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Even though concrete is not combustible, it is susceptible to the effects of intense heat and has limitations in terms of maintaining its strength and integrity at high temperatures. The cement in cured concrete is bound into a rock-like substance by water molecules.
Intense heat causes cured cement to dehydrate and revert back to dehydrated powdery cement. Heat, by dehydrating the cement, reduces the strength and the modulus of elasticity of concrete. And the water, released as steam, sometimes violently, causes chipping and other physical structural damage.
Concrete cannot burn, but it can fail structurally due to the effects of heat. Liquid granite is much less susceptible to structural failure due to intense heat. Because of its ability to stand up much longer to heat, it can provide more precious time to evacuate burning buildings in which it has been used structurally.
Liquid granite is also more eco-friendly than concrete. Therefore, its carbon footprint is smaller than concrete. Engineers may now have to make decisions based on comparing two materials, concrete and liquid granite.
Structure comparative experiments involving two samples as hypothesis tests. Perform hypothesis tests and construct confidence intervals on the difference in means of two normal distribtions.
Perform hypothesis tests and construct confidence intervals on the ratio of the variances of two normal distributions. Perform hypothesis tests and construct confidence intervals on the difference in two population proportions. Compute power and type II error, and make sample size selection decisions for hypothesis tests and confidence intervals.
Understand how the analysis of variance can be used in an experiment to compare several means. Understand the blocking principle and how it is used to isolate the effect of nuisance factors in an experiment. Design and conduct experiments using a randomized complete block design.
This chapter extends those results to the case of two independent populations. The general situation is shown in Fig. Inferences will be based on two random samples of c05DecisionMakingforTwoSamples. That is, X11, X12,.
The assumptions for this section are summarized next. Assumptions 1. X11, X12,. X21, X22,. The two populations represented by X1 and X2 are independent. Both populations are normal, or if they are not normal, the conditions of the central limit theorem apply.
This is exactly what we did in the one-sample z-test of Section This would give a test with level of significance. P-values or critical regions for the one-sided alternatives would be determined similarly. Formally, we summarize these results for the two-sample z-test in the following display.
Two formulations of the paint are tested; formulation 1 is the standard chemistry, and formulation 2 has a new drying ingredient that should reduce the drying time. From experience, it is known that the standard deviation of drying time is 8 minutes, and this inherent variability should be unaffected by the addition of the new ingredient. Ten specimens are painted with formulation 1, and another 10 specimens are painted with formulation 2; the 20 specimens are painted in random order.
What conclusions can the product developer draw about the effectiveness of the new ingredient? We apply the seven-step procedure to this problem as follows: 1.
We want to reject H0 if the new ingredient reduces mean drying time. Note that because the P-value for this test is 0. The practical engineering conclusion is that adding the new ingredient to the paint significantly reduces the drying time. EXAMPLE Paint Drying Time To illustrate the use of these sample size equations, consider the situation described in Example , and suppose that if the true difference in drying times is as much as 10 minutes, we want to detect this with probability at least 0.
What sample size is appropriate? Recall that X11, X12,. For nonnormal populations, the confidence level is approximately valid for large sample sizes. EXAMPLE Aircraft Spars Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft.
From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. The data obtained are shown in Table The required sample size from each population is as follows. A computer program has produced the following output for a hypothesis testing problem: Difference in sample means: 2. A computer program has produced the following output for a hypothesis testing problem: Difference in sample means: Two machines are used for filling plastic bottles with a net volume of A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is A random sample of 10 bottles is taken from the output of each machine.
Machine 1 Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi.
Based on the sample information, should it use plastic 1? Use the P-value approach in reaching a decision. The burning rates of two different solid-fuel propellants used in aircrew escape systems are being studied. What is the practical meaning of this interval? Two machines are used to fill plastic bottles with dishwashing detergent.
