This is file Q8Intl-IM13E.doc - The fifth of 5 files for solutions to this chapter. This file also includes the solutions to the end-of-chapter cases.

30.

Thirty samples of 75 items each were inspected at the Yummy Candy Company and 75 items were found to be defective. Compute control limits for a p-chart for this process.

Answer 30.

Control limits for Yummy Candy Company are as follows: CL p = 75/2250 = 0.0333 sp = sp =

[( p)(1 p)] / n

(0.0333)(0.9667)/75 = 0.0207

Control limits: UCLp = p + 3 s p UCLp = 0.0333+ 3(0.0207) = 0.0954 LCLp = p - 3 s p LCLp = 0.0333 - 3(0.0207) = - 0.0288, use 0 31.

Twenty-five samples of 50 orders each at the Dixie Ice Co. were inspected, and 56 items were found to be defective. Compute control limits for a p-chart.

Answer 31.

Control limits for Dixie Ice Company orders can be calculated using:

CL p = 56/(25)(50) = 56/1250 = 0.045 1 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

sp = sp =

2

[( p)(1 p)] / n

(0.045)(0.955) / 50 = 0.0293

Control limits: UCLp = p + 3 s p UCLp = 0.045 + 3(0.0293) = 0.1329 LCLp = p - 3 s p LCLp = 0.045 - 3(0.0293) = - 0.0429, use 0

32.

The fraction defective for a folding process in a Quality Printing plant is given in the worksheet Prob. 13-32 for 25 samples. Fifty units are inspected each shift. a. Construct a p-chart and interpret the results. b. After the process was determined to be under control, process monitoring began, using the established control limits. The results of 25 more samples are shown in the second part of the worksheet. Is there a problem with the process? If so, when should the process have been stopped, and steps taken to correct it?

Answer 32.

See data and control chart below for Quality Printing Company’s plant. See spreadsheet Prob13-32P.xls for details. CL p = 0.06 sp =

[( p)(1 p)] / n =

[(0.06) (0.94)] / 50

= 0.0336

Control limits: UCLp = p + 3 sp = 0.06 + 3(0.0336) = 0.1608 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

3

LCLp = p - 3 sp = 0.06 - 3(0.0336) = - 0.0408 use 0 The process appears to be under control.

b) When the additional data is added, while the process is being monitored using the previously calculated control limits, the process starts to go out of control, with samples 29, 30, 31 being the first indicators. Two out of three of these are more than 2 away from the mean, p . Later, four out of five samples between 37-41 are more than 1 away from the mean, p . Finally, sample 48 exceeds the upper control limit. See Part B tabs on spreadsheet Prob13-32P.xls for details. The process should have been stopped and corrected when the first indications were seen. If these were missed, it is certain that sample 48, which was above the control limit should have been spotted, and the process stopped.

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Statistical Process Control

33.

4

The fraction defective of automotive pistons made by the Hasty Piston Co. is given in the worksheet Prob.13-33 for 20 samples. Two hundred units are inspected each day. Construct a p-chart and interpret the results.

Answer 33.

See data and control charts for Hasty Piston Company, below. See spreadsheets Prob1333P.xls, sheets for details. a) Initially, CL p = 0.134 (see printout) sp =

[( p)(1 p)] / n

sp =

[ (0.134)(0.866)]/200

= 0.024

Control limits: UCLp = p + 3 s p UCLp = 0.134 + 3(0.024) = 0.206 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

5

LCLp = p - 3 s p LCLp = 0.134 - 3(0.024) = 0.062 See initial data and control chart, below.

b) Hasty Piston Company (Continued) Revised CL p = 0.1296 (after sample 12, with a fraction defective of 0.23, was removed). sp =

[( p)(1 p)] / n =

[(0.1296)(0.8704)]/200

= 0.0237

Control limits: UCLp = p + 3 s p UCLp = 0.1296 + 3(0.0237) = 0.2007 LCLp = p - 3 s p LCLp = 0.1296 - 3(0.0237) = 0.0585 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

6

See data and control chart, below, after sample 12 was removed. The conclusion is, tht the process is now in control.

Problem 33-B Final Revised Control Chart

34.

One hundred insurance claim forms are inspected daily at Full Life Insurance Co. over 25 working days, and the number of forms with errors have been recorded in the worksheet Prob. 13-34. Construct a p-chart. If any points occur outside the control limits, assume that assignable causes have been determined. Then construct a revised chart.

Answer 34.

