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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
=+a. Construct normal quantile plots to verify the plausibility of both samples having been selected from normal population distributions.
=+fabric (H) and poor-quality fabric (P) specimens:H: 1.2 .9 .7 1.0 1.7 1.7 1.1 .9 1.7 1.9 1.3 2.1 1.6 1.8 1.4 1.3 1.9 1.6.8 2.0 1.7 1.6 2.3 2.0 P: 1.6 1.5 1.1 2.1 1.5 1.3 1.0 2.6
=+31. Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article “Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments” (Textile Res. J., 1997: 137–142) gave the
=+concluding that the true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles?Carry out a test using a significance level of .01.(Note: The cited paper reported P-values of t tests for comparing means of the various types
=+20 natural latex condoms of a certain type, the sample mean and sample standard deviation of the number of cycles to break were 4358 and 2218, respectively, whereas a sample of 20 polyisoprene condoms gave a sample mean and sample standard deviation of 5805 and 3990, respectively. Is there strong
=+of dimensions and lubricant quality. The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break.The article reported that for a sample of
=+30. According to the article “Fatigue Testing of Condoms” (Polymer Testing, 2009: 567–571), “tests currently used for condoms are surrogates for the challenges they face in use,” including a test for holes, an inflation test, a package seal test, and tests
=+Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of .01.
=+reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection
=+29. Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome(CTS). The article “A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome” (Ergonomics, 1995: 2588–2601)
=+ Assuming that both zinc mass distributions are at least approximately normal, carry out a test at significance level .05 to decide whether true average zinc mass is different for the two types of batteries.
=+28. Urban storm water can be contaminated by many sources, including discarded batteries. When ruptured, these batteries release metals of environmental significance. The article “Urban Battery Litter”(J. of Environ. Engr., 2009: 46–57) presented summary data for characteristics of a
=+27. Determine the number of degrees of freedom for the two-sample t test in each of the following situations:a. n1 5 10, n2 5 10, s1 5 5.0, s2 5 6.0b. n1 5 10, n2 5 15, s1 5 5.0, s2 5 6.0c. n1 5 10, n2 5 15, s1 5 2.0, s2 5 6.0d. n1 5 12, n2 5 24, s1 5 5.0, s2 5 6.0
=+N Mean Std Dev T Prob>|T|15 0.0453333 0.0899100 1.9527887 0.0711(Note: SAS explicitly tests H0: 5 0, so to test H0: 5 .60, the null value .60 must be subtracted from each xi; the reported mean is then the average of the (xi 2 .60) values. Also, SAS’s P-value is always for a two-tailed
=+26. The relative conductivity of a semiconductor device is determined by the amount of impurity“doped” into the device during its manufacture.A silicon diode to be used for a specific purpose requires an average cut-on voltage of .60 V, and if this is not achieved, the amount of impurity must
=+measured by DSC is at least approximately normal.The sample mean and standard deviation are 181.4 and .7242, respectively. Is there compelling evidence for concluding that true average melting point exceeds 181°C? Carry out a test of hypotheses using a significance level of .05.
=+ A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the Unless otherwise noted, all content on this page is © Cengage Learning.population distribution of PHB melting points as
=+DSC to measure the melting point (in °C) of the polymer, which is the temperature for 99% completion of the fusion process.180.5 181.7 180.9 181.6 182.6 181.6 181.3 182.1 182.1 180.3 181.7 180.5
=+25. Poly(3-hydroxybutyrate) (PHB), a semicrystalline polymer that is fully biodegradable and biocompatible, is obtained from renewable resources. From a sustainability perspective, PHB offers many attractive properties though it is more expensive to produce than standard plastics. The authors
=+24. Reconsider the sample observations introduced in Exercise 15 in Chapter 2 on the required force (N)to cause initial cracks in a thin enclosure for a subdermally implanted biotelemetry device:2006.1 2065.2 2118.9 1686.6 1966.9 1792.5 Suppose the device will not be used unless the true average
=+23. Exercise 5 in Chapter 2 gave n 5 12 observations on daily energy demand readings (kW h) for remote telecommunications stations throughout Cameroon, from which the sample mean and sample standard deviation are 32.59 and 10.66, respectively. Suppose the investigators had believed a priori that
=+b. A normal quantile plot of the data was quite straight. Use the descriptive output to test the appropriate hypotheses.
=+a. What does the boxplot suggest about the status of the specification for true average coating weight?
