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Last 1/3 of Chapter 13 problem solutions
Chapter 13 - Statistical Process Control 2
This is file Q8IM13E - 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.
Ans
30.Control limits for Yummy Candy Company are as follows:
CL
= 75/2250 = 0.0333
s =
s = = 0.0207
Control limits:
UCLp =
+ 3 s
UCLp = 0.0333+ 3(0.0207) = 0.0954
LCLp =
- 3 s
LCLp = 0.0333 - 3(0.0207) = - 0.0288, use 0
31.Thirty-five samples of 25 orders each at the Bakery Bread, Inc. were inspected, and 25 items were found to be defective. Compute control limits for a p-chart.
Ans.31. Control limits for Bakery Bread, Inc. orders can be calculated using:
CL
= 25/(35)(25) = 25/875 = 0.029 s =
s = = 0.034
Control limits:
UCLp =
+ 3 s
UCLp = 0.029 + 3(0.034) = 0.131
LCLp =
- 3 s
LCLp = 0.029 - 3(0.034) = - 0.073, use 0
32.Samples of packages orders were taken at the R.A. Treinta Package Co. to determine if the orders were prepared correctly. The percent defectives for each sample are given in the worksheet Prob.13-32 for 25 samples. Five hundred orders are inspected each day for each sample. Construct a p-chart and interpret the results.
Ans.32. Control limits for the R.A. Treinta Package Co. orders can be calculated using:
CL
= 0.012 s =
= = 0.0049
Control limits:
UCLp =
+ 3 sp = 0.012 + 3(0.0049) = 0.0267
LCLp =
- 3 sp = 0.012 - 3(0.0049) = - 0.0027 use 0
The process appears to be under control.
33.The fraction defective for a folding process in an Out-of-the-Box plant is given in the worksheet Prob.13-33 for 25 samples. One hundred 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?Ans
33. See data and control chart below for Out-of-the-Box Companys plant. See spreadsheet Prob13-33P.xls for details.
CL
= 0.024 s =
= = 0.0153
Control limits:
UCLp =
+ 3 sp = 0.024 + 3(0.0153) = 0.0699
LCLp =
- 3 sp = 0.024 - 3(0.0153) = - 0.0219 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,
, and sample 30 is outside the control limits. Later, four out of five samples between 37-41 are more than 1 ( away from the mean,
, and sample 39 is outside the control limits. The overall pattern seems to show an upward shift in the process proportion. See Part B tabs on spreadsheet Prob13-33P.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 39, which was above the control limit, should have been spotted, and the process stopped and readjusted.
34.The fraction defective of automotive pistons made by the Precision Piston Co. is given in the worksheet Prob.13-34 for 25 samples. One hundred units are inspected each day. Construct a p-chart and interpret the results.
Ans
34.See data and control charts for Precision Piston Company, below. See spreadsheets Prob13-34P.xls, sheets for details.
a) Initially, CL
= 0.096 (see printout)
s =
s = = 0.029
Control limits:
UCLp =
+ 3 s
UCLp = 0.096 + 3(0.029) = 0.183
LCLp =
- 3 s
LCLp = 0.096 - 3(0.029) = 0.009
See initial data and control chart, below.
34.b. Precision Piston Company (Continued)
Revised CL
= 0.092 (after sample 12, with a fraction defective
of 0.19, was removed). s =
= = 0.029
Control limits:
UCLp =
+ 3 s
UCLp = 0.092 + 3(0.029) = 0.179
LCLp =
- 3 s
LCLp = 0.092 - 3(0.029) = 0.005
See data and control chart, below, after sample 12 was removed. The conclusion is, that the process is now in control.Problem 34-B Final Revised Control Chart
35.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. 13-35. Construct a p-chart. If any points occur outside the control limits, assume that assignable causes have been determined. Then construct a revised chart.
