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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

Euro Coins Consider a process of minting coins with a value of one euro. Listed below are the numbers of defective coins in successive batches of 10,000 coins randomly selected on consecutive days of production.

32 21 25 19 35 34 27 30 26 33

Short Answer

Expert verified

The following p chart is constructed for the proportions of defective coins:

The process appears to be within statistical control since none of the threeout-of-control criteria is met.

Step by step solution

01

Given information

Data are given on the proportion of defective coins in 10 samples.

The sample size of each sample is 10,000.

02

Important values of p chart

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

It is computed as follows:

\(\begin{array}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{samples}}}}\\ = \frac{{32 + 21 + 25 + ..... + 33}}{{10\left( {10000} \right)}}\\ = \frac{{282}}{{100000}}\\ = 0.00282\end{array}\)

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

\(\begin{array}{c}\bar q = 1 - 0.00282\\ = 0.99718\end{array}\)

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

\(\begin{array}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.00282 - 3\sqrt {\frac{{\left( {0.00282} \right)\left( {0.99718} \right)}}{{10000}}} \\ = 0.001229\end{array}\)

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

\(\begin{array}{c}UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.00282 + 3\sqrt {\frac{{\left( {0.00282} \right)\left( {0.99718} \right)}}{{10000}}} \\ = 0.004411\end{array}\)

03

Computation of the fraction defective

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

\({p_i} = \frac{{{d_i}}}{{10000}}\)

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

32

0.0032

2

21

0.0021

3

25

0.0025

4

19

0.0019

5

35

0.0035

6

34

0.0034

7

27

0.0027

8

30

0.0030

9

26

0.0026

10

33

0.0033

04

Construction of the p chart

Follow the given steps to construct the p chart:

  • Mark the values 1, 2, ...,10 on the horizontal axis and label it 鈥淪ample.鈥
  • Mark the values 0.0010, 0.0015, 0.0020, 鈥︹, 0.0045 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 = 0.00282\).鈥
  • Plot a horizontal line corresponding to the value 0.001229 on the vertical axis and label it鈥淟CL=0.001229.鈥
  • Similarly, plot a horizontal line corresponding to the value 0.004411 on the vertical axis and label it UCL=0.004411.鈥
  • Mark all ten sample points (fraction defective of the ith lot) on the graph and join the dots using straight lines.

The following p chart is obtained:

05

Analysis of the p chart

None of the three out-of-control criteria is depicted in the plotted p chart.

As all the points lie within upper and lower control limits, it can be concluded that the process is within statistical control.

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Most popular questions from this chapter

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