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

#### Doug I. Jones

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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

$$\frac{1}{N} \sum N_h\left(\frac{y_{h 1}+y_{h 2}}{2}\right)=\sum W_h \bar{y}h=\bar{y}{s t} \text {, say }$$

$$V\left(\bar{y}{s t}\right)=\sum W_h^2 S_h^2 / 2$$ 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
$$v=\frac{1}{4} \sum W_h^2 d_h^2$$

$$y_j=Y_{i_j}$$

$$y=\left(y_1, y_2, \ldots, y_n\right)^{\prime}$$

${ }^$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$y_1=\left(y_2, y_2, y_4, y_2\right)$。

$$\underline{y}2, \underline{y}_3, \ldots, \underline{y}_B$$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}

$$\widehat{\theta}\left(y_b\right)-\widehat{\theta}(y), b=1,2, \ldots, B$$

$$\widehat{\theta}(y)-\theta(Y)$$

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