Assume normality. Interpret this interval. Compare and comment on the width of this interval to the width of the interval in part a. Reconsider the situation described in Exercise Two different formulations of an oxygenated motor fuel are being tested to study their road octane numbers. Formulate and test an appropriate hypothesis using the P-value approach. Consider the situation described in Exercise Consider the road octane test situation described in Exercise A polymer is manufactured in a batch chemical process.
Fifteen batch viscosity measurements are given as follows: , , , , , , , , , , , , , , A process change is made that involves switching the type of catalyst used in the process. Following the process change, eight batch viscosity measurements are taken: , , , , , , , Assume that process variability is unaffected by the catalyst change. The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process.
The standard deviation of active concentration is known to be 3 grams per liter, regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data are shown here: Catalyst 1: Catalyst 2: Base your answer on the results of part a.
Consider the polymer batch viscosity data in Exercise If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability.
For the laundry detergent problem in Exercise , test the hypothesis that the mean active concentrations are the same for both types of catalyst. What is the P-value for this test? Compare your answer to that found in part b of Exercise , and comment on why they are the same or different. If the sample sizes n1 and n2 exceed 40, the normal distribution procedures in Section could be used.
However, when small samples are taken, we will assume that the populations are normally distributed and base our hypothesis tests and CIs on the t distribution. This nicely parallels the case of inference on the mean of a single sample with unknown variance. A t-statistic will be used to test these hypotheses. As noted above and in Section , the normality assumption is required to develop the test procedure, but moderate departures from normality do not adversely affect the procedure.
Two different situations must be treated. Let X1, X2, S 12, S 22 be the sample means and sample variances, respectively. Now the expected value c05DecisionMakingforTwoSamples.
The determination of P-values in the location of the critical region for fixed-level testing for both two- and one-sided alternatives parallels those in the one-sample case. This procedure is often called the pooled t-test. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Because catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table Is there any difference between the mean yields?
Assume equal variances. Since these are all separate runs of the pilot plant, it is reasonable to assume that we have two independent populations and random samples from each population. The solution using the seven-step hypothesis testing procedure is as follows: 1. When the sample sizes are the same from both populations, the t-test is very robust or insensitive to the assumption of equal variances.
The practical conclusion is that at the 0. This was obtained from Minitab computer software. Checking the Normality Assumption Notice that the numerical results are essentially the same as the manual computations in Example We will give the computing formula for the CI in Section Figure shows the normal probability plot of the two samples of yield data and comparative box plots.
The normal probability plots indicate that there is no problem with the normality assumption. Furthermore, both straight lines have similar slopes, providing some verification of the assumption of equal variances. The comparative box plots indicate that there is no obvious difference in the two catalysts, although catalyst 2 has slightly greater sample variability.
However, the following test statistic is used. If v is not an integer, round down to the nearest integer. Arsenic concentration in public drinking water supplies is a potential health risk.
An article in the Arizona Republic May 27, reported drinking water arsenic concentrations in parts per billion ppb for 10 metropolitan Phoenix communities and 10 communities in rural Arizona. For our illustrative purposes, we are going to assume that these two data sets are representative random samples of the two types of communities. Figure shows a normal probability plot for the two samples of arsenic concentration. The assumption of normality appears quite reasonable, but because the slopes of the two straight lines are very different, it is unlikely that the population variances are the same.
Applying the seven-step procedure gives the following: 1. Therefore, the P-value is less than 0. Practical engineering conclusion: There is evidence to conclude that mean arsenic concentration in the drinking water in rural Arizona is different from the mean arsenic concentration in metropolitan Phoenix drinking water.
Furthermore, the mean arsenic concentration is higher in rural Arizona communities. We will discuss its computation in c05DecisionMakingforTwoSamples. For the one-sided alternative hypothesis, we use Charts Vc and Vd and define d and as in equation Suppose that if catalyst 2 produces a mean yield that differs from the mean yield of catalyst 1 by 4. What sample size is required? Reduced levels of calcium would indicate that the hydration mechanism in the cement is blocked and would allow water to attack various locations in the cement structure.