=+22. The article “The Foreman’s View of Quality Control” (Quality Engr., 1990: 257–280) described an investigation into the coating weights for large pipes resulting from a galvanized coating process.Production standards call for a true average weight of 200 lb per pipe. The accompanying
=+21. The true average diameter of ball bearings of a certain type is supposed to be .5 in. A one-sample t test will be carried out to see whether this is the case.What conclusion is appropriate in each of the following situations?a. n 5 13, t 5 1.6, 5 .05b. n 5 13, t 5 21.6, 5 .05c. n 5 25,
=+d. What should be concluded if the hypotheses of part (a) are tested and t 5 23.6?
=+c. What conclusion is appropriate if the hypotheses of part (a) are tested, t 5 21.8, and 5 .01?
=+b. What conclusion is appropriate if the hypotheses of part (a) are tested, t 5 22.3, and 5 .05?
=+a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?
=+20. A certain pen has been designed so that true average writing lifetime under controlled conditions(involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal quantile plot of the resulting data
=+19. The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let denote the true average reflectometer reading for a new type of paint under consideration. A test of H0: 5 20 versus Ha: . 20 will be based on a random sample of size n from a normal
=+18. Give as much information as you can about the P-value of a t test in each of the following situations:a. Upper-tailed test, df 5 8, t 5 2.0b. Lower-tailed test, df 5 11, t 5 22.4c. Two-tailed test, df 5 15, t 5 21.6d. Upper-tailed test, df 5 19, t 5 2.4e. Upper-tailed test, df 5 5, t 5 5.0f.
=+c. The authors commented that in most cases the ALD is better than or on the order of 1.0. Does the data in fact provide strong evidence for concluding that true average ALD under these circumstances is less than 1.0? Carry out an appropriate test of hypotheses.
=+b. Is it plausible that ALD is at least approximately normally distributed? Must normality be assumed prior to testing hypotheses about true average ALD? Explain.
=+a. Summarize and describe the data.
=+A measure of the accuracy of the automatic region is the average linear displacement (ALD). The paper gave the following ALD observations for a sample of 49 kidneys (units of pixel dimensions).1.38 0.44 1.09 0.75 0.66 1.28 0.51 0.39 0.70 0.46 0.54 0.83 0.58 0.64 1.30 0.57 0.43 0.62 1.00 1.05 0.82
=+17. Automatic identification of the boundaries of significant structures within a medical image is an area of ongoing research. The article “Automatic Segmentation of Medical Images Using Image Registration: Diagnostic and Simulation Applications” (J. of Medical Engr. and Tech., 2005:
=+deviation of 4.8. The conduits were manufactured with the specification that true average penetration be at most 50 mils. Does the sample data indicate that specifications have not been met? State the relevant hypotheses, calculate the value of the appropriate z statistic, determine the
=+16. To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 35 specimens are buried in soil for an extended period. The maximum penetration (in mils) is then measured for each specimen, yielding a sample mean penetration of 52.7 and a sample standard
=+of the appropriate z statistic, determine the P-value, and state the conclusion for a significance level of .01.
=+15. A sample of 40 speedometers of a particular type is selected, and each speedometer is calibrated for accuracy at 55 mph, resulting in a sample mean and sample standard deviation of 53.87 and 1.36, respectively.Does this data suggest that the true average reading when speed is 55 mph is in
=+ What significance level and conclusion would you recommend?
=+b. What conclusion would be appropriate for a significance level of .05? A significance level of .01?
=+a. How can you tell from the output that the alternative hypothesis was not Ha: . 750?
=+unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using Minitab, resulting in the accompanying output:Variable N
=+14. Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement
=+b. What conclusion would be reached for a significance level of .05, and why? Answer the same question for a significance level of .10.
=+a. What hypotheses were tested?
