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HW#9 - 260 CHAPTER 8 STATISTICAL INTERVALS FOR A SINGLE...

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Unformatted text preview: 260 CHAPTER 8 STATISTICAL INTERVALS FOR A SINGLE SAMPLE LargeSample Confidence Interval for a Parameter V The large-sample confidence interval for p. in Equation 8—11 is a special case of a more general result. Suppose that 3 is a parameter of a probability distribution, and let 6) be an estimator of - 0. If (:9 (1) has an approximate normal distribution, (2) is approximately unbiased for 6, and (3) has standard deviation (n23. that can be estimated from the sample data, then the quantity (0) ‘— 0)/U'é has an approximate standard normal distribution. Then a large~sarnp1e approximate CI for 0 is given by 7 ' Large-Sample Approximate Confidence Interval 0 — VZu/Z'U‘é) S g + ZQ/zfl‘é Maximum likelihood estimators usually satisfy the three conditions listed above, so Equation 8-12 is often used when (3) is the maximum likelihood estimator of 0. Finally, note that Equation 8—12 can be used even when 09 is a function of other unknown parameters (or of 0). Essentially, all one does is to use the sample data to compute estimates of the unknown parameters and substitute those estimates into the expression for erg. EXERCISES FOR SECTION 8-1 8—1. For a normal population with known variance oz, answer the following questions: 7 (a) What is the confidence level for the interval 2 — 2140/ Vii , s ,1. if + 2.14a/Vfi? (b) Whatis the confidence level for the interval)? * 2.49o/V17t‘ ,- s a s r + awn/Vi? (c) What is the confidence level for the interval 3? - 1.850/W s a 52 + ism/Vi? 8,2. For a normal population with known variance 01: (a) What value of 2m in Equation 8—5 gives 98% confidence? (b) What value of 2.1,; in Equation 8-5 gives 80% confidence? (c) What value of zap in Equation 8-5 gives 75% confidence? 8%. Consider the one-sided confidence interval expressions for a mean of a normal population. (a) What value of 2“ would result in a 90% CI? (b) What value of 2., would result in a 95% Cl? (C) What value ofzu would result in a 99% CI? 8-4. A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is nor- mally distributed with stande deviation 0' : 20. (a) Find a 95% CI for u when n = 10 and J”: = 1000. (b) Find a 95% CI for u. when n = 25 and E = 1000. (c) Find a 99% CI for p. when )1 =10 and} = 1000. ’ ((1) Find a 99% CI for p. when n = 25 and i = 1000. (e) How does the length of the CIS computed above change with the changes in sample size and confidence level? 85. A random sample has been taken from a normal distri— bution and the following confidence intervals constructed us- ing the same data: (38.02, 61.98) and (39.95, 60.05) . (a) What is the value of the sample mean? (b) One of these intervals is a 95% CI and the other is a 90% CI. Which one is the 95% CI and why? A random sample has, been taken from a normal dishi- ution and the following confidence intervals constructed us- ing the same data: (37.53, 49.87) and (35.59, 51.81) (a) What is the value of the sample mean? (5) One of these intervals is a 99% CI and the other is a 95% CI. Which one is the 95% CI and why? 8-7. Consider the gain estimation problem in Exercise 8—4. (a) How large must it be if the length of the 95% Cl is to be 40? (b) How large must n be if the length of the 99% CI is to he 40? 