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SampleVariability
Considerthesmallpopula5onofintegers{0,2,4,6,8}
Itisclearthatthemean,μ=4.Supposewedidnotknowthe
popula5onmeanandwantedtoes5mateitwithasamplemean
withsamplesize2.(Wewillusesamplingwithreplacement)
Wetakeonesampleandgetsamplemean,ū1=(0+2)/2=1and
takeanothersampleandgetasamplemeanū2=(4+6)/2=5.
Whyarethesesamplemeansdifferent?
Aretheygoodes5matesofthetruemeanofthepopula5on?
Whatistheprobabilitythatwetakearandomsampleandgeta
samplemeanthatwouldexactlyequalthetruemeanofthe
popula5on?
Sec5on7.1,Page137
1
SamplingDistribu5on
Eachofthesesampleshasasamplemean,ū.
Thesesamplemeansrespec5velyareas
follows:
P(ū=1)=2/25=.08
P(ū=4)=5/25=.20
Sec5on7.1,Page138
2
SamplingDistribu5on
Shapeisnormal
Meanofthesampling
distribu5on=4,themean
ofthepopula5on
Sec5on7.1,Page138
3
SamplingDistribu5onsand
CentralLimitTheorem
If all possible random samples, each of size n, are
taken from any population with a mean µ, and standard
deviation σ, then the sampling distribution will have
the following:
1.  A mean
µx equal to the population mean µ
2.  A standard deviation:
€
σx =
σ
= SE(x ) alsocalled"S tan dard Error"
n
3.  If the population is normal, then the sampling
distribution will be normal for all sample sizes
€
4.  And further, if the population is not normal, then
the the sampling distribution can be assured to be
normal for all samples sizes ≥ 30.
5.  It is clear from 2 above that as the sample size
increases, the standard deviation of the sampling
distribution gets smaller.
6.  Sampling Distributions and the Central Limit
Theorem make possible statistical inference.
Sec5on7.2,Page141
4
CentralLimitTheorem
Sec5on7.2,Page144
5
CentralLimitTheorem
Sec5on7.2,Page145
6
Calcula5ngProbabili5esforthe
Mean
Kindergartenchildrenhaveheightsthatareapproximately
normallydistributedaboutameanof39inchesanda
standarddevia5onof2inches.Arandomsampleof25is
taken.Whatistheprobabilitythatthesamplemeanis
between38.5and40inches?
P(38.5<samplemean<40)=
NORMDIST1
LOWERBOUND=38.5
UPPERBOUND=40
MEAN=39
SE(x ) = 2/ 25 = 0.4
ANSWER:0.8881
€
€
Sec5on7.3,Page147
7
Calcula5ngMiddle90%
Kindergartenchildrenhaveheightsthatareapproximately
normallydistributedaboutameanof39inchesanda
standarddevia5onof2inches.Arandomsampleof25is
taken.Findtheintervalthatincludesthemiddle90%ofall
samplemeansforthesampleofkindergarteners.
Sampling Distribution
σx =
€
2
25
x
ux = 39
NORMDIST2
€
AREAFROMLEFT=0.05
MEAN=39
SE(x ) = 2 / 25 = 0.4
ANSWER:38.3421
€
NORMDIST2
AREAFROMLEFT=.95
€
MEAN=39
SE(x ) = 2 / 25 = 0.4
ANSWER:39.6579
Theinterval(38.3inches,39.7inches)containsthe
€
€middle90%ofallsamplemeans.Ifwechoosea
€
randomsample,thereisa90%probabilitythatitwillbe
intheinterval.
Sec5on7.3,Page147
8
Problems
Problems,Page149
9
Problems
Problems,Page150
10
Problems
Problems,Page151
11
Problems
Problems,Page151
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