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Introduction To Statistical Quality Control 7th Edition Douglas C Montgomery - Solutions

A normally distributed process has specifications of LSL = 75 and USL = 85 on the output. A random sample of 25 parts indicates that the process is centered at the middle of the specification band, and the standard deviation is s = 1.5. (a) Find a point estimate of Cp. (b) Find a 95% confidence
A company has been asked by an important customer to demonstrate that its process capability ratio Cp exceeds 1.33. It has taken a sample of 50 parts and obtained the point estimate Cp = 152. Assume that the quality characteristic follows a normal distribution. Can the company demonstrate that Cp
Suppose that a quality characteristic has a normal distribution with specification limits at USL = 100 and LSL = 90. A random sample of 30 parts results in x = 97 and s = 1.6. (a) Calculate a point estimate of Cpk. (b) Find a 95% confidence interval on Cpk.
The molecular weight of a particular polymer should fall between 2,100 and 2,350. Fifty samples of this material were analyzed with the results x = 2,275 and s = 60. Assume that molecular weight is normally distributed. (a) Calculate a point estimate of Cpk.. (b) Find a 95% confidence interval on
A normally distributed quality characteristic has specification limits at LSL = 10 and USL = 20. A random sample of size 50 results in x =16 and s =1.2 . (a) Calculate a point estimate of Cpk. (b) Find a 95% confidence interval on Cpk.
A normally distributed quality characteristic has specification limits at LSL = 50 and USL = 60. A random sample of size 35 results in x  55.5 and s  0.9. (a) Calculate a point estimate of Cpk. (b) Find a 95% confidence interval on Cpk. (c) Is this a 6-sigma process?
Consider a simplified version of equation 8.19 (below). Note that this was obtained by assuming that the term 9n in equation 8.19 will probably be large. Rework Exercise 8.24 using this equation and compare your answer to the original answer obtained from equation 8.19. How good is the
An operator-instrument combination is known to test parts with an average error of zero; however, the standard deviation of measurement error is estimated to be 3. Samples from a controleld process were analyzed, and the total variability was estimated to be σ = 5. What is the true process
Consider the piston ring data in Table 6.3. Estimate the process capability assuming that specifications are 74.00 ±0.035 mm.
Consider the situation in Example 8.7. A new gauge is being evaluated for this process. The same operator measures the same 20 parts twice using the new gauge and obtains the data shown in Table 8E.8.(a) What can you say about the performance of the new gauge relative to the old one? (b) If
Ten parts are measured three times by the same operator in a gauge capability study. The data are shown in Table 8E.9(a) Describe the measurement error that results from the use of this gauge. (b) Estimate the total variability and product variability. (c) What percentage of total variability is
In a study to isolate both gauge repeatability and gauge reproducibility, two operators use the same gauge to measure ten parts three times each. The data are shown in Table 8E.10.(a) Estimate gauge repeatability and reproducibility. (b) Estimate the standard deviation of measurement error. (c) If
The data in Table 8E.11 were taken by one operator during a gauge capability study.(a) Estimate gauge capability. b) Does the control chart analysis of these data indicate any potential problem in using the gauge?
A measurement systems experiment involving 20 parts, three operators, and two measurements per part is shown in Table 8E.12.(a) Estimate the repeatability and reproducibility of the gauge (b) What is the estimate of total gauge variability? (c) If the product specifications are at LSL = 6 and USL =
Reconsider the gauge R&R experiment in Exercise 8.34. Calculate the quantities SNR and DR for this gauge. Discuss what information these measures provide about the capability of the gauge.
Three parts are assembled in series so that their critical dimensions x1, x2, and x3 add. The dimensions of each part are normally distributed with the following parameters: 1 100,1  4, 2  75, 2  4, 3 75, 3 2 . What is the probability that an assembly chosen at
Two parts are assembled as shown in the figure. The distributions of x1 and x2 are normal, with ï1 ï€½20,ï³1 ï€½ 0.3ï2 ï€½19.6, ï³2 ï€½ 0.4 . The specifications of the
A product is packaged by filling a container completely full. This container is shaped as shown in the figure. The process that produces these containers is examined, and the following information collected on the three critical dimensions:
A rectangular piece of metal of width W and length L is cut from a plate of thickness T. If W, L, and T are independent random variables with means and standard deviations as given here and the density of the metal is 0.08 g/cm3, what would be the estimated mean and standard deviation of the
Perform a process capability analysis using x and R charts for the data in Exercise 6.7.