8-8. Following are two- confidence interval estimates of the mean u. of the cycles to failure of an automotive door latch mechanism (the test was conducted at an elevated stress level to accelerate the failure): 3124.9 2 p 5 3215.7 3110.5 5 p. 5 3230.1 (a) What is the value of the sample mean cycles to failure? (b) The confidence level for one of these C15 is 95% and the confidence level for the other is 99%. Both Cls are calcu— lated from the same sample data. Which is the 95% CI? Explain why. 8—9. Suppose that n = 100 random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A 95% CI on the mean cal- cium concentration $70.49 5 u. E 0.82. (a) Would a 99% C1 calculated from the same sample data be longer or shorter? ISTRIBUTION, VARIANCE UNKNOWN 261 l 8—2 CONFIDENCE INTERVAL ON THE MEAN OF A NORMAL D (a) Construct a 95% two—sided confidence interval on mean (19) Consider the following statement: There is a 95% chance that p. is between 0.49 and 0.82. ls this statement correct? compressive strength. (b) Construct a 99% two— Explain your answer. der the following statement: If n = 100 random compressive strength. Compare the width of this confi- t‘ the one found in part (a). (c) COflSl samples of water from the lake were taken and the 95% C1 .o ence interval with the width 0 use that in Exercise 8-14 we wanted the error in on p. computed, and this process were repeated 1000 Supp times! 950 Gfme C15 would 9011mm the true Value Of ll» 15 mating the mean life from the two-sided confidence inter— this statement correct? Explain your answer. va1 to be five hours at 9 8.10. Past experience has indicated that the breaking Sho‘fld be “sad? 8—17. Suppose that in Exercise 8—14 we wanted the total strength of yarn used in manufacturing drapery material is normally distributed and that a = 2 psi, A random sample of width of the two-sided confidence interval on mean life to be nine specimens is tested. and the average breaking strength is sixhours at 95% confidence. What sample size should be used? found to be 98 psi. Find a 95% two-sided confidence interval 8—18. Suppose that in Exercise 8-15 it is desired to estimate an error that is less than 15 psi on the true mean breaking strength. the compressive strength with at 99% confidence. What sample size is required? 8—11. The yield of a chemical process is being studied. From I _ _ previous emafleme, yield is know to be normally dismbuted _.8-19. By how much must the sample size n be increased if 7' the length of the CI on u. in Equation 8—5 is to be halved? and 0' = 3. The past five days of plant operation have resulted in 840. If the sample size n is doubled, by how much is the ‘. 91.6, 88.75, 90.8, 89.95, and 91.3. film—val on the true mean yield. length of the CI on p. in Equation 8—5 reduced? What happens to the length of the interval if the sample size is increased by a 8—12. The diameter of holes for a cable harness is known to . . . . . factor of four? . ve anormal distribution With 0' = 0.01 111011. A random sample _ . . I 8’21. An article in the Journal of Agricultural SCLBI’ICEL’ [“The r' 10 'lds d‘ t £15045; as d 990/ ° “26 We ma‘mge 13mm “1° m a “ Use ofResidual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen two-sided confidence interval on the mean hole diameter. 8‘13: A manufacture produces piston rings for an auto- Fertilizer Eifects” (1997, Vol. 123, pp. 1357142)} investigated “109113 engme- It 15 known that ring dlametef 13 normally (115' means of wheat grain crude protein content (CP) and Hagberg tributed with o i 0.001 millimeters. A random sample of 15 falling number (HFN) surveyed in the UK The analysis used a [51185 has a mean diameter 0f? :. 74-036 millimeters- variety of nitrogen fertilizer applications (kg N/ha), tempera- (a) Construct a 99% two-sided confidence interval on the mammal-1d total monthly ramfau (mm) The dam shown be- mean piston ring diameter. for wheat grown at Harper Adams 0’) Constmct a 99% lower-00 Agricultural College between 1982 and 1993. The tempera— tures measured in June Were obtained as follows: sided confidence interval on mean 5% confidence. What sample size nfidence bound on the mean e the lower bound of this piston ring diameter. Compar confidence 1nterva1w1ti1 the one in part (a). 15-2 142 140 p 12.2 14.4 125 8-14. The life in hours of a 75—watt light bulb is known to be 143 142 13.5 11.8 15.2 hams“ A random sample 0f Assume that the standard deviation is known to be u = 0.5. ' annually distributed with o = 25 two-sided confidence interval on the Egbglbs has a mgapl/life of 2-:(1101igicpurs. . 