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抽样调查Survey sampling是主流统计的边缘。这里的特殊之处在于，我们有一个具有某些特征的有形物体集合，我们打算通过抓住其中一些物体并试图对那些未被触及的物体进行推断来窥探它们。这种推论传统上是基于一种概率论，这种概率论被用来探索观察到的事物与未观察到的事物之间的可能联系。这种概率不被认为是在统计学中，涵盖其他领域，以表征我们感兴趣的变量的单个值之间的相互关系。但这是由调查抽样调查人员通过任意指定的一种技术从具有预先分配概率的对象群体中选择样本而创建的。

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统计代写|抽样调查作业代写sampling theory of survey代考|BALANCED REPEATED REPLICATION

Suppose a finite population of $N$ units is divided into $L$ strata of $N_1, N_2, \ldots, N_L$ units, respectively. From each stratum let SRSWORs be independently selected, making $n_h$ draws from the $h$ th, $h=1, \ldots, L$. Let $L$ be sufficiently large and $n_h$ be taken as 2 for each $h=1, \ldots, L$. Let us write $\left(y_{h 1}, y_{h 2}\right)$ as the vector of variable values on the variable of interest $y$ observed for the sample from the $h$ th stratum. Then, with $W_h=N_h / N$,
$$
\frac{1}{N} \sum N_h\left(\frac{y_{h 1}+y_{h 2}}{2}\right)=\sum W_h \bar{y}h=\bar{y}{s t} \text {, say }
$$
is taken as the usual unbiased estimator for $\bar{Y}=\sum W_h \bar{Y}h$, the population mean. Neglecting $n_h / N_h=f_h$, that is, ignoring the finite population correction $1-f_h$ for every $h$, we have the variance of $\bar{y}{s t}$ as
$$
V\left(\bar{y}_{s t}\right)=\sum W_h^2 S_h^2 / 2
$$
where
$$
S_h^2=\frac{1}{N_h-1} \sum_1^{N_h}\left(Y_{h i}-\bar{Y}h\right)^2, $$ writing $Y{h i}$ as the value of $i$ th unit of $h$ th stratum and $\bar{Y}h$ for their mean. This $V\left(\bar{y}{s t}\right)$ is unbiasedly estimated by
$$
v=\frac{1}{4} \sum W_h^2 d_h^2
$$
where $d_h=\left(y_{h 1}-y_{h 2}\right)$. Let us now form two half-samples by taking into the first half-sample one of $y_{h 1}$ and $y_{h 2}$ for every $h=1, \ldots, L$ leaving the other ones, which together, over $h=$ $1, \ldots, L$, form the second half-sample. We denote the first halfsample by I and the second by II. There are, in all, $2^L$ possible ways of forming these half-samples. For the $j$ th $\left(j=1, \ldots, 2^L\right)$ such formation, let $\delta_{h j}=1(0)$ if $y_{h 1}$ appears in I (II). Then,
$$
\begin{aligned}
t_{h 1} & =\sum W_h\left[\delta_{h j} y_{h 1}+\left(1-\delta_{h j}\right) y_{h 2}\right] \
t_{h 2} & =\sum W_h\left[\left(1-\delta_{h j}\right) y_{h 1}+\delta_{h j} y_{h 2}\right]
\end{aligned}
$$
form two unbiased estimators of $\bar{Y}$ based respectively on I and II. Then, $\bar{t}j=\frac{1}{2}\left(t{j 1}+t_{j 2}\right)=\sum W_h \bar{y}h$ for every $j=1, \ldots, 2^L$. Also $$ v_j=\left(t{j 1}-t_{j 2}\right)^2 / 4
$$
may be taken as an estimator for
$$
V\left(\bar{t}j\right)=V\left(\sum W_h \bar{y}_h\right)=V\left(\bar{y}{s t}\right) .
$$

