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applied statistics and probability for engineers
Applied Statistics For Engineers And Scientists 3rd Edition Jay L. Devore, Nicholas R. Farnum, Jimmy A. Doi - Solutions
=+b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p?
=+a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who have watched streamed programming.
=+23. TV advertising agencies face growing challenges in reaching audience members because viewing TV programs via digital streaming is increasingly popular. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV
=+22. The American Taxpayer Relief Act of 2012 was passed by the U.S. Congress on January 1, 2013.This act helped address what became famously known as the “fiscal cliff” crisis. However, during the last months of 2012, heated debates concerning the crisis were ongoing in Congress, and there
=+21. The article “Ultimate Load Capacities of Expansion Anchor Bolts” (J. of Energy Engr., 1993:139–158) gave the following summary data on shear strength (kip) for a sample of 3y8-in. anchor bolts: n 5 78, x 5 4.25, s 5 1.30. Calculate a lower confidence bound using a confidence level of
=+20. A Brinell hardness test involves measuring the diameter of the indentation made when a hardened steel ball is pressed into material under a standard test load. Suppose that the Brinell hardness is determined for each specimen in a sample of size 32, resulting in a sample mean hardness of
=+18. Determine the confidence level for each of the following large-sample one-sided confidence bounds:a. Upper bound: x 1 .84sy1nb. Lower bound: x 2 2.05sy1nc. Upper bound: x 1 .67sy1n
=+17. When the population distribution is normal and n is large, the statistic s has approximately a normal distribution with s , s y12n. Use this fact to develop a large-sample two-sided confidence interval formula for . Then calculate a 95% confidence interval for the true standard
=+b. Suppose the investigators had believed a priori that the population standard deviation was about 4 MPa. Based on this supposition, how large a sample would have been required to estimate to within .5 MPa with 95% confidence?
=+a. Calculate a two-sided confidence interval for true average fracture strength using a confidence level of 95%. Does it appear that true average fracture strength has been precisely estimated?
=+16. The article “Evaluating Tunnel Kiln Performance”(Amer. Ceramic Soc. Bull., August 1997: 59–63)gave the following summary information for fracture strengths (MPa) of n = 169 ceramic bars fired in a particular kiln: x 5 89.10, s 5 3.73.
=+b. Suppose the investigators had made a rough guess of 320 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 50 ml for a confidence level of 95%?
=+a. Calculate and interpret a 95% (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected.
=+ Burning coal inside houses can lead to increased levels of indoor air toxins that may have negative effects on lung function. Among the children in the study, 514 came from households that use coal for cooking or heating or both. Their FEV1 mean was 1427 with standard deviation 325. (Using a
=+indoor air pollution metrics and lung function growth among children ages 6–13 years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in 1 second. Higher FEV1
=+15. The negative effects of ambient air pollution on children’s lung function has been well established, but less research is available about the effects of indoor air pollution. The authors of “Indoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities” (Indoor
=+b. Suppose the investigators had made a rough guess of .16 for the value of s before collecting data.What sample size would be necessary to obtain an interval width of .05 for a confidence level of 95%?
=+a. Calculate a 95% two-sided CI for the true average dye-layer density for all such trees.
=+14. The article “Extravisual Damage Detection? Defining the Standard Normal Tree” (Photogrammetric Engr.and Remote Sensing, 1981: 515–522) discusses the use of color infrared photography in identification of normal trees in Douglas fir stands. Among data reported were summary statistics
=+c. The American Academy of Orthopedic Surgeons recommends that backpack weight be at most 10% of body weight. What does your calculation of part (b) suggest and why?
=+a. Calculate and interpret a 99% CI for population mean backpack weight.b. Obtain a 99% CI for population mean weight as a percentage of body weight.
=+13. Young people may feel they are carrying the weight of the world on their shoulders when in reality they are too often carrying an excessively heavy backpack. The article “Effectiveness of a School-Based Backpack Health Promotion Program” (Work, 2003: 113–123) reported the following
=+b. Calculate a 99% two-sided confidence interval for population mean concentration for the Pogonias cromis species. Why is this interval wider than the interval of part (a) even though it is based on a somewhat larger sample size?