1 h (a) Construct a 99% moralitrlppct a o tWO—Sl e co ence interva on t e_ mean temperatme' r e l e. Co tru ta 95°/ low — o fiden e b and on the mea 0)) Construct a-95% lower-confidence bound on the mean life. (b) tefisemzna 0 at C n c 0 n anted to be 95% confident that the error Compare the lower bound or this confidence interval wrth (c) Suppose that we W in estimating the mean temperature is less than 2 degrees the one in part (a). Celsius. What sample size should be used? Ev15. A civil engineer is analyz 0f Concrete. Compressive strength is ndrrnally distributed with ((1) Suppose that we wanted the total width of the two-sided con— 02 = 1000(psi)2. A random sample of .12 specimens has a fidence interval on mean temperature to be 1.5 degrees Celsius at 95% confidence. What sample size shouldbe used? mean compressive strength of i = 3250 psi. ing the compressive strength 8'2 CONFIDENCE INTERVAL ON THE MEAN OF A NORIVIAL ; DISTRIBUTION, VARIANCE UNKNOWN ' intervals on the mean u of a normal population when e in Section 8-1.1. This C1 is also approximately valid ss of whether or not the underlying population When we are constructing confidence 02 is known, we can use the procedur (because of the central limit theorem) regardle 8-2 CONFIDENCE lNTERVAL ON MEAN OF A NORMAL DISTRIBUTION, VARIANCE UNKNOWN 265 EXERCISES FOR SECTION 8-2 7 p 8.22. Find the values of the following percentiles: {0325115, ; -. a 10.1030: 30.00125: and rename- '3.: Determine the r-percentile that is required to construct of the following two—sided confidence intervals: (3) Confidence level = 95%, degrees of freedom = 12 (b) Confidence level = 95%, degrees of freedom = 24 (c) Confidence level = 99%, degrees of freedom = 13 (d) Confidence level = 99.9%, degrees of freedom = 15 8’24. Determine the t—percentile that is required to construct each of the following one—sided confidence intervals: (a) Confidence level = 95%, degrees of freedom = 14 (b) Confidence level = 99%, degrees of freedom = 19 i (c) Confidence level = 99.9%, degrees of freedom = 24 825. A random sample has been taken from a normal dis- tribution. Output from a software package is given below: Variable N Mean SEMean StDev Variance Sum x 10 ? 0.507 1.605 7 251.848 (a) Fill in the missing quantities. (b) Find a 95% C1 on the population mean. 8-26. A random sample has been taken from a normal dis- tribution. Output from a software package is given below: Variable N Mean SE Mean StDev Variance Sum x ? ‘? _ 1.58 6.11 7 751.40 (3.) Fill in the missing quantities. (b) Find a 95% CI on the population mean. 8-27. A research engineer for a tire manufacturer is investi- gating tire life for a new rubber compound and has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are 60,1397 and 3645 .94 kilometers. Find a 95% confidence interval on mean the life. . 8—28. An Izod impact test was performed on 20 specimens of PVC pipe. The sample mean is it = 1.25 and the sample stan- dard deviation is s = 0.25. Find a 99% lower confidence bound on Izod impact strength. 8-29. A postmix beverage machine is adjusted to release a certain amount of syrup into a chamberwhe‘re it is mixed with i » carbonated water. A random sample of 25 beverages was . found to have a mean syrup content of E = 1.10 fluid ounce and a standard deviation of s = 0.015 fluid ounce. Find a 95% CI on the mean volume of syrup dispensed. 8'30‘ An article in Medicine and Science in Sports and Exercise [“Maximal Leg-Strength Training Improves Cycling Economy in Previously Untrained Men” (2005, Vol. 37, PP. 131—136)] studied cycling performance before and after Eight Weeks of leg-strength training. Seven previously untrained males performed leg-strength training three days per week for eight Weeks (with four sets of five replications at 85% of one Eepfitition maximum). Peak power during incremental cycling increased to a mean of 315 watts with a standard deviation of 16 watts. Construct a 95% confidence interval for the mean peak power after training. 