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Solution Manual For Managing, Controlling, and Improving Quality, 1st Edition

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    C HAPTER 1 Introduction to Quality Learning Objectives After completing this chapter you should be able to: 1. Define and discuss quality and quality improvement 2. Discuss the different dimensions of quality 3. Discuss the evolution of modern quality improvement methods 4. Discuss the role that variability and statistical methods play in controlling and improving quality 5. Explain the links between quality and productivity and between quality and cost 6. Discuss product liability Important Terms and Concepts Acceptance - sampling Appraisal costs Critical - to - quality (CTQ) Dimensions of quality Fitness for use Internal and external failure costs Nonconforming product or service Prevention costs Product liability Quality assurance Quality characteristics Quality control and improvement Quality engineering Quality of conformance Quality of design Quality planning Specifications Variability Exercises 1.1. Why is it difficult to define quality? Even the American Society for Quality describes “quality” as a subjective term for which each person or sector has its own definition. Given a large set of customers considering purchasing the same product or service, it is likely that each customer evaluate s it in terms of a completely different set of desirable characteristics. As a result, it is extremely difficult to come up with a single definition of quality that could meet the expectations of all customers for all products or services. ( For further details refer to page 2 ) 1.7. What are the three primary technical tools used for quality control and improvement? The three primary statistical technical tools include: statistical process control , design of experiments , and acceptance - sampling . 1.10. Discuss the statement “Quality is the responsibility of the quality department.” The responsibility for quality spans the entire organization . Quality improvement must be a total, company - wide activity, and that every organizational unit must actively participate. Because quality improvement activities are so broad, successful efforts require, as an initial step, senior management

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    commitment. This commitment involves emphasis on the importance of quality, identification of the respective quality responsibilities of the various organizational units, and explicit accountability for quality improvement of all managers and employees in the company. Nevertheless, the quality department takes care of the quality planning and analyses in order to make sure all quality efforts are being successfully implemented throughout the organization. ( For further details refer to pages 26 - 27 ) 1.12. Explain why it is necessary to consider variability around the mean or nominal dimension as a measure of quality. The primary objective of quality engineering efforts is the systematic reduction of variability in the key quality characteristics of the product. The introduction of statistical process control will help to stabilize process es and reduce the ir variability. However, it is not satisfactory just to meet requirements - F f urther reduction of variability around the mean or nominal dimension often also leads to better product performance and enhanced competitive position. ( For further details refer to page 17 ) .

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    C HAPTER 2 Managing aspects of Quality Learning Objectives After completing this chapter you should be able to: 1. Describe the quality management philosophies of W. Edwards Deming, Joseph M. Juran, and Armand V. Feigenbaum 2. Discuss total quality management, six - sigma, the Malcolm Baldrige National Quality Award, and quality systems and standards 3. Understand the importance of selecting good projects for improvement activities 4. Explain the five steps of DMAIC 5. Know when and when not to use DMAIC Important Terms and Concepts Analyze step Control step Define step Design for Six - Sigma (DFSS) DMAIC Failure modes and effects analysis (FMEA) Improve step Key process input variables (KPIV) Key process output variables (KPOV) Measure step Project charter SIPOC diagram Six - Sigma Tollgate Exercises 2.12. Explain the importance of tollgates in the DMAIC process. At a tollgate, a project team presents its work to managers and “owners” of the process. In a six - sigma organization, the tollgate participants also would include the project champion, master black belts, and other black belts not working directly on the project. Tollgates are where the project is reviewed to ensure that it is on track and they provide a continuing opportunity to evaluate whether the team can successfully complete the project on schedule. Tollgates also present an opportunity to provide guidance regarding the use of specific technical tools and other information about the problem. Organization problems and other barriers to success — and strategies for dealing with them — also often are identified during tollgate reviews. Tollgates are critical to the overall problem - solving process; It is important that these reviews be conducted very soon after the team completes each step. 2.18. Suppose that your business is operating at the four - and - a - half - sigma quality level. If projects have an average improvement rate of 50% annually, how many years will it take to achieve six - sigma quality? Assuming the a 1 . 5  shift in the mean that is customary with six - sigma applications [ 𝑋 ′ ~ 𝑁 ( 𝜇 ′ = 1 . 50 , 𝜎 = 1 ) ] , the percentage within the − 4 . 5 𝜎 and 4 . 5 𝜎 limits is:

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    𝑃 ( − 4 . 5 ≤ 𝑋 ′ ≤ 4 . 5 ) = 𝑃 ( 𝑋 ′ ≤ 4 . 5 ) − 𝑃 ( 𝑋 ′ ≤ − 4 . 5 ) = 0.9987 Then, the 𝑝𝑝𝑚 defective is: ( 1 − 0 . 9987 ) ∗ 10 6 ≈ 1 , 350 . Using the equation in Example 2.1: 3 . 4 = 1 , 350 ∗ ( 1 − 0 . 5 ) 𝑥 𝑥 = 𝑙𝑛 ( 3 . 4 1 , 350 ) 𝑙𝑛 ( 1 − 0 . 5 ) ⁄ 𝑥 = 8 . 6332 ` It will take the business nearly 8 years and 7 months to achieve 6 𝜎 quality . 2.19. Explain why it is important to separate sources of variability into special or assignable causes and common or chance causes. Common or chance causes are due to the inherent variability in the system and cannot generally be controlled. Special or assignable causes can be discovered and removed, thus reducing the variability in the overall system. It is important to distinguish between the two types of variability, because the strategy to reduce variability depends on the source. Chance cause variability can only be removed by changing the system, while assignable cause variability can be addressed by finding and eliminating the assignable causes. 2.21. Suppose that during the analyze phase an obvious solution is discovered. Should that solution be immediately implemented and the remaining steps of DMAIC abandoned? Discuss your answer. The answer is generally NO . The advantage of completing the rest of the DMAIC process is that the solution will be documented, tested, and it’s applicability to other parts of the business will be evaluated. An immediate implementation of an “obvious” solution may not lead to an appropriate control plan. Completing the rest of DMAIC process can also lead to further refinements and improvements to the solution. Also the transition plan developed in the control phase includes a validation check several months after project completion, if the DMAIC process is not completed; there is a danger of the original results not being sustained.

