Question 10.P.4: p-chart and c-chart. Using the appropriate control chart, de......
p-chart and c-chart. Using the appropriate control chart, determine two-sigma control limits for each case:
a. An inspector found an average of 3.9 scratches in the exterior paint of each of the automobiles being prepared for shipment to dealers.
b. Before shipping lawn mowers to dealers, an inspector attempts to start each mower and notes any that do not start on the first try. The lot size is 100 mowers, and an average of 4 did not start (4 percent).
The choice between these two types of control charts relates to whether two types of results can be counted (p – chart) or whether only occurrences can be counted (c – chart).
a. The inspector can only count the scratches that occurred, not the ones that did not occur. Consequently, a c – chart is appropriate. The sample average is 3.9 scratches per car. Two sigma control limits are found using the formulas
\mathrm{UCL}\,=\,\overline{{{c}}}\,+\,z\,\sqrt{\overline{{{c}}}}\\ \mathrm{LCL}\,=\,\overline{{{c}}}\,-\,z\,\sqrt{\overline{{{c}}}}where \overline{c} = 3.9\ \text{and}\ z = 2. Thus,
\mathrm{UCL}\,=\,3.9\,+\,2{\sqrt{3.9}}\,=\,7.85\,\,{\mathrm{scratches}}\\ \mathrm{{{LCL}}}\,=\,3.9\,-\,2{\sqrt{3.9}}\,=\,-.05,\;{\mathrm{so~the~lower~limit~is~0~scratches}}( Note: Round to zero only if the computed lower limit is negative.)
b. The inspector can count both the lawn mowers that started and those that did not start. Consequently, a p – chart is appropriate. Two-sigma control limits can be computed using the following:
\mathrm{UCL}\;=\;\overline{{{p}}}\;+\;z\sqrt{\frac{\overline{{{p}}}(1-\overline{{{p}}})}{n}}\\ \mathrm{LCL}\,=\,{\overline{{p}}}\,-\,z\sqrt{\frac{{\overline{{p}}}(1-{\overline{{p}}})}{n}}where
\begin{array}{l}{{\overline{{{p}}}\,=\,.04}}\\ {{n\,=\,100}}\\ {{z\,=\,2}}\end{array}Thus,
\mathrm{UCL}\,=\,.04\,+\,2\sqrt{\frac{.04(96)}{100}}\,=\,.079\\ \mathrm{LCL}\,=\,.04\,-\,2{\sqrt{\frac{.04(.96)}{100}}}\,=\,.001