=+a. Calculate a 95% two-sided confidence interval for population mean concentration for the Mugil liza species.
=+zinc concentration ( g@g) in the liver of fish:Species n x s Mugil liza 56 9.15 1.27 Pogonias cromis 61 3.08 1.71
=+12. Heavy-metal pollution of various ecosystems is a serious environmental threat, in part because of the potential transference of hazardous substances to humans via food. The article “Cadmium, Zinc, and Total Mercury Levels in the Tissues of Several Fish Species from La Plata River Estuary,
=+d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding 95% interval is repeated 100 times, 95 of the resulting intervals will include .Is this statement correct? Why or why not?
=+c. Consider the following statement: We can be highly confident that 95% of all bottles of this type of cough syrup have an alcohol content that is between 7.8 and 9.4. Is this statement correct?Why or why not?
=+a. Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.
=+11. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected, and the alcohol content of each bottle is determined. Let denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence
=+b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is 90% and for the other is 99%.Which of the intervals has the 90% confidence level, and why?
=+10. Each of the following is a confidence interval for 5 true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type:(114.4, 115.6) (114.1, 115.9)
=+9. Discuss how each of the following factors affects the width of the large-sample two-sided confidence interval for :a. Confidence level (for fixed n and s)b. Sample size n (for fixed confidence level and s)c. Sample standard deviation s (for fixed confidence level and n)
=+8. What z critical value in the large-sample two-sided confidence interval for should be used to obtain each of the following confidence levels?a. 98%b. 85%c. 75%d. 99.9%
=+7. Assuming that n is large, determine the confidence level for each of the following two-sided confidence intervals:a. x 6 3.09sy1nb. x 6 2.81sy1nc. x 6 1.44sy1nd. x 6 sy2n
=+b. What is the standard error of the estimator in part (a)? Hint: The variance of a Poisson random variable also equals .
=+a. Find an unbiased estimator for and compute the estimate using the data. Hint: The mean of a Poisson random variable equals .
=+6. Each of 150 newly manufactured items is examined, and the number of surface flaws per item is recorded, yielding the following data:Number of flaws: 0 1 2 3 4 5 6 7 Observed frequency: 18 37 42 30 13 7 2 1 Let x denote the number of flaws on a randomly chosen item, and assume that x has a
=+c. Plot the sample sizes found in parts (a) and (b)versus their corresponding probabilities. What can you conclude from this graph?
=+b. Repeat the calculations in part (a) for areas of 80%, 95%, and 99%.
=+99%, and 90%. The higher the confidence level, the more strongly we believe that the value of the parameter being estimated lies within the interval.Information about the precision of an interval estimate is conveyed by the width of the interval. If the confidence level is high and the resulting
=+the 95% confidence level, any value of between 9162.5 and 9482.9 is plausible. A confidence level of 95% implies that 95% of all samples would give an interval that includes , or whatever other parameter is being estimated, and only 5% of all samples would yield an erroneous interval. The most
=+virtually never the case that x 5 . The point estimate says nothing about how close it might be to . An alternative to reporting a single most plausible value of the parameter being estimated is to calculate and report an entire interval of plausible values—an interval estimate or confidence
=+A point estimate, because it is a single number, by itself provides no information about the precision and reliability of estimation. Consider, for example, using the statistic x to calculate a point estimate for the true average breaking strength (g) of paper towels of a certain brand, and
=+7.2 Large-Sample Confidence Intervals for a Population Mean
=+a. Suppose you want 90% of the area under the sampling distribution of x to lie within 61 unit of a population mean . Find the
=+5. Random samples of size n are selected from a normal population whose standard deviation is known to be 2.
=+b. Repeat the probability calculation in part (a) for samples of size n550, n5100, and n51000.(Use the normal approximation to the binomial.)
=+ 1 .10. That is, find the probability that the sample proportion lies within6.10 (i.e., 10%)of the population proportion. Use the formula for the upper bound on the standard error of p(see Section 5.6) in your calculations.