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    C HAPTER 3 Tools and Techniques for Quality Control and Improvement Learning Objectives After completing this chapter you should be able to: 1. Understand chance and assignable causes of variability in a process 2. Explain the statistical basis of the Shewhart control chart 3. Understand the basic process improvement tools of SPC: the histogram or stem - and - leaf plot, the check sheet, the Pareto chart, the cause - and - effect diagram, the defect concentration diagram, the scatter diagram, and the control chart 4. Explain how sensitizing rules and pattern recognition are used inconjunction with control charts Important Terms and Concepts Action limits Assignable causes of variation Cause - and - effect diagram Chance causes of variation Check sheet Control chart Control limits Defect concentration diagram Designed experiments Flow charts, operations process charts, and value stream mapping Factorial experiment In - control process Magnificent seven Out - of - control - action plan (OCAP) Out - of - control process Pareto chart Patterns on control charts Scatter diagram Shewhart control charts Statistical control of a process Statistical process control (SPC) Three - sigma control limits Exercises 3.6 . Consider the control charts shown here. Does the pattern appear random? No, the last four samples are located at a distance greater than 1  from the center line.

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    3.7 . Consider the control charts shown here. Does the pattern appear random? Yes, the pattern is random. 3.8 . Consider the control charts shown here. Does the pattern appear random? Yes, the pattern is random.

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    C HAPTER 4 Statistical Inference about Product and Process Quality Learning Objectives After completing this chapter you should be able to: 1. Construct and interpret visual data displays, including the stem - and - leaf plot, the histogram, and the box plot 2. Compute and interpret the sample mean, the sample variance, the sample standard deviation, and the sample range 3. Explain the concepts of a random variable and a probability distribution 4. Understand and interpret the mean, variance, and standard deviation of a probability distribution 5. Determine probabilities from probability distributions 6. Construct and interpret confidence intervals on a single mean and on the difference in two means 7. Construct and interpret confidence intervals on a single variance and the ratio of two variances 8. Construct and interpret confidence intervals on a single proportion and on the difference in two proportions 9. Understand how the analysis of variance (ANOVA) is used to compare more than two samples Important Terms and Concepts Alternative hypothesis Analysis of variance (ANOVA) Approximations to probability distributions Binomial distribution Box plot Central limit theorem Checking assumptions for statistical inference procedures Chi - square distribution Confidence interval Confidence intervals on means, known variance(s) Confidence intervals on means, unknown variance(s) Confidence intervals on proportions Confidence intervals on the variance of a normal distribution Confidence intervals on the variances of two normal distributions Continuous distribution Control limit theorem Critical region for a test statistic Descriptive statistics Discrete distribution Exponential distribution F - distribution Histogram Hypothesis testing Interquartile range Mean of a distribution Median Negative binomial distribution Normal distribution Normal probability plot Parameters of a distribution Percentile Poisson distribution Population Power of a statistical test Probability distribution Probability plotting P - value Quartile Random sample Random variable Residual analysis Run chart Sampling distribution Standard deviation Standard normal distribution Statistic Statistics Stem - and - leaf display t - distribution Test statistic Tests of hypotheses on means, known variance(s) Tests of hypotheses on means, unknown variance(s) Tests of hypotheses on proportions Tests of hypotheses on the variance of a normal distribution Tests of hypotheses on the variances of two normal distributions Time series plot Type I error Type II error Uniform distribution Variance of a distribution

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    Exercises 4 . 1 . The fill amount of liquid detergent bottles is being analyzed. Twelve bottles, randomly selected from the process, are measured, and the results are as follows (in fluid ounces): 16.05, 16.03, 16.02, 16.04, 16.05, 16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07. a. Calculate the sample average. ∑ 𝑥 𝑖 𝑛 𝑖 = 1 𝑛 = 16 . 05 + ⋯ + 16 . 07 12 = 16 . 0292 𝑓𝑙𝑢𝑖𝑑 𝑜𝑢𝑛𝑐𝑒𝑠 b. Calculate the sample standard deviation. √ ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) 2 𝑛 𝑖 = 1 𝑛 − 1 = √ ( 16 . 05 − 16 . 0255 ) 2 + ⋯ + ( 16 . 07 − 16 . 0255 ) 2 11 = 0 . 0202 𝑓𝑙𝑢𝑖𝑑 𝑜𝑢𝑛𝑐𝑒𝑠 4.3 . Waiting times for customers in an airline reservation system are (in seconds) 953, 955, 948, 951, 957, 949, 954, 950, 959 . a. Calculate the sample average. ∑ 𝑥 𝑖 𝑛 𝑖 = 1 𝑛 = 953 + ⋯ + 959 9 = 952 . 8889 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 b. Calculate the sample standard deviation. √ ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) 2 𝑛 𝑖 = 1 𝑛 − 1 = √ ( 96 − 121 . 25 ) 2 + ⋯ + ( 156 − 121 . 25 ) 2 7 = 3 . 7231 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 4 . 5 . The time to complete an order (in seconds) is as follows: 96, 102, 104, 108, 126, 128, 150, and 156. a. Calculate the sample average. ∑ 𝑥 𝑖 𝑛 𝑖 = 1 𝑛 = 96 + ⋯ + 156 8 = 121 . 25 𝑠𝑒𝑐𝑜𝑛𝑑𝑠