Furthermore, assume that both normal populations have the same standard deviation. Can you answer this question without doing any additional calculations? This off-center operation was ultimately traced to an oversized wax tool. Changing the tooling resulted in a substantial improvement in the process. The fractions of nonconforming output or fallout below the lower specification limit and above the upper specification limit are often of interest.
Suppose that the measurement from a normally distributed process in statistical control is denoted as X. Estimate Cp, Cpk, and the probability of not meeting specification. Departures from normality can seriously affect the results. The calculation should be interpreted as an approximate guideline for process performance. Montgomery b provides guidelines on appropriate values of the Cp and a table relating fallout for a normally distributed process in statistical control to the value of Cp.
Many U. Assuming a normal distribution, the calculated fallout for this process is 0. The reason that such a large process capability is often required is that it is difficult to maintain a process mean at the center of the specifications for long periods of time. A common model that is used to justify the importance of a six-sigma process is illustrated in Fig. If the process mean shifts off-center by 1. Assuming a normally distributed process, the fallout of the shifted process is 3. Consequently, the mean of a six-sigma process can shift 1.
We repeat that process capability calculations are meaningful only for stable processes; that is, processes that are in control. A process capability ratio indicates whether or not the natural or chance variability in a process is acceptable relative to the specifications. Six standard deviations of a normally distributed process use It is centered at the nominal dimension, located halfway between the upper and lower specification limits.
Interpret these ratios. Reconsider Exercise Use the revised control limits and process estimates. Reconsider Exercise , where the specification limits are Estimate the percentage of defective items that will be produced. Estimate Cp and interpret this ratio. What is the fallout level?
Estimate Cp and Cpk and interpret these ratios. Given that the specifications are at 6. What are the natural tolerance limits of this process? This classification is usually done to achieve economy and simplicity in the inspection operation. For example, the diameter of a ball bearing may be checked by determining whether it passes through a gauge consisting of circular holes cut in a template. This kind of measurement is much simpler than directly measuring the diameter with a device such as a micrometer.
Control charts for attributes are used in these situations. Attributes control charts often require a considerably larger sample size than do their measurements counterparts. In this section, we will discuss the fraction-defective control chart, or P chart.
Sometimes the P chart is called the control chart for fraction nonconforming. Suppose D is the number of defective units in a random sample of size n. We assume that D is a binomial random variable with unknown parameter p. Suppose that m preliminary samples each of size n are available, and let di be the number of defectives in the ith sample.
These control limits are based on the normal approximation to the binomial distribution. When p is small, the normal approximation may not always be adequate. In such cases, we may use control limits obtained directly from a table of binomial probabilities. If p is small, the lower control limit may be a negative number.
If this should occur, it is customary to consider zero as the lower control limit. Construct a fraction-defective control chart for this ceramic substrate production line. Assume that the samples are numbered in the sequence of production.
All samples are in control. If they were not, we would search for assignable causes of variation and revise the limits accordingly. This chart can be used for controlling future production. We should take appropriate steps to investigate the process to determine why such a large number of defective units are being produced.
Defective units should be analyzed to determine the specific types of defects present. Once the defect types are known, process changes should be investigated to determine their impact on defect levels.
Designed experiments may be useful in this regard. The points, center line, and control limits for this chart are multiples times n of the corresponding elements of a P chart.
The use of an nP chart avoids the fractions in a P chart. The number of defectives in Table would be plotted on such a chart and the conclusions would be identical to those from the P chart. Suppose that in the production of cloth it is necessary to control the number of defects per yard or that in assembling an aircraft wing the number of missing rivets must be controlled.
In these situations, we may use the control chart for defects per unit, or the U chart. Many defects-per-unit situations can be modeled by the Poisson distribution. A U chart may be constructed for such data. These control limits are based on the normal approximation to the Poisson distribution. In such cases, we may use control limits obtained directly from a table of Poisson probabilities. If u is small, the lower control limit may be a negative number.