The surface tension of a chemical product, measured on a coded scale, is given by the relationship (3 + 0.05x)2 x where x is a component of the product with probability distribution Find the mean and variance of s.
Two resistors are connected to a battery as shown in the figure. Find approximate expressions for the mean and variance of the resulting current (I). E, R1, and R2 are random variables with means E ,R 1 ,R2 and variances
Two mating parts have critical dimensions x1 and x2 as shown in the figure. Assume that x1 and x2 are normally distributed with means 1 and 2 and standard deviations 1 = 0.400and 2 = 0.300. If it is desired that the probability of a smaller clearance (i.e., x1 − x2) than 0.09 should
An assembly of two parts is formed by fitting a shaft into a bearing. It is known that the inside diameters of bearings are normally distributed with mean 2.010 cm and standard deviation 0.002 cm, and that the outside diameters of the shafts are normally distributed with mean 2.004 cm and standard
We wish to estimate a two-sided natural tolerance interval that will include 99% of the values of a random variable with probability 0.80. If nothing is known about the distribution of the random variable, how large should the sample be?
A sample of ten items from a normal population had a mean of 300 and standard deviation of 10. Using these data, estimate a value for the random variable such that the probability is 0.95 that 90% of the measurements on this random variable will lie below the value.
Sample of 25 measurements on a normally distributed quality characteristic has a mean of 85 and a standard deviation of 1. Using a confidence probability of 0.95, find a value such that 90% of the future measurements on this quality characteristic will lie above it.
A sample of 20 measurements on a normally distributed quality characteristic had x = 350 and s =10 . Find an upper natural tolerance limit that has probability 0.90 of containing 95% of the distribution of this quality characteristic.
How large a sample is required to obtain a natural tolerance interval that has probability 0.90 of containing 95% of the distribution? After the data are collected, how would you construct the interval?
A random sample of n= 40 pipe sections resulted in a mean wall thickness of 0.1264 in. and a standard deviation of 0.0003 in. We assume that wall thickness is normally distributed. (a) Between what limits can we say with 95% confidence that 95% of the wall thicknesses should fall? (b) Construct a
Estimate process capability using the x and R charts for the power supply voltage data in Exercise 6.8 (note that early printings of the 7th edition indicate Exercise 6.2). If specifications are at 350 ± 50 V, calculate Cp, Cpk, and Cpkm. Interpret these capability ratios.
Find the sample size required to construct an upper nonparametric tolerance limit that contains at least 95% of the population with probability at least 0.95. How would this limit actually be computed from sample data?
Consider the hole diameter data in Exercise 6.9. Estimate process capability using x and R charts. If specifications are 0 ± 0.01, calculate Cp, Cpk, and Cpkm. Interpret these ratios.
A process is in control with x = 100.s = 1.05, and n = 5. The process specifications are at 95 ±10. The quality characteristic has a normal distribution.(a) Estimate the potential capability. (b) Estimate the actual capability. (c) How much could the fallout in the process be reduced if the
A process is in statistical control with x = 199 and R = 3.5. The control chart uses a sample size of n= 4. Specifications are at 200 ± 8. The quality characteristic is normally distributed.USL = 200 + 8 = 208; LSL = 200 €“ 8 = 192 (a) Estimate the potential capability of the
A process is in statistical control with x = 39.7 and R = 2.5. The control chart uses a sample size of n= 2. Specifications are at 40 ± 5. The quality characteristic is normally distributed.USL = 40 + 5 = 45; LSL = 40 €“ 5 = 35 (a) Estimate the potential capability of the
The data in Table 9E.1 represent individual observations on molecular weight taken hourly from a chemical process. The target value of molecular weight is 1,050 and the process standard deviation is thought to be about Ïƒ = 25.(a) Set up a tabular CUSUM for the mean of this process.