See spreadsheets for Full Life Insurance Co. in file Prob13-34P.xls for details. a) Initially, based on the sum of the p values for the 25 samples, p1 + p2 + p3 + … CL p = ----------------------N CL p = 0.63 / 30 = 0.0210

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Statistical Process Control

sp =

[( p)(1 p)] / n =

7

[(0.0210)(0.979)]/100

= 0.0143

UCLp = p + 3 s p = 0.0210 + 3(0.0143) = 0.0640 LCLp = p - 3 s p = 0.0210 - 3(0.0143) = -0.0220, use 0 Throw out #9 and #23, out-of-control values, revise. See initial data and control chart, below.

b) Revised CL p = 0.480 / 28 = 0.0171 sp =

[( p)(1 p)] / n =

[(0.0171)(0.9829)]/100

= 0.0130

Control limits: UCLp = p + 3 s p = 0.0171 + 3(0.0130) = 0.0561 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

8

LCLp = p - 3 s p = 0.0171 - 3(0.0130) = -0.0219, use 0 The conclusion is that the process is now in control. Problem 34-b Final Revised Control Chart

35.

Samples of size 200 have been randomly selected during each shift of 25 shifts in a production process at Lean Manufacturing, Inc. The data are given in the worksheet Prob.13-35. Construct a p-chart and determine whether the process is in control. If not, eliminate any data points that appear to be due to assignable causes and construct a new chart.

Answer 35. a) See spreadsheets Prob13-35P.xls with various worksheets, for details. Based on the average number of defects per sample, the average proportion can be calculated as: p1 + p2 + p3 + … CL p = ----------------------N © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

9

CL p = 3.610 / 25 = 0.1444 sp =

[( p)(1 p)] / n =

[(0.1444)(0.8556)]/200

= 0.0249

Control limits: UCLp = p + 3 s p = 0.1444 + 3(0.0249) = 0.2191 LCLp = p - 3 s p = 0.1444 - 3(0.0377) = 0.0697

See initial p-chart, below. It is highly likely that this process is not in control and that major changes need to be made. However, we can approximate our results by throwing out all points containing 43 defectives and over, and revising the chart.

b) (Continued) This results in: CL p = 2.245 / 20 = 0.1123

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Statistical Process Control

sp =

[( p)(1 p)] / n =

10

[(0.1123)(0.8877)]/200

= 0.0223

Control limits: UCLp = p + 3 s p = 0.1123 + 3(0.0223) = 0.1793 LCLp = p - 3 s p = 0.1123 - 3(0.0343) = 0.0453 Throw out all points containing 35 defectives and over, and revise again.

First revised p-chart

c) (Continued) Final Revised CL p = 1.49/ 16 = 0.0931 sp =

[( p)(1 p)] / n =

[ (0.0931)(0.9069)]/200

= 0.0205

Control limits: © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

11

UCLp = p + 3 s p = 0.0931 + 3(0.0205) = 0.1546 LCLp = p - 3 s p = 0.0931 - 3(0.0205) = 0.0316

Problem 35c - Final revised p-chart

36.

AtYourService.com, an Internet service provider (ISP), is concerned that the level of access of customers is decreasing, due to heavier use. The proportion of peak period time when a customer is likely to receive busy signals is considered a good measure of service level. The percentage of times a customer receives a busy signal during peak periods varies. Using a sampling process, the ISP set up control charts to monitor the service level, based on proportion of busy signals received. Construct the p-chart using on the sample data in the table in the worksheet Prob.13-36. What does the chart show? Is the service level good or bad, in your opinion?

Answer 36.

See spreadsheet Prob13-36P.xls for details about AtYourService.com.

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Statistical Process Control

12

The average sample size = 15755/30 = 525.17 CL p = 202 / 15755 = 0.0128 sp =

[( p)(1 p)] / n =

[(0.0128)(0.9872)]/525.17

= 0.0049

UCLp = p + 3 s p = 0.0128 + 3(0.0049) = 0.0275 LCLp = p - 3 s p = 0.0128 - 3(0.0049) = - 0.0019, use 0 All points fall within the control limits. Problem 13-36: p-charts

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Statistical Process Control

37.

13

Edgewater Hospital surveys all outgoing patients by means of a patient satisfaction questionnaire. The number of patients surveyed each month varies. Control charts that monitor the proportion of unsatisfied patients for key questions are constructed and studied. Construct a p-chart for the data in the worksheet Prob.13-37, which represent responses to a question on satisfaction with hospital meals.

Answer 37.