=+a. For random samples of size n510, calculate the area under the sampling distribution curve for p between the points 2 .10 and
=+4. Random samples of n trees are taken from a large area of forest, and the proportion of diseased trees in each sample is determined. The actual proportion of diseased trees, , is unknown.
=+n. What can you conclude from this graph?
=+c. Graph the probabilities you found in parts (a)and (b) versus their corresponding sample sizes,
=+b. Repeat the probability calculation in part (a) for samples of size n 5 50, n 5 100, and n 5 1000.
=+a. For random samples of size n510, calculate the area under the sampling distribution curve for x between the values 2 1 and 1 1. That is, find the probability that the sample mean lies within 61 unit of the population mean.
=+3. Random samples of size n are taken from a normal population whose standard deviation is known to be 5.
=+b. Use an unbiased estimator to compute a point estimate of , the proportion of all homes that use over 100 therms.
=+a. Use an unbiased estimator to compute a point estimate of , the average amount of gas used by all houses in the area.
=+2. A random sample of ten homes in a particular area, each heated with natural gas, is selected, and the amount of gas (therms) used during January is determined for each home. The resulting observations are 103, 156, 118, 89, 125, 147, 122, 109, 138, and 99.
=+1. A single plastic part is randomly selected from a large population of such parts. Can the length of the chosen part be considered an unbiased estimator of the average length of all the parts?
=+64. A small system contains three components that are connected according to the following diagram. Assuming that the components all function independently of one another, find the general expression for the system reliability R(t)in terms of the component reliabilities R1(t), R2(t), and R3(t).
=+b. Then prove it in the more general case, where the two components may or may not function independently of one another.1 2 3
=+a. Prove this under the assumption that the two components function independently of one another.
=+63. Show that any series system consisting of two components with reliabilities R1(t) and R2(t) can never have a system reliability R(t) that exceeds the reliability of its weakest link, that is, R(t) #min{R1(t), R2(t)}.
=+the two components do not necessarily function independently of one another. Show in this case that min5R1(t), R2(t)6 # R1(t)R2(t) 1 1 4.
=+62. Suppose that two components with reliabilities R1(t) and R2(t) are connected in series, but that
=+another t2 hours is the same as its initial probability of lasting t2 hours. Prove that the exponential distribution is memoryless. That is, prove that P(X . t1 1 t2uX . t1) 5 P(X . t2) for a random variable X that has an exponential distribution with parameter .
=+61. A “memoryless” system or component is one that satisfies the following property: If it has already lasted for t1 hours, then the probability it lasts for
=+60. For a fixed value of p, how large does the subgroup size n have to be to yield a positive lower control limit on a p chart?
=+59. For a certain process, x and R charts based on subgroups of size 5 have centerlines of 14.5 and 1.163, respectively. Given that the process has specification limits of 12 and 16, calculate Cp, Cpu, Cpl, and Cpk.
=+58. Explain why it is possible for all the measurements in a given sample to lie within the specification limits and for the same data to yield a nonzero estimate of the proportion of the process data that exceeds the specification limits.
=+b. Construct x and R charts of the transformed data in part (a). Evaluate the charts, and comment on the milling process.
=+a. Using the nominal lengths given, convert this data into the deviations from nominal format.
=+Subgroup x1 x2 x3 x4 Part type 15 .252 .250 .251 .247 P2 16 .251 .249 .250 .250 P2 17 .126 .127 .122 .125 P1 18 .123 .123 .123 .128 P1 19 .252 .250 .247 .248 P2 20 .502 .496 .502 .502 P4
=+milled steel bars of various sizes, denoted P1, P2, P3, and P4. The nominal length for bars of type P1 is .125;for bars of type P2, .250; for P3, .375; and for P4, .500.Subgroup x1 x2 x3 x4 Part type 1 .251 .252 .250 .249 P2 2 .372 .378 .379 .375 P3 3 .247 .249 .254 .251 P2 4 .248 .247 .250 .252
=+hour to hour, so there may be insufficient data to create control charts on any particular bar size. However, by subtracting the nominal value from each batch of bars, the resulting subgroups of data are sufficient to create a control chart for the milling process itself. The following table
=+is common to all the parts. For example, consider a milling process in which metal bars of various sizes are machined to specified lengths. The size of the bars submitted to the machining process may vary from
=+57. The deviations from nominal transformation in Exercise 21 can be used in so-called short-run processes.Even though small numbers of different-size parts are created by such processes, the deviations from the various nominal values of these parts provide information about the particular
=+b. Check the chart in part (a) for any out-of-control points. If there are any, eliminate them from the data and reconstruct the s chart. Repeat this process, if necessary, until there are no out-ofcontrol signals in the s chart.
