分层抽样

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Ni k k Ni 1 k Ni 2 2 2 ⇒S = [ ∑∑(Xij − Xi ) ] +∑(Xi − X )[∑(Xij − Xi )]+∑∑ (Xi − X ) N −1 i=1 j=1 i=1 j=1 i=1 j=1
k k 1 k N i − 1 Ni 2 2 2 ⇒S = ∑[ ∑ ( X ij − X i ) ] +∑ ( X i − X )0 +∑ Ni ( X i − X ) N − 1 i=1 Ni − 1 j =1 i =1 i =1
2 are unbiased estimators of the stratum variances S i (i = 1,2, L , k )
We obtain an unbiased estimator of var(X st ) as
si2 si2 1 k 2 2 = 2 ∑ N i ( N i − ni ) s ( x st ) = ∑ Wi (1 − f i ) ni N i =1 ni i =1
1− f 2 S = S , (i = 1,2,L, k ) var(xxt ) = SW n
2 i 2 W
Directly analogous(相似的) results hold for the estimation of the population total X T Thus from the stratified fandom sample we obtain an unbiased estimator
k
k
provided that cov(xi , x j ) = 0
Some special cases of this are (a)sampling fractions f i = ni / N i are negligible忽略
Si2 var(xst ) = ∑ Wi 2 ni i =1
k
∑x
j =1
ni
ij
and
1 ni s i2 = ∑ xij − xi ni − 1 j =1
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2
In each stratum we have a sampling fraction f i = ni / N i
(i = 1,2,L, k )
The estimator of X commonly employed is the so-called stratifieБайду номын сангаас sample mean. This is defined as
k
i
=X
x′ will be unbiased only in the
case of proportional allocation. Th2, for the variance of x st ,we have
si2 1 k si2 var(xst ) = ∑Wi 2 var(xi ) = ∑Wi 2 (1 − fi ) = 2 ∑ Ni ( Ni − ni ) ni N i=1 ni i =1 i =1
X T = N xst = ∑i =1 N i xi
k
with variance Var ( X T ) =
N i2 (1 − f i ) S i2 / ni ∑i =1
k
In practical situations the stratum variance S i2 will not be known and we need to estimate the S i2
. The strata are just sub-populations. The sampled values in each stratum constitute a s.r. sample. Thus, as
1 ni 2 2 si = ∑ ( xij − xi ) , (i = 1,2,L, k ) ni − 1 j =1
st
means
:
xi
are unbiased.
k k
Ni 1 Proof E ( xst ) = ∑ Wi X i = ∑ Xi = N i =1 i =1 N
1 k Note that E( x′) = ∑ ni X i so that n i =1
1 ∑ Ni X i = N i =1
k
∑X
i =1
.
x st
,is not the same as the arithmetic mean
this would imply that the sampling fractions fi =n / N are identical for i i all strata; A special form of stratified sampling where the stratum sample sizes, ni are said to be chosen by proportional allocation. （按比例配置）
STARTIFIED POPULSTIONS（总体分层概念） Suppose we have a finite population of 20 members in which X takes values 5 3 4 4 5 3 6 2 3 2 2 6 5 3 5 2 4 6 4 5 Its mean is X = 4 its variance S X = 40 / 19
k
In some situations the practical circumstances may suggest that all stratum variances are equal. If this so, then it is desirable to combine the data from the different strata to obtain a “pooled” unbiased estimator of 2 2 the common variance SW in the form k ni
k 1 k (Ni − 1)Si2 + ∑ Ni ( X i − X )2 S2 = ∑ N − 1 i =1 i =1
2 2 1 k Ni 1 k Ni 2 2 S = ∑∑ X ij − X ⇒ S = ∑∑ X ij − X i + X i − X N − 1 i=1 j =1 N − 1 i=1 j =1 2 1 k Ni 2 ⇒S = ∑∑ ( X ij − X i ) + ( X i − X ) N − 1 i =1 j =1 1 k Ni 2 2 2 ⇒S = ∑∑ ( Xij − Xi ) + 2( Xij − Xi )( Xi − X ) + ( Xi − X ) N −1 i=1 j=1
N1，N2， ，Nk L k ∑ Ni = N 层容量 i=1
with members X ij (i = 1,2, L , k ; j = 1,2, L , N i )数字指标 1、总体平均数、总和及比例的估計 、总体平均数 平均 Denote the stratum means and variance by X 1 , X 2 ,L, X k 2 S12 , S 2 , L S k2 respectively. and
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The s.r. sample from the ith-stratum has members
xi1 , xi 2 ,L xini (i = 1,2,L , k )
and we denote the sample mean and variance in the ith-stratum by
1 xi = ni
(b) proportional allocation (比例配置）ni
= nW, fi = f = n / N i
1− f var(xst ) = n
Wi Si2 ∑
i =1
k
(c) proportional allocation, and constant within strata variance 方差为常 数
xst = ∑Wi xi
i =1
k
Note that this assumes that we know the stratum size N i precisely in order to determinethe stratum weights Wi = N i / N The stratified sample mean
2
If we take a s.r. sample of size 5 we could obtain quite a range of different values for x in different samples: from 2.2 to 5.8 But notice the structure of the population. It could be rearranged as
Stratified Random Sampling（分层随机抽样）
Suppose we wish to estimate the mean X ，of the set of values
X 1 , X 2 , X 3 ,L, X N , in a finite population.
We shall assume that the population is stratified, that is to say it has been divided into k non-overlapping groups, or strata of sizes（把总体 分为K个互不重迭的子总体，每个子总体称为层） 层
The population mean and variance X and S 2 will of course have the forms k 1 k X = ∑ N i X i = ∑Wi X i 总体平均数(加权平均） 总体平均 加权平均 平均数 加权平均） N i =1 i =1
Ni (where Wi = N is termed the weight of the ith-stratum),and
1 k x ′ = ∑ ni x i n i =1
except where
ni Ni = n N
We first consider the mean and variance of xst Th1, x is unbiased estimator of X since the stratum sample
2222333344445555666 6
It consists of 5 groups in each of which all 4 X-values are the same. Suppose we had some mechanism by which we could choose one member at random from each group to constitute our sample of size 5. We must inevitably obtain on all occasions 2 3 4 5 6 with sample mean 4. Thus our estimate is always equal to the population mean X ,it is estimating.
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k 1 k 2 2 = ∑ ( N i − 1)S i + ∑ N i ( X i − X ) N − 1 i =1 i =1
We shall assume that a sample of size n is chosen by taking a s.r. sample of predetermined size from each stratrum. The stratum sample size will be denoted by k n1 , n2 , L, nk ∑i =1 ni = n