Download Interval estimation

IntroductiontoEstimation
Basicdefinitionsandconcepts
Theassignmentofvalue(s)toapopulationparameter
basedonavalueofthecorrespondingsamplestatisticis
called estimation
calledestimation.
POINT&INTERVALESTIMATION
AND
(INTRODUCTIONTOTESTING)
Thevalue(s)assignedtoapopulationparameterbasedon
The
value(s) assigned to a population parameter based on
thevalueofasamplestatisticiscalledanestimate.
Thesamplestatisticusedtoestimateapopulation
parameteriscalledanestimator.
2
Estimationsteps
PointEstimators
y APointEstimation
Theestimationprocedureinvolvesthefollowingsteps:
1
1.
2.
3.
4.
Thevalueofasamplestatisticthatisusedtoestimatea
The
value of a sample statistic that is used to estimate a
populationparameteriscalledapointestimate.Usually,
wheneverweusepointestimation,wecalculatethemarginof
error associated with that point estimation which is
errorassociatedwiththatpointestimation,whichis
calculatedasfollows: Margin of error r1.96V x or r 1.96s x
Pointestimateisbasedonjustonesample,wecannot
expectittobeequaltothecorrespondingpopulation
parameter.Indeed,eachsamplewillhaveadifferent,
non of them is equal to P. But they are all unbiased
nonofthemisequaltoP.Buttheyareallunbiased
estimatesofP.(Recallthatunbiased meanstheir
expectedvalueisequaltoP.)
Selectasample.
Select
a sample
Collecttherequiredinformationfromthemembersof
the sample.
thesample.
Calculatethevalueofthesamplestatistic.
Assignvalue(s)tothecorrespondingpopulation
g
()
p
gp p
parameter.
3
4
Statistics
PointEstimates
y Statistics
y Parameters
¾ Astatisticisapropertyofasamplefromthe
¾ Instatisticalinference,thetermparameter
isusedtodenote
aquantity,say,thatisapropertyofanunknownprobability
T
distribution.
¾ Forexample,themean,variance,oraparticularquantile
p ,
,
,
p
q
of
theprobabilitydistribution
¾ Parametersareunknown,andoneofthegoalsofstatistical
inference is to estimate them
inferenceistoestimatethem.
y Estimation
¾ Aprocedureof“guessing”propertiesofthepopulationfrom
which data are collected
whichdataarecollected.
¾ Apointestimateofanunknownparameterisastatisticthat
representsa“guess”oftheparameterofinterest.
¾ Theremaybemorethanonesensiblepointestimateofa
There may be more than one sensible point estimate of a
parameter.
population
population.
¾ Astatisticisdefinedtobeanyfunctionofrandom
variables.So,itisalsoarandomvariable.For
example,thesamplemean,samplevariance,ora
particularsamplequantile.
¾ Theobservedvalueofthestatisticcanbecalculated
The observed value of the statistic can be calculated
fromtheobserveddatavaluesofrandomvariables.
Examples of statistics:
X1 X 2 X n
sample mean X
n
sample variance S 2
5
¦
n
i 1
( X i X )2
n 1
6
Therelationshipbetweenanunknown
parameteranditspointestimator
PropertiesofEstimatorsthatWeDesire
y Unbiasedness:
U bi d
ˆ
E( T)
T
Inotherwordswewouldwishthattheexpectedvalueof
theestimatoristhesameasitstruevalue.
Wedefinebiasofanestimator asthedifferencebetween
theexpectedvalueoftheestimatorandthetruevaluein
p p
thepopulation:
y Efficiency:wewishtominimizethemeansquareerror
aroundthetruevalue.Theefficiencytellsushowwellthe
estimatorperformsinpredicting.Amongunbiased
estimatorstherefore,wewanttheonewiththesmallest
variance.
y
7
Consistency.Assamplesizeincreases,variationofthe
estimatorfromthetruepopulationvaluedecreases.