统计代写|抽样调查作业代写sampling theory of survey代考|BOOTSTRAP

Consider a population $U=(1,2, \ldots, N)$ and unknown values $Y_1, Y_2, \ldots, Y_N$ associated with the units $1,2, \ldots, N$. Let $\theta=$ $\theta(Y)$ be a population parameter, for example, the population mean $\bar{Y}$, or some not necessarily linear function $f(\bar{Y})$ of $\bar{Y}$, or the median of the values $Y_1, \ldots, Y_N$, etc. Suppose a sample $s=\left(i_1, \ldots, i_n\right)$ is drawn by SRSWR, write for $j=1,2, \ldots, n$
$$
y_j=Y_{i_j}
$$
and define
$$
y=\left(y_1, y_2, \ldots, y_n\right)^{\prime}
$$
Let $\widehat{\theta}=\widehat{\theta}(y)$ be an estimator of $\theta$; in the special case $\theta=f(\bar{Y})$, for example, it suggests itself to choose $\widehat{\theta}=f(\bar{y})$, where $\bar{y}$ is the sample mean. To calculate confidence intervals for $\theta$ we need some information on the distribution of $\widehat{\theta}$ relative to $\theta$.
Now, choose a sample $s^$ of size $n$ from $s$ by SRSWR, denote the observed values by $$ \stackrel{}{y}{11}, \stackrel{}{y}{21}, \ldots, \stackrel{}{y}{n 1} $$ and define $$ \stackrel{}{y}_1=\left(\stackrel{}{y}{11}, \stackrel{}{y}{21}, \ldots, \stackrel{}{y}{n 1}\right)^{\prime}
$$
and ${ }^$ is called a bootstrap sample. If, for example, $s=(4,2,4$, $5)$, then $\stackrel{}{s}=(2,2,4,2)$ would be possible, and in this case $y_1=\left(y_2, y_2, y_4, y_2\right)$.
Repeat the selection of a bootstrap sample independently to obtain
$$
\underline{y}2, \underline{y}_3, \ldots, \underline{y}_B $$ where $B=500,1000$, or even larger, and calculate $$ \begin{aligned} \widehat{\theta}_0 & =\frac{1}{B} \sum{b=1}^B \widehat{\theta}\left(yb^\right) \ v_B & =\frac{1}{B-1} \sum{b=1}^B\left[\widehat{\theta}\left(_b^*-\widehat{\theta}_0\right]^2\right.
\end{aligned}
$$
It may be shown that the empirical distribution of
$$
\widehat{\theta}\left(y_b\right)-\widehat{\theta}(y), b=1,2, \ldots, B
$$
for large $n$ and $B$ approximates closely the distribution of
$$
\widehat{\theta}(y)-\theta(Y)
$$
and that $v_B$ approximates the variance of $\hat{\theta}(y)$. For details, good references are RAO and WU $(1985,1988)$.

抽样调查代考

统计代写|抽样调查作业代写sampling theory of survey代考|BALANCED REPEATED REPLICATION

假设一个有限的人口$N$个单位分别被分成$L$个单位的地层$N_1, N_2, \ldots, N_L$个单位。从每个地层中独立选择SRSWORs，使$n_h$从$h$ th、$h=1, \ldots, L$中抽取。设$L$足够大，$n_h$等于2对应$h=1, \ldots, L$。让我们将$\left(y_{h 1}, y_{h 2}\right)$作为从$h$层的样本中观察到的感兴趣变量$y$的变量值向量。然后，通过$W_h=N_h / N$，
$$
\frac{1}{N} \sum N_h\left(\frac{y_{h 1}+y_{h 2}}{2}\right)=\sum W_h \bar{y}h=\bar{y}{s t} \text {, say }
$$
作为总体均值$\bar{Y}=\sum W_h \bar{Y}h$的通常无偏估计量。忽略$n_h / N_h=f_h$，即忽略每个$h$的有限总体修正$1-f_h$，我们得到$\bar{y}{s t}$ as的方差
$$
V\left(\bar{y}{s t}\right)=\sum W_h^2 S_h^2 / 2 $$ 在哪里 $$ S_h^2=\frac{1}{N_h-1} \sum_1^{N_h}\left(Y{h i}-\bar{Y}h\right)^2, $$用$Y{h i}$表示$h$层的单位$i$的值，用$\bar{Y}h$表示其平均值。这个$V\left(\bar{y}{s t}\right)$是由
$$
v=\frac{1}{4} \sum W_h^2 d_h^2
$$
在哪里$d_h=\left(y_{h 1}-y_{h 2}\right)$。现在让我们形成两个半样本，将一个$y_{h 1}$和$y_{h 2}$放到前半样本中，每个$h=1, \ldots, L$留下其他的，它们一起，除以$h=$$1, \ldots, L$，形成第二个半样本。我们用I表示前半样本，用II表示后半样本。总之，有$2^L$种可能的方法来形成这些半样品。对于$j$和$\left(j=1, \ldots, 2^L\right)$这种形式，如果$y_{h 1}$出现在I (II)中，则设$\delta_{h j}=1(0)$。
$$
\begin{aligned}
t_{h 1} & =\sum W_h\left[\delta_{h j} y_{h 1}+\left(1-\delta_{h j}\right) y_{h 2}\right] \
t_{h 2} & =\sum W_h\left[\left(1-\delta_{h j}\right) y_{h 1}+\delta_{h j} y_{h 2}\right]
\end{aligned}
$$
分别基于I和II形成$\bar{Y}$的两个无偏估计量。然后，$\bar{t}j=\frac{1}{2}\left(t{j 1}+t_{j 2}\right)=\sum W_h \bar{y}h$对应每一个$j=1, \ldots, 2^L$。还有$$ v_j=\left(t{j 1}-t_{j 2}\right)^2 / 4
$$
可以作为一个估计
$$
V\left(\bar{t}j\right)=V\left(\sum W_h \bar{y}_h\right)=V\left(\bar{y}{s t}\right) .
$$

