Service Times The Newport Diner records the times (min) it takes before customers are asked for their order. Each day, 50 customers are randomly selected, and the order is considered to be defective if it takes longer than three minutes. The numbers of defective orders are listed below for consecutive days. Construct an appropriate control chart and determine whether the process is within statistical control. If not, identify which criteria lead to rejection of statistical stability.

3 2 3 5 4 6 7 9 8 10 11 9 12 15 17

The number of defective ordersis given for 15 samples with a sample size of 50 orders each.

Here, the sample values show the number of defectives (which is an attribute) in each sample.

Thus, the appropriate control chart for depicting the proportion of defectives will be the p-chart which is an attribute chart.

It is computed as follows:

\(\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\),

where\(\bar p\)isthe proportion of defective orders in all the samples and

\(\bar q = 1 - \bar p\).

The value of the lower control limit (LCL) and upper control limit (UCL) is computed below:

\[\begin{aligned}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \end{aligned}\]

It is computed as follows:

\(\begin{aligned}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\\ = \frac{{3 + 2 + 3 + ..... + 17}}{{15\left( {50} \right)}}\\ = 0.1613\end{aligned}\)

The value of\(\bar q\)is computed as shown below:

\(\begin{aligned}{c}\bar q = 1 - \bar p\\ = 1 - 0.161\\ = 0.839\end{aligned}\)

The value of the lower control limit (LCL) is computed below:

\[\begin{aligned}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.161 - 3\sqrt {\frac{{\left( {0.161} \right)\left( {0.839} \right)}}{{50}}} \\ = 0.0053\end{aligned}\]

The value of the upper control limit (UCL) is computed below:

\[\begin{aligned}{c}LCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.161 + 3\sqrt {\frac{{\left( {0.161} \right)\left( {0.839} \right)}}{{50}}} \\ = 0.3174\end{aligned}\]

The sample fraction defective for the ith sample or lot can be computed as:

\[{p_i} = \frac{{{d_i}}}{{50}}\],

where\[{p_i}\]isthe sample fraction defective for the ith lot, and

\[{d_i}\]isthe number of defective orders in the lot.

The computation of fraction defective for the ith lot is given as follows:

Follow the given steps to construct the p-chart:

• Mark the values 1, 2, ...,15 on the horizontal axis and label the axis as “Day.”
• Mark the values 0.00, 0.05, 0.10, ……,0.35 on the vertical axis and label the axis as “Proportion.”
• Plot a straight line corresponding to the value “0.1613” on the vertical axis and label the line (on the left side) as “\(\bar P\;{\rm{or}}\;\bar p = 0.1613\).”
• Plot a horizontal line corresponding to the value “0.0053” on the vertical axis and label the line as “LCL=0.0053.”
• Similarly, plot a horizontal line corresponding to the value “0.3174” on the vertical axis and label the line as “UCL=0.3174.”
• Mark the given15 sample points (fraction defective of the ith lot) on the graph and join the dots using straight lines.

The following p-chart is plotted:

The following features can be observed:

Here, the order times are increasing, so there is an upward trend in the proportion of defectives over the 15 days.

There is one point that lies beyond the upper control limit.

As these characteristics indicate the instability of the process, it can be concluded that the process is not within statistical control.