A flow solder machine is used to make the mechanical and electrical connections of the leaded components to the board. The boards are run through the flow solder process almost continuously, and every hour five boards are selected and inspected for process-control purposes. The number of defects in each sample of five boards is noted. Results for 20 samples are shown in Table Construct a U chart.
Because LCL is negative, it is set to zero. Practical interpretation: From the control chart in Fig. An investigation should be made of the specific types of defects found on the printed circuit boards to suggest potential avenues for process improvement. This is simply a control chart of C, the total number of defects in a sample.
The use of a C chart avoids the fractions that can occur in a U chart. The number of defects in Table would be plotted on such a chart. Suppose the following number of defects has been found in successive samples of size 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.
If not, assume that assignable causes can be found and outof-control points eliminated. Revise the control limits. Using an injection molding process, a plastics company produces interchangeable cell phone covers. After molding, the covers are sent through an intricate painting process.
Quality control engineers inspect the covers and record the paint blemishes. The number of blemishes found in 20 samples of 5 covers are as follows: 2, 1, 5, 5, 3, 3, 1, 3, 4, 5, 4, 4, 1, 5, 2, 2, 3, 1, 4, 4.
If not, assume that assignable causes can be found, list points, and revise the control limits. The following represent the number of defects per feet in rubber-covered wire: 1, 1, 3, 7, 8, 10, 5, 13, 0, 19, 24, 6, 9, 11, 15, 8, 3, 6, 7, 4, 9, 20, 11, 7, 18, 10, 6, 4, 0, 9, 7, 3, 1, 8, Do the data come from a controlled process? Consider the data in Exercise Set up a C chart for this process.
Compare it to the U chart in Exercise Comment on your findings. The following are the numbers of defective solder joints found during successive samples of solder joints.
Day No. By moving the control limits farther from the center line, we decrease the risk of a type I error—that is, the risk of a point falling beyond the control limits, indicating an outof-control condition when no assignable cause is present. However, widening the control limits will also increase the risk of a type II error—that is, the risk of a point falling between the control limits when the process is really out of control. If we move the control limits closer to the center line, the opposite effect is obtained: The risk of type I error is increased, whereas the risk of type II error is decreased.
The control limits on a Shewhart control chart are customarily located a distance of plus or minus three standard deviations of the variable plotted on the chart from the center line; that is, the constant k in equation should be set equal to 3. These limits are called three-sigma control limits. A way to evaluate decisions regarding sample size and sampling frequency is through the average run length ARL of the control chart.
That is, even if the process remains in control, an out-of-control signal will be generated every points, on the average. Consider the piston ring process discussed earlier, and suppose we are sampling every hour. Thus, we will have a false alarm about every hours on the average. Then the probability that X falls between the control limits of Fig. Suppose this approach is unacceptable because production of piston rings with a mean diameter of How can we reduce the time needed to detect the out-of-control condition?
One method is to sample more frequently. For example, if we sample every half hour, only 1 hour will elapse on the average between the shift and its detection. The second possibility is to increase the sample size.
The probability of X falling between the control limits when the process mean is Table provides average run lengths for an X chart with three-sigma control limits. The average run lengths are calculated for shifts in the process mean from 0 to 3.
Consider the X control chart in Fig. Suppose that the mean shifts to An X chart uses samples of size 6. The center line is at , and the upper and lower three-sigma control limits are at and 94, respectively. Find the probability that this shift will be detected on the next sample.
Suppose that the mean shifts to 0. Suppose that the mean shifts to 6. In any problem involving measurements, some of the observed variability will arise from the experimental units that are being measured and some will be due to measurement error. Two types of error associated with a gauge or measurement device are precision and accuracy. These two components of measurement error are illustrated in Fig. Accuracy refers to the ability to measure the true value of the characteristic correctly on average, and precision reflects the inherent variability in the measurements.