Note that the values obtained for Edgewater Hospital’s data with a hand calculator will differ slightly from those obtained from the Excel spreadsheets, due to rounding differences. See spreadsheets in the file Prob13-37P.xls for details. Initial Calculation a) The average sample size = 10525/ 24 = 438.54, so the “approximate” control limits, based on total defects over total items samples, are: CL p = 305 / 10525 = 0.0290 sp =

[( p)(1 p)] / n =

[(0.0290)(0.9710)]/438.54

= 0.0080

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Statistical Process Control

14

UCLp = p + 3 s p = 0.029 + 3(0.0080) = 0.0530 LCLp = p - 3 s p = 0.029 - 3(0.0080) = 0.0050 Below are charts showing “exact” and “approximate” control limits. Problem 13-37: p-charts

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Statistical Process Control

15

Points #12 and #19 are out of limits. After deleting these points, we get: b) Revised The revised average sample size = 9695/22 = 440.68 CL p = 256 / 9695 = 0.0264 sp =

[( p)(1 p)] / n =

[(0.0264)(0.9736)]/440.68

= 0.0076

UCLp = p + 3 s p = 0.0264 + 3(0.0076) = 0.0492 LCLp = p - 3 s p = 0.0264 - 3(0.0076) = 0.0036

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Statistical Process Control

16

Problem 37b - Final revised p-charts

The process is now considered to be under control. 38.

Construct an np-chart for the data in Problem 34, the Full Life Insurance Company. What

does the chart show? © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

17

Answer 38.

See spreadsheets Prob13-38NP.xls for details. Using data from problem 13-34, we get: CL n p = n p = 100 (0.021) = 2.1 snp =

[n( p)(1 p)] =

100 (0.021) (0.979)

= 1.434

Control limits: UCL n p = n p + 3 s n p = 2.1 + 3 (1.434) = 6.402 LCL n p = n p - 3 s n p = 2.1 - 3 (1.434) = -2.202, use 0 As was shown in the previous control chart for problem 13-34, values for samples 9 and 23 are out of limits. Eliminating these points, we get revised control limits shown for the final control chart, below. Note that the two values of 6 or over were dropped.

Problem 13-38- Revised So, CL n p = n p = 100 (0.0171) = 1.71 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

snp =

[n( p)(1 p)] =

18

100 (0.0171)(0.9829)

= 1.296

Control limits: UCL n p = n p + 3 s n p = 1.71 + 3 (1.296) = 5.598 LCL n p = n p - 3 s n p = 1.71 - 3 (1.296) = -2.178, use 0

Problem 13-38 Final np-control chart

39.

Construct an np-chart for the data in Problem 13-35 from Lean Manufacturing, Inc. What

does the chart show? Answer 39.

See spreadsheet Prob13-39NP.xls for details Using data from problem 13-35, we get results for the np chart that are similar to the p-chart: Initial So, CL n p = n p = 200 (0.1444) = 28.88

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Statistical Process Control

snp =

[n( p)(1 p)] =

19

200 (0.1444)(0.8556)

= 4.971

Control limits: UCL n p = n p + 3 s n p = 28.88 + 3 (4.971) = 43.793 LCL n p = n p - 3 s n p = 28.88 - 3 (4.971) = 13.967

As was shown in the previous control charts for problem 13-35, values for 9 out of limits samples had to be eliminated, leaving 16 usable data points. After eliminating the unusable points, we get revised control limits shown for the final control chart, below. Final Revised Using the slightly more accurate spreadsheet value, p = 0.0928 So, CL n p = n p = 200 (0.0928) = 18.56 snp =

[n( p)(1 p)] =

200 (0.0928)(0.9072)

= 4.103

Control limits: © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

20

UCL n p = n p + 3 s n p = 18.56 + 3 (4.103) = 30.869 LCL n p = n p - 3 s n p = 18.56 - 3 (4.103) = 6.251

Problem 13-39: Final revised np control chart

40.

Construct a c-chart for a situation involving 40 samples and having a total of 1,000 defects and interpret the results.

Answer 40.

Center line for the c-chart: c = 1000/40 = 25 c ±3

41.

c = 30 ± 3

25 = 30 ± 15 = 15 to 45

Construct a c-chart for a situation involving 30 samples and having a total of 340 defects and interpret the results.

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Statistical Process Control

21

Answer 41.

Center line for the c-chart: c = 340/30 = 11.33

c = 11.33 ± 3 11.33 = 11.33 ± 10.1 = 1.23 to 21.33

c ±3

42.

Consider the sample data for defects per pizza in a new store being opened by Rob’s Pizza Palaces in the worksheet Prob.13-42. Construct a c-chart for these data. What does the chart show?

Answer 42.

See spreadsheet Prob13-42CC.xls for details. For the c-chart: Number defective = 84; number of samples = 25 Center Line for the c-chart: c = 84/25 = 3.36 c ±3

c

= 3.36 ± 3

3.36 = 3.36 ± 5.50 = -2.14 to 8.86, use 0 for lower control limit.