统计代写|抽样调查作业代写sampling theory of survey代考|BOOTSTRAP

考虑一个总体$U=(1,2, \ldots, N)$和与单位$1,2, \ldots, N$相关的未知值$Y_1, Y_2, \ldots, Y_N$。设$\theta=$$\theta(Y)$为总体参数，例如，总体均值$\bar{Y}$，或$\bar{Y}$的某些不一定是线性的函数$f(\bar{Y})$，或值的中位数$Y_1, \ldots, Y_N$，等等。假设SRSWR绘制了一个示例$s=\left(i_1, \ldots, i_n\right)$，为$j=1,2, \ldots, n$编写
$$
y_j=Y_{i_j}
$$
然后定义
$$
y=\left(y_1, y_2, \ldots, y_n\right)^{\prime}
$$
设$\widehat{\theta}=\widehat{\theta}(y)$为$\theta$的估计值;例如，在特殊情况$\theta=f(\bar{Y})$中，它建议自己选择$\widehat{\theta}=f(\bar{y})$，其中$\bar{y}$是样本均值。为了计算$\theta$的置信区间，我们需要关于$\widehat{\theta}$相对于$\theta$的分布的一些信息。
现在，通过SRSWR从$s$中选择大小为$n$的示例$s^$，用$$ \stackrel{}{y}{11}, \stackrel{}{y}{21}, \ldots, \stackrel{}{y}{n 1} $$表示观测值并定义$$ \stackrel{}{y}_1=\left(\stackrel{}{y}{11}, \stackrel{}{y}{21}, \ldots, \stackrel{}{y}{n 1}\right)^{\prime}
$$
${ }^$被称为bootstrap样本。例如，如果是$s=(4,2,4$, $5)$，那么可能是$\stackrel{}{s}=(2,2,4,2)$，在本例中是$y_1=\left(y_2, y_2, y_4, y_2\right)$。
重复选择一个独立的bootstrap样本来获得
$$
\underline{y}2, \underline{y}_3, \ldots, \underline{y}_B $$其中$B=500,1000$，甚至更大，并计算$$ \begin{aligned} \widehat{\theta}_0 & =\frac{1}{B} \sum{b=1}^B \widehat{\theta}\left(yb^\right) \ v_B & =\frac{1}{B-1} \sum{b=1}^B\left[\widehat{\theta}\left(_b^*-\widehat{\theta}_0\right]^2\right.
\end{aligned}
$$
的经验分布可以证明
$$
\widehat{\theta}\left(y_b\right)-\widehat{\theta}(y), b=1,2, \ldots, B
$$
对于较大的$n$和$B$近似的分布
$$
\widehat{\theta}(y)-\theta(Y)
$$
这个$v_B$近似于$\hat{\theta}(y)$的方差。详细信息请参考RAO和WU $(1985,1988)$。

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