Set up a tabular CUSUM scheme for the flow width data used in Example 6.1 (see Tables 6.1 and 6.2). When the procedure is applied to all 45 samples, does the CUSUM react more quickly than the x chart to the shift in the process mean? Use Ïƒ = 0.14 in setting up the CUSUM, and design the
Consider the loan processing cycle time data in Exercise 8.15. Set up a CUSUM chart for monitoring this process. Does the process seem to be in statistical control?x ï‚»p50 ï€½18.05; p84 ï€½20.53;
Consider the loan processing cycle time data in Exercise 8.15. Set up an EWMA control chart for monitoring this process using ï¬ï€ = 0.1. Does the process seem to be in statistical control?x ï‚»p50 ï€½18.05; p84
Consider the hospital emergency room waiting time data in Exercise 8.16. Set up a CUSUM chart for monitoring this process. Does the process seem to be in statistical control?x ï‚»p50 ï€½4.55; p84 ï€½7.34;
Consider the hospital emergency room waiting time data in Exercise 8.16. Set up an EWMA control chart for monitoring this process using ï¬ï€ = 0.2. Does the process seem to be in statistical control? ï‚»p50 ï€½4.55; p84
Consider the €œminute clinic€ waiting time data in Exercise 6.66. These data may not be normally distributed. Set up a CUSUM chart for monitoring this process. Does the process seem to be in statistical control?ïï€ ï€½
Consider the €œminute clinic€ waiting time data in Exercise 6.66. These data may not be normally distributed. Set up a EWMA control chart using ï¬ï€ = 0.1 for monitoring this process. Does the process seem to be in statistical control?
Consider the hospital emergency room waiting time data in Exercise 8.16. Set up an EWMA control chart for monitoring this process using ï¬ï€ = 0.1. Compare this EWMA chart to the one from Exercise 9.14.x ï‚»p50 ï€½4.55; p84
Consider the €œminute clinic€ waiting time data in Exercise 6.66. Set up an EWMA control chart for monitoring this process using ï¬ï€ = 0.4. Compare this EWMA chart to the one from Exercise 9.16.
Apply the scale CUSUM discussed in Section 9.1.8 to the data in Exercise 9.1Exercise 9.1
Rework Exercise 9.1 using a standardized CUSUM.Exercise 9.1
Apply the scale CUSUM discussed in Section 9.1.8 to the concentration data in Exercise 9.8.Exercise 9.8
Consider a standardized two-sided CUSUM with k = 0.2 and h=8. Use Siegmund’s procedure to evaluate the in-control ARL performance of this scheme. Find ARL1 for * = 0.5. In control ARL performance:
Consider the viscosity data in Exercise 9.9. Suppose that the target value of viscosity is µ0 = 3,150 and that it is only important to detect disturbances in the process that result in increased viscosity. Set up and apply an appropriate one-sided CUSUM for this process. µ0 = 3150, s = 5.95238, k
Consider the velocity of light data introduced in Exercises 6.69 and 6.70. Use only the 20 observations in Exercise 6.69 to set up a CUSUM with target value 734.5. Plot all 40 observations from both Exercises 6.69 and 6.70 on this CUSUM. What conclusions can you draw?
Rework Exercise 9.23 using an EWMA with = 0.10. = 0.1; L = 2.7; CL = 0 = 734.5; = 108.7
Rework Exercise 9.1 using an EWMA control chart with = 0.1 and L = 2.7. Compare your results to those obtained with the CUSUM. = 25, CL = 0 = 1050, UCL = 1065.49, LCL = 1034.51
Consider a process with 0 = 10 and = 1. Set up the following EWMA control charts: (a) = 0.1, L = 3 (b)  = 0.2, L = 3 (c)  = 0.4, L = 3
Reconsider the data in Exercise 9.4. Set up an EWMA control chart with  = 0.2 and L = 3 for this process. Interpret the results.. Assume = 0.05. CL = 0 = 8.02, UCL = 8.07, LCL = 7.97
Reconstruct the control chart in Exercise 9.27 using  = 0.1 and L = 2.7. Compare this chart with the one constructed in Exercise 9.27. Assume = 0.05. CL = 0 = 8.02, UCL = 8.05, LCL = 7.99
Reconsider the data in Exercise 9.7. Apply an EWMA control chart to these data using  = 0.1 and L = 2.7. 12.16 , CL = 0 = 950, UCL = 957.53, LCL = 942.47.
