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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 “Defective Aircraft Altimeters” in this section. Compare the np chart to the control chart for p given in this section

Short Answer

Expert verified

The np chart constructed is shown below:

From the constructed np chart, the process is not within statistical control because

  • there is at least one point beyond the upper control limit, and
  • there seems to be an upward trend in the number of defects.

For the given data, the p chart and the np chart are nearly identical and have the same structure. The only variation between the two graphs is in the vertical scale values.

Step by step solution

01

Given information

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

The size of each sample is 100.

02

np Chart

The np chart is an attribute control chart that depicts the number of defects in the individual samples as compared to the p chart, which shows the proportion of defects in each sample.

The following data is utilized to construct the np chart, which shows the number of defective altimeters in each sample:

Sample number

Number of defects

1

2

2

0

3

1

4

3

5

1

6

2

7

2

8

4

9

3

10

5

11

12

12

7

03

Important values of thenp chart

Let\(n\bar p\)be the estimated number of defectivealtimetersin all the samples.

It is computed as follows:

\(\begin{array}{c}n\bar p = n \times \left( {\frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{samples}}}}} \right)\\ = \left( {100} \right) \times \left( {\frac{{2 + 0 + 1 + ..... + 7}}{{12\left( {100} \right)}}} \right)\\ = \left( {100} \right) \times \left( {\frac{{42}}{{1200}}} \right)\\ = \left( {100} \right) \times \left( {0.035} \right)\end{array}\)

\( = 3.5\)

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

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.035\\ = 0.965\end{array}\)

The value ofthe central line is calculated below:

\(\begin{array}{c}CL = n\bar p\\ = 3.5\end{array}\)

The lower control limit (LCL) is computed below:

\(\begin{array}{c}LCL = n\bar p - 3\sqrt {n\bar p\bar q} \\ = 3.5 - 3\sqrt {\left( {100} \right)\left( {0.035} \right)\left( {0.965} \right)} \\ = - 2.01339\\ \approx 0\end{array}\)

The upper control limit (UCL) is computed below:

\(\begin{array}{c}UCL = n\bar p + 3\sqrt {n\bar p\bar q} \\ = 3.5 + 3\sqrt {\left( {100} \right)\left( {0.035} \right)\left( {0.965} \right)} \\ = 9.01\end{array}\)

04

Tabulation of the number of defectives

The following table shows the number of defective altimeters corresponding to the sample number:

Serial number

Number of defectives (d)

1

2

2

0

3

1

4

3

5

1

6

2

7

2

8

4

9

3

10

5

11

12

12

7

05

Construction

Follow the given steps to construct the p chart:

  • Mark the values 1, 2, 3 ...,12 on the horizontal axis and label it “Sample.”
  • Mark the values 0, 2, 4 ...,14 on the vertical axis and label it “Sample Count.”
  • Plot a horizontal line parallel to the horizontal axis corresponding to the value “3.5” on the vertical axis and label the line (on the left side) “\(n\bar p\)= 3.5.”
  • Plot a horizontal line parallel to the horizontal axis corresponding to the value “9.01” on the vertical axis and label the line (on the left side) “UCL= 9.01.”
  • Plot a horizontal line parallel to the horizontal axis corresponding to the value “0” on the vertical axis and label the line (on the left side) “LCL= 0.”
  • Mark the 12 samplepoints on the graph and join the dots using straight lines.

The following np chart is obtained:

06

Analysis of the np chart

The following characteristics can be observed from the plotted chart:

  • There is at least one point beyond the upper control limit.
  • There appears to be an upward trend in the number of defects.

Since the above criteria point towards the violation of the stability of the given process, the process is not under statistical control.

07

Comparison of np chart and p chart

Referring to Example 1, the centerline and control limits (LCL and UCL) are given as follows:

\(\begin{array}{c}CL = 0.035\\UCL = 0.090\\LCL = - 0.020\\ \approx 0\end{array}\)

The p chart and the np chart plotted for the given data are approximately identical and have the same structure. The only difference in the two charts is in the values marked on the vertical scale.

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Most popular questions from this chapter

Notation The control chart for Exercise 1 shows a value of \(\bar p\) = 0.0975. What does that value denote, and how is it obtained? What do UCL and LCL indicate?

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

Quarters. In Exercises 9–12, 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: Notation Find the values of \({\bf{\bar \bar x}}\)and\({\bf{\bar R}}\). Also find the values of LCL and UCL for an R chart, then find the values of LCL and UCL for an \({\bf{\bar x}}\) chart

Energy Consumption. Exercises 1–5 refer to the amounts of energy consumed in the author’s home. (Most of the data are real, but some are fabricated.) Each value represents energy consumed (kWh) in a two-month period. Let each subgroup consist of the six amounts within the same year. Data are available for download atwww.TriolaStats.com.


Jan.-Feb.

Mar.-April

May-June

July-Aug.

Sept.-Oct.

Nov.-dec.

Year 1

3637

2888

2359

3704

3432

2446

Year 2

4463

2482

2762

2288

2423

2483

Year 3

3375

2661

2073

2579

2858

2296

Year 4

2812

2433

2266

3128

3286

2749

Year 5

3427

578

3792

3348

2937

2774

Year 6

4016

3458

3395

4249

4003

3118

Year 7

4016

3458

3395

4249

4003

3118

Year 8

4016

3458

3395

4249

4003

3118

Energy Consumption: R Chart Let each subgroup consist of the 6 values within a year. Construct an R chart and determine whether the process variation is within statistical control. If it is not, identify which of the three out-of-control criteria lead to rejection of statistically stable variation

\(\bar x\)- Chart Based on Standard Deviations An x chart based on standard deviations (instead of ranges) is constructed by plotting sample means with a centerline at x and control limits at x + A3s and x - A3s, where A3 is found in Table 14-2 on page 660 and s is the mean of the sample standard deviations. Use the data in Table 14-1 on page 655 to construct an xchart based on standard deviations. Compare the result to the x chart based on sample ranges in Example 5 “x Chart of Altimeter Errors.”

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