Ans.35. See spreadsheets for Full Life Insurance Co. in file Prob13-35P.xls for details.
a) Initially, based on the sum of the p values for the 25 samples,
p1 + p2 + p3 +
CL
= -----------------------
NCL
= 0.55 / 25 = 0.022
s =
= = 0.0147
UCLp =
+ 3 s = 0.022 + 3(0.0147) = 0.0661
LCLp =
- 3 s = 0.022 - 3(0.0147) = -0.0221, use 0
Throw out #9 and #23, out-of-control values, revise.
See initial data and control chart, below.
35.b) Revised
CL
= 0.400 / 23 = 0.0174 s =
= = 0.0131
Control limits:
UCLp =
+ 3 s = 0.0174 + 3(0.0131) = 0.0567
LCLp =
- 3 s = 0.0174 - 3(0.0131) = -0.0219, use 0
The conclusion is that the process is now in control.
Problem 35-b Final Revised Control Chart
36.Samples of size 50 have been randomly selected during each shift of 30 shifts in a production process at Delgado Manufacturing, Inc. The data are given in the worksheet Prob.13-36. 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.Ans.36.a)See spreadsheets Prob13-36P.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
= -----------------------
N
CL
= 4.12 / 30 = 0.1373
s =
= = 0.0487
Control limits:
UCLp =
+ 3 s = 0.1373 + 3(0.0487) = 0.2834
LCLp =
- 3 s = 0.1373 - 3(0.0487) = - 0.0088 Use 0
See initial p-chart, above. It appears that this process is not in control and that major changes need to be made. However, we can assume that the out of control points #20 and #25 (and most likely, point #21) are due to assignable causes. By throwing out all points containing 13 defectives and over, and revising the chart, we can come up with the results shown in part 36 b), below.
36. b)(Continued)
This results in:
CL
= 3.22 / 27 = 0.1193
s =
= = 0.0458
Control limits:
UCLp =
+ 3 s = 0.1193 + 3(0.0458) = 0.2568
LCLp =
- 3 s = 0.1193 - 3(0.0458) = - 0.0181 Use 0
Revised p-chart
After eliminating the outlying points, the process appears to be under control.37.SpeedyNetService.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.13-37. What does the chart show? Is the service level good or bad, in your opinion?
Ans.37.See spreadsheet Prob13-37P.xls for details about SpeedyNetService.com.
The average sample size = 15755/30 = 525.17
CL
= 202 / 15755 = 0.0128
s =
= = 0.0049UCLp =
+ 3 s = 0.0128 + 3(0.0049) = 0.0275
LCLp =
- 3 s = 0.0128 - 3(0.0049) = - 0.0019, use 0
All points fall within the control limits, so the service level appears to be good.Problem 13-37: p-charts
38.Ellswater 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.13-38, which represent responses to a question on satisfaction with hospital housekeeping services.
Ans.38. Note that the values obtained for Ellswater Hospitals data with a hand calculator will differ slightly from those obtained from the Excel spreadsheets, due to rounding differences. See spreadsheets in the file Prob13-38P.xls for details.
Initial Calculationa) The average sample size = 10468/ 30 = 348.93, so the approximate control limits, based on total defects over total items samples, are:
CL
= 259 / 10468 = 0.0247
s =
= = 0.0083
UCLp =
+ 3 s = 0.0247 + 3(0.0083) = 0.0496
LCLp =
- 3 s = 0.0247 - 3(0.0083) = - 0.0002 Use 0 Below are charts showing exact and approximate control limits.
Problem 13-38: p-charts
Points #12 and #19 are above the control limits. After deleting these points, we get:
38.b) Revised
The revised average sample size = 9804/28 = 350.14
CL
= 223 / 9804 = 0.0228
s =
= = 0.0080
UCLp =
+ 3 s = 0.0228 + 3(0.0080) = 0.0468
LCLp =
- 3 s = 0.0228 - 3(0.0080) = - 0.0012 Use 0Problem 38b - Final revised p-charts
The process is now considered to be under control.
39.Construct an np-chart using the data in Prob.13-35, the Full Life Insurance Company. What does the chart show?