=+Variable N Mean StDev SE Mean Z P-Value sil cont 32 0.8228 0.1894 0.0335 -0.81 0.42
=+13. It is specified that a certain type of iron should contain .85 gm of silicon per 100 gm of iron(.85%). The silicon content of each of 32 randomly selected iron specimens was determined, and the accompanying Minitab output resulted from a test of the appropriate hypotheses:
=+12. Newly purchased automobile tires of a certain type are supposed to be filled to a pressure of 34 psi. Let denote the true average pressure. A test of H0: 5 34 versus Ha: Þ 34 will be based on a large sample of tires so that the test statistic z 5 (x 2 34)y(s1n) will have approximately a
=+11. Let denote the true average reaction time to a certain stimulus. A test of H0: 5 5 versus Ha: . 5 will be based on a large sample size so that when H0 is true, the test statistic z 5 (x 2 5)y(sy1n)has approximately a standard normal distribution(the z curve). Determine the value of z
=+10. For each of the given pairs of P-values and significance levels, state whether H0 should be rejected.a. P@value 5 .084, 5 .05b. P@value 5 .003, 5 .001c. P@value 5 .048, 5 .05d. P@value 5 .084, 5 .10e. P@value 5 .039, 5 .01f. P@value 5 .017, 5 .10
=+9. For which of the given P-values will the null hypothesis be rejected when using a test with a significance level of .05?a. .001b. .021c. .078d. .047e. .156
=+8.a. Use the definition of a P-value to explain why H0 would certainly be rejected if P-value 5.0003.b. Use the definition of a P-value to explain why H0 would definitely not be rejected if P-value 5.350.
=+for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate.State the relevant hypotheses, and describe the type I and type II errors in the context of
=+7. A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed in an attempt to reduce warpage.The regular laminate will be used on one sample of specimens and the special laminate on another sample; the amount of warpage will
=+6. A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a true average compressive strength of more than 1300 KN/m2.The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. State the
=+if sample data strongly indicates a reduction in true average braking distance for the new design. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.
=+5. A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be 120 ft. It is proposed that the new design be implemented only
=+4. Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure, oil-filled submarine power cable, a company wants to see conclusive evidence that the population standard deviation of sheath thickness is less than .05 mm.What hypotheses should be
=+40, purchasers will complain because the fuses will have to be replaced too frequently, whereas if the average exceeds 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. After obtaining data from a sample of fuses, what null and alternative hypotheses
=+3. Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-amp fuses wants to make sure that the true average amperage at which its fuses burn out is indeed 40. If the average amperage is lower than
=+ is the dividing line between welds meeting specification or not doing so. Explain why it might be better to test the hypotheses H0: 5 100 versus Ha: . 100 rather than H0: 5 100 versus Ha: , 100.
=+2. To decide whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is to be selected and the strength of each weld (force required to break the weld) determined. Suppose a population mean strength of 100 lb/in2
=+1. State whether each of the following assertions is a legitimate statistical hypothesis and why:a. H: . 100b. H: x 5 45c. H: Þ 2.0d. H: s # .50e. H: 1 y 2 , 1f. H: x1 2 x2 5 25.0 g. H: , .01, where is the parameter of an exponential distribution used to model component lifetime h. H:
=+b. For 5 6.58 and df 5 15,t.025 and t.975 are(from Minitab) 4.1690 and 10.9684, respectively.Use this information to obtain a 95% CI for the 5th percentile of the modulus of elasticity distribution considered in Example 7.10.
=+a. Use the given information to obtain a formula for a 95% confidence interval for some particular percentile of a normal population distribution.
=+.025 and upper tail area .025, respectively, under the noncentral t curve with v df and noncentrality parameters (when 5 0,t.025 5 2t.975, since central t distributions are symmetric about 0).
=+the value of the noncentrality parameter ( 5 0 gives the central t distribution). The key result is that the variable t 5 x 2 y1n 2 (zpercentile)1n sy has a noncentral t distribution with df 5 n 2 1 and 5 (2zpercentile)1n. Let t.025.v, and t.975,v, denote the critical values that capture
=+92. The one-sample CI for a normal mean and PI for a single observation from a normal distribution were both based on the central t distribution. A CI for a particular percentile (e.g., the 1st percentile or the 95th percentile) of a normal population distribution is based on the noncentral t
=+b. A 1998 survey based on people’s own assessments revealed that 20% of all adult Americans consider themselves obese. Does the estimate of part (a) suggest that the 2002 percentage is more than 1.5 times the 1998 percentage? Explain.
=+a. Estimate the proportion of all American adults who are obese in a way that conveys information about the reliability and precision of the estimate.