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    b. Calculate the sample standard deviation. √ ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) 2 𝑛 𝑖 = 1 𝑛 − 1 = √ ( 96 − 121 . 25 ) 2 + ⋯ + ( 156 − 121 . 25 ) 2 7 = 22 . 6258 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 4.9 . Construct and interpret a normal probability plot of the volumes in Exercise 4.1 . Either its question 4.8 or its exercise 4.3 . If its Question 4.9, the data is different, the whole solution changes. Data: 16.05, 16.03, 16.02, 16.04, 16.05, 16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07 MTB > Graph > Probability Plot > Single It can be assumed that the data follows a normal distribution. 4.11 . Construct a normal probability plot of the chemical process yield data in Exercise 4.7. Does the assumption that process yield is well modeled by a normal distribution seem reasonable? 13.3 14.3 14.9 15.2 15.8 14.2 16 14 14.5 16.1 13.7 15.2 13.7 16.9 14.9 14.4 15.3 13.1 15.2 15.9 15.1 14.9 13.6 13.7 15.3 15.5 14.5 16.5 13.4 15.2 15.3 13.8 14.3 12.6 15.3 14.8 14.1 14.4 14.3 15.6 14.8 14.6 15.6 15.1 14.8 15.2 15.6 14.5 15.2 14.3 15.8 17 14.3 14.6 16.1 12.8 14.5 15.4 13.3 14.9 14.3 16.4 13.9 16.1 14.6 15.2 14.1 14.8 16.4 14.2 15.2 16.6 14.1 16.8 15.4 14 16.9 15.7 14.4 15.6 970 965 960 955 950 945 940 99 95 90 80 70 60 50 40 30 20 10 5 1 C1 Percent Mean 952.9 StDev 3.723 N 9 AD 0.166 P-Value 0.908 Probability Plot of C1 Normal - 95% CI Formatted: Font: Bold

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    MTB > Graph > Probability Plot > Single It can be assumed that the data follows a normal distribution. 4.13 . Consider the chemical process yield data in Exercise 4.12. Calculate the sample average and standard deviation. 94.1 87.3 94.1 92.4 84.6 85.4 93.2 84.1 92.1 90.6 83.6 86.6 90.6 90.1 96.4 89.1 85.4 91.7 91.4 95.2 88.2 88.8 89.7 87.5 88.2 86.1 86.4 86.4 87.6 84.2 86.1 94.3 85 85.1 85.1 85.1 95.1 93.2 84.9 84 89.6 90.5 90 86.7 87.3 93.7 90 95.6 92.4 83 89.6 87.7 90.1 88.3 87.3 95.3 90.3 90.6 94.3 84.1 86.6 94.1 93.1 89.4 97.3 83.7 91.2 97.8 94.6 88.6 96.8 82.9 86.1 93.1 96.3 84.1 94.4 87.3 90.4 86.4 94.7 82.6 96.1 86.4 89.1 87.6 91.1 83.1 98 84.5 a. Calculate the sample average. ∑ 𝑥 𝑖 𝑛 𝑖 = 1 𝑛 = 94 . 1 + ⋯ + 84 . 5 90 = 89 . 4756 19 18 17 16 15 14 13 12 11 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 C3 Percent Mean 14.90 StDev 0.9804 N 80 AD 0.249 P-Value 0.740 Probability Plot of C3 Normal - 95% CI

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    b. Calculate the sample standard deviation. √ ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) 2 𝑛 𝑖 = 1 𝑛 − 1 = √ ( 94 . 1 − 89 . 4756 ) 2 + ⋯ + ( 90 − 89 . 4756 ) 2 89 = 4 . 1578 4.17 . Suppose that two fair dice are tossed and the sum of the dice is observed. Determine the probability distribution of 𝑥 , the sum of the dice. Let 𝑥 = 𝑑 1 + 𝑑 2 represent the sum of two dice. First, let u s determine the possible values for 𝑥 , and then, let us determine how likely it is to obtain each value. For example, 𝑥 = 3 can occur when 𝑑 1 = 1 and 𝑑 2 = 2 or when 𝑑 1 = 2 and 𝑑 2 = 1 . Hence, there are two possible scenarios that lead to 𝑥 = 3 . Similarly, let us list all possible scenarios for 𝑥 = 𝑑 1 + 𝑑 2 : Scenario 𝑑 1 𝑑 2 𝑥 Scenario 𝑑 1 𝑑 2 𝑥 1 1 1 2 19 4 1 5 2 1 2 3 20 4 2 6 3 1 3 4 21 4 3 7 4 1 4 5 22 4 4 8 5 1 5 6 23 4 5 9 6 1 6 7 24 4 6 10 7 2 1 3 25 5 1 6 8 2 2 4 26 5 2 7 9 2 3 5 27 5 3 8 10 2 4 6 28 5 4 9 11 2 5 7 29 5 5 10 12 2 6 8 30 5 6 11 13 3 1 4 31 6 1 7 14 3 2 5 32 6 2 8 15 3 3 6 33 6 3 9 16 3 4 7 34 6 4 10 17 3 5 8 35 6 5 11 18 3 6 9 36 6 6 12 From the previous table, we can conclude 𝑥 can assume integer values between 2 and 12. Now, let us define the likelihood of each scenario by counting how many times each scenario occurs and dividing it by the total number of possible scenarios ( 36 ).

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    𝑥 Number of Scenarios 𝑃 ( 𝑑 1 + 𝑑 2 = 𝑥 ) 2 1 1/36 3 2 2/36 4 3 3/36 5 4 4/36 6 5 5/36 7 6 6/36 8 5 5/36 9 4 4/36 10 3 3/36 11 2 2/36 12 1 1/36 4.18 . Find the mean and standard deviation of 𝑥 in Exercise 4.17 . 𝑥 𝑝 ( 𝑥 ) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 Mean: 𝐸 ( 𝑥 ) = ∑ 𝑥 ∗ 𝑝 ( 𝑥 ) = 2 ∗ 1 36 + ⋯ + 12 ∗ 1 36 ⁄ ⁄ = 7 𝑛 𝑖 = 1 Standard Deviation: √ 𝑉 ( 𝑥 ) = ∑ 𝑥 2 ∗ 𝑝 ( 𝑥 ) − [ 𝐸 ( 𝑥 ) ] 2 = 2 2 ∗ 1 36 + ⋯ + 12 2 ∗ 1 36 ⁄ ⁄ − 7 2 = 5 . 8333 𝑛 𝑖 = 1 4.19 . The tensile strength of a metal part is normally distributed with mean 40 lb and standard deviation 5 lb. If 50,000 parts are produced, how many would you expect to fail to meet a minimum specification limit of 35 lb tensile strength? How many would have a tensile strength in excess of 48 lb? 𝑋 ~ 𝑁 ( 𝜇 = 40 , 𝜎 = 5 ) Probability that a part fails to meet 35 lb specification (look up probability i n a normal probability table) : 𝑃 ( 𝑋 < 35 ) = 𝑃 ( 𝑍 < 35 − 40 5 ) = 𝑃 ( 𝑍 < − 1 ) = 0 . 1587

