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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

Smartphone Batteries TheSmartBatt company manufactures batteries for smartphones. Listed below are numbers of defects in batches of 200 batteries randomly selected in each of 12 consecutive days of production. What action should be taken?

5 7 4 6 3 10 10 13 4 15 4 19

Short Answer

Expert verified

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

The process is not within statistical control as one sample point lies above the upper control limit.

One sample proportion is exceptionally high, above the upper control limit, and the proportion of defects appears to be generally increasing. As a result, the manufacturer should make the necessary changes to limit the number of defects and ensure that high-quality products are manufactured.

Step by step solution

01

Given information

Data are given on the number of defective batteries in 12 samples of smartphones.

The sample size of each of the 12 samples is equal to 200.

02

Important values of p chart

Let\(\bar p\)be the estimated proportion of defectivebatteriesin all the samples.

It is computed as follows:

\(\begin{array}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{samples}}}}\\ = \frac{{5 + 7 + 4 + ..... + 19}}{{12\left( {200} \right)}}\\ = \frac{{100}}{{2400}}\\ = 0.041667\end{array}\)

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

\(\begin{array}{c}\bar q = 1 - 0.041667\\ = 0.958333\end{array}\)

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

\(\begin{array}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.041667 - 3\sqrt {\frac{{\left( {0.041667} \right)\left( {0.958333} \right)}}{{200}}} \\ = - 0.000723\\ \approx 0\end{array}\)

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

\(\begin{array}{c}UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.041667 + 3\sqrt {\frac{{\left( {0.041667} \right)\left( {0.958333} \right)}}{{200}}} \\ = 0.0840567\end{array}\)

03

Computation of the fraction defective

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

\({p_i} = \frac{{{d_i}}}{{200}}\)

Where,

\({p_i}\)be the sample fraction defective for the ith batch;

\({d_i}\)be 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

5

0.025

2

7

0.035

3

4

0.020

4

6

0.030

5

3

0.015

6

10

0.050

7

10

0.050

8

13

0.065

9

4

0.020

10

15

0.075

11

4

0.020

12

19

0.095

04

Construction of the p chart

Follow the given steps to construct the p chart:

  • Mark the values 1, 2, ...,12 on the horizontal axis and label the axis as 鈥淪ample.鈥
  • Mark the values 0.00, 0.01, 0.02, 鈥︹, 0.1 on the vertical axis and label the axis as 鈥淧roportion.鈥
  • Plot a straight line corresponding to the value 鈥0.041667鈥 on the vertical axis and label the line (on the left side) as\(CL = 0.041667\).鈥
  • Plot a horizontal line corresponding to the value 鈥0鈥 on the vertical axis and label the line as 鈥淟CL=0.鈥
  • Similarly, plot a horizontal line corresponding to the value 鈥0.08405623鈥 on the vertical axis and label the line as 鈥淯CL=0.08405623.鈥
  • Mark the given12 sample points (fraction defective of the ith lot) on the graph and join the dots using straight lines.

The following p chart is plotted:

05

Analysis of the p chart

It can be observed from the plotted chart that there is at least one point that lies beyond the upper control limit.

Since the above criterion violates the stability of the given process, it can be concluded that the process is not within statistical control.

06

Corrective action

One value of the sample proportion is extremely high and lies beyond the upper control limit, and overall, there appears to be an increase in the proportion of defects. Therefore, the manufacturer should make appropriate changes, wherever required, to reduce the number of defects and ensure that good quality products are produced.

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Most popular questions from this chapter

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

Quarters. In Exercises 9鈥12, refer to the accompanying table of weights (grams) of quarters minted by the U.S. government. This table is available for download at www.TriolaStats.com.

Day

Hour 1

Hour 2

Hour 3

Hour 4

Hour 5

\(\bar x\)

s

Range

1

5.543

5.698

5.605

5.653

5.668

5.6334

0.0607

0.155

2

5.585

5.692

5.771

5.718

5.72

5.6972

0.0689

0.186

3

5.752

5.636

5.66

5.68

5.565

5.6586

0.0679

0.187

4

5.697

5.613

5.575

5.615

5.646

5.6292

0.0455

0.122

5

5.63

5.77

5.713

5.649

5.65

5.6824

0.0581

0.14

6

5.807

5.647

5.756

5.677

5.761

5.7296

0.0657

0.16

7

5.686

5.691

5.715

5.748

5.688

5.7056

0.0264

0.062

8

5.681

5.699

5.767

5.736

5.752

5.727

0.0361

0.086

9

5.552

5.659

5.77

5.594

5.607

5.6364

0.0839

0.218

10

5.818

5.655

5.66

5.662

5.7

5.699

0.0689

0.163

11

5.693

5.692

5.625

5.75

5.757

5.7034

0.0535

0.132

12

5.637

5.628

5.646

5.667

5.603

5.6362

0.0235

0.064

13

5.634

5.778

5.638

5.689

5.702

5.6882

0.0586

0.144

14

5.664

5.655

5.727

5.637

5.667

5.67

0.0339

0.09

15

5.664

5.695

5.677

5.689

5.757

5.6964

0.0359

0.093

16

5.707

5.89

5.598

5.724

5.635

5.7108

0.1127

0.292

17

5.697

5.593

5.78

5.745

5.47

5.657

0.126

0.31

18

6.002

5.898

5.669

5.957

5.583

5.8218

0.185

0.419

19

6.017

5.613

5.596

5.534

5.795

5.711

0.1968

0.483

20

5.671

6.223

5.621

5.783

5.787

5.817

0.238

0.602

Quarters: R Chart Treat the five measurements from each day as a sample and construct an R chart. What does the result suggest?

p Chart A variation of the control chart for p is the np chart, in which the actual numbers of defects are plotted instead of the proportions of defects. The np chart has a centerline value of \(n\bar p\), and the control limits have values of \(n\bar p + 3\sqrt {n\bar p\bar q} \)and\(n\bar p - 3\sqrt {n\bar p\bar q} \). The p chart and the np chart differ only in the scale of values used for the vertical axis. Construct the np chart for Example 1 鈥淒efective Aircraft Altimeters鈥 in this section. Compare the np chart to the control chart for p given in this section

Sunspots and the DJIA Use the data from Exercise 5 and find the equation of the regression line. Then find the best predicted value of the DJIA in the year 2004, when the sunspot number was 61. How does the result compare to the actual DJIA value of 10,855?

Listed below are annual sunspot numbers paired with annual high values of the Dow Jones Industrial Average (DJIA). Sunspot numbers are measures of dark spots on the sun, and the DJIA is an index that measures the value of select stocks. The data are from recent and consecutive years. Use a 0.05 significance level to test for a linear correlation between values of the DJIA and sunspot numbers. Is the result surprising?

Sunspot

DJIA

45

10941

31

12464

46

14198

31

13279

50

10580

48

11625

56

12929

38

13589

65

16577

51

18054

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