Question

Control Charts for p. In Exercises 5–12, use the given process data to construct a control chart for p. In each case, use the three out-of-control criteria listed near the beginning of this section and determine whether the process is within statistical control. If it is not, identify which of the three out-of-control criteria apply

Cola Cans In each of several consecutive days of production of cola cans, 50 cans are tested and the numbers of defects each day are listed below. Do the proportions of defects appear to be acceptable? What action should be taken?

8 7 9 8 10 6 5 7 9 12 9 6 8 7 9 8 11 10 9 7

Short answer

The following p chart is constructed for the given number of defects:

None of the features in the chart indicates that the process is unstable.

But the values of the proportions of defectives seem to be very high.

As a result, the producer must respond quickly to improve the quality of goods produced.

Step-by-step solution

Given information

Data are given on the number of defects in 20 randomly selected samples.

The size of each sample is 50.

Important values of p chart

Let\(\bar p\)be the estimated proportion of defective cans in all samples.

It is computed as follows:

\(\begin{array}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\\ = \frac{{8 + 7 + 9 + ..... + 7}}{{20\left( {50} \right)}}\\ = \frac{{165}}{{1000}}\\ = 0.165\end{array}\)

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

\(\begin{array}{c}\bar q = 1 - 0.165\\ = 0.835\end{array}\)

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

\(\begin{array}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.165 - 3\sqrt {\frac{{\left( {0.165} \right)\left( {0.835} \right)}}{{50}}} \\ = 0.0075\end{array}\)

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

\(\begin{array}{c}UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.165 + 3\sqrt {\frac{{\left( {0.165} \right)\left( {0.835} \right)}}{{50}}} \\ = 0.3225\end{array}\)

Computation of the fraction defective

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

\({p_i} = \frac{{{d_i}}}{{50}}\)

Here,

\({p_i}\)is the sample fraction defective for the ith batch, and

\({d_i}\)is the number of defective orders in the ith batch.

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

S.No.	Defectives (d)	Sample fraction defective (p)
1	8	0.16
2	7	0.14
3	9	0.18
4	8	0.16
5	10	0.20
6	6	0.12
7	5	0.10
8	7	0.14
9	9	0.18
10	12	0.24
11	9	0.18
12	6	0.12
13	8	0.16
14	7	0.14
15	9	0.18
16	8	0.16
17	11	0.22
18	10	0.20
19	9	0.18
20	7	0.14

Construction of the p chart

Follow the given steps to construct the p chart:

• Mark the values 2, 4, ...,20on the horizontal axis and label it “Sample.”
• Mark the values 0.00, 0.05, 0.10, ……, 0.35 on the vertical axis and label it “Proportion.”
• Plot a straight line corresponding to the value 0.165 on the vertical axis and label it (on the left side) “\(\bar P\;or\;\bar p = 0.165\).”
• Plot a horizontal line corresponding to the value 0.0075 on the vertical axis and label it “LCL=0.0075.”
• Similarly, plot a horizontal line corresponding to the value 0.3225 on the vertical axis and label it “UCL=0.3225.”
• Mark all 20 sample points (fraction defective of the ith lot) on the graph and join the dots using straight lines.

The following p chart is obtained:

Analysis of the p chart

The chart has no feature that indicates the violation of the stability of the process. Thus, the process is within statistical control.

But the values of the proportions of defectives seem to be very high.

Thus, the manufacturer should take quick action to correct the quality of the goods produced.