8’3 1. An article in Obesity Research [“Impaired Pressure Nah'iuresis in Obese Youths” (2003, Vol. 11, pp. 745—75 1)] de- scribed a study in which all meals were provided for 14 lean boys for three days followed by one stress (with a video-game task). The average systolic blood pressure (SBP) during the test was 118.3 mm HG with a standard deviation of 9 .9 mm HG. Construct a 99% one-sided upper confidence interval for mean 8131?. 8-32. An article in the Journal of Composite Materials - (December 1989, Vol. 23, p. 1200) describes the effect of delam— ination on the natural frequency of beams made from composite laminates. Five such delarnjnated beams were subjected to I loads, and the resulting frequencies were as follows (in hertz): 230.66, 233.05, 232.58, 229.48, 232.58 Check the assumption of normality in the population. Calculate a ' % two— sided confidence interval on mean natural frequency. @ The Bureau of Meteorology of the Australian Government provided the mean annual rainfall (in milli- meters) inAustralia 19834002 as follows (http://wwwbomgov. . au/climate/change/rain03.txt): 7 499.2, 555.2, 398.8, 391.9, 453.4, 459.8, 483.7, 417.6, 469.2, 452.4, 499.3, 340.6, 522.8, 469.9, 527.2, 565.5, 584.1, 727.3, 558.6, 338.6 Check the assumption of normality in the population. Construct a 95% confidence interval for the mean annual rainfall. 8-34. The solar energy consumed (in trillion BTU) in the U.S. by year from 1989 to 2004 (source: U.S. Department of Energy Web site, hfipzlfwmv.eia.doe.gov/emeu) is shown in the table below. Read down, then right for year. ' are ._ 63.886 70.833 66.3 88 63.287 _______________——-——-— Check the assumption of normality in the population. Construct a 95% confidence interval for the mean solar energy consumed. 8-35. The brightness of a television picture tube can be eval- uated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in :7: = 317.2 and s = 15.7. Find (in microarnps) a 99% confi- dence interval 011, mean current required. State any necessary assumptions about the underlying distribution of the data. 836. A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percent- ages). A sample of six packages resulted in the following data: 16.8,17.2,17.4,16.9,16.5,17.1. ‘ (21) Check the assumption that the level of polyunsaturated fatty acid is normally distributed. 5.3 CONFIDENCE INTERVAL ON THE VARIANCE AND STANDARD DEVIATION OF A NORMAL DISTRIBUTION =_ EXAMPLE 8-6 Detergent Filling An automatic filling machine is used to fill bottles with liquid ,3 detergent. A random sample of 20 bottles results in a sample 5 variance of fill volume of s2 = 0.0153 (fluid ounce)? If the variance of fill volume is too large, an unacceptable proportion of bottles will be under— or overfilled. We will assume that the fill volume is approximately normally distributed. A 95% up- per confidenCe bound is found from Equation 8-26 as follows: i: _, 2 02 S (n 2 1).? ; Xo.95,l9 or 19 0.0153 02 s Q— = 0.0287 (fluid ounce)2 10.117 269 This last expression may be converted into a confidence inter- val on the standard deviation 0' by taking the square root of both sides, resulting in (r S 0.17 Practical Interpretation: Therefore, at the 95% level of confi- dence, the data indicate that the process standard deviation could be as large as 0.17 fluid ounce. The process engineer or manager now needs to determine if a standard deviation this large could lead to an operational problem with under-or over filled bottles. EXERCISES FOR SECTION 86 @ Determine the values of the following percentiles: X0.05,10! Xhozsass Xinmz: X03520: Kiwis, X0.995,§6= 311d thosas- 8'43. Determine the x2 percentile that is required to construct each of the following 01s: (a) Confidence level = 95%, degrees of freedom = 24, one—sided (upper) ' (b) Confidence level = 99%, degrees of freedom = 9, one- sided (lower) (c) Confidence level = 90%, degrees of freedom : 19, two— sided. 