8
Unbiasedness
Efficiency
P(X)
Unbiased
P(X) Sampling
Biased
Distribution
of Median
P
Sampling
Distribution of
Mean
P
9
10
Consistency
Intervalestimation:
General approach
Generalapproach
Larger
sample size
P(X)
B
Smaller
sample size
A
P
11
12
IntervalEstimation
ConfidenceIntervalEstimation
Definition
In interval estimation an interval is constructed around
Inintervalestimation,anintervalisconstructedaround
thepointestimate,anditisstatedthatthisintervalislikely
tocontainthecorrespondingpopulationparameter.
y Outline:
y Procedure:
y 1. Sample o point estimator (
X or p )
y 2.
2 Confidence level and Table o Z or tn-11
y 3. Formulas o compute UCL and LCL:
y
Px
13
P
$1130
x
point estimator r margin
p
g of error
$1370
$1610
IntervalEstimationofthePopulationMean
Intervalestimationofapopulation
mean: The case of known mean:Thecaseofknown
Eachintervalisconstructedwithregardtoagivenconfidencelevel
andiscalledaconfidenceinterval.
The confidence level associated with a confidence interval states how
Theconfidencelevelassociatedwithaconfidenceintervalstateshow
muchconfidencewehavethatthisintervalcontainsthetrue
populationparameter.Theconfidencelevelisdenotedby(1–
)100%.
)
%
The(1– )100%confidenceintervalfor (populationmean)is:
x r zV x if V is known and x r zs x if V is not known,
where Vx V / n and sx s / n
Thevalueofz usedherecanbefoundfromthestandardnormal
distributiontable,forthegivenconfidencelevel.
Themaximumerrorofestimatefor,denotedbyE,isthequantity
that is subtracted from and added to the value of x toobtaina
thatissubtractedfromandaddedtothevalueofx
to obtain a
confidenceintervalfor.Thus, E zV x or zs x
17
16
IntervalEstimationofthePopulationMean
when isknown:Example
when
is known: Example
Apublishingcompanyhasjustpublishedanewcollegetextbook.
p y
p
Beforethecompanydecidesthepriceatwhichtosellthis
textbook,itwantstoknowtheaveragepriceofallsuch
textbooksinthemarket.Theresearchdepartmentatthe
company took a sample of 36 comparable textbooks and
companytookasampleof36comparabletextbooksand
collectedinformationontheirprices.Thisinformationproduces
ameanpriceof$70.50forthissample.Itisknownthatthe
standarddeviationofthepricesofallsuchtextbooksis$4.50.
d dd i i
f h
i
f ll
h
b k i $4 50
(a) Whatisthepointestimateofthemeanpriceofallsuch
textbooks? What is the margin of error for the estimate?
textbooks?Whatisthemarginoferrorfortheestimate?
(b) Constructa90%confidenceintervalforthemeanpriceofall
suchcollegetextbooks.
18
IntervalEstimationofthePopulationMeanwhen is
known:AnswerstotheExample
x
Herewetakeadvantageofourknowledgeondistributionofto
developaconfidenceintervalforP.
a)
n=36,x =$70.50,and =$4.5,thus: V x V 4.50 $.75
n
36
Pointestimateof =x =$70.50
( 75) r$1.47
Margin of error = r 1.96V x r1.96(.
Marginoferror=
b)
Confidencelevelis90%or.90;andz =1.65.
x r zV x
70.50 r 1.65(.
( 75)
70.50 r 1.24
(70.50 - 1.24) to (70.50 1.24)
$69.26 to $71.74
Basedonourresults,wecansaythatweare90%confidentthat
the mean price of all such college textbooks is between
themeanpriceofallsuchcollegetextbooksisbetween
$69.26and$71.74.
19
1. Interval Estimation for Population
p
Mean
Example 1: ( known case)
In an effort to estimate the mean amount spent per customer
for dinner at a major Atlanta restaurant. Data were collected
for a sample of 49 customers over a three-week period.
Assume a population standard deviation of $5.