Ans.39.See spreadsheets Prob13-39NP.xls for details. Using data from Prob.13-35, we get:
CL n
= n
= 100 (0.022) = 2.2
s n
=
= = 1.467
Control limits:
UCL n
= n
+ 3 s n
= 2.2 + 3 (1.467) = 6.601
LCL n
= n
- 3 s n
= 2.2 - 3 (1.467) = -2.201, use 0
As was shown in the previous control chart for Prob. 13-35, 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 13-39- Revised
So, CL n
= n
= 100 (0.0174) = 1.74 s n
=
= = 1.308
Control limits:
UCL n
= n
+ 3 s n
= 1.74 + 3 (1.308) = 5.664
LCL n
= n
- 3 s n
= 1.74 - 3 (1.308) = -2.184, use 0
Note: Does not conform to spreadsheet values due to rounding errors.
Problem 13-39 Final np-control chart
40.Construct an np-chart using the data in Prob. 13-36 from Delgado Manufacturing Co. What does the chart show?
Ans.40. See spreadsheet Prob13-40NP.xls for details. Note that values obtained from a hand calculator are slightly different from those from an Excel spreadsheet, due to rounding errors. Using data from Prob.13-36, we get results for the np chart that are similar to the p-chart:
Initial
So, CL n
= n
= 50 (0.1373) = 6.865 s n
=
= = 2.434
Control limits:
UCL n
= n
+ 3 s n
= 6.865 + 3 (2.434) = 14.167
LCL n
= n
- 3 s n
= 6.865 - 3 (2.434) = -0.437
As was shown in the previous control charts for Prob.13-36, values for 3 out of limits samples (actually sample 20 would have been out of limits after re-calculation) had to be eliminated, leaving 27 usable data points. After eliminating the unusable points, we get revised control limits shown for the final control chart, below.
Final Revised
So, CL n
= n
= 50 (0.1193) = 5.965 s n
=
= = 2.292
Control limits:
UCL n
= n
+ 3 s n
= 5.965 + 3 (2.292) = 12.841
LCL n
= n
- 3 s n
= 5.965 - 3 (2.292) = -0.911Problem 13-40: Final revised np control chart
41.Calculate the centerline and control limits for a c-chart involving 40 samples and having a total of 1,000 defects and interpret their meaning.
Ans.41.Center line for the c-chart:
= 1000/40 = 25
3
= 25 3 = 25 15 = 10 to 4042.Find 3 ( control limits for a c-chart with an average number of defects equal to 18.
Ans.42.For the c-chart: Center Line:
(average number of defects) = 18
3
= 18 3 (4.24) = 18 12.72 = 5.28 to 30.72
43.Calculate the centerline and control limits for a c-chart involving 35 samples and having a total of 350 defects and interpret the results.
Ans.
43. Center line for the c-chart:
= 350/35 = 10.0
3
= 10.0 3 = 10.0 9.49 = 0.513 to 19.4944.Consider the sample data for defects per pizza in a new store being opened by Robs Pizza Palaces in the worksheet Prob.13-44. Construct a c-chart for these data. What does the chart show?
Ans.
44. See spreadsheet Prob13-44CC.xls for details.
For the c-chart: Number defective = 84; number of samples = 25
Center Line for the c-chart:
= 84/25 = 3.36
3
= 3.36 3 = 3.36 5.50 = -2.14 to 8.86, use 0 for lower control limit.
The process appears to be in control.
45.FarmaSuitica, Inc., a mail-order prescription drug vendor, measured the number of defects per standard 200 line order being picked in their distribution center. Construct a c-chart for data in the table in the worksheet Prob. 13- 45 and interpret the results.
Ans.
45.See spreadsheet Prob13-45CC.xls for details.
For the c-chart: Number defective = 52; number of samples = 30
Center Line for the c-chart:
= 52/30 = 1.73
3
= 1.73 3 = 1.73 3.95 = -2.22 to 5.68, use 0 for lower control limit.
The process appears to be in control.
46.A quality consultant was asked to analyze the data from order errors at the Audubon Books, Inc., distribution center as shown in the table in the worksheet Prob. 13-46. The data show the number of orders processed per month and the error found in those orders. 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 distribution center manager to do about the errors?
Ans.
