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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The attributes $A$ and $\bar{A}$ may both be sensitive, for example, affiliation to two rival political blocks. An alternative RR device for estimating $\pi_A$ in this dichotomous case is described below.
Suppose $B$ is another innocuous character unrelated to the sensitive attribute $A$, for example, $B$ may mean preference for fish over chicken and $\bar{B}$ its complement. Assume further that the proportion of persons bearing $B$ is a known number $\pi_B$. Then, for an SRSWR in $n$ draws a sampled respondent is requested to report Yes or No truthfully about bearing $A$ with a probability $P$ and about bearing $B$ with the complementary probability $Q=1-P$. The sample proportion $p_{y r}$ of Yes responses is an unbiased estimator for
$$
\pi_y=P \pi_A+(1-P) \pi_B .
$$
Since $\pi_B$ is supposed known and $P$ is preassigned, an unbiased estimator for $\pi_A$ is
$$
\widehat{\pi}A=\left[p{y r}-(1-P) \pi_B / P,\right.
$$
provided $P \neq 0$.
One way to have $\pi_B$ known is to adopt the following modified device where a respondent is asked to (1) report Yes or No truthfully about bearing $A$ with probability $P_1$, (2) report Yes with a probability $P_2$ and (3) report No with a probability $P_3$, choosing numbers $P_1, P_2, P_3$ such that $0<P_1, P_2, P_3<1$ and $P_1+P_2+P_3=1$, using a pack of cards of three types mixed in proportions $P_1: P_2: P_3$. Then,
$$
\pi_y=P_1 \pi_A+P_2=P_1 \pi_A+\left(\frac{P_2}{P_2+P_3}\right)\left(1-P_1\right)
$$
and the known quantity $\frac{P_2}{P_2+P_3}$ may be supposed to play the role of $\pi_B$.
统计代写|抽样调查代考Survey sampling代考|Polychotomous Populations
Many alternative devices are available for the purpose we are discussing. We will mention selectively a few more. Suppose a population may be classified into several mutually exclusive and exhaustive categories according to a sensitive characteristic. For example, women may be classified according to the number of self-induced abortions so far implemented. Suppose, in general $\pi_i, i=1, \ldots, k, \sum_k^k \pi_i=1$, denote the unknown proportions of individuals belonging to $k$ disjoint and exhaustive categories according to a stigmatizing character. In order to estimate $\pi_i$ on taking an SRSWR of a given size $n$, let us apply the following device. Suppose small marbles or beads of $k$ distinct colors numbering $m_i, i=1,2, \ldots, k, \sum_k^k m_i=m$ are put into a flask with a long neck marked $1, \ldots, m$ spaced apart to accommodate one bead each when turned upside down with the mouth tightly closed. Each color represents a category and a sampled person is requested to shake the flask thoroughly, unobserved by the investigator, and to record on the questionnaire the number on the flask-neck accomodating the bottommost bead of the color of his/her category when turned upside down. Writing $\lambda_j$ as the probability of reporting the value $j, P_{i j}$ as the probability of reporting $j$ when the true category is $i$, and $p_j$ as the sample proportion of RR as $j$, we have $p_j$ as an estimator for $\lambda_j$ given by
$$
\lambda_j=\sum_{i=1}^k P_{i j} \pi_i, j=1, \ldots, J, \text { where } J=m-\min _{1 \leq i \leq k} m_i+1 .
$$
Here $P_{i j}$ is easy to calculate for the given $m_i$ ‘s, $i=1, \ldots, k$. For example,
$$
\begin{aligned}
& P_{11}=\frac{m_1}{m}, \
& P_{21}=\frac{m_2}{m}, \
& P_{12}=\frac{m-m_1}{m} \cdot \frac{m_1}{m-1}, \
& P_{23}=\frac{m-m_2}{m} \cdot \frac{m-m_2-1}{m-1} \cdot \frac{m_2}{m-2}
\end{aligned}
$$
and so on. The values of $m_i$ should be kept small and distinct for simplicity. Yet $J>k$. One good choice is $m_i=i ; i=1, \ldots, k$, in which case $J=m=k(k+1) / 2$. So, $\pi_i$ is to be estimated as $\hat{\pi}i$ on solving $$ p_j=\sum_1^p P{i j} \widehat{\pi}_i
$$
but a unique solution is not possible. One procedure recommended in the literature is to apply the theory of linear models. The solution requires evaluation of generalized inverses and is complicated and unlikely in practice to yield $\hat{\pi}_i$ within the permitted range $[0,1]$.