8-44. A rivet is to be inserted into a hole. A random sample of n = 1.5 parts is selected, and the hole diameter is measured. The sample standard deviation of the. hole diameter measure- ments is s = 0.008 millimeters. Construct a 99% lower confi— dence bound for cl. ' 8—45. Consider the situation in Exercise 8-44. Find a 99% 1 er confidence bound on the standard deviation. The sugar content of the syrup in canned peaches is a sample standard deviation of s = 4.8 milligrams. Calculate a 95% two-sided confidence interval for a. 8-47. The percentage of titanium in an alloy used in aero- space castings is measured in 51 randomly selected parts. The Sample standard deviation is s = 0.37. Construct a 95% two- sided confidence interval for or. 8—48. An article in Medicine and Science in Sports and Exercise [“Electrostimulation Training Effects on the Physical Performance of Ice Hockey Players” (2005, Vol. 37, pp. 455-460)] considered the use of electromyostimulation (EMS) as a method to train healthy skeletal muscle. EMS ses- Sions cdnsisted of 30 contractions (4-second duration, 85 Hz) and Were carried out three times per week for three weeks on 17 ice hockey players. The 10-meter skating performance test Showed a standard deviation of 0.09 seconds. Construct a 95% B ‘ ally distributed. A random sample of n = 10 cans yields ' confidence interval of the standard deviation of the skating performance test". ' 8-49. An article in Urban Ecosystem, “Urbanization and Warming of Phoenix (Arizona, USA): Impacts, Feedbacks and Mitigation” (2002, Vol. 6, pp. 183—203), mentions that Phoenix is ideal to study the effects of an urban heat island because it has grown from a population of 300,000 to approximately 3 million over the last 50 years and this is a period with a continuous, de— tailed climate record. The 50-year averages of the mean annual temperatures at eight sites in Phoenix are shown below. Check the assumption of normality in the population with a probabil- ity plot. Construct a 95% confidence interval for the standard deviation over the sites of the mean annual temperatures. Average Mean Site Temperature (°C) HarborAirport _' ' 23.3 Phoenix Greenway __ ‘ 21 ._7 Phoenix-Encanto . 21.6 H Weddell _ _ ' 21.7 Litchfield ‘ r a ' 21.3 Laveen W _ _ 20.7 Maricopa r i _ ' '7 20.9 Harlquahala 20.1 —_______.-———-—— 8-50. An article in Cancer Research [“Analyses of Litter- Matched Time-to-Response Data, with Modifications for Recovery of Interlitter Information” (1977, Vol. 37, pp. 3863—3868)] tested the tumorigenesis of a drug. Rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows: 101, 104, 104, 77, 89, 88, 104, 96, 82, 70, 89, 91, 39, 103, 93, 85, 104, 104, 81, 67, 104, 104, 104, 87, 104, 89, 78, 104, 86, 76, 103, 102, 80, 45, 94, 104, 104, 76, 80, 72, 73 8'5 GUlDELlNES FOR CONSTRUCTING CONFIDENCE lNTERVALS 273 Approximate One— a 1 -i - Sided Confidence . Bounds on a -' Binomial Proportion ‘ Thahpproxhnate or)%:rlower' and'upperlcdnfidence bpundg are: 7' r 1 EXERCISES FOR SECTION 8-4 8.53 . The fraction of defective integrated circuits produced 7 in aphotolithography process is being studied. A random sam- ple of 300 circuits is tested, revealing l3 defectives. (3.) Calculate a 95% two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a 95% upper confidence bound on the fraction of defective circuits. 8-54. An article in Knee Surgery, Sports Tmumatology, Arthroscopy [“Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique” (2005, Vol. 13, pp. 273—279)] showed that only 25 out of 37 tears (67.6%) located between 3 and 6 mm from the meniscus rim Were healed. (a) Calculate a two-sided 95% confidence interval on the proportion of such tears that will heal. (b) Calculate a 95% lower confidence bound on the propor- tion of such tears that will heal. 8—55 . The 2004 presidential election exit polls from the crit- ical state of Ohio provided the following r...
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