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    Expected number of parts that fail to meet 35 lb: 𝑛𝑃 ( 𝑋 < 35 ) = 50 , 000 ∗ 0 . 1587 ≈ 7 , 933 7935 𝑝𝑎𝑟𝑡𝑠 4.23 . A light bulb has a normally distributed light output with mean 5,000 end foot - candles and standard deviation of 50 end foot - candles. Find a lower specification limit such that only 0.5% of the bulbs will not exceed this limit. 𝑋 ~ 𝑁 ( 𝜇 = 5000 , 𝜎 = 50 ) Lower specification limit (x): 𝑃 ( 𝑋 < 𝑥 ) = 𝑃 ( 𝑍 < 𝑥 − 5000 50 ) = 𝑃 ( 𝑍 < 𝑧 ) = 0 . 005 Look up inverse value of 𝑧 in probability table : 𝑧 = 𝑥 − 5000 50 = − 2 . 5758 𝑥 = 5000 + 50 𝑧 = 5000 − 2 . 5758 ∗ 50 = 4 , 871 . 2085 4.25 . The tensile strength of fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi. a. Build a 95% lower confidence interval in the mean tensile strength. 𝜇 ≥ 𝑥 ̅ − 𝑧 0 . 05 ∗ 𝜎 √ 𝑛 ⁄ = 127 − 1 . 6449 ∗ 2 √ 8 = 125 . 8369 ⁄ psi b. What can you conclude? For 95% of the time CIs constructed , the mean tensile strength is expected to be equal to or exceed 125.8 369 psi. 4.27 . The service life of a battery used in a cardiac pacemaker is assumed o be normally distributed. A random sample of 10 batteries is subjected to an accelerated life test by running the, continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7. The manufacturer wants to be certain that the mean battery life exceeds 25 hours in accelerated life testing. a. Construct a 90% two - sided confidence interval on mean life in the accelerated test. 𝑥 ̅ − 𝑡 𝛼 2 , 𝑛 − 1 ⁄ 𝑠 √ 𝑛 ≤ 𝜇 ≤ ⁄ 𝑥 ̅ + 𝑡 𝛼 2 , 𝑛 − 1 ⁄ 𝑠 √ 𝑛 ⁄ 26 − 𝑡 0 . 1 2 , 10 − 1 ⁄ ∗ 1 . 6248 √ 10 ≤ 𝜇 ≤ ⁄ 26 + 𝑡 0 . 1 2 , 10 − 1 ⁄ ∗ 1 . 6248 √ 10 ⁄ 26 − 1 . 8331 ∗ 1 . 6248 √ 10 ≤ 𝜇 ≤ ⁄ 26 + 1 . 8331 ∗ 1 . 6248 √ 10 ⁄ 25 . 0581 ≤ 𝜇 ≤ 26 . 9419

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    b. Construct a normal probability plot of the battery life data. What conclusions can you draw? MTB > Graph > Probability Plot > Single Normality assumption is reasonable. Also, the manufacturer can be certain that the mean exceeds 25 hours , 95% of the CIs constructed will not contain a mean of 25 hrs. time , given the lower limit for the confidence interval exceeds 25 hrs . NOTE: Conclusions on the confidence interval are highly dependent on the number of significant digits used throughout the calculations. 4.29 . A company has just purchased a new billboard near the freeway. Sales for the past 10 days have been 483, 532, 444, 510, 467, 461, 450, 444, 540, and 499. Build a 95% two - sided confidence interval on sales. Is there any evidence that the billboard has increased sales from its previous average of 475 per day? 𝑥 ̅ − 𝑡 𝛼 2 , 𝑛 − 1 ⁄ 𝑠 √ 𝑛 ≤ 𝜇 ≤ ⁄ 𝑥 ̅ + 𝑡 𝛼 2 , 𝑛 − 1 ⁄ 𝑠 √ 𝑛 ⁄ 483 − 𝑡 0 . 05 2 , 10 − 1 ⁄ ∗ 35 . 7553 √ 10 ≤ 𝜇 ≤ ⁄ 483 + 𝑡 0 . 05 2 , 10 − 1 ⁄ ∗ 35 . 7553 √ 10 ⁄ 483 − 2 . 2622 ∗ 35 . 7553 √ 10 ≤ 𝜇 ≤ ⁄ 483 + 2 . 2622 ∗ 35 . 7553 √ 10 ⁄ 457 . 4217 ≤ 𝜇 ≤ 508 . 5783 Since 475 is within the confidence interval, there is no evidence that the billboard has increased sales from its previous average of 475 per day. NOTE: Confidence intervals endpoints are highly dependent on the number of significant digits used throughout the calculations . 4.31. A company is evaluating the quality of aluminum rods received in a recent shipment. Diameters of aluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001 in. A random sample of 25 rods has an average diameter of 0.5046 in. Test whether or not the mean rod diameter is 0.5025 using a two - sided confidence interval. 𝑥 ̅ − 𝑧 0 . 05 2 ⁄ ∗ 𝜎 √ 𝑛 ⁄ ≤ 𝜇 ≤ 𝑥 ̅ + 𝑧 0 . 05 2 ⁄ ∗ 𝜎 √ 𝑛 ⁄ 0 . 5046 − 1 . 96 ∗ 0 . 0001 √ 25 ⁄ ≤ 𝜇 ≤ 0 . 5046 − 1 . 96 ∗ 0 . 0001 √ 25 ⁄ 0 . 504561 ≤ 𝜇 ≤ 0 . 504639 32 30 28 26 24 22 20 99 95 90 80 70 60 50 40 30 20 10 5 1 C1 Percent Mean 26 StDev 1.625 N 10 AD 0.114 P-Value 0.986 Probability Plot of C1 Normal - 95% CI

