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Unformatted text preview: 260 CHAPTER 8 STATISTICAL INTERVALS FOR A SINGLE SAMPLE LargeSample Conﬁdence Interval for a Parameter V The largesample 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 ' LargeSample
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
812 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 81 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 85 gives 80% conﬁdence?
(c) What value of zap in Equation 85 gives 75% conﬁdence? 8%. Consider the onesided 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? 84. 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? 87. 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? 88. 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 814 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 twosided 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 twosided 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% twosided conﬁdence interval 8—18. Suppose that in Exercise 815 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 _.819. 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 twosided 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? :. 74036 millimeters variety of nitrogen fertilizer applications (kg N/ha), tempera (a) Construct a 99% twosided conﬁdence interval on the mammal1d total monthly ramfau (mm) The dam shown be
mean piston ring diameter. for wheat grown at Harper Adams 0’) Constmct a 99% lower00 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). 152 142 140 p 12.2 14.4 125
814. 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
twosided 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 a95% lowerconﬁdence 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 twosided 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 81.1. 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 82 CONFIDENCE lNTERVAL ON MEAN OF A NORMAL DISTRIBUTION, VARIANCE UNKNOWN 265 EXERCISES FOR SECTION 82 7 p 8.22. Find the values of the following percentiles: {0325115, ; . a 10.1030: 30.00125: and rename
'3.: Determine the rpercentile 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. 826. 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. 827. 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 endoflife 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. 829. 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 LegStrength Training Improves Cycling
Economy in Previously Untrained Men” (2005, Vol. 37,
PP. 131—136)] studied cycling performance before and after
Eight Weeks of legstrength training. Seven previously untrained
males performed legstrength 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 videogame 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% onesided upper conﬁdence interval for mean 8131?. 832. 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. 834. 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. 835. 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 86 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 826 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 underor 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. 844. 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 844. 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% twosided conﬁdence interval for a. 847. 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.
455460)] considered the use of electromyostimulation
(EMS) as a method to train healthy skeletal muscle. EMS ses
Sions cdnsisted of 30 contractions (4second duration, 85 Hz)
and Were carried out three times per week for three weeks on
17 ice hockey players. The 10meter 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". ' 849. 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 50year 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
PhoenixEncanto . 21.6 H
Weddell _ _ ' 21.7
Litchﬁeld ‘ r a ' 21.3
Laveen W _ _ 20.7
Maricopa r i _ ' '7 20.9
Harlquahala 20.1 —_______.————— 850. An article in Cancer Research [“Analyses of Litter
Matched TimetoResponse 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 84 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% twosided 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. 854. 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 twosided 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 results. There were 2020 respondents in the: exit polls and 768 were college grad— uates. 0f the college graduates, 412 voted for George Bush. (3.) Calculate a 95% conﬁdence interval for the proportion of
college graduates in Ohio who voted for George Bush. (h) Calculate a 95% lower conﬁdence bound for the proportion
of college graduates in Ohio who voted for George Bush. 856. Of 1000 randomly selected cases of lung cancer, 823
resulted in ‘death within 10 years. ’
(3.) Calculate a 95% twosided conﬁdence interval on the death rate from lung cancer. _
(b) Using the point estimate of p obtained from the prelimi nary sample, what sample size is needed to be 95% conﬁ dent that the error in estimating the true value of p is less (0) How large must the sample be if we wish to be at least
95% conﬁdent that the error in estimating p is less than
0.03, regardless of the true value of p? 8—57. An article in the Journal of the American Statistical Association (1990, Vol. 85, pp. 972—985) measured the weight of 30 rats under experiment controls. Suppose that there are 12 underweight rats. (3) Calculate a 95% twosided conﬁdence interval on the true
proportion of rats that would show underweight from the
experiment. . (b) Using the point estimate of p obtained from the prelimi—
nary sample, what sample size is needed to be 95% conﬁ
dent that the error in estimating the true value of p is less
than 0.02? (c) How large must the sample be if we wish to be at least
95% conﬁdent that the error in estimating p is less than
0.02, regardless of the true value of p? A random sample of 50 suspension helmets used by otorcycle riders and automobile racecar drivers was sub jected to an impact test, and on 18 of these helmets some dam age was observed. (a) Find a 95% twosided conﬁdence interval on the true pro
portion of helmets of this type that would show damage
from this test. (b) Using the point estimate of p obtained from the prelimi
nary sample of 50 helmets, how many helmets must be
tested to be 95% conﬁdent that the error in estimating the
true value of p is less than 0.02? (c) How large must the sample be if we Wish to be at least
95% conﬁdent that the error in estimating p is less than
0.02, regardless of the true value of p? 8:59. The Arizona Department of Transportation
wishes to survey state residents to determine what propor—
tion of the population would like to increase statewide
highway speed limits to 75 mph from 65 mph. How many