The process appears to be in control.

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Statistical Process Control

22

43.

Smoothsoft, Inc., a software developer, measured the number of defects per 1,000 lines of code in software modules being developed by the company. Construct a c-chart for data in the table in the worksheet Prob. 13- 43 and interpret the results. Answer 43.

See spreadsheet Prob13-43CC.xls for details. For the c-chart: Number defective = 178; number of samples = 30 Center Line for the c-chart: c = 178/30 = 5.93 c ±3

c = 5.93 ± 3

5.93 = 5.93 ± 7.31 = -1.38 to 13.27, use 0 for lower control limit.

The process appears to be in control.

44.

Find 3

control limits for a c-chart with an average number of defects equal to 18.

Answer 44.

For the c-chart: Center Line: c (average number of defects) = 18 c ±3

c = 18 ± 3 (4.24) = 18 ± 12.72 = 5.28 to 30.72

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Statistical Process Control 45.

23

Tom Pyzdek, a noted quality consultant, presented data on falls at the Great Falls Hospital (fictitious name), for patients as shown in the table in the worksheet Prob. 13-45. Note that the “sample size” is in 100s of Patient Care Days (PCDs). Develop a run chart, a frequency histogram, and a u-chart for these data. What insights do you get from each chart? What would you advise the administration of the hospital to do about falls?

Answer 45.

The run chart, frequency distribution, and u-chart all provide different insights on the problem. The run chart shows that there were as many as 6 falls recorded in four different months at the hospital.

7 6 5 4 3 2 1 0

25

23

21

19

17

15

13

11

9

7

5

Number

3

1

Falls

Number of Falls

Samples

The frequency histogram shows that the most common number of monthly falls was 4 per month. Looking back at the run chart, one can see the months in which each number of falls happened.

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Statistical Process Control

24

Frequency of Occurrance

Histogram 8 7 6 5 4 3 2 1 0

Frequency

0

1

2

3

4

5

6

Monthly Falls

The u-chart shows what the rate of falls was, based on the number of patient care days (PCD’s), a variable “sample size”. These data permit an analyst to develop a control chart, with variable control limits. The chart shows that the process was under control, although it doesn’t mean that falls were at an acceptable level! The acceptable level of patient falls is 0. Therefore, the hospital administration should be encouraged to use these data to analyze and find the root causes of falls and make every effort to reduce the frequency to 0.

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Statistical Process Control 46.

25

Determine, using Figure 13.40, the appropriate sample size for detecting: a.

A 1-sigma shift in the mean with a 0.80 probability.

b.

A 2-sigma shift with 0.95 probability

c.

A 2.5-sigma shift with 0.90 probability

Answer 46.

This is simply an exercise in reading values from the curves to fit required conditions. a. For a 1 shift and a 0.80 probability, use n = 15 (if rounded to next higher value). b. For a 2 shift and a 0.95 probability, use n = 8 (rounded to next higher value). c. For a 2.5 shift and a 0.90 probability, use n = 3 (rounded to next higher value).

47.

Find 3 control limits for a u-chart with u = 16 and n = 4. What do the limits show?

Answer 47.

For the u-chart conditions: Number of samples = 4, number of defectives = 16 Center Line for the u-chart: u = 16/4 = 4 u ±3

u / n = 4 ± 3 = 4 ± 3 = 1 to 7

This calculation provides the control limits on one sample with a size of 4.

For problems 48 through 51, see the Advanced Control Charts Section in the Bonus Materials folder on the CD-ROM.

48.

Develop a stabilized control chart for the post office example (Figure 13.29). What does the chart show?

Answer 48.

The stabilized p-chart below, based on the post office example, plots the “transformed z statistic” instead of p, and it shows the process is in control. (See spreadsheet Prob1348PS.xls for details). To verify calculations, for example, the first data point is: p = 0.022

process =

p(1 p) =

0.022(1 0.022) = 0.1467

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Statistical Process Control

26

Note that p (1 - p ) is not divided by n, since this is the estimated process standard deviation, not the sample standard deviation. Thus variations in sample and lot sizes can be tolerated here, where they might cause problems with the standard p-chart. z =

p- p = p

49.

0.03 - 0.022 = 0.0545 0.1467

Develop a stabilized control chart for the silicon wafer example (Figure 13.6). What does the chart show?

Answer 49.