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    Since 0.5025 is not within the confidence interval, we do not have evidence to state the mean rod diameter is 0.5025 in. 4.33 . Last month, a large national bank’s average payment time was 33 days with a standard deviation of 4 days . This month, the average payment time was 33.5 days with a standard deviation of 4 days. They had 1,000 customers both months. a. Build a 95% confidence interval on the difference between this month’s and last month’s payment times. Is there evidence that payment times are increasing? Using a one - sided confidence interval: 𝜇 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝜇 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≥ 𝑥 ̅ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝑥 ̅ 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 − 𝜎 0 . 05 √ 𝜎 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 2 𝑛 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 + 𝜎 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 2 𝑛 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝜇 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝜇 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≥ 33 . 5 − 33 − 1 . 6449 √ 4 2 1000 + 4 2 1000 = 0 . 2058 Since the lower limit for the difference is positive, there is evidence to state that the payment times are increasing. U s ing a two - sided confidence interval : 𝑥 ̅ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝑥 ̅ 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 − 𝜎 0 . 05 2 ⁄ √ 𝜎 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 2 𝑛 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 + 𝜎 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 2 𝑛 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≤ 𝜇 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝜇 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≤ 𝑥 ̅ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 + 𝑥 ̅ 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 + 𝜎 0 . 05 2 ⁄ √ 𝜎 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 2 𝑛 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 + 𝜎 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 2 𝑛 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 33 . 5 − 33 − 1 . 96 √ 4 2 1000 + 4 2 1000 ≤ 𝜇 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝜇 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≤ 33 . 5 − 33 + 1 . 96 √ 4 2 1000 + 4 2 1000 0 . 149 ≤ 𝜇 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 − 𝜇 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 ≤ 0 . 851 We still arrive at the same conclusion that payment times have increased given the entire confidence interval is greater than zero. 4.35 . A supplier received results on the hardness of metal from two different hardening process (1) salt - water quenching and (2) oil quenching . The results are shown in Table 4E.4.

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    Salt Quench Oil Quench 145 152 150 150 153 147 148 155 141 140 152 146 146 158 154 152 139 151 148 143 a. Construct a 95% confidence interval on the difference in mean hardness. Using a two - sided confidence interval assuming unequal variances :  = ( 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ) 2 ( 𝑠 1 2 𝑛 1 ) 2 𝑛 1 − 1 + ( 𝑠 2 2 𝑛 2 ) 2 𝑛 2 − 1 = ( 24 . 7111 10 + 29 . 8222 10 ) 2 ( 24 . 7111 10 ) 2 10 − 1 + ( 29 . 8222 10 ) 2 10 − 1 = 17 . 3992 ⇒ 17 𝑥 ̅ 𝑆𝑄 − 𝑥 ̅ 𝑂𝑄 − 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 𝑆𝑄 2 𝑛 𝑆𝑄 + 𝑠 𝑂𝑄 2 𝑛 𝑂𝑄 ≤ 𝜇 𝑆𝑄 − 𝜇 𝑂𝑄 ≤ 𝑥 ̅ 𝑆𝑄 − 𝑥 ̅ 𝑂𝑄 + 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 𝑆𝑄 2 𝑛 𝑆𝑄 + 𝑠 𝑂𝑄 2 𝑛 𝑂𝑄 147 . 6 − 149 . 4 − 2 . 1098 √ 24 . 7111 10 + 29 . 8222 10 ≤ 𝜇 𝑆𝑄 − 𝜇 𝑂𝑄 ≤ 33 . 5 − 33 + 2 . 1098 √ 24 . 7111 10 + 29 . 8222 10 − 6 . 7269 ≤ 𝜇 𝑆𝑄 − 𝜇 𝑂𝑄 ≤ 3 . 1269 b. Construct a 95% confidence interval on the ratio of variances. 𝑠 1 2 𝑠 2 2 𝐹 1 − 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ ≤ 𝜎 𝑆𝑄 2 𝜎 𝑂𝑄 2 ⁄ ≤ 𝑠 1 2 𝑠 2 2 𝐹 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ 24 . 7111 29 . 8222 𝐹 1 − 0 . 05 2 , 10 − 1 , 10 − 1 ⁄ ≤ 𝜎 𝑆𝑄 2 𝜎 𝑂𝑄 2 ⁄ ≤ 24 . 7111 29 . 8222 𝐹 0 . 05 2 , 10 − 1 , 10 − 1 ⁄ 24 . 7111 29 . 8222 ∗ 0 . 2484 ≤ 𝜎 𝑆𝑄 2 𝜎 𝑂𝑄 2 ⁄ ≤ 24 . 7111 29 . 8222 ∗ 4 . 0260 0 . 2058 ≤ 𝜎 𝑆𝑄 2 𝜎 𝑂𝑄 2 ⁄ ≤ 3 . 3360 c. Does this assumption of normality seem appropriate for this data?

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    MTB > Graph > Probability Plot > Single The normality assumption for both samples looks reasonable. d. What can you conclude? For the confidence interval based on the mean difference, zero is included with in the interval. For the confidence interval based on the ratio of the variances, one is included with in the interval . Hence, there is no significance significant difference in the sample means or variances . 4.37 . A random sample of 500 connecting rod pins contains 65 nonconforming units. Estimate the process fraction nonconforming. Construct a 90% upper confidence interval on the true process fraction nonconforming. Is it possible that the true fraction nonconforming defective is 10%? Process fraction nonconforming estimate: 𝑝 ̂ = 65 500 = 0 . 13 90% upper confidence interval: 𝑝 ≤ 𝑝 ̅ + 𝑧 0 . 1 √ 𝑝 ̅ ( 1 − 𝑝 ̅ ) 𝑛 𝑝 ≤ 0 . 13 + 1 . 2816 √ 0 . 13 ∗ ( 1 − 0 . 13 ) 500 𝑝 ≤ 0 . 1493 For 90% of the time CIs constructed , the true defective fraction is less than 14 . 93 % (evidence suggests p might be larger than 0.10) 4.41 . A new purification unit is installed in a chemical process. Before and after installation data was collected regarding the percentage of impurity: Before (1) : Sample mean = 9.85, Sample variance = 6.79, Number of samples = 10 After (2) : Sample mean = 8.08, Sample variance = 6.18, Number of samples = 8 170 160 150 140 130 99 95 90 80 70 60 50 40 30 20 10 5 1 170 160 150 140 130 Salt Quench Percent Oil Quench Mean 147.6 StDev 4.971 N 10 AD 0.218 P-Value 0.779 Salt Quench Mean 149.4 StDev 5.461 N 10 AD 0.169 P-Value 0.906 Oil Quench Probability Plot of Salt Quench, Oil Quench Normal - 95% CI