=+a. Construct an s chart based on this data.
=+56. A manufacturer of dustless chalk monitors the consistency of chalk by running an s chart on the density of chalk in subgroups of size 8. The most recent 24 such subgroups had the accompanying sample standard deviations (read across):.204 .315 .096 .184 .230 .212 .322 .287.145 .211 .053 .145
=+55. A tool that drills holes in metal parts eventually wears out and periodically must be replaced. If the hole diameters drilled by this machine are monitored on a control chart, describe the type of pattern you would expect to see on the chart as the drill wears out.
=+54. Instead of constructing x and R charts for 30 subgroups of size 4, a friend suggests the simpler alternative of calculating the standard deviation s of the 30 means to establish 3-sigma limits for a control chart. That is, it is suggested that the 30 means be plotted on a chart with
=+the sensor, thereby meeting the required fill specification; these bottles are then shipped to customers.Describe the shape of the distribution of fill volumes for the bottles that pass this inspection.
=+53. In a bottling process, a beam of light is passed through the necks of bottles passing by on a conveyor system. Underfilled bottles, which allow the beam of light to pass through, trip a sensor that routes the bottles off the conveyor system. Bottles with liquid levels above the level of the
=+52. When affixed to an object, each piece of paper in a pad of adhesive notepaper must stay in place but must also be easily removable. The strength of the adhesive used is a critical quality characteristic of such pads. For this type of product, does adhesive strength have a one- or a two-sided
=+b. Suppose any disk has an exponential lifetime(in months) with parameter 5 .025. Calculate the reliability of this system.
=+a. Draw a diagram of this RAID system.
=+corresponding mirror disk and that the three such pairs of disks are connected in series.
=+system consists of three disks (A, B, and C) and three “mirror” disks that contain complete copies of the data in the first three disks. Suppose each A, B, and C disk is connected in parallel to its
=+51. RAID (Redundant Arrays of Inexpensive Disks)structures consist of various combinations of computer disks that use parallel design elements to achieve high reliability. Suppose that a RAID
=+b. Compare the estimates from part (a) with those obtained in Example 7.18 (page 338).
=+va. Using the data of Example 2.18 (page 93), estimate the parameters of the Weibull distribution that fit this data.
=+ with the [(i 2 .5)yn]th sample quantile of the Weibull distribution. That is, we use pi 5 (i 2 .5)yn in place of F(x) and perform a regression of ln[ln(1y1 2 pi)] on ln(xi) to find estimates of and .
=+, and it is easy to show algebraically that ln3ln (1y1 2 F(x))4 5 ln(x) 2 ln() (see page 93). For an ordered set of data x1 # x2 # x3 ## xn, we associate xi
=+50. Estimates of Weibull parameters can be obtained using simple linear regression (see Section 3.3). Denoting the lifetime of a product by t, the Weibull cumulative area function can be written as F(x) 5 1 2 e 2(xy)
=+b. Based on your result in part (a), what type of failure rate (decreasing, constant, or increasing) do products with normal failure laws have?
=+49.a. Assume that a certain product can be modeled with a normal failure law having a mean lifetime of 5 10 years and standard deviation 5 2 years. Use a spreadsheet program or other software to create a graph of the failure rate, Z(t), for such a product.
=+d. Answer part (c) using only the “memoryless”property of the exponential distribution.
=+c. Based on results in (a) and (b), do you think the exponential distribution is a good one for modeling human lifetimes?
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