46. The run chart, frequency distribution, and u-chart all provide different insights on the problem. a. The run chart shows that there were as many as 10 errors recorded in the monthly samples taken at the distribution center.
b. The frequency histogram shows that the most common number of errors was 5 per monthly sample. Looking back at the run chart, one can see the months in which each number of errors happened.
c. The u-chart shows what the rate of errors was, based on the orders processed, 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 doesnt mean that errors were at an acceptable level! The acceptable level of errors is 0. Therefore, the distribution center manager should be encouraged to use these data to analyze and find the root causes of errors and make every effort to reduce the frequency to 0. In addition, she/he should investigate the special causes on 3/7, 4/7 & 11/7. The manager should determine what is different about these days. Different operators, in their ability to handle capacity of large orders with current manpower, boredom, etc.; who was working on the day when there were few, or no errors? What were they doing?
47.Find 3( control limits for a small u-chart with the following errors per sample unit. What do the limits show?
ErrorsSample Unit
592
3136
470
878
7165
Ans.
47. For the u-chart conditions: Number of samples = range from 70 to 165 and the number of errors from 3 to 8.
Center Line for the u-chart:
= (5+3+4+8+7) / (92+136+70+78+165) = 27/541 = 0.05
So, su =
= . The limits will therefore vary for each sample. For example, the first sample will be:
3
= 0.05 3 = 0.05 3 (0.0233) = - 0.0199 to 0.1199 Use 0 for the lower limit.
This calculation provides the control limits on one sample with a size of 92.
Limits for the samples are, respectively:
SampleDefects Standard
ErrorsSizeper unitDeviationLCLuCLUCLu
5920.05430.023300.04990.1198
31360.02210.019200.04990.1074
4700.05710.026700.04990.1300
8780.10260.025300.04990.1258
71650.04240.017400.04990.1021
The u-chart (see spreadsheet Prob13-47U.xls for details) looks like this:
48.Determine, using Figure 13.42, p. 704 in the text, gives 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
Ans.
48.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).
For problems 49 through 51, see the Statistical Foundations of Control Charts Section in the Bonus Materials folder on the CD-ROM.
49.What are the probability limits corresponding to a Type I error of ( = 0.10?
Ans.
49. ( = 0.10 = 0.05; From the normal probability table, P(z) = 0.4500
2 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.
50.If control limits for a project are based on 2.75 standard deviations, what percentage of observations will be expected to fall beyond the limits?Ans.
50. 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 %
51.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?Ans.
51. Using the binomial formula:
n
n
Probability (acceptance) = ( f(x) = and f(x) = ( x ) px (1-p)n-x
x=0
11 in a row = (0.5)11
= 0.049%
11
10 of 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).
ANSWERS TO CASE QUESTIONS
Case - MORELIA MORTGAGE 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 R-chart 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 Shaun) 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 not good means that the company should be concerned, and 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.
Assignment I - Company Case
1. Interpret the data in the MMC Case worksheet in the Excel workbook C13CaseData (available on the Premium website), establish a state of statistical control, and evaluate the capability of the process to meet specifications. Consider the following questions: What do the initial control charts tell you? Do any out-of-control conditions exist? If the process is not in control, what might be the likely causes, based on the information that is available? What is the process capability? What do the process capability indexes tell the company? Is MMC facing a serious problem that it needs to address? How might the company eliminate the problems of slow loan processing?
Ans.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 Morelia Mortgage Co. - 1 (from the Excel spreadsheet called MMCxbar&R1) 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
:
= 16.021; CLR :
= 8.424
Control limits for the
- chart are:
A2
= 16.021 0.577 (8.424) = 20.882 to 11.160
For the R-chart: UCLR = D4
= 2.114 (8.424) = 17.808
LCLR = D3
= 0
Figure Morelia Mortgage Co. - 1 shows the range and averages charts with these control limits. The range chart does not have any obvious out of control points. However, the
- chart has two points above the upper control limit, point 9 and point 21. Inspecting the production records, we see that when each of these samples were taken, a different operator, "Shaun" was running the cutting operation. Apparently he was a substitute for operator "Carmen" who was absent. Hence, these data may be construed as a special cause. (The company should determine if "Shaun" was knowledgeable about the operation and equipment or needs additional training.)