(a) Add a headstart feature to the CUSUM in Exercise 9.1. (b) Use a combined Shewhart-CUSUM scheme on the data in Exercise 9.1. Interpret the results of both charts.
Reconstruct the control chart in Exercise 9.29 using = 0.4 and L = 3. Compare this chart to the one constructed in 9.29 12.16 , CL = 0 = 950, UCL = 968.24, LCL = 931.76.
Reconsider the data in Exercise 9.8. Set up and apply an EWMA control chart to these data using  = 0.05 and L = 2.6. 5.634, CL = 0 = 175, UCL = 177.30, LCL = 172.70.
Reconsider the homicide data in Exercise 7.76. Set up an EWMA control chart for this process µ = 0.1 and L = 2.7. Does potential non-normality in the data pose a concern here?
Reconsider the data in Exercise 9.9. Set up and apply an EWMA control chart to these data using  = 0.1 and L = 2.7.
Analyze the data in Exercise 9.1 using a moving average control chart with w = 6. Compare the results obtained with the cumulative sum control chart in Exercise 9.1. w = 6, 0 = 1050, = 25, CL = 1050, UCL = 1080.6, LCL = 1019.4
Analyze the data in Exercise 9.4 using a moving average control chart with w = 5. Compare the results obtained with the cumulative sum control chart in Exercise 9.4. w = 5, 0 = 8.02, = 0.05, CL = 8.02, UCL = 8.087, LCL = 7.953
Analyze the homicide data in Exercise 7.76 using a moving average control chart with w = 5. Does potential non-normality in the data pose a concern here?
Show that if the process is in control at the level µ, the exponentially weighted moving average is an unbiased estimator of the process mean.
Derive the variance of the exponentially weighted moving average zi.
Show that if  = 2/(w + 1) for the EWMA control chart, then this chart is equivalent to a w-period moving average control chart in the sense that the control limits are identical in the steady state.
A machine is used to fill cans with motor oil additive. A single sample can is obtained. Since the filling process is automated, it has very stable variability, and long experience indicates that Ïƒ = 0.05 oz. The individual observations for 24 hours of operation are shown in Table
Show that if  = 2/(w + 1), then the average “ages” of the data used in computing the statistics zi and Mi are identical.
Show how to modify the control limits for the moving average control chart if rational subgroups of size n > 1 are observed every period, and the objective of the control chart is to monitor the process mean.
A Shewhart x chart has center line at 10 with UCL = 16 and LCL = 4. Suppose you wish to supplement this chart with an EWMA control chart using  = 0.1 and the same control limit width in σ-units as employed on the x chart. What are the values of the steady-state upper and lower control limits on
An EWMA control chart uses ï¬ = 0.4. How wide will the limits be on the Shewhart control chart, expressed as a multiple of the width of the steady-state EWMA limits?For EWMA, steady-state limits are
Consider the valve failure data in Example 7.6. Set up a CUSUM chart for monitoring the time between events using the transformed variable approach illustrated in that example. Use standardized values of h = 5 and k = ½.
Consider the valve failure data in Example 7.6. Set up a one-sided CUSUM chart for monitoring and detecting an increase in failure rate of the valve. Assume that the target value of the mean time between failures is 700 hr. µ0 = 700, h = 5, k = 0.5, estimate σ using the average moving range
Set up an appropriate EWMA control chart for the valve failure data in Example 7.6. Use the transformed variable approach illustrated in that example.
Discuss how you could set up one-sided EWMA control charts.
Rework Exercise 9.4 using the standardized CUSUM parameters of h = 8.01 and k = 0.25. Compare the results with those obtained previously in Exercise 9.4. What can you say about the theoretical performance of those two CUSUM schemes?
Reconsider the data in Exercise 9.4. Suppose the data there represent observations taken immediately after a process adjustment that was intended to reset the process to a target of µ0 = 8.00. Set up and apply an FIR CUSUM to monitor this process.
