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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 r...
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