Q8BSC Control Charts for p. In Exercis... [FREE SOLUTION]

Chapter 14: Q8BSC (page 654)

Control Charts for p. In Exercises 5鈥�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
Aspirin Tablets Bottles of aspirin tablets typically have labels indicating that each tablet contains 325 mg of aspirin, but the actual amounts can vary between 315 mg and 335 mg. A tablet is defective if it has less than 315 mg of aspirin or more than 335 mg of aspirin. Listed below are numbers of defects found in batches of 1000 tablets.
16 18 13 9 10 8 6 5 5 3

Short Answer

Expert verified

The following p chart is constructed for the given proportion of defective tablets:

The sample proportions appear to be following a downward trend, as shown in the graph. Here, the occurrence of a trend is one of the three out-of-control conditions that appears in the chart, so the process cannot be considered to be within control.

Step by step solution

Given information

Data are given forthe number of defective tablets in 10 samples.

The size of each sample is 1000.

Important values of p chart

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

It is computed as follows:

\(\begin{array}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\\ = \frac{{16 + 18 + 13 + ..... + 3}}{{10\left( {1000} \right)}}\\ = \frac{{93}}{{10000}}\\ = 0.0093\end{array}\)

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

\(\begin{array}{c}\bar q = 1 - 0.0093\\ = 0.9907\end{array}\)

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

\(\begin{array}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.0093 - 3\sqrt {\frac{{\left( {0.0093} \right)\left( {0.9907} \right)}}{{1000}}} \\ = 0.00019\end{array}\)

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

\(\begin{array}{c}UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.0093 + 3\sqrt {\frac{{\left( {0.0093} \right)\left( {0.9907} \right)}}{{1000}}} \\ = 0.01841\end{array}\)

Computation of the fraction defective

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

\({p_i} = \frac{{{d_i}}}{{1000}}\)

Here,

\({p_i}\)isthe 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	16	0.016
2	18	0.018
3	13	0.013
4	9	0.009
5	10	0.010
6	8	0.008
7	6	0.006
8	5	0.005
9	5	0.005
10	3	0.003

Construction of the p chart

Follow the given steps to construct the p chart:

Mark the values 1, 2, ...,10on the horizontal axis and label it 鈥淪ample.鈥�
Mark the values 0.000, 0.005, 0.010, 鈥︹€�, 0.020 on the vertical axis and label it 鈥淧roportion.鈥�
Plot a straight line corresponding to the value 0.0093 on the vertical axis and label it (on the left side) 鈥淺(\bar P\;or\;\bar p = 0.0093\).鈥�
Plot a horizontal line corresponding to the value 0.00019 on the vertical axis and label it 鈥淟CL=0.00019.鈥�
Similarly, plot a horizontal line corresponding to the value 0.01841 on the vertical axis and label it 鈥淯CL=0.01841.鈥�
Mark the givensample 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 shows a downward trend in the sample proportions of defects. Since the presence of a trend is one of the three out-of-control criteria,it can be concluded that the process is not within statistical control. Here, the number of defects in the aspirin tablets is decreasing. It means there is an improvement in the statistical process.