residents do they need to survey if they want to be at least
99% conﬁdent that the sample proportion is within 0.05 of
the true proportion? 8—60. A study is to be conducted of the percentage of home—
owners who own at least two television sets. How large a
sample is required if we wish'to be 99% conﬁdent that the
error in estimating this quantity is less than 0.017? 857 GUIDELINES FOR CONSTRUCTING CONFIDENCE INTERVALS The most difﬁcult step in constructing a co
priate calculation to the objective of the study. nﬁdence interval is often the match of the appro—
Common cases are listed in Table 81 along with the reference to the section that covers the appropriate calculation for a conﬁdence interval test. ‘31.; 4,... ..__. axial“;  Lugre 9" ' set remerﬁnaetneriesesmrn a—az .W'ﬁr': rue raIr —r— 
I . r v‘ Ar. r «Pt—fi .4352, mm iv"éwrﬁﬁeﬁqwmrimrwﬂbs; ewerm. “NW—‘1‘ "ertv‘rﬂt~=;=vv.—m:1nw—.rta—p. 35*: _—, :n':"'—.—.: 'i'l_'  the next specimen of PVC pipe tested. Compare the length of
the prediction interval with the length of the 99% Cl on the
population mean. 8’63. Consider the syrupdispensing measurements de
scribed in Exorcise 829. Compute a 95% prediction interval
on the syrup volume in the next beverage dispensed. Compare
the length of the prediction interval with the length of the 95%
CI on the population mean. 864. Consider the natural frequency of beams described
in Exercise 8—32. Compute a 90% prediction interval on the
diameter of the natural frequency of the next beam of this
type that will be tested. Compare the length of the prediction
interval with the length of the 90% CI on the population 1116 an. 7 Consider the rainfall in Exercise 8—33. Compute a
9 prediction interval on the rainfall for the next year. Compare the length of the prediction interval with the length
. of the 95% Cl on the population mean. 866. Consider the margarine test described in Exercise 836. Compute a 99% prediction interval on the polyunsaturated fatty acid in the next package of margarine that is tested. Compare the length of the prediction interval with the length ' of the 99% or on the population mean. 867. Consider the television tube brightness test described
in Exercise 835. Compute a 99% prediction interval 011 the
brightness of the next tube tested. Compare the length of the
prediction interval with the length of the 99% Cl on the popu
lation mean. 868. Consider the suspension rod diameter measurements
described in Exercise 838. Compute a 95% prediction inter
val on the diameter of the next rod tested. Compare the length
of the prediction interval with the length of the 95% CI on the
population mean. 869. Consider the test on the compressive strength of con
crete described in Exercise 8—37. Compute a 90% prediction
interval on the next specimen of concrete tested. 8—70. Consider the bottlewall thickness measurements
described in Exercise 840. Compute a 90% prediction interval
011 the wall thickness of the next bottle tested. 8—71. Consider the fuel rod enrichment data described
in Exercise 841. Compute a 90% prediction interval on the
enrichment ofthe next rod tested. Compare the length of
the prediction interval with the length of the 99% CI on the'
population mean. 8—72. How would'you obtain a onesided prediction bound
on a future observation? Apply this procedure to obtain a 95%
Onesided prediction bound on the wall'thickness of the next bottle for the situation described in Exercise 840.
8‘73. Consider the tiretesting data in Exercise 827.
 Compute a 95% tolerance interval on the life of the tires that has conﬁdence level 95%. Compare the length of the toler
ance interval with the length of the 95% CI on the population 8.6 TOLERANCE AND PREDICTION INTERVAis 27 7 mean. Whichinterval is shorter? Discuss the difference in
interpretation of these two intervals. 8’74. Consider the Izod impact test described in Exercise
8—28. Compute a 99% tolerance interval on the impact
strength of PVC pipe that has conﬁdence level 90%. Compare
the length ofthe tolerance interval with the length of the 99%
Cl on the population mean. Which interval is shorter? Discuss
the difference in interpretation of these two intervals. 8—75. Consider the syrupvolume data in Exercise 829.
Compute a 95% tolerance interval on the syrup volume that
has conﬁdence level 90%. Compare the length of the tolerance
interval with the length of the 95% CI on the population mean. 876. Consider the margarine test described in Exercise 836.
Compute a 99% tolerance interval on the polyunsaturated
fatty acid in this particular type of margarine that has conﬁ
dence level 95%. Compare the length of the tolerance inter
val with the length of the 99% CI on the population mean.
Which interval is shorter? Discuss the difference in inter— pretation of these two intervals. , 8—77. Consider the rainfall data in Exercise 8—33. Compute a
95% tolerance interval that has conﬁdence level 95%. Compare
thelength of the tolerance interval with the length of the 95% CI
on the population mean. Discuss the difference in interpretation
of these two intervals; ' 8—7 8. Consider the suspension rod diameter data in Exercise 838. Compute a 95% tolerance interval on the diameter of the '
rods described that has'90% conﬁdence. Compare the length
of the tolerance interval with thelength of the 95% C1 on the population mean. Which interval is shorter? Discuss the dif ference in interpretation of these two intervals. 879. Consider the television tube brightness data in
Exercise 835. Compute a 99% tolerance interval on the
brightness of the television tubes that has conﬁdence level
95%. Compare the length of the tolerance interval with the
length of the 99% Cl on the population mean. Which interval
is shorter? Discuss the difference in interpretation of these two intervals.
8430. Consider the strengthofconcrete data in Exercise 837. Compute a 90% tolerance interval on the compressive
strength of the concrete that has 90% conﬁdence. 8:81. Consider the fuel rod enrichment data described in
Exercise 841. Compute a 99% tolerance interval on rod
enrichment that has confidence level 95%. Compare the
length of the tolerance interval with the length of the 95% CI on the population mean. Consider the bottlewall thickness measurements
ascribed in Exercise 840. (a) Compute a 90% tolerance interval on bottle—wall thick ness that has conﬁdence level 90%. (b) Compute a 90% lower tolerance bound on bottlewall
thickness that has conﬁdence level 90%. Why would a lower tolerance bound likely be of interest here? ...
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This note was uploaded on 03/13/2012 for the course ISE 130 taught by Professor Patel,n during the Spring '08 term at San Jose State University .
 Spring '08
 Patel,N

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