FIGURE Morelia Mortgage Co. - 1
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, Figure Morelia Mortgage Co. - 2 (from the Excel spreadsheet called MMCxbar&R2.xls) shows the mean, range and control limits for the revised set of 28 samples. Using these data, we find that the mean and the average range are as follows:
CL
:
= 15.532; CLR :
= 8.147These lead to the new control limits:
Control limits for the
- chart are:
A2
= 15.532 0.577 (8.147) = 20.233 to 10.831
For the R-chart: UCLR = D4
= 2.114 (8.147) = 17.223
LCLR = D3
= 0
FIGURE Morelia Mortgage Co. - 2
The new control charts are shown in Figure Morelia Mortgage Co.-2. The process now appears to be in control (note that samples 9 and 21 were removed, and prior and succeeding points were connected).
Process capability may now be evaluated. An estimate of the standard deviation from the revised control chart statistics is
Estimated ( =
/ d2 = 8.147/2.326 = 3.503The six-standard deviation spread is 15.532 3 (3.503), or 26.041 to 5.023. (If one computes the standard deviation of the raw data after the two samples are deleted, the actual standard deviation is found to be 3.470, so the estimate is very close.) The six standard deviation spread does not fall within the specifications of 20.5 and 10.5. This can also be seen by computing the process capability indexes:
Cp = 10 / [6 (3.503)] = 0.476Since the process is almost exactly centered, Cpl = Cpu = Cpk = approximately 0.47 or 0.48, also. (See spreadsheet MMCxbar&R2.xls for exact figures). This means that the process is not capable of meeting the specifications. Thus, the process is in control, but not capable.Assignment II Morelia Mortgage Company Case
2. The process manager who initiated the trial project implemented the recommendations that resulted from the initial study. Because of her success in using control charts, MMC made a decision to continue using them on that process. After establishing control, one additional sample was taken over the next 20 shifts, shown in second part of the table in the MMC Case worksheet. Evaluate whether the process remains in control, and suggest any actions that should be taken. Consider the following issues: Does any evidence suggest that the process has changed relative to the established control limits? If any out-of-control patterns are suspected, what might be the cause? What should the company investigate?
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 48 samples (excluding the bad earlier samples) are shown below in Figure Morelia Mortgage-3 (from the Excel spreadsheet called MMCxbar&R3.xls). The R-chart is in control. Sample 46 shows that a point is out of control on the
- chart. In addition, it appears that values are hugging the centerline, except for the bad sample.The company should investigate the process, including training of workers, to determine if adjustments should be made. It should also investigate the possible causes of sample data hugging the centerline. In addition, the company should work to reduce overall variation in the process in order to meet the specification limits which have been set up.FIGURE Morelia Mortgage Co. 3
Case - Murphy Trucking, Inc.The Billing Study - Part IThe 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
= 252 / 400 = 0.63
sp =
=
= 0.108
Control limits:
UCLp =
+ 3 sp = 0.63 + 3 (0.108) = 0.954
LCLp =
- 3 sp = 0.63 - 3 (0.108) = 0.306Perhaps 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 C13MTIP1.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.
Figure MTI-A
The Billing Study - Part IIThe 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.)
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,
, must be used to calculate the individual control limits.
Center Line for the u-chart:
= 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:
3
= 0.6270 3 = 0.6270 3 (0.1078) =
0.304 to 0.950
Figure MTI-B
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.
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-CMurphy Trucking, Inc
Error Category
% of TotalCumulative
ErrorsErrorsPercent
230024.35%24.35%
826421.43%45.78%
115912.91%58.69%
615112.26%70.94%
414711.93%82.87%
5877.06%89.94%
7625.03%94.97%
3625.03%100.00%
Totals1232100.00%
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.
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:
ViscosityPercent SolidsLb./Gal.
Mean74.14663.05613.261
Std. Dev.4.5690.5110.080
Spec. Limits60-8060-6512.6-13.5
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:
Solids
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.