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    a. Can you conclude that the two variances are equal using a two - sided 95% confidence interval? 𝑠 1 2 𝑠 2 2 𝐹 1 − 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 𝑠 1 2 𝑠 2 2 𝐹 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ 6 . 79 6 . 18 𝐹 1 − 0 . 05 2 , 8 − 1 , 10 − 1 ⁄ ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 6 . 79 6 . 18 𝐹 0 . 05 2 , 8 − 1 , 10 − 1 ⁄ 24 . 7111 29 . 8222 ∗ 0 . 2073 ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 24 . 7111 29 . 8222 ∗ 4 . 6113 0 . 2278 ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 4 . 6113 Since one is included in the interval, we can conclude that the variance s are equal. b. Can you conclude that the new purification device has reduced the mean percentage of impurity using a two - sided 95% confidence interval? Using a two - sided confidence interval assuming equal variances:  = ( 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ) 2 ( 𝑠 1 2 𝑛 1 ) 2 𝑛 1 − 1 + ( 𝑠 2 2 𝑛 2 ) 2 𝑛 2 − 1 = ( 6 . 79 10 + 6 . 18 8 ) 2 ( 6 . 79 10 ) 2 10 − 1 + ( 6 . 18 8 ) 2 8 − 1 = 15 . 4373 ⇒ 15 𝑥 ̅ 1 − 𝑥 ̅ 2 − 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ≤ 𝜇 1 − 𝜇 2 ≤ 𝑥 ̅ 1 − 𝑥 ̅ 2 + 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 9 . 85 − 8 . 08 − 2 . 1315 √ 6 . 79 10 + 6 . 18 8 ≤ 𝜇 1 − 𝜇 2 ≤ 9 . 85 − 8 . 08 + 2 . 1315 √ 6 . 79 10 + 6 . 18 8 − 0 . 7980 ≤ 𝜇 1 − 𝜇 2 ≤ 4 . 3380 Since zero is within the interval, we cannot conclude that the new device has reduced th e mean percentage of impurity .

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    4.43 . The diameter of a metal rod is measured by 12 inspectors, each using both a micrometer caliper and a vernier caliper. The results are shown in Table 4E.5. Is there a difference between the mean measurements produced by the two types of caliper? Use alpha = 0.01. Inspector Micrometer Caliper Vernier Caliper 1 0.15 0.151 2 0.151 0.15 3 0.151 0.151 4 0.152 0.15 5 0.151 0.151 6 0.15 0.151 7 0.151 0.153 8 0.153 0.155 9 0.152 0.154 10 0.151 0.151 11 0.151 0.15 12 0.151 0.152  = ( 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ) 2 ( 𝑠 1 2 𝑛 1 ) 2 𝑛 1 − 1 + ( 𝑠 2 2 𝑛 2 ) 2 𝑛 2 − 1 = ( 6 . 97 ∗ 10 − 7 12 + 2 . 63 ∗ 10 − 6 12 ) 2 ( 6 . 97 ∗ 10 − 7 12 ) 2 12 − 1 + ( 2 . 63 ∗ 10 − 6 12 ) 2 12 − 1 = 16 . 4498 ⇒ 16 𝑥 ̅ 1 − 𝑥 ̅ 2 − 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ≤ 𝜇 1 − 𝜇 2 ≤ 𝑥 ̅ 1 − 𝑥 ̅ 2 + 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 0 . 1512 − 0 . 1516 − 2 . 9208 √ 6 . 97 ∗ 10 − 7 12 + 2 . 63 ∗ 10 − 6 12 ≤ 𝜇 1 − 𝜇 2 ≤ 0 . 1512 − 0 . 1516 + 2 . 9208 √ 6 . 97 ∗ 10 − 7 12 + 2 . 63 ∗ 10 − 6 12 − 0 . 0020 ≤ 𝜇 1 − 𝜇 2 ≤ 0 . 0011 Since zero is within the confidence interval, no evidence suggests the mean measurements produced by the two types of calipers are different.

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    4.45 . Rehab, Inc. is evaluating patient success results for its Northbrook and Southbrook locations. Each successfully treated 10 patients within the last year following elbow surgery. Total recovery times and % range of motion achieved are listed in Table 4E.6. Northbrook (1) Southbrook (2) Recovery Time % ROM Achieved Recovery Time % ROM Achieved 148.81 0.69 135.25 0.98 188.72 0.89 174.99 0.47 186.77 0.65 144.15 0.85 152.72 0.73 161.81 0.71 197.8 0.79 151.35 0.94 162.78 0.81 149.69 0.56 192.18 0.64 136.17 0.57 200.17 0.88 146.25 0.2 181.32 0.67 162.88 0.84 193.03 0.74 183.95 0.62 a. Build a 95% two - sided confidence interval for the differences of the average of the recovery times. Is there evidence to support that the facilities are different?  = ( 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ) 2 ( 𝑠 1 2 𝑛 1 ) 2 𝑛 1 − 1 + ( 𝑠 2 2 𝑛 2 ) 2 𝑛 2 − 1 = ( 353 . 0122 10 + 258 . 3751 10 ) 2 ( 353 . 0122 10 ) 2 10 − 1 + ( 258 . 3751 10 ) 2 10 − 1 = 17 . 5788 ⇒ 17 𝑥 ̅ 1 − 𝑥 ̅ 2 − 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ≤ 𝜇 1 − 𝜇 2 ≤ 𝑥 ̅ 1 − 𝑥 ̅ 2 + 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 180 . 43 − 154 . 649 − 2 . 1098 √ 353 . 0122 10 + 258 . 3751 10 ≤ 𝜇 1 − 𝜇 2 ≤ 180 . 43 − 154 . 649 + 2 . 1098 √ 353 . 0122 10 + 258 . 3751 10 9 . 2841 ≤ 𝜇 1 − 𝜇 2 ≤ 42 . 2779 The confidence interval is positive; thus, there is evidence that recovery times are larger at Northbrook. b. Build a 95% two - sided confidence interval for the ratio of the variance of recovery times. Is there evidence to support that the facilities are different? 𝑠 1 2 𝑠 2 2 𝐹 1 − 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 𝑠 1 2 𝑠 2 2 𝐹 𝛼 2 , 𝑛 2 − 1 , 𝑛 1 − 1 ⁄ 353 . 0122 258 . 3751 𝐹 1 − 0 . 05 2 , 10 − 1 , 10 − 1 ⁄ ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 353 . 0122 258 . 3751 𝐹 0 . 05 2 , 10 − 1 , 10 − 1 ⁄ 353 . 0122 258 . 3751 ∗ 0 . 2484 ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 353 . 0122 258 . 3751 ∗ 4 . 0260