$5
a. At the 95% confidence, what is the margin error?
b. If the sample mean is $24.80, What is the 95%
confidence interval for the population mean?
Example 1:
Answer:
•
n = 49
X
$24.8
V=$5
V
$5
• Z: (1-D)/2 = 0.95/2 = 0.475 o Table 1: Z = 1.96
V
5
• 1. ZD / 2 x (1.96)
1.4
2. UCL
n
49
X ZD /2
X ZD /2
LCL
P: [23.4, 26.2]
V
x
n
V
x
n
24 . 8 1 . 4
26 . 2
24 . 8 1 . 4
23 . 4
IntervalEstimationofthePopulationMean
when isunknown
Intervalestimationofapopulation
mean: The case of unknown mean:Thecaseofunknown
y InsteadofpopulationstandarddeviationVwehave
sample standard deviation s.
samplestandarddeviations.
y Insteadofnormaldistribution,wehavetdistribution
,
Thetdistribution isusedtoconstructaconfidence
intervalabout if:
1. Thepopulationfromwhichthesampleisdrawnis
Th
l ti f
hi h th
l i d
i
(approximately)normallydistributed;
2. Thesamplesizeissmall(thatis,n<30);
p
(
,
);
3. Thepopulationstandarddeviation,,isnotknown.
22
23
Thet Distribution
Thetdistribution isaspecifictypeofbellshapeddistributionwith
alowerheightandawiderspreadthanthestandardnormal
p
g ,
distribution.Asthesamplesizebecomeslarger,thet
distributionapproachesthestandardnormaldistribution.A
specifictdistributiondependsononlyoneparameter,calledthe
degrees of freedom (df) The mean of the t distribution is equal
degreesoffreedom(df).Themeanofthetdistributionisequal
to0anditsstandarddeviationisfoundby.The
df /(df 2)
graphbelowdepictsthecaseofdf=3.
Thestandarddeviationofthe
standardnormaldistributionis1.0
24
Thet Distribution:Example
Findthevalueoft for16degreesoffreedomand.05areain
th i ht t il f t distributioncurve.
therighttailofa
di t ib ti
Area in the Right Tail Under the t Distribution Curve
Thestandarddeviationofthet
distributionis 9 /(9 2) 1.134
=0
25
df
.10
10
.05
05
.025
025
…
.001
001
1
2
3
.
16
.
3.078
1.886
1.638
…
1 337
1.337
…
6.314
2.920
2.353
…
1 746
1.746
…
12.706
4.303
3.182
…
2 120
2.120
…
…
…
…
…
…
…
318.309
22.327
10.215
…
3 686
3.686
…
Therequiredvalueoft for16 df and.05areaintherighttail.
ConfidenceIntervalforPopulationmean Using
the t Distribution
thet
Thet Distribution(continued)
Thet distributionwith16degreesoffreedom,areasunderthe
right and the left tails
rightandthelefttails.
The (1 – )100%confidenceinterval
The(1
)100% confidence interval for
for is
x r ts x where s x
s
n
Thevalueoft isobtainedfromthet distributiontableforn
– 1degreesoffreedomandthegivenconfidencelevel.
d
ff d
d h
fd
l l
.05
26
-1.746
1.746
0
27
ConfidenceIntervalforPopulationmean Using
thet Distribution:Example
Dr.Moorewantedtoestimatethemeancholesterollevelfor
all adult men living in Hartford. He took a sample of 25 adult
alladultmenlivinginHartford.Hetookasampleof25adult
menfromHartfordandfoundthatthemeancholesterol
levelforthissampleis186withastandarddeviationof12.
Assumethatthecholesterollevelsforalladultmenin
h h h l
ll l f
ll d l
Hartfordare(approximately)normallydistributed.Construct
a 95% confidence interval for the population mean .
a95%confidenceintervalforthepopulationmean.