In this section we describe some methods for evaluating the precision of a measurement device or system. Determining accuracy often requires the use of a standard, for which the true value of the measured characteristic is known. Often the accuracy feature of a measurement system or device can be modified by making adjustments to the device or by using a properly constructed calibration curve. The regression methods of Chapter 6 can be used to construct calibration curves.
Data from a measurement system study in the semiconductor industry are shown in Table An electronic tool was used to measure the resistivity of 20 randomly selected silicon wafers following a process step during which a layer was deposited on the wafer surface.
The technician who was responsible for the setup and operation of the measurement tool measured each wafer twice. Each measurement on all 20 wafers was made in random order. Figure shows X and R charts from Minitab for the data in Table The X chart indicates that there are many out-of-control points because the control chart is showing the discriminating capability of the measuring instrument—literally the ability of the device to distinguish between different units of product.
Notice that this is a somewhat different interpretation for an X control chart. The R chart directly reflects the magnitude of measurement error because the range values are the difference between measurements made on the same wafer using the same measurement tool. The R chart is in control, indicating that the operator is not experiencing any difficulty making consistent measurements. Nor is there any indication that measurement variability is increasing with time. We may also consider the measurement system study in Table as a single-factor completely randomized experiment with parts as treatments.
Notice that the F-statistic for wafers is significant, implying that there are differences in the parts used in the study. This yields This is a desirable situation. It can be extended to more complex types of experiments. Table presents an expanded study of the tool for measuring resistivity of silicon wafers. In the original study, the 20 wafers were measured on the first shift, and an operator from that shift was responsible for setup and operation of the measurement tool.
In the expanded study, the 20 wafers were measured on two additional shifts, and operators from those shifts did the setup and ran the measurement tool. Both parts and shifts are considered as random factors. The other component of measurement variability is called reproducibility, and it reflects the variability associated with the shifts which arise from the setup procedure for the tool, drift in the tool over time, and different operators.
In the Minitab analysis, we have specified that both factors, wafers and shifts, are random factors. The estimates of the variance components were obtained by solving the equations derived from the expected mean squares essentially as we did with the single-factor experiment. The Minitab output also contains the expected mean squares using the 1, 2, 3, 4, notation for the variance components. Notice that the estimate of one of the variance components is negative. The ANOVA method of variance component estimation sometimes produces negative estimates of one or more variance components.
This is usually taken as evidence that the variance component is really zero. Therefore, the only significant component of overall gauge variability is the component due to repeatability, and the operators are very consistent about how the tool is set up and run over the different shifts.
Consider the wafers measured in example data given in Table for two measurements on 20 different wafers. Assume that a third measurement is recorded on these 20 wafers, respectively. The measurements are as follows: , , , , , , , , , , , , , , , , , , , Do we have a desirable situation with respect to the variability of the gauge?
The process engineer is concerned with the device used to measure purity level of a steel alloy. To assess the device variability he measures 10 specimens with the same instrument twice. The resultant measurements follow: Measurement 1 Measurement 2 2. An implantable defibrillator is a small unit that senses erratic heart signals from a patient. The quality control department of the manufacturer of these defibrillators is responsible for checking the output voltage of these assembled devices.
To check the variability of the voltmeters, the department performed a designed experiment, measuring each of eight units twice. The data collected are as follows: Measurement 1 Measurement 2 Comment on your results. Asphalt compressive strength is measured in units of psi. To test the repeatability and reproducibility of strength c08StatisticalProcessControl.
The data are as follows: Operator 1 Operator 2 Meas. A handheld caliper is used to measure the diameter of fuse pins in an aircraft engine. A repeatability and reproducibility study is carried out to determine the variability of the gauge and the operators. Two operators measure five pistons three times. The data are as follows equals operator 1, measurement 1; equals operator 1, measurement 2, and so forth : Consider a gauge study in which two operators measure 15 parts six times.
The result of their analysis is given in the following Minitab output. Consider a gauge study in which two operators measure 15 parts five times. The diameter of fuse pins used in an aircraft engine application is an important quality characteristic.
