts.

The first step in a process capability study is to determine if the process is in statistical control using control charts.

Now we plot the sample values, $ \overline{X} $ and $ R $ on the charts.

We appear to have a process that is in statistical control, i.e., there are no signals on the charts that would suggest that the process is not subject to only a constant system of common cause variation.

At this point, since no individual process measurements are available, to proceed we will assume that the process can be approximated by the normal distribution, with a mean estimated to be $ \overline{\overline{X}} $ = 2.9984 and a standard deviation is estimated by $ \overline{R} $/d_{2} = 0.447/2.326 = 0.01922. We can find the percent conforming to specifications in the following manner:

Clearly, we have a process here that may be showing good statistical control, but at the same time is not very capable of meeting the specifications.

- Log in to post comments