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    0 . 3394 ≤ 𝜎 1 2 𝜎 2 2 ⁄ ≤ 5 . 5006 Since one is included in the interval, we can conclude the variance s are equal. c. Build a 95% two - sided confidence interval for the differences on the percentage of motion restored. Is there evidence to support that the facilities are different? 𝑝 ̅ 1 − 𝑝 ̅ 2 − 𝑧 0 . 05 2 ⁄ √ 𝑝 ̅ 1 ( 1 − 𝑝 ̅ 1 ) 𝑛 1 + 𝑝 ̅ 2 ( 1 − 𝑝 ̅ 2 ) 𝑛 2 ≤ 𝑝 1 − 𝑝 2 ≤ 𝑝 ̅ 1 − 𝑝 ̅ 2 + 𝑧 0 . 05 2 ⁄ √ 𝑝 ̅ 1 ( 1 − 𝑝 ̅ 1 ) 𝑛 1 + 𝑝 ̅ 2 ( 1 − 𝑝 ̅ 2 ) 𝑛 2 0 . 7490 − 0 . 674 − 1 . 96 √ 0 . 7490 ( 1 − 0 . 7490 ) 10 + 0 . 674 ( 1 − 0 . 674 ) 10 ≤ 𝑝 1 − 𝑝 2 ≤ 0 . 0909 − 0 . 2397 + 1 . 96 √ 0 . 7490 ( 1 − 0 . 7490 ) 10 + 0 . 674 ( 1 − 0 . 674 ) 10 − 0 . 3208 ≤ 𝑝 1 − 𝑝 2 ≤ 0 . 4708 Since zero is within the interval, there is not enough evidence to c onclude the facilities are different in terms of the percentage of motion restored. 4.47. An article in the ACI Materials Journal (Vol. 84, 1987, pp. 213 - 216) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3 - in diameter cylinder was used, and the number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in Table 4E.8. a. Is there any difference in compressive strength due to rodding level? Answer the question by using analysis of variance with alpha = 0.05. Rodding level Compressive strength 10 1530 10 1530 10 1440 15 1610 15 1650 15 1500 20 1560 20 1730 20 1530 25 1500 25 1490 25 1510

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    MTB > Stat > ANOVA > One - way One - way ANOVA: Compressive strength versus Rodding level Source DF SS MS F P Rodding level_1 3 28633 9544 1.87 0.214 Error 8 40933 5117 Total 11 69567 S = 71.53 R - Sq = 41.16% R - Sq(adj) = 19.09% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---- + --------- + --------- + --------- + ----- 10 3 1500.0 52.0 ( ----------- * ---------- ) 15 3 1586.7 77.7 ( ----------- * ----------- ) 20 3 1606.7 107.9 ( ----------- * ----------- ) 25 3 1500.0 10.0 ( ----------- * ---------- ) ---- + --------- + --------- + --------- + ----- 1440 1520 1600 1680 Pooled StDev = 71.5 Since p - value = 0.214 > 0.05, there is no evidence that suggests rodding level influences compressive strength. NOTE: Let us now assume only data on levels 10 and 15 is available, and that we will use a 95% two - sided confidence interval to evaluate whether there are any differences in terms of compressive strength between the two rodding levels.  = ( 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ) 2 ( 𝑠 1 2 𝑛 1 ) 2 𝑛 1 − 1 + ( 𝑠 2 2 𝑛 2 ) 2 𝑛 2 − 1 = ( 2700 3 + 6033 . 3333 3 ) 2 ( 2700 3 ) 2 3 − 1 + ( 6033 . 3333 3 ) 2 3 − 1 = 3 . 4914 ⇒ 3 𝑥 ̅ 1 − 𝑥 ̅ 2 − 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 ≤ 𝜇 1 − 𝜇 2 ≤ 𝑥 ̅ 1 − 𝑥 ̅ 2 + 𝑡 0 . 05 2 ,  ⁄ √ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 1500 − 1586 . 6667 − 3 . 1825 √ 2700 3 + 6033 . 3333 3 ≤ 𝜇 1 − 𝜇 2 ≤ 1500 − 1586 . 6667 + 3 . 1825 √ 2700 3 + 6033 . 3333 3 − 258 . 3748 ≤ 𝜇 1 − 𝜇 2 ≤ 85 . 0415 Since zero is within the confidence interval, we cannot conclude there is a difference in the compressive strength associated with the two different rodding levels. b. Construct box plots of compressive strength by rodding level. Provide a practical interpretation of the plots.