The stabilized control charts for x and R show the same out-of-control condition for the Silicon Wafer example as discussed earlier in chapter. Only the first x control chart is produced here, and it shows sample 17 out of control, as in body of the chapter. It would be a simple matter to remove sample 17, which created the out-of- control condition, since control limits on the stabilized chart remain constant for a given sample size. See spreadsheet Prob13-49ST.xls for details. Note that slightly different values for the control chart averages are calculated versus those in the text problem, due to rounding the data in the original problem to whole number values.

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Statistical Process Control

27

To verify calculations, for example, the first data point is: x = 48.973

R = 27.040

CL x = 0; CL R = 1

UCL x = A2 = 1.023 LCL x = - A2 = - 1.023 UCL R = D4 = 2.574 LCL R = D3 = 0 Note that the limits on x are not multiplied by 3, which is the estimated process range variation, not the sample range variation. The following example shows how the transformed values for the first sample was obtained. z

x

= x - x = 44.333 - 48.973 = - 0.172 R

z= R = R

27.040 48 = 1.775 27.04

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Statistical Process Control

50. Using a value of

28

= 0.8, construct an EWMA chart for the data shown in the worksheet

Prob. 13-50 as applied to the means of the samples. What does the chart show? Answer 50.

The control chart for the EMWA versus observed values is shown below. (See spreadsheet Prob13-50EW.xls for details). With an = 0.8, we see that the process is under control, and the EMWA estimate fairly closely anticipates the next observed value. The conclusion is that a better “forecast” of future values may be obtained for volatile values such as these if a larger value is used to give greater weight to more recent values.

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Statistical Process Control

51.

Repeat problem 50 for a value of

29

= 0.4. What does the chart show, now?

Answer 51.

The control chart for the EMWA versus observed values is shown below. (See spreadsheet Prob13-51EW.xls for details). With an = 0.4, we see that the process is under control, and the EMWA estimate less closely anticipates the next observed value. The conclusion is that a better “forecast” of future values was obtained (as shown by those in problem 13-50) for volatile values such as these by using a larger value, which gave greater weight to more recent values. Thus, a larger is preferable for more volatile values, although there is less forecasting certainty available, for the long term.

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Statistical Process Control

30

For problems 52 through 54, see the Statistical Foundations of Control Charts Section in the Bonus Materials folder on the CD-ROM. 52.

What are the probability limits corresponding to a Type I error of

= 0.10?

Answer 52. 2

= 0.10 = 0.05; From the normal probability table, P(z) = 0.4500 2

Therefore, z0.02 = 1.64 or 1.65, since it is equidistant (0.4495 and 0.4505, respectively) between the closest table values to 0.4500.

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Statistical Process Control 53.

31

If control limits for a project are based on 2.5 standard deviations, what percentage of observations will be expected to fall beyond the limits?

Answer 53.

For z = ± 2.75, P (z > 2.75) = P (z < -2.75) From the normal probability table: 0.5000 - 0.4970 = 0.003; Therefore, the % outside = 100 x 2 (0.003) = 0.6 %

54.

What is the probability of observing 11 consecutive points on one side of the center line if the process is in control? 10 of 11 points? 9 of 11 points? How many points out of 11 on one side of the center would indicate lack of control?

Answer 54.

Using the binomial formula: n

Probability (acceptance) = 11 in a row = (0.5)11

10 of 11

n f(x) = and f(x) = ( x ) px (1-p)n-x x=0 = 0.049%

11 = ( 10 ) (0.5)10 (0.5)1 = 11 (0.5)11 = 0.539%

11 9 of 11 = ( 9 ) (0.5) 9 (0.5)2 = 55 (0.5)11 = 2.695% 11 8 of 11 = ( 8 ) (0.5) 8 (0.5)3 = 165 (0.5)11 = 8.085% 11 7 of 11 = ( 7 ) (0.5) 7 (0.5)4 = 330 (0.5)11 = 16.17%

9 out of 11 points are statistically significant (p < 0.05). © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

32

ANSWERS TO CASE QUESTIONS I. La Ventana Window Company This case is designed to test the students' abilities to apply SPC principles, to interpret the results effectively, and see "beyond the data." The key points are summarized below: 1. The student should be able to apply the formulas from this chapter to construct an x-bar and Rchart and to determine the state of control, remove out-of-control points, and compute new control limits. 2. A key aspect of the case is to recognize potential differences among operators. This is going beyond the computations and using the data for diagnosis. The astute student might even take a different approach and stratify the data by operator to study differences among them. Although the key result (concerning operator Shane) can be gleaned by cross-comparing the control chart with the table in the database, such an analysis would clearly show the source of the problem. 3. It is critical that process capability calculations be performed after the process is brought into control by removing out-of-control points. 4. The fact that process capability is extremely good means that the company should not be concerned, but should devote its attention to improved training of any substitute operators and use SPC as more of an audit tool. 5. The additional data indicate that the process tends to drift upward after some time. The student should speculate about potential reasons for this. This also provides a clue as to why “interference” occurs in windows, occasionally. Assignment I - La Ventana Window Company Case Since the data are variables data, the first step is to construct x-bar and R-charts and determine if the process is in control. Figure DDC-1 (from the Excel spreadsheet) shows the mean, range and control limits for the original set of 30 samples. Using these data, we find that the mean and the average range are as follows: For the Center Lines, CL x : x = 25.4992; CLR : R = 0.0136 Control limits for the x - chart are: x ± A2 R = 25.4992± 0.577 (0.0136) = 25.491 to 25.507 © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