ConfidenceIntervalforPopulationmean Using
thet Distribution:ExampleAnswered
y Confidencelevelis95%or.95,withdf =n – 1=25– 1=24
y Areaineachtail=.5– (.95/2)=.5 .4750=.025
s
12
y Thevalueoft intherighttailis2.064,and s x
n
25
2.40
df = 24
.025
.025
.4750
.4750
x r tsx 186 r 2.064(2.40) 186 r 4.95 181.05 to 190.95
y Thus,wecanstatewith95%confidencethatthemean
28
Example 2: ( known case)
The mean flying time for pilots at Continental
Airlines is 49 hours p
per month. This mean was
based on a sample of 100 pilots and the sample
standard deviation was 8.5 hours.
a. At 95% confidence, what is the margin of error?
b What is the 95% confidence interval estimate of
b.
the population mean flying time?
c. The mean flying time for pilots at United Airlines
is 36 hours per month. Discuss difference
between the flying times at two airlines.
29
cholesterollevelforalladultmenlivinginHarfordliesbetween
h l t ll lf
ll d lt
li i i H f d li b t
181.05and190.95.
Example 2:
Given: n = 100,,X = 49,, S = 8.5,, 1-D = .95
Think: What to estimate? Use Z or t?
Answer:
• Sample info (given): n = 100,X = 49, S = 8.5
• t: 1- D=0.95, so D/2=0.025, d.f.=n-1=99 o
Table 2: d.f.=100,
d f =100 D/2=0.025
/2=0 025 o t=1.984
t=1 984
d.f.=80, D/2=0.025 o t=1.990
Interpolation: t 1.984 (1.990 1.984) 100 99 1.9843
*Interpolation:
100 80
S
8 .5
a. m.o.e.: m .o .e . t
1 . 9843
1 . 69
D /2
n
b. UCL = 49+1.69 =50.69
P: [47.31,
[47 31 50
50.69]
69]
100
LCL = 49 – 1.69 = 47.31
c. 36 < LCL. The mean flying time is lower at United.
Example
Solution
Twentyfiverandomlyselectedadultswhobuybooks
forgeneralreadingwereaskedhowmuchthey
usually spend on books per year.
usuallyspendonbooksperyear.
y Confidence level is 99% or .99
y
sx
s
300
n
$60
25
y df = n – 1 = 25 – 1 = 24
Thesampleproducedameanof$1450anda
standarddeviationof$300forsuchannualexpenses.
d dd i i
f $300 f
h
l
y Area in each tail = .5
5 – (.99/2)
( 99/2) = .5
5 - .4950
4950 = .005
005
y The values of t are 2.797 and -2.797
Assume that such expenses for all adults who buy
Assumethatsuchexpensesforalladultswhobuy
booksforgeneralreadinghaveanapproximate
normaldistribution.
y The 99% confidence interval for is
x r ts x
Determinea99%confidenceintervalforthe
correspondingpopulationmean.
p
gp p
32
$1450 r 2.797(60)
$1450 r $167.82
$1282.18 to $1617.82
33
Intervalestimationofapopulationproportion:
g
p
Thecaseoflargesamples
Intervalestimationofapopulation
proportion: The case of large samples
proportion:Thecaseoflargesamples
EstimatoroftheStandardDeviationof p̂
Thevalueof,whichgivesapointestimateof,is
The
value of s pˆ which gives a point estimate of V p̂ is
calculatedas
pˆ qˆ
s pˆ
n
The(1– )100%confidenceintervalforthepopulation
proportion,p,is
pˆ r zs
pˆ
Thevalueofz usedhereisobtainedfromthestandard
normaldistributiontableforthegivenconfidencelevel,and
pˆ qˆ/n
s pˆ
34
35
Intervalestimationofapopulationproportionin
caseoflargesamples:Example
Intervalestimationofapopulationproportionin
caseoflargesamples:Examplesolved
Accordingtoa2002surveybyFindLaw.com,20%of
Americansneededlegaladviceduringthepastyearto
resolvesuchthornyissuesasfamilytrustsandlandlord
disputes(CBS.MarketWach.com,August6,2002).Suppose
a recent sample of 1000 adult Americans showed that 20%
arecentsampleof1000adultAmericansshowedthat20%
ofthemneededlegaladviceduringthepastyeartoresolve
suchfamilyrelatedissues.
a) Whatisthepointestimateofthepopulation
proportion?Whatisthemarginoferrorofthis
estimate?
b) Find,witha99%confidencelevel,thepercentageofall
adultAmericanswhoneededlegaladviceduringthe
pastyeartoresolvesuchfamilyrelatedissues.
p̂
y n =1000,=.20,and,=.80
q̂
36
y Notethatandarebothgreaterthan5.