抽样调查代写
属性$A$和$\bar{A}$可能都是敏感的，例如，隶属于两个敌对的政治集团。下面描述了用于估计这种二分情况下$\pi_A$的另一种RR装置。
假设 $B$ 是否另一个无害的字符与敏感属性无关 $A$例如， $B$ 可能意味着更喜欢吃鱼而不是鸡肉和 $\bar{B}$ 它是互补的。进一步假设人口生育比例 $B$ 是已知的数 $\pi_B$． 然后，对于SRSWR $n$ 抽样调查对象被要求如实回答“是”或“否” $A$ 有概率地 $P$ 关于轴承 $B$ 具有互补概率 $Q=1-P$． 样本比例 $p_{y r}$ 是的无偏估计量
$$
\pi_y=P \pi_A+(1-P) \pi_B .
$$
自从 $\pi_B$ 应该是已知的 $P$ 是预分配的无偏估计量 $\pi_A$ 是
$$
\widehat{\pi}A=\left[p{y r}-(1-P) \pi_B / P,\right.
$$
提供 $P \neq 0$．
一种方法是 $\pi_B$ 在要求被调查者如实回答“是”或“否”的情况下，采用以下改进的方法 $A$ 有概率地 $P_1$，(2)有概率报告Yes $P_2$ (3)以概率报告No $P_3$，选择数字 $P_1, P_2, P_3$ 这样 $0<P_1, P_2, P_3<1$ 和 $P_1+P_2+P_3=1$使用一副按比例混合的三种纸牌 $P_1: P_2: P_3$． 然后，
$$
\pi_y=P_1 \pi_A+P_2=P_1 \pi_A+\left(\frac{P_2}{P_2+P_3}\right)\left(1-P_1\right)
$$
已知的量 $\frac{P_2}{P_2+P_3}$ 可能应该扮演的角色 $\pi_B$．
统计代写|抽样调查代考Survey sampling代考|Polychotomous Populations
对于我们正在讨论的目的，有许多可供选择的装置。我们将有选择地再提几个。假设一个种群可以根据一个敏感特征被分为几个相互排斥和详尽的类别。例如，可根据迄今实施的自行流产次数对妇女进行分类。假设，一般情况下$\pi_i, i=1, \ldots, k, \sum_k^k \pi_i=1$表示根据污名化特征属于$k$不相交和详尽类别的个人的未知比例。为了在给定大小$n$的SRSWR上估计$\pi_i$，让我们应用以下设备。假设将颜色为$k$、编号为$m_i, i=1,2, \ldots, k, \sum_k^k m_i=m$的小弹珠或小珠子放入一个长颈瓶中，长颈瓶中标有$1, \ldots, m$，长颈瓶与长颈瓶之间间隔一段，当瓶口紧闭时，每个长颈瓶可容纳一颗珠子。每种颜色代表一个类别，被采样的人被要求在调查员不注意的情况下彻底摇动烧瓶，并在问卷上记录当倒扣时，他/她类别颜色的最底部的头所在的烧瓶颈上的数字。将$\lambda_j$作为报告值的概率$j, P_{i j}$作为真实类别为$i$时报告$j$的概率，将$p_j$作为RR的样本比例$j$，我们得到$p_j$作为$\lambda_j$的估计量，由下式给出
$$
\lambda_j=\sum_{i=1}^k P_{i j} \pi_i, j=1, \ldots, J, \text { where } J=m-\min _{1 \leq i \leq k} m_i+1 .
$$
对于给定的$m_i$和$i=1, \ldots, k$, $P_{i j}$很容易计算。例如，
$$
\begin{aligned}
& P_{11}=\frac{m_1}{m}, \
& P_{21}=\frac{m_2}{m}, \
& P_{12}=\frac{m-m_1}{m} \cdot \frac{m_1}{m-1}, \
& P_{23}=\frac{m-m_2}{m} \cdot \frac{m-m_2-1}{m-1} \cdot \frac{m_2}{m-2}
\end{aligned}
$$
等等……为了简单起见，$m_i$的值应该保持小而不同。然而$J>k$。一个不错的选择是$m_i=i ; i=1, \ldots, k$，在这种情况下$J=m=k(k+1) / 2$。因此，求解$$ p_j=\sum_1^p P{i j} \widehat{\pi}_i
$$时，$\pi_i$被估计为$\hat{\pi}i$
但唯一的解是不可能的。文献中推荐的一个程序是应用线性模型理论。该解决方案需要评估广义逆，是复杂的，不太可能在实践中产生$\hat{\pi}_i$在允许范围$[0,1]$。
统计代写|抽样调查代考Survey sampling代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