33

For the R-chart: UCLR = D4 R = 2.114 (0.0136) = 0.029 LCLR = D3 R = 0 Figure La Ventana-1 shows the range and averages charts with these control limits. The range chart shows sample 24 is out of control. Also, the x - chart has three points below the lower control limit. Inspecting the production records, we see that when each of these samples were taken, a different operator, "Shane" was running the cutting operation. Apparently this was a substitute for operator "Juanita" who was absent. Hence, these data may be construed as a special cause. (The company should determine if "Shane" was knowledgeable about the operation and equipment or needs additional training.) These data that show defective products must be removed from consideration and new control limits must be computed before capability can be assessed. After deleting these samples, the new averages are: CL x : x = 25.5007; CLR : R = 0.0124 These lead to the new control limits: Control limits for the x - chart are: x ± A2 R = 25.501± 0.577 (.0124) = 25.494 to 25.508

For the R-chart: UCLR = D4 R = 2.114 (.0124) = 0.026 LCLR = D3 R = 0

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Statistical Process Control

34

FIGURE La Ventana-1 Averages Lower Control Limit Center Line Upper Control Limit

X-bar Chart 1 La Ventana Window Co. Averages

25.510 25.505 25.500 25.495 25.490 25.485

28

25

22

19

16

13

10

7

4

1

25.480 Sample number

R-Chart 1 La Ventana Window Co.

R Values Lower Control Limit Center Line Upper Control Limit

0.035

Ranges

0.030 0.025 0.020 0.015 0.010 0.005

29

27

25

23

21

19

17

15

13

11

9

7

5

3

1

0.000 Sample number

The new control charts are shown in Exhibit La Ventana-2. The process now appears to be in control.

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Statistical Process Control

35

Process capability may now be evaluated. An estimate of the standard deviation from the revised control chart statistics is Estimated

= R / d2 = .0124/2.326 = .0.0053

The six-standard deviation spread is 25.501 ± 3 (.0.0053), or 25.485 to 25.517. (If one computes the standard deviation of the raw data after the three samples are deleted, the actual standard deviation is found to be 0.0052, so the estimate is very close.) The variation is well within the specifications of 25.470 and 25.530. This can also be seen by computing the process capability indexes: Cp = 0.060 / [6 (.0053)] = 1.88 Since the process is almost exactly centered, Cpl = Cpu = Cpk = approximately 1.88 also. (See spreadsheet C13lvwc2.xls for exact figures). This means that the process is very capable of meeting the specifications. FIGURE La Ventana-2 Averages Lower Control Limit Center Line Upper Control Limit

X-bar Chart 2 La Ventana Window Co. Averages

25.510 25.505 25.500 25.495

27

25

23

21

19

17

15

13

11

9

7

5

3

1

25.490 Sample number

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Statistical Process Control

36

R Values Lower Control Limit Center Line Upper Control Limit

R-Chart 2 La Ventana Window Co. 0.030

Ranges

0.025 0.020 0.015 0.010 0.005 27

25

23

21

19

17

15

13

11

9

7

5

3

1

0.000 Sample number

Assignment II - La Ventana Window Company Case The additional 20 samples must be plotted using established control limits. It is incorrect to use the data to find new control limits. The first set of samples established the state of control and no process changes were made. Sample means and ranges for all 50 samples are shown in Figure La Ventana-3. The R-chart is in control. The last 9 samples show consecutive points are above the center line on the x - chart. (This meets the rule of 8 consecutive points cited in the text, showing that the process has drifted out of control.) The company should investigate the fixtures and tools to determine if adjustments should be made.

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Statistical Process Control

37

FIGURE La Ventana-3

X-bar Chart 3 La Ventana Window Co.

Averages Lower Control Limit Center Line Upper Control Limit

Averages

25.510 25.505 25.500 25.495

45

41

37

33

29

25

21

17

13

9

5

1

25.490 Sample number

R-Chart 3 La Ventana Window Co.