Note that npˆ and nqˆ are both greater than 5
s pˆ
pˆ qˆ
n
(.20)(.80)
1000
.01264911
a)Pointestimateofp ==.20
p̂
s pˆ
M i f
Marginoferror=±1.96=±1.96(.01264911)=±
±1 96
±1 96( 01264911) ± .025or±2.5%
025 ±2 5%
b)Theconfidencelevelis99%,or.99.
Thez valuefor.4950isapproximately2.58.
pˆ r zss pˆ .20
20 r 2.58(.0126
2 58( 01264911) .20
20 r .033
033
37
.167 to .233 or 16.7% to 23.3%
Determiningthesampleforestimationofthe
mean
y Giventheconfidencelevelandthestandarddeviationofthe
population,thesamplesizethatwillproducea
population,
the sample size that will produce a
predeterminedmaximumerrorE oftheconfidenceinterval
estimateof is:
V
z 2V 2
E zV x z.
n
2
,with
n
E
y Example:
Analumniassociationwantstoestimatethemeandebtofthisyear’s
collegegraduates.Itisknownthatthepopulationstandard
deviation of the debts of this year’sscollegegraduatesis$11,800.
deviationofthedebtsofthisyear
college graduates is $11 800
Howlargeasampleshouldbeselectedsothattheestimatewitha
99%confidenceleveliswithin$800ofthepopulationmean?
z 2V 2 ( 2 .58 ) 2 (11,800 ) 2
n
1448 .18 | 1449
38
E2
(800 ) 2
Determiningthesamplesizeforestimatingthe
p p
proportion
y Given the confidence level and the values of p and q, the
sample size that will produce a predetermined maximum
error E of the confidence interval estimate of p is:
n
makesapartthatisusedinclocks.Thecompanywantstoestimatethe
proportionofthesepartsproducedbythismachinethataredefective.
Thecompanymanagerwantsthisestimatetobewithin.02ofthe
populationproportionfora95%confidencelevel.Whatisthemost
conservativeestimateofthesamplesizethatwilllimitthemaximum
errortowithin.02ofthepopulationproportion?
y Answer:
y Thevalueofzfora95%confidencelevelis1.96; p =.5andq =.5
y n
2
(1.96) (.50)(.50)
(.02) 2
and calculate
to find n.
p̂p and qq̂
from this sample
sample. Then use them
39
Determiningthesamplesizeforestimatingthe
p p
proportion:Example(continued)
p (
)
y ConsiderthepreviousExampleagain.Supposeapreliminary
sampleof200partsproducedbythismachineshowedthat
sample
of 200 parts produced by this machine showed that
7%ofthemaredefective.Howlargeasampleshouldthe
p y
p is
companyselectsothatthe95%confidenceintervalfor
within.02ofthepopulationproportion?
y Answer:
p̂
n
y Thus,ifthecompanytakesasampleof2401parts,thereis95%chance
thattheestimateofp willbewithin.02ofthepopulationproportion.
41
q̂
=.07and=.93
,thus:
z 2 pˆ qˆ
E2
(1.96) 2 (.07)(.93)
(.02) 2
(3.8416)(.
)( 07)(.
)( 93)
.0004
625.22 | 626
INTERVALESTIMATIONOFAPOPULATIONMEAN:
LARGESAMPLES
AnIntervalEstimationcont.