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    MTB > Graph > Boxplot > One Y > With Groups There does not seem to be a significant difference in terms of the mean compressive strength for rodding levels 10, 15, and 25 20 and between , 10 and 25 . Nevertheless, rodding Rodding level 25 shows remarkably smaller variability compared to other levels . On the other hand, the boxplot s also suggests there a significant difference in terms of the mean compressive strength and variance for rodding levels 20 and 25. c. Construct a normal probability plot of the residuals from this experiment. Does the assumption of a normal distribution for compressive strength seem reasonable? MTB > Graph > Probability Plot > Single Normality assumption appears reasonable. 25 20 15 10 1750 1700 1650 1600 1550 1500 1450 1400 Rodding level_1 Compressive strength_1 Boxplot of Compressive strength_1 150 100 50 0 -50 -100 -150 99 95 90 80 70 60 50 40 30 20 10 5 1 Residual Percent Normal Probability Plot (response is Compressive strength_1)

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    C HAPTER 5 Control Charts for Variables Learning Objectives After completing this chapter you should be able to: 1. Understand chance and assignable causes of variability in a process 2. Explain the statistical basis of the Shewhart control chart, including choice of sample size, control limits, and sampling interval 3. Explain the rational subgroup concept 4. Explain how sensitizing rules and pattern recognition are used in conjunction with control charts 5. Know how to design variables control charts 6. Know how to set up and use and R control charts 7. Know how to set up and use and S control charts 8. Know how to set up and use control charts for individual measurements 9. Understand the importance of the normality assumption for individuals control charts and know how to check this assumption 10. Set up and use CUSUM control charts for monitoring the process mean 11. Set up and use EWMA control charts for monitoring the process mean 12. Understand the difference between process capability and process potential 13. Calculate and properly interpret process capability ratios 14. Understand the role of the normal distribution in interpreting most process capability ratios Important Terms and Concepts Assignable causes of variation Control chart Control limits CUSUM control chart EWMA control chart In - control process Individuals control chart One - sided process - capability ratios Out - of - control process Out - of - control - action plan (OCAP) Pareto chart PCR C p PCR C pk PCR C pm Process capability R control chart Rational subgroups s control chart Sampling frequency for control charts Shewhart control charts Signal resistance of a control chart Specification limits Statistical control of a process Statistical process control (SPC) Three - sigma control limits Variable sample size on control charts Variables control charts x control chart

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    Exercises 5.14 . Consider the control chart shown here. Does the pattern appear random? No. The last four runs appear to plot at a distance of one - sigma or beyond from the center line. 5.15 . Consider the control chart shown here. Does the pattern appear random? Yes, the p attern appears random. 5.16 . Consider the control chart shown here. Does the pattern appear random? Yes, the p attern appears random.

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    5.17 . Apply the Western Electric rules to the control chart in Exercise 5. 17. Are any of the criteria for declaring the process out of control satisfied? Check: 1. Any point outside the 3 - sigma control limits? NO. 2. 2 of 3 beyond 2 sigma of centerline? NO. 3. 4 of 5 at 1 sigma or beyond of centerline? YES. Points 17, 18, 19, and 20 are outside the lower 1 - sigma area. 4. 8 consecutive points on one side of centerline? NO. The process can be declared out - of - control because the last four runs appear to plot at a distance of one - sigma or beyond from the center line .

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    5.19 . Consider the time - varying process behavior shown below. Match each of these several patterns of process performance to the corresponding and R charts shown in Figures (a) to (e) below.

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    Behavior Control Chart ( a ) ( 2 ) ( b ) ( 4 ) ( c ) ( 5 ) ( d ) ( 1 ) ( e ) ( 3 ) 5.21. The net weight of a soft drink is to be monitored by 𝑥 ̅ and R control charts using a sample size of 𝑛 = 5 . Data for 20 preliminary samples are shown in Table 5E.2. Sample Number 𝑥 1 𝑥 2 𝑥 3 𝑥 4 𝑥 5 1 15.8 16.3 16.2 16.1 16.6 2 16.3 15.9 15.9 16.2 16.4 3 16.1 16.2 16.5 16.4 16.3 4 16.3 16.2 15.9 16.4 16.2 5 16.1 16.1 16.4 16.5 16 6 16.1 15.8 16.7 16.6 16.4 7 16.1 16.3 16.5 16.1 16.5 8 16.2 16.1 16.2 16.1 16.3 9 16.3 16.2 16.4 16.3 16.5 10 16.6 16.3 16.4 16.1 16.5 11 16.2 16.4 15.9 16.3 16.4 12 15.9 16.6 16.7 16.2 16.5 13 16.4 16.1 16.6 16.4 16.1 14 16.5 16.3 16.2 16.3 16.4 15 16.4 16.1 16.3 16.2 16.2 16 16 16.2 16.3 16.3 16.2 17 16.4 16.2 16.4 16.3 16.2 18 16 16.2 16.4 16.5 16.1 19 16.4 16 16.3 16.4 16.4 20 16.4 16.4 16.5 16 15.8

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    a. Set up 𝑋 ̅ and R control charts using these data. Does the process exhibit statistical control? MTB > Stat > Control Charts > Variables Charts for Subgroups > Xbar - R 𝑥 𝑋 ̅ ̅ ̅ ̅ chart : 𝑈𝐶𝐿 = 16 . 5420 , 𝐶𝐿 = 16 . 268 , 𝐿𝐶𝐿 = 15 . 9940 𝑅 𝑐 ℎ 𝑎𝑟𝑡 : 𝑈𝐶𝐿 = 1 . 004 , 𝐶𝐿 = 0 . 475 , 𝐿𝐶𝐿 = 0 The process appears to exhibit statistical control. b. Estimate the process mean and standard deviation. 𝜇 ̂ = 𝑥 ̿ = 16 . 268 , 𝜎 ̂ = 𝑅 ̅ 𝑑 2 ⁄ = 0 . 475 2 . 326 ⁄ = 0 . 2042 c. Does fill weight seem to follow a normal distribution? MTB > Stat > Basic Statistics > Normality Test Normality assumption is reasonable. 19 17 15 13 11 9 7 5 3 1 16.6 16.4 16.2 16.0 Sample Sample M ean _ _ X=16.268 U C L=16.5420 LC L=15.9940 19 17 15 13 11 9 7 5 3 1 1.00 0.75 0.50 0.25 0.00 Sample Sample Range _ R=0.475 U C L=1.004 LC L=0 Xbar-R Chart of Net Weight 17.00 16.75 16.50 16.25 16.00 15.75 15.50 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 Net Weight Percent Mean 16.27 StDev 0.2014 N 100 AD 1.257 P-Value <0.005 Probability Plot of Net Weight Normal - 95% CI

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