R Values Lower Control Limit Center Line Upper Cotrol Limit

0.030

Ranges

0.025 0.020 0.015 0.010 0.005 46

43

40

37

34

31

28

25

22

19

16

13

10

7

4

1

0.000 Sample number

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Statistical Process Control

38

II. Murphy Trucking, Inc. The Billing Study - Part I The first assignment requires the construction of a p-chart, since we are interested in the proportion of bills in error. (Summary data is in spreadsheet C13MTISUMS.xls)The calculations are shown below. The average proportion of bills in error is 0.63 and the standard deviation is 0.108. Using the formulas for a p-chart, the lower and upper control limits are, respectively, 0.306 and 0.954. The control chart is shown in Figure MTI-A. CL p = 252 / 400 = 0.63 sp =

[( p)(1 p)] / n =

[(0.63) (0.37)] / 20

= 0.108

Control limits: UCLp = p + 3 sp = 0.63 + 3 (0.108) = 0.954 LCLp = p - 3 sp = 0.63 - 3 (0.108) = 0.306 Perhaps the most surprising finding is that except for points hugging the center line, the process appears to be under control! However, improvements definitely need to be made. Although the process may be in control, an error rate of 63 percent is clearly unacceptable. The capability of the process is specified by the control limits, since they are 3 standard deviations on either side of the average. This can be interpreted to mean that error rates only as low as 31 percent and as high as 95 percent might be reasonably expected. Thus, further analysis is warranted. See spreadsheet C14MTIP1.xls for further details. Note that a u-chart has also been constructed for comparison purposes with the Billing Study, Part II. See spreadsheet C13MTIUPt1.xls for further details.

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Statistical Process Control

39

Figure MTI-A p Values Lower Control Limit Center Line Upper Control Limit

Attribute (p) Chart 1.00

% Non conforming

0.800 0.600 0.400 0.200 0.00

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Sample number

U Values Low er Control Limit Center Line Upper Control Limit

1.0000 0.8000 0.6000 0.4000

25

23

21

19

17

15

13

11

9

7

5

3

0.2000

1

Defects per unit

Attribute (u) Chart

Sample number

The Billing Study - Part II The second part of the study is to analyze the distribution of actual errors identified by the management team. The analysis consists of two phases. First, a u-chart should be constructed to study the total number of errors per bill. The calculations are shown below. The control chart, shown in Figure MTI-B, is clearly in control. No special causes of variation are apparent; thus, management must attack the common causes. (See spreadsheet C13MTIUPt2.xls for further details.) © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

40

For the u-chart: 25 samples are available. Because the number of bills each day varies considerably, individual control limits are established for each day. For the Day 1 sample, n = 54 total bills, number of defects = 36 for all categories. The overall average, u , must be used to calculate the individual control limits. Center Line for the u-chart: u = 1232 / 1965 = 0.6270 Control limits must be calculated for each sample, since each sample size is different, hence the "ragged" appearance of the upper control limit on the u-chart below. For the first sample, where n = 54: u ± 3 u / n = 0.6270 ± 3 0.304 to 0.950

(0.6270 )/54

= 0.6270 ± 3 (0.1078) =

Figure MTI-B U Values Low er Control Limit Center Line Upper Control Limit

1.0000 0.8000 0.6000 0.4000

25

23

21

19

17

15

13

11

9

7

5

3

0.2000

1

Defects per unit

Attribute (u) Chart

Sample number

Note that a p-chart has also been constructed with these data for comparison purposes with the initial billing study, Part I, above. See spreadsheet C13MTIP2.xls for further details.

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Statistical Process Control

41

Attribute (p) Chart

p Values Lower Control Limit Center Line Upper Control Limit

0.700 0.600

% Non conforming

0.500 0.400 0.300 0.200 0.100 0.00

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 Sample number

Finally, if each error category is summed, we may construct a Pareto diagram of the distribution of errors by category as shown in Figure MTI-C. (See C13MTIPARETO.xls for details.) By examining the nature of the errors, you should realize that many of the errors, specifically categories 1, 2, 3, 6 and 7, can easily be recognized by the driver, while the other categories are "true" billing errors. Nearly seventy percent of the errors fall into this category. This suggests that an increased focus on driver training and awareness could reduce a majority of the errors. The customers also should be informed of their role in providing correct information to reduce the scope of the problem. Figure MTI-C Murphy Trucking, Inc Error Category

2 8 1 6 4 5 7 3 Totals

% of Total Cumulative Errors Errors Percent 300 24.35% 24.35% 264 21.43% 45.78% 159 12.91% 58.69% 151 12.26% 70.94% 147 11.93% 82.87% 87 7.06% 89.94% 62 5.03% 94.97% 62 5.03% 100.00% 1232