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 specification62.5Average63.0561Cp1.6311
Upper tolerance limit65Standard deviation0.5109Cpl1.9939
Lower tolerance limit60Cpu1.2683
Cpk1.2683
Process Capability- Viscosity
The process capability for viscosity is not good, as shown here:Nominal specification70Average74.1463Cp0.72951
Upper tolerance limit80Standard deviation4.5693Cpl1.03200
Lower tolerance limit60Cpu0.42703
Cpk0.42703
Process Capability- Pounds/Gallon
The process capability for pounds per gallon is not as good as it might be, as shown here:Nominal specification13.05Average13.2610Cp1.8693
Upper tolerance limit13.5Standard deviation0.0802Cpl2.7457
Lower tolerance limit12.6Cpu0.9929
Cpk0.9929
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.
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.
1
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_1316722333.unknown
_1316723269.unknown
_1316723864.unknown
_1316722685.unknown
_1288179408.unknown
_1288179740.unknown
_1288167282.unknown
_1288130088.unknown
_1288130683.unknown
_1288045205.unknown
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_1287760413.unknown
_1287761594.unknown
_1287692854.unknown
_1287518688.unknown
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_1287514647.unknown
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_1139120847.unknown
_1141423177.xlsChart1
0.66666666670.30409236850.6274809160.9508694636
0.55263157890.35488790330.6274809160.9000739287
0.62686567160.33715613020.6274809160.9178057018
0.73033707870.37558187610.6274809160.879379956
0.63157894740.35488790330.6274809160.9000739287
0.58333333330.36819324150.6274809160.8867685905
0.59016393440.3232126930.6274809160.9317491391
0.64383561640.34934307090.6274809160.9056187611
0.67777777780.37698522420.6274809160.8779766078
0.52040816330.38742717540.6274809160.8675346567
0.65853658540.36505024590.6274809160.8899115862
0.656250.33042956720.6274809160.9245322648
0.54166666670.34741821860.6274809160.9075436135
0.57954545450.37415467460.6274809160.8808071575
0.60465116280.37122595020.6274809160.8837358818
0.73118279570.38105859640.6274809160.8739032357
0.62962962960.36343527270.6274809160.8915265594
0.67045454550.37415467460.6274809160.8808071575
0.63636363640.33496496870.6274809160.9199968634
0.71739130430.37972296430.6274809160.8752388678
0.59649122810.31271761750.6274809160.9422442146
0.63235294120.33929877780.6274809160.9156630542
0.58823529410.36972297570.6274809160.8852388564
0.55696202530.36011381330.6274809160.8948480188
0.63541666670.38493950540.6274809160.8700223267
U Values
Lower Control Limit
Center Line
Upper Control Limit
Sample number
Defects per unit
Attribute (u) Chart
Data and calculations
Murphy Trucking, Inc.
Initial U-Chart (Base Case)
Average Number of Defects Per Unit (u) Chart
Click on the sheet tab to display the control chart (some rescaling may be needed).