The data in Table 9E.3 are temperature readings from a chemical process in ï‚°C, taken every two minutes. (Read the observations down, from left.) The target value for the mean is ï0 = 950.(a) Estimate the process standard deviation. (b) Set up and apply a tabular
Bath concentrations are measured hourly in a chemical process. Data (in ppm) for the last 32 hours are shown in Table 9E.4 (read down from left). The process target is µ0 = 175 ppm.(a) Estimate the process standard deviation. (b) Construct a tabular CUSUM for this process, using standardized
Viscosity measurements on a polymer are made every 10 minutes by an on-line viscometer. Thirty-six observations are shown in Table 9E.5 (read down from left). The target viscosity for this process is µ0 = 3,200.(a) Estimate the process standard deviation. (b) Construct a tabular CUSUM for
Use the data in Table 10E.1 to set up short-run charts using the DNOM approach. The nominal and x R dimensions for each part are TA = 100, TB = 60, TC = 75, and TD = 50.
A manufacturing process operates with an in-control fraction of nonconforming production of at most 0.1%, which management is willing to accept 95% of the time; however, if the fraction nonconforming increases to 2% or more, management wishes to detect this shift with probability 0.90. Design an
Consider a modified control chart with center line at µ = 0 and σ = 1.0 (known). If n = 5, the tolerable fraction nonconforming  = 0.00135, and the control limits are at three-sigma, sketch the OC curve for the chart. On the same set of axes sketch the OC curve corresponding to the chart with
Specifications on a bearing diameter are established at 8.0 ± 0.01 cm. Samples of size n = 8 are used, and a control chart for s shows statistical control, with the best current estimate of the population standard deviation S = 0.001. If the fraction of nonconforming product that is barely
An x chart is to be designed for a quality characteristic assumed to be normal with a standard deviation of 4. Specifications on the product quality characteristics are 50 ± 20. The control chart is to be designed so that if the fraction nonconforming is 1%, the probability of a point falling
An x chart is to be designed for a quality characteristic assumed to be normal with a standard of 4. Specifications on the quality characteristic are 800 ± 20. The control chart is to be designed so that if the fraction nonconforming is 1%, the probability of a point falling inside the control
A normally distributed quality characteristic is controlled by x and R charts having the following parameters (n = 4, both charts are in control):(a) What is the estimated standard deviation of the quality characteristic x? (b) If specifications are 610 ±15, what is your estimate of the
The data in Table 10E.8 are molecular weight measurements made every two hours on a polymer (read down, then across from left to right).(a) Calculate the sample autocorrelation function and provide an interpretation. (b) Construct an individuals control chart with the standard deviation estimated
Consider the molecular weight data in Exercise 10.16. Construct a CUSUM control chart on the residuals from the model you fit to the data in part (c) of that exercise.Let ï0 = 0, ï¤ = 1 sigma, k = 0.5, h = 5.In Exercise 10.16
Consider the molecular weight data in Exercise 10.16. Construct a EWMA control chart on the residuals from the model you fit to the data in part (c) of that exercise. Let  = 0.1 and L = 2.7 (approximately the same as a CUSUM with k = 0.5 and h = 5).
Set up a moving center-line EWMA control chart for the molecular weight data in Exercise 10.16. Compare it to the residual control chart in Exercise 10.16, part (c). To find the optimal  , fit an ARIMA (0,1,1) (= EWMA = IMA(1,1)):
Use the data in Table 10E.2 to set up appropriate short-run x and R charts, assuming that the standard deviations of the measured characteristic for each part type are not the same. The nominal dimensions for each part are TA = 100, TB = 200, and TC = 2000.
The data shown in Table 10E.9 are concentration readings from a chemical process, made every 30 minutes (read down, then across from left to right).(a) Calculate the sample autocorrelation function and provide an interpretation. (b) Construct an individuals control chart with the standard deviation
Consider the concentration data in Exercise 10.20. Construct a CUSUM chart on the residuals from the model you fit in part (c) of that exercise.
Consider the concentration data in Exercise 10.20. Construct a EWMA chart on the residuals from the model you fit in part (c) of that exercise.
Set up a moving center line EWMA control chart for the concentration data in Exercise 10.20. Compare it to the residuals control chart in Exercise 10.20, part (c). To find the optimal  , fit an ARIMA (0,1,1) (= EWMA = IMA(1,1)):
The data shown in Table 10E.10 are temperature measurements made every 2 minutes on an intermediate chemical product (read down, then across from left to right).(a) Calculate the sample autocorrelation function. Interpret the results that you have obtained. (b) Construct an individuals control

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