100.00%

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Statistical Process Control

42

Frequency of Billing Errors

Murphy Trucking, Inc. Pareto Diagram 350

120.00%

300

100.00%

250

80.00%

200 60.00% 150 40.00%

100

20.00%

50 0

0.00% 2

8

1

6

4

Error Categories

5

7

3 Errors Cumulative Percent

III. Day Industries Day Industries’ data can be analyzed in three ways. The measures are interrelated, so they may be looked at individually, in groups, and by control charts. (See spreadsheets labeled: C13DayScatter, C13DayViscos, C13DaySolids, and C13DayProcCap for details). Visual inspection of the individual data for each variable reveals one conclusion about the “pounds per gallon” variable. That is, there is very little variation over the 41 samples.

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Statistical Process Control

43

Day Industries - Scatter Diagram 80

Measure

75 Viscosity Percent Solids

70

65

31

28

25

22

19

16

13

10

7

4

1

60

Batches

On the surface, it seems that this variable would not need to be charted continuously. A reading could be taken, at random and infrequently, to evaluate whether further testing should be done. There is one concern – the process mean is not centered on the nominal value of 13.05 pounds per gallon. Because of the obvious stability of the pounds per gallon measure, it will not be placed on a control chart, but its process capability will be calculated. The other two variables, viscosity and solids, show much more variability than the pounds per gallon measure. (See Scatter Diagram, above) They should be closely watched and carefully controlled. Values for the sample means and standard deviations versus their specification limits are:

Mean Std. Dev. Spec. Limits

Viscosity Percent Solids Lb./Gal. 74.146 63.056 13.261 4.569 0.511 0.080 60-80

60-65

12.6-13.5

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Statistical Process Control

44

Statistical Control of Viscosity and Solids Analysis of viscosity and solids required use of charts for individuals, since each of the measures was from a chemical process, from which each sample was taken individually. Results show:

Individuals (X) Chart - Viscosity

Individuals Upper control limit Center line Lower control limit

100

90

80

70

Value

60

50

40

30

20

10

0 1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Observation number

Moving Range Chart - Viscosity 18

Moving ranges Lower control limit Center line Upper control limit

16

Moving ranges

14

12

10

8

6

4

2

0 1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

Observation number

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Statistical Process Control

45

Solids

Individuals (X) Chart - Solids

Individuals Upper control limit Center line Lower control limit

65.0

64.5

64.0

63.5

Value

63.0

62.5

62.0

61.5

61.0

60.5

60.0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Observation number

Moving Range Chart - Solids

Moving ranges Lower control limit Center line Upper control limit

1.8

1.6

Moving ranges

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

Observation number

Although there was a sample on each of the charts that was just inside the control limits, the measures for both variables appear to be stable and under control. © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

46

Process Capability- Solids The process capability for solids is good, although not quite at the ideal level of 2.0 for all measures as shown here: Nominal specification Upper tolerance limit Lower tolerance limit

Average 62.5 65 Standard deviation 60

63.0561 Cp 0.5109 Cpl Cpu Cpk

1.6311 1.9939 1.2683 1.2683

Process Capability- Viscosity The process capability for viscosity is not good, as shown here: Nominal specification Upper tolerance limit Lower tolerance limit

70 80 60

Average Standard deviation

74.1463 4.5693

Cp Cpl Cpu Cpk

0.72951 1.03200 0.42703 0.42703

Process Capability- Pounds/Gallon The process capability for pounds per gallon is not as good as it might be, as shown here: Nominal specification Upper tolerance limit Lower tolerance limit

13.05 13.5 12.6

Average Standard deviation

13.2610 Cp 0.0802 Cpl Cpu Cpk

1.8693 2.7457 0.9929 0.9929

The report to the plant manager would include the analysis presented above. It should also include the recommendations that he/she: Work to center the part of the process that controls solids, so that the measure averages its nominal value of 62.5. Attempt to center the part of the process that involves pounds per gallon, so that its average approximates its nominal value of 13.05. Perform a study of the part of the process that involves viscosity and work to center its average and reduce its variation so as to make the process capable, as well as continuing in control. © 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Statistical Process Control

47

Continue to monitor both solids and viscosity using charts for individuals (x and Moving Range charts), and take action if the process appears to be heading out of control. Monitor the part of the process that affects pounds per gallon until it is clear that there is no tendency for the process average to “drift.” Afterward, it may be possible to discontinue regular control charting and periodically sample the pounds per gallon. If process instability occurs, step up monitoring and control activities.

© 2011 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.