Average (u-bar)0.6275
NumberSampleDefectsStandard
Sampleof DefectsSizeper unitDeviationLCLuCLUCLu
136540.66670.10779618250.30409236850.6274809160.9508694636
242760.55260.09086433760.35488790330.6274809160.9000739287
342670.62690.09677492860.33715613020.6274809160.9178057018
465890.73030.08396634670.37558187610.6274809160.879379956
548760.63160.09086433760.35488790330.6274809160.9000739287
649840.58330.08642922480.36819324150.6274809160.8867685905
736610.59020.1014227410.3232126930.6274809160.9317491391
847730.64380.0927126150.34934307090.6274809160.9056187611
961900.67780.08349856390.37698522420.6274809160.8779766078
1051980.52040.08001791360.38742717540.6274809160.8675346567
1154820.65850.08747689010.36505024590.6274809160.8899115862
1242640.65630.09901711630.33042956720.6274809160.9245322648
1339720.54170.09335423250.34741821860.6274809160.9075436135
1451880.57950.08444208050.37415467460.6274809160.8808071575
1552860.60470.08541832190.37122595020.6274809160.8837358818
1668930.73120.08214077320.38105859640.6274809160.8739032357
1751810.62960.08801521450.36343527270.6274809160.8915265594
1859880.67050.08444208050.37415467460.6274809160.8808071575
1942660.63640.09750531580.33496496870.6274809160.9199968634
2066920.71740.08258598390.37972296430.6274809160.8752388678
2134570.59650.10492109950.31271761750.6274809160.9422442146
2243680.63240.09606071270.33929877780.6274809160.9156630542
2350850.58820.08591931350.36972297570.6274809160.8852388564
2444790.55700.08912236760.36011381330.6274809160.8948480188
2561960.63540.08084713690.38493950540.6274809160.8700223267
260.00000000
270.00000000
280.00000000
290.00000000
300.00000000
310.00000000
320.00000000
330.00000000
340.00000000
350.00000000
360.00000000
370.00000000
380.00000000
390.00000000
400.00000000
410.00000000
420.00000000
430.00000000
440.00000000
450.00000000
460.00000000
470.00000000
480.00000000
490.00000000
500.00000000
u-chart
u-chart
0.66666666670.30409236850.6274809160.9508694636
0.55263157890.35488790330.6274809160.9000739287
0.62686567160.33715613020.6274809160.9178057018
0.73033707870.37558187610.6274809160.879379956
0.63157894740.35488790330.6274809160.9000739287
0.58333333330.36819324150.6274809160.8867685905
0.59016393440.3232126930.6274809160.9317491391
0.64383561640.34934307090.6274809160.9056187611
0.67777777780.37698522420.6274809160.8779766078
0.52040816330.38742717540.6274809160.8675346567
0.65853658540.36505024590.6274809160.8899115862
0.656250.33042956720.6274809160.9245322648
0.54166666670.34741821860.6274809160.9075436135
0.57954545450.37415467460.6274809160.8808071575
0.60465116280.37122595020.6274809160.8837358818
0.73118279570.38105859640.6274809160.8739032357
0.62962962960.36343527270.6274809160.8915265594
0.67045454550.37415467460.6274809160.8808071575
0.63636363640.33496496870.6274809160.9199968634
0.71739130430.37972296430.6274809160.8752388678
0.59649122810.31271761750.6274809160.9422442146
0.63235294120.33929877780.6274809160.9156630542
0.58823529410.36972297570.6274809160.8852388564
0.55696202530.36011381330.6274809160.8948480188
0.63541666670.38493950540.6274809160.8700223267
U Values
Lower Control Limit
Center Line
Upper Control Limit
Sample number
Defects per unit
Attribute (u) Chart
_1141423481.xlsChart1
3000.2433090024
2640.4574209246
1590.5863746959
1510.7088402271
1470.8280616383
870.898621249
630.9497161395
621
Errors
Cumulative Percent
Error Categories
Frequency of Billing Errors
Murphy Trucking, Inc. Pareto Diagram
Data
Murphy Trucking, Inc
Error Category
% of TotalCumulative
ErrorsErrorsPercent
0
230024.33%24.33%
826421.41%45.74%
115912.90%58.64%
615112.25%70.88%
414711.92%82.81%
5877.06%89.86%
7635.11%94.97%
3625.03%100.00%
Totals1233100.00%
&A
Page &P
Chart
&A
Page &P
Chart
3000.2433090024
2640.4574209246
1590.5863746959
1510.7088402271
1470.8280616383
870.898621249
630.9497161395
621
&A
Page &P
Errors
Cumulative Percent
Error Categories
Frequency of Billing Errors
Murphy Trucking, Inc. Pareto Diagram
_1141424372.unknown
_1141423282.xlsChart1
0.300.3050.6138506921
0.300.3050.6138506921
0.3500.3050.6138506921
0.3500.3050.6138506921
0.200.3050.6138506921
0.300.3050.6138506921
0.3500.3050.6138506921
0.400.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.2500.3050.6138506921
0.200.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.3500.3050.6138506921
p Values
Lower Control Limit
Center Line
Upper Control Limit
Sample number
% Non - conforming
Attribute (p) Chart
Data and calculations
Murphy Trucking, Inc.