Twenty-five samples of three pins each are as follows in mm. Sample Number Diameter 1 2 3 4 5 6 7 If necessary, revise limits so that no observations are out of control. Calculate an estimate of Cp. Use this ratio to draw conclusions about process capability. What should this new variance value be? What is the probability that this shift will be detected on the next sample? What is the ARL after the shift? Plastic bottles for liquid laundry detergent are formed by blow molding. The data are as follows: 9, 11, 10, 8, 3, 8, 8, 10, 3, 5, 9, 8, 8, 8, 6, 10, 17, 11, 9, Is the process in statistical control?
Use the data given to set up a P chart for this process. Explain why they differ. Also explain why your assessment about statistical control differs for the two sizes of n. Cover cases for a personal computer are manufactured by injection molding. Samples of five cases are taken from the process periodically, and the number of defects is noted. The results for 25 samples follow: 3, 2, 0, 1, 4, 3, 2, 4, 1, 0, 2, 3, 2, 8, 0, 2, 4, 3, 5, 0, 2, 1, 9, 3, 2.
If necessary, revise your control limits. Repeat parts a and b. Explain how this change alters your answers to parts a and b. Explain how this alters your answers to parts a and b. Suppose that a process is in control and an X chart is used with a sample size of 4 to monitor the process.
Suddenly there is a mean shift of 1. Also, which limits would you recommend using and why? Consider the control chart for individuals with threesigma limits. Consider a control chart for individuals, applied to a continuous hour chemical process with observations taken every hour. How many false alarms would occur each day month, on the average, with this chart?
Recall that the ARL for detecting this shift with three-sigma limits is How many false alarms would occur each month with this chart? Is this in-control ARL performance satisfactory? The depth of a keyway is an important part quality characteristic. Observation Sample a Using all the data, find trial control limits for X and R charts. Is the process in control? Then estimate the process standard deviation. Using the results from part b , what statements can you make about process capability?
Compute estimates of the appropriate process capability ratios. A process is controlled by a P chart using samples of size The center line on the chart is 0. Also explain why a shift to 0. Suppose the average number of defects in a unit is known to be 8. Suppose the average number of defects in a unit is known to be Suppose that an X control chart with two-sigma limits is used to control a process.
Find the probability that a false out-of-control signal will be produced on the next sample. Compare this with the corresponding probability for the chart with three-sigma limits and discuss. Comment on when you would prefer to use two-sigma limits instead of three-sigma limits. Consider an X control chart with k-sigma control limits. Consider the X control chart with two-sigma limits in Exercise What is the probability of making a product outside the specification limits?
Week No. If not, assume assignable causes can be found and out-ofcontrol points eliminated. Obtain time-ordered data from a process of interest. Use the data to construct appropriate control charts and comment on the control of the process. Can you make any recommendations to improve the process? If appropriate, calculate appropriate measures of process capability. Ferris, F. Grubbs, and C. Bowker and G. Lieberman, Prentice-Hall, A very well-written presentation of graphical methods in statistics.
Freedman, D. An excellent introduction to statistical thinking, requiring minimal mathematical background. Hoaglin, D. Good discussion and illustration of techniques such as stem-and-leaf displays and box plots. Tanur, J. Contains a collection of short nonmathematical articles describing different applications of statistics. A comprehensive treatment of probability at a higher mathematical level than this book. Hoel, P. A wellwritten and comprehensive treatment of probability theory and the standard discrete and continuous distributions.
Mosteller, F. A precalculus introduction to probability with many excellent examples. Ross, S. More mathematically sophisticated than this book but has many excellent examples and exercises. A more comprehensive book on engineering statistics at about the same level as this one. More tightly written and mathematically oriented than this book but contains some good examples. An excellent reference containing many insights on data analysis. Draper, N. A comprehensive book on regression written for statistically oriented readers.
The first part of the book is an introduction to simple and multiple linear regression. The orientation is to business and economics. Montgomery, D. A comprehensive book on regression written for engineers and physical scientists.
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