ConfidenceIntervalfor forLargeSamples
The(1– )100%confidenceintervalfor is
y Definition
y Eachintervalisconstructedwithregardtoagiven
g
g
x r zV x if V is known
confidencelevel andiscalledaconfidenceinterval.The
confidencelevelassociatedwithaconfidenceinterval
stateshowmuchconfidencewehavethatthisinterval
containsthetruepopulationparameter.Theconfidence
level is denoted by (1 )100%.
levelisdenotedby(1–
)100%
42
pq
n
y Take a preliminary sample of arbitrarily determined size
2401
40
zu
by using p = .5 and q = .5. For a given E, these values of p
and q will give the largest sample size in comparison to
any other pair of values of p = .5 and q = .5 since their
product is greater than the product of any other pair
pair.
2
z pq
E2
zV pˆ
y In case the values of p and q are not known:
y Take the most conservative estimate of the sample size n
Determiningthesamplesizeforestimatingthe
proportion:Example
y LombardElectronicsCompanyhasjustinstalledanewmachinethat
z 2 pq
,with E
E2
where
x r zs x if V is not known
V x V / n and s x s / n
Thevalueofz usedhereisreadfromthestandard
normaldistributiontableforthegivenconfidence
level.
43
Repeat, again:
Finding z for a 95% confidence level.
Area in the tails – what this means?
Total shaded area is
.9500 or 95%
D
D
2
.4750
.4750
x
-z
-1.96
2
(1 – )
0
1.96
44
45
Interval Estimation for Population Proportion
Example 3:
Answer:
a. Point estimate of p:
Example
E
l 3
3:
A survey asked 346 job seekers. The answer selected most
(152 times) was “higher compensation.”
a. What is the point estimate of the proportion of job seekers
who would select “higher
higher compensation
compensation” as the reason of
changing jobs?
a What is the 95% confidence inter
a.
interval
al estimate of the
population proportion?
z
0
z
z
p
x
n
152
346
.4393
b. Confidence interval:
• Z:
Z (1-D)/2=0.475,
(1 )/2 0 475 T
Table
bl 1 o Z=1.96
Z 1 96
• Margin of error =
Z
p (1 p )
n
(0.4393)(1 0.4393)
= 0.0523
346
(1.96)
• Confidence interval
UCL
p m .o .e .
. 4393 . 0523
. 4916
LCL
p m .o .e .
. 4393 . 0523
. 3870
p: [.3870,
[ 3870 .4916]
4916]
Sample Size and Precision
Intervalestimation:
Sample size and precision
Samplesizeandprecision
Quality of estimation:
• Confidence level: 1 - D
• Precision: margin of error
Confidence level: 1 - D is guaranteed by procedure.
sampling distribution for sample mean.
Probability that
X falls between
P ZD
V
/2
x
n
and P Z D
V
/2
x
n
is 1 - D. In general, any sample mean that is within this range
will provide an interval that contains the population mean P.
Margin of error:
Given n, then (1 - D)np o margin of error np.
Given 1 - D, then n np o margin of error pn.
pn
48
Determine sample size to meet requirements for both
g of error:
confidence level and margin
1. Determine sample size for estimation of P
n
ª Z D2 / 2V 2 º
«
»
2
»¼
¬« E
E: desired margin of error
[ ]: round up (“Be conservative”)
p size for estimation of p
2. Determine sample
n
ª Z D2 / 2 p (1 p ) º
»
«
E2
¬
¼
p = .5 (“Be conservative”)
desired
ed margin
ag o
of e
error
o
E: des
[ ]: round up (“Be conservative”)
Example 4:
Bride’s magazine reported that the mean cost of a wedding is
$19,000. Assume that the population standard deviation is
$9,400. Use 95% confidence,
a. What
a
at iss the
t e recommended
eco
e ded sa
sample
p e ssize
e if tthe
e des
desired
ed margin
ag
of error is $1,000?
b. What is the recommended sample size if the desired margin
of error is $500?