Billing Study Part II
Fraction Nonconforming (p) Chart
Click on the sheet tab to display the control chart (some rescaling may be needed).
Average (p-bar)0.305
SampleFractionStandard
SampleValueSizeNonconformingLCLpCLUCLpDeviation
16200.3000.0000.3050.6140.103
26200.3000.0000.3050.6140.103
37200.3500.0000.3050.6140.103
47200.3500.0000.3050.6140.103
54200.2000.0000.3050.6140.103
66200.3000.0000.3050.6140.103
77200.3500.0000.3050.6140.103
88200.4000.0000.3050.6140.103
97200.3500.0000.3050.6140.103
105200.2500.0000.3050.6140.103
115200.2500.0000.3050.6140.103
124200.2000.0000.3050.6140.103
137200.3500.0000.3050.6140.103
145200.2500.0000.3050.6140.103
157200.3500.0000.3050.6140.103
165200.2500.0000.3050.6140.103
177200.3500.0000.3050.6140.103
185200.2500.0000.3050.6140.103
197200.3500.0000.3050.6140.103
207200.3500.0000.3050.6140.103
210.00000000
220.00000000
230.00000000
240.00000000
250.00000000
260.00000000
270.00000000
280.00000000
290.00000000
300.00000000
310.00000000
320.00000000
330.00000000
340.00000000
350.00000000
360.00000000
370.00000000
380.00000000
390.00000000
400.00000000
410.00000000
420.00000000
430.00000000
440.00000000
450.00000000
460.00000000
470.00000000
480.00000000
490.00000000
500.00000000
p-chart
p-chart
0.300.3050.6138506921
0.300.3050.6138506921
0.3500.3050.6138506921
0.3500.3050.6138506921
0.200.3050.6138506921
0.300.3050.6138506921
0.3500.3050.6138506921
0.400.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.2500.3050.6138506921
0.200.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.2500.3050.6138506921
0.3500.3050.6138506921
0.3500.3050.6138506921
p Values
Lower Control Limit
Center Line
Upper Control Limit
Sample number
% Non - conforming
Attribute (p) Chart
_1139120897.unknown
_1035445302.unknown
_1035700461.unknown
_1035711628.unknown
_1137678807.xlsChart1
7486.579852101774.146341463461.7128308251
6986.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7586.579852101774.146341463461.7128308251
6286.579852101774.146341463461.7128308251
6986.579852101774.146341463461.7128308251
7386.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
6886.579852101774.146341463461.7128308251
6986.579852101774.146341463461.7128308251
7786.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7486.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7386.579852101774.146341463461.7128308251
7686.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7086.579852101774.146341463461.7128308251
6986.579852101774.146341463461.7128308251
7486.579852101774.146341463461.7128308251
7586.579852101774.146341463461.7128308251
7486.579852101774.146341463461.7128308251
7086.579852101774.146341463461.7128308251
7586.579852101774.146341463461.7128308251
6586.579852101774.146341463461.7128308251
6986.579852101774.146341463461.7128308251
7486.579852101774.146341463461.7128308251
7286.579852101774.146341463461.7128308251
7786.579852101774.146341463461.7128308251
7386.579852101774.146341463461.7128308251
7786.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
6686.579852101774.146341463461.7128308251
7786.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
7686.579852101774.146341463461.7128308251
7986.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
86.579852101774.146341463461.7128308251
Individuals
Upper control limit
Center line
Lower control limit
Observation number
Value
Individuals (X) Chart - Viscosity
Data and calculations
Day Industries
Viscosity Measures
X and Moving Range Chart
This spreadsheet is designed for up to 50 observations and a moving range from 2 to 5. Enter data ONLY in yellow-shaded cells.
Enter the number of samples in cell E6 and the sample size in cell E7. Then enter your data in the grid below.
Click on sheet tabs to display the control charts (some rescaling may be needed).
Number of samples (