如果你也在 怎样代写抽样调查Survey sampling 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。抽样调查Survey sampling是数学工程这一广泛新兴领域中的一个自然组成部分。例如，我们可以断言，数学工程之于今天的数学系，就像数学物理之于一个世纪以前的数学系一样;毫不夸张地说，数学在诸如语音和图像处理、信息理论和生物医学工程等工程学科中的基本影响。
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|NONRESPONSE
To cite a simple example, suppose that unit $i$, provided it is included in a sample $s$, responds with probability $q_i, q_i$ not depending on $s$ or $Y=\left(Y_i, \ldots, Y_N\right)$. Suppose $n$ units are drawn by SRSWOR and define
$$
M_i= \begin{cases}1 & \text { if unit } i \text { is sampled and responds } \ 0 & \text { otherwise }\end{cases}
$$
Consider the arithmetic mean
$$
\bar{y}=\frac{\sum_1^N M_i Y_i}{\sum_1^N M_i}
$$
of all observations as an estimator of $\bar{Y}$. Then
$$
E M_i=\frac{n}{N} q_i
$$
and $E \bar{y}$ is asymptotically equal to
$$
\frac{\sum q_i Y_i}{\sum q_i}
$$
The bias
$$
\sum\left(\frac{q_i}{\sum q_i}-\frac{1}{N}\right) Y_i
$$
is negligible only if approximately
$$
q_i=\frac{1}{N} \sum q_i .
$$
统计代写|抽样调查作业代写sampling theory of survey代考|CALLBACKS
HANSEN and HURWITZ (1946) gave an elegant procedure for callbacks to tackle nonresponse problems later modified by SRINATH (1971) and J. N. K. RAO (1973), briefly described below. The population is conceptually dichotomized with $W_1\left(W_2=1-W_1\right)$ and $\widehat{Y}_1\left(\widehat{Y}_2=\left[\widehat{Y}-W_1 \widehat{Y}_1\right] / W_2\right)$ as the proportion of respondents (nonrespondents) and mean of respondents (nonrespondents) and an SRSWOR of size $n$ yields proportions $w_1=n_1 / n$ and $w_2=1-w_1=1-n_1 / n=n_2 / n$ of respondents and nonrespondents, respectively. Choosing a suitable number $K>1$ an SRSWOR of size $m_2=n_2 / K$, assumed to be an integer, is then drawn from the initial $n_2$ sample nonrespondents. Supposing that more expensive and persuasive procedures are followed in this second phase so that each of the $m_2$ units called back now responds, let $\bar{y}1$ and $\bar{y}{22}$ denote the first-phase and second-phase sample means based respectively on $n_1$ and $m_2$ respondents. Then, $\bar{Y}$ may be estimated by $\bar{y}d=w_1 \bar{y}_1+w_2 \bar{y}{22}$, and the variance
$$
V\left(\bar{y}d\right)=(1-f) \frac{S^2}{n}+W_2 \frac{(K-1)}{n} S_2^2 $$ by $$ \begin{aligned} v_d= & (1-f)\left(\frac{n_1-1}{n-1}\right) w_1 \frac{s_1^2}{n_1} \ & +\frac{(N-1)\left(n_2-1\right)-(n-1)\left(m_2-1\right)}{N(n-1)} w_2 \frac{s{22}^2}{m_2} \
& +\frac{N-n}{N(n-1)}\left[w_1\left(\bar{y}1-\bar{y}_d\right)^2+w_2\left(\bar{y}{22}-\bar{y}_d\right)^2\right] .
\end{aligned}
$$
抽样调查代考
统计代写|抽样调查作业代写sampling theory of survey代考|NONRESPONSE
举一个简单的例子，假设单元$i$(如果它包含在示例$s$中)的响应概率为$q_i, q_i$，而不取决于$s$或$Y=\left(Y_i, \ldots, Y_N\right)$。假设$n$单位由SRSWOR绘制并定义
$$
M_i= \begin{cases}1 & \text { if unit } i \text { is sampled and responds } \ 0 & \text { otherwise }\end{cases}
$$
考虑算术平均值
$$
\bar{y}=\frac{\sum_1^N M_i Y_i}{\sum_1^N M_i}
$$
作为$\bar{Y}$的估计值。然后
$$
E M_i=\frac{n}{N} q_i
$$
$E \bar{y}$渐近等于
$$
\frac{\sum q_i Y_i}{\sum q_i}
$$
偏见
$$
\sum\left(\frac{q_i}{\sum q_i}-\frac{1}{N}\right) Y_i
$$
只有近似时才可忽略
$$
q_i=\frac{1}{N} \sum q_i .
$$
统计代写|抽样调查作业代写sampling theory of survey代考|CALLBACKS
HANSEN和HURWITZ(1946)给出了一个优雅的回调程序来处理无响应问题，后来由SRINATH(1971)和j.n. K. RAO(1973)修改，下面简要介绍。人口在概念上分为$W_1\left(W_2=1-W_1\right)$和$\widehat{Y}_1\left(\widehat{Y}_2=\left[\widehat{Y}-W_1 \widehat{Y}_1\right] / W_2\right)$作为受访者的比例(非受访者)和受访者的平均值(非受访者)，大小为$n$的SRSWOR分别产生受访者和非受访者的比例$w_1=n_1 / n$和$w_2=1-w_1=1-n_1 / n=n_2 / n$。选择一个合适的数字$K>1$，然后从初始的$n_2$样本非受访者中提取大小为$m_2=n_2 / K$的SRSWOR(假设为整数)。假设在第二阶段中遵循更昂贵和更有说服力的程序，以便每个回调的$m_2$单位现在都做出响应，让$\bar{y}1$和$\bar{y}{22}$分别表示基于$n_1$和$m_2$受访者的第一阶段和第二阶段样本平均值。然后，$\bar{Y}$可以用$\bar{y}d=w_1 \bar{y}_1+w_2 \bar{y}{22}$来估计，方差
$$
V\left(\bar{y}d\right)=(1-f) \frac{S^2}{n}+W_2 \frac{(K-1)}{n} S_2^2 $$ by $$ \begin{aligned} v_d= & (1-f)\left(\frac{n_1-1}{n-1}\right) w_1 \frac{s_1^2}{n_1} \ & +\frac{(N-1)\left(n_2-1\right)-(n-1)\left(m_2-1\right)}{N(n-1)} w_2 \frac{s{22}^2}{m_2} \
& +\frac{N-n}{N(n-1)}\left[w_1\left(\bar{y}1-\bar{y}_d\right)^2+w_2\left(\bar{y}{22}-\bar{y}_d\right)^2\right] .
\end{aligned}
$$
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