=+91. Recent information suggests that obesity is an increasing problem in America among all age groups.The Associated Press (October 9, 2002) reported that 1276 individuals in a sample of 4115 adults were found to be obese (a body mass index exceeding 30;this index is a measure of weight relative
=+three different brands of tires with identical lifetime ratings—a store brand (1) and two national brands(2 and 3)—are selected, and the lifetime of each tire is determined, resulting in the following data:Sample Sample Sample standard Brand size mean deviation 1 40 38,376 1522 2 32 41,569
=+An estimated variance s 2n results from replacing the 2>s; by the s 2>s; n can then be standardized to obtain a z variable from which the confidence interval n 6 (z crit)s n is obtained. Suppose that samples of
=+are specified numerical constants. A point estimate of is n 5 a1x1 1 a2x2 1 a3x3. When the sample sizes are all large, n has approximately a normal distribution with variance 2 n 5 a2 1 ? 2 1n1 1 a2 2 ? 2 2n2 1 a2 3 ? 2 3n3
=+90. Suppose that samples of size n1, n2, and n3 are independently selected from three different populations. Let i and i(i 5 1, 2, 3) denote the population means and standard deviations, and consider estimating 5 a1 1 1 a2 2 1 a3 3, where the ai>s
=+c. Compare your results in parts (a) and (b).
=+b. Use the bootstrap method to find a 95% bootstrap interval for the population mean.
=+a. Compute the t-based confidence interval of Section 7.4.
=+the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk (“Neural Network Training on Human Body Core Temperature Data,” Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999):38.4 38.7
=+89. Nine Australian soldiers were subjected to extreme conditions that involved a 100-min walk with a 25-lb pack when the temperature was 40°C (104°F). One of them overheated (above 39°C) and was removed from
=+c. Predict the total coating layer thickness for a single electrode in a way that conveys information about precision and reliability.
=+b. Calculate and interpret a 95% CI for true average total coating layer thickness in all such electrodes.
=+ The article “High-Performance Wire Electrodes for Wire Electrical-Discharge Machining—A Review”(J. of Engr. Manuf., 2012: 1757–1773) gave the following sample observations on total coating layer thickness (in m) of eight wire electrodes used for WEDM:21 16 29 35 42 24 24 25
=+88. Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coated wires have been used to substantially increase the cutting speed and precision of the process. Coating on the
=+Would you use the one-sample t confidence interval to estimate the population mean and median? Estimate the population median percentage of oil left using the interval suggested in Exercise 84, and determine the corresponding confidence level.
=+87. The article cited in Exercise 86 also included the following data on percentage of oil remaining for the commutator bearings:71.02 86.49 81.14 84.89 87.42 84.49 82.09 80.97 69.80 89.29 86.10 86.80 83.41 60.56 88.80 86.41 86.19
=+bearing, and do so in a way that conveys information about the reliability and precision of the estimate. (Note: A normal quantile plot validates the necessary normality assumption.) Would you say that the population mean difference has been precisely estimated? Does it look as though
=+Pinion: 278 208 281 274 268 Calculate an estimate of the population mean difference between penetration for the commutator armature bearing and penetration for the pinion
=+for the commutator armature bearing, resulting in the following data:Motor: 1 2 3 4 5 6 Commutator: 211 273 305 258 270 209 Pinion: 226 278 259 244 273 236 Motor: 7 8 9 10 11 12 Commutator: 223 288 296 233 262 291 Pinion: 290 287 315 242 288 242 Motor: 13 14 15 16 17 Commutator: 278 275 210 272
=+86. The derailment of a freight train due to the catastrophic failure of a traction motor armature bearing provided the impetus for a study reported in the article “Locomotive Traction Motor Armature Bearing Life Study” (Lubrication Engr., Aug. 1997: 12–19). A sample of 17 high-mileage
=+d. What is P(y1 , xn11 , yn), and what does this say about the prediction level associated with the interval (y1, yn)? Determine the interval and associated prediction level for the curing time data given in Exercise 83.
=+c. What is P(xn11 , y1)? What is P(xn11 . yn)?
=+b. What is P(xn11 , x1 and xn11 , x2), that is, the probability that xn11 is the smallest of these three observations?
=+a. What is P(xn11 , x1)?
=+85. Suppose we have obtained a random sample x1,…, xn from a continuous distribution and wish to use it as a basis for predicting a single new observation xn11 without assuming anything about the shape of the distribution. Let y1 and yn denote the smallest and largest, respectively, of the n
=+f. Using the results of parts (d) and (e), what is P(y2 , , yn21)? What does this imply about the confidence level associated with the interval (y2, yn21)? Determine the interval and associated confidence level for the data given in Exercise 83.
=+? Hint: , y2 occurs if either all n of the observations exceed the median or all but one of the xi’s does.e. With yn21 denoting the second largest xi, what is P( . yn21)?
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