Answer:
1D
a. Z D / 2 :
. 475 o Z D / 2 1 . 96
2
ª (1 . 96 ) 2 ( 9400 ) 2 º
n «
» >339 . 44 @ | 340
1000 2
¬
¼
b.
n
Example 6:
>1357 .78 @ | 1358
Example
The League of American Theatres and Producers uses an
ongoing audience tracking survey that provides up-to-date
i f
information
ti about
b t Broadway
B d
th
theater
t audiences.
di
E
Every week,
k th
the
League distributes a one-page survey on random theater seats
at a rotation roster of Broadway shows.
a. How large a sample should be taken if the desired margin of
error on any proportion is 0.04? Use 95% confidence.
Answer:
1D
. 475 , Table 1 o Z D / 2 1 . 96
2
ª (1 . 96 ) 2 . 5 (1 . 5 ) º
«
» >600 . 25 @ | 601
(. 04 ) 2
¬
¼
According to the analysis of a CNN–USA TODAY–Gallup
poll conducted in October 2002, “Stress has become a
common part of everyday life in the United States
States. The
demands of work, family, and home place an increasing
burden on the average American.”
According to this poll, 40% of Americans included in the
survey indicated that they had a limited amount of time to
relax (Gallup
(Gallup.com,
com November 8
8, 2002)
2002).
ZD /2 :
The poll was based on a randomly selected national
sample of 1502 adults aged 18 and older
older.
n
Construct a 95% confidence interval for the corresponding
population
l ti proportion.
ti
53
Solution
Example
y Confidence level = 95% or .95
According to a report by the Consumer Federation of
America,, National Credit Union Foundation,, and the Credit
Union National Association, households with negative
assets carried
g of $15,528 in debt in 2002
an average
(CBS.MarketWatch.com, May 14, 2002).
y The value of z for .95 / 2 = .4750 is 1.96.
s pˆ
pˆ r zs pˆ
pˆ qˆ
n
(.40)(.60)
1502
.01264069
Assume that this mean was based on a random sample
of 400 households and that the standard deviation of
debts for households in this sample was $4200.
.40 r 1.96(.01264069)
.40
40 r .025
025
Make a 99% confidence interval for the 2002 mean debt
for all such households.
.375 to .425 or 37.5% to 42.5%
54
ª (1.96 ) 2 (9400 ) 2 º
«
»
500 2
¬
¼
55
Solution
Example
y Confidence level 99% or .99
y
sx
s
4200
n
400
Lombard Electronics Company has just installed a new
machine that makes a part that is used in clocks. The
company wants to estimate the proportion of these parts
produced by this machine that are defective.
$210
y The sample is large (n > 30)
¾ Therefore,
f
x r zs x
we use the normal distribution z = 2.58
2 8
The company manager wants this estimate to be within
.02 of the population proportion for a 95% confidence
level.
15,528 r 2.58(210) 15,528 r 541.80
$14,986.20 to $16,069.80
What is the most conservative estimate of the sample size
that will limit the maximum error to within .02
02 of the
population proportion?
Thus, we can state with 99% confidence that the 2002
mean debt for all households with negative assets was
between $14,986.20
$14 986 20 and $16
$16,069.80.
069 80
56
57
Example
Solution
y The value of z for a 95% confidence level is 1.96.
Consider previous example again.
y p = .50 and q = .50
n
y
z 2 pq
E2
(1.96) 2 (.50)(.50)
((.02) 2
Suppose a preliminary sample of 200 parts produced by
this machine showed that 7% of them are defective.
2401
y Thus, if the company takes a sample of 2401 parts, there is
How large a sample should the company select so that the
% confidence interval for p is within .02 of the p
population
p
95%
proportion?
95% chance that the estimate of p will be within .02 of the
population proportion.
58
59
Solution
pp̂
n
60
= .07
z 2 pˆ qˆ
E2
and
q̂
q
= .93
(1.96) 2 (.
( 07)(.
)( 93)
2
(.02)
(3.8416)(.07)(.93)
.0004
625.22 | 626