# 统计代写|抽样调查作业代写sampling theory of survey代考|PSY279

## 统计代写|抽样调查作业代写sampling theory of survey代考|Estimation of Population Proportion

Sometimes we need to estimate the proportion of the population that possesses a certain attribute $A$, such as smoking, drug addiction, or unemployment. In such a situation, we take $y_i=1$ if the ith unit belongs to the group $A$ and $\gamma_i=0$ otherwise (if the ith unit does not belong to the group $A$ ). So in this case, $Y=N_A=$ total number of units possessing the attribute $A$ and $\bar{Y}=N_A / N=\pi_A=$ proportion of units in the population belonging to the group $A ; \bar{\gamma}(s)=n_A / n=\widehat{\pi}_A=$ proportion of units in the sample that belong to the group $A$ and $n_A$ is the total number of units in the sample that belong to the group $A$. Now noting

(i) $S_Y^2=\frac{1}{(N-1)} \sum_{i=1}^N\left(y_i-\bar{Y}\right)^2=\frac{1}{(N-1)}\left(\sum_{i=1}^N y_i-N \bar{Y}^2\right)$ $=\frac{N\left{\pi_A\left(1-\pi_A\right)\right}}{N-1}\left(\right.$ since $\gamma_i=0$ or 1$)$,
(ii) $s_y^2=\sum_{i \in s}\left{y_i-\bar{y}(s)\right}^2 /(n-1)=\frac{n}{n-1} \widehat{\pi}_A\left(1-\widehat{\pi}_A\right)$, and using Theorem 3.2.2, we have the following.
Theorem 3.2.8
For an SRSWOR sampling of size $n$,
(i) $\hat{\pi}_A$ is an unbiased estimator for the population proportion $\pi_A$.
(ii) Variance of $\widehat{\pi}_A$ is
$$V\left(\widehat{\pi}_A\right)=\frac{N-n}{n(N-1)} \pi_A\left(1-\pi_A\right)$$
and
(iii) An unbiased estimator of $V\left(\widehat{\pi}_A\right)$ is
$$\widehat{V}\left(\widehat{\pi}_A\right)=\frac{N-n}{N(n-1)} \widehat{\pi}_A\left(1-\widehat{\pi}_A\right) .$$

## 统计代写|抽样调查作业代写sampling theory of survey代考|Estimation of Domain Mean and Total

Suppose we want to estimate $Y_D=\sum_{i \in D} \gamma_i$, the domain total of the $\gamma$-values of a section $D$ of the population $U$. Let $N_D$ be the population domain size $=$ total number of units in the domain $D$, then $\bar{Y}D=Y_D / N_D=$ population domain mean of $\gamma \quad$ and $S{\gamma D}^2=$ $\sum_{i \in D}\left(y_i-\bar{Y}D\right)^2 /\left(N_D-1\right)=$ population domain variance of $\gamma$. Let $s_D(\subset s)$ be the set of units in the sample $s$, which belong to $D$, and $n_D(\leq n)$ be the total numbers of units in $s_D, \bar{y}\left(s_D\right)=\sum{i \in s_D} y_i / n_D=$ sample domain mean and $s_{\gamma D}^2=\sum_{i \in s_D}\left{\gamma_i-\bar{\gamma}\left(s_D\right)\right}^2 /\left(n_D-1\right)$ be the sample domain variance.

Let us define $z_i=d_i y_i$, where $d_i=1$ if the $i$ th unit belongs to $D$ and $d_i=0$ otherwise. Then, $Z=\sum_{i=1}^N z_i=Y_D=$ population domain total, $\bar{Y}_D=Y_D / N_D=$ population domain mean, $\pi_D=N_D / N=$ population domain proportion, and $\bar{Z}=Z / N=\pi_D \bar{Y}_D$; the sample domain mean $=\bar{\gamma}\left(s_D\right)=\sum_{i \in s_D} \gamma_i / n_D$ and the sample domain proportion $=$ $\widehat{\pi}D=n_D / n$. Furthermore, $S_z^2=\sum{i \in U}\left(z_i-\bar{Z}\right)^2 /(N-1)$
$=\left[\sum_{i \in D} \gamma_i^2-N\left(\pi_D \bar{Y}D\right)^2\right] /(N-1)$ $=\left(N_D-1\right) S{\gamma D}^2 /(N-1)+N_D \bar{Y}D^2\left(1-\pi_D\right) /(N-1)$ $=\left[\left(N \pi_D-1\right) S{\gamma D}^2+N \pi_D\left(1-\pi_D\right) \bar{Y}D^2\right] /(N-1)$ $\bar{z}(s)=\sum{i \in s} z_i / n=\sum_{i \in s_D} \gamma_i / n=\widehat{\pi}D \bar{\gamma}\left(s_D\right)$ and $s_z^2=\sum{i \in s}\left[z_i-\bar{z}(s)\right]^2 /(n-1)$
$=\left(n \widehat{\pi}D-1\right) s{y D}^2 /(n-1)+n \widehat{\pi}_D\left(1-\widehat{\pi}_D\right)\left{\bar{\gamma}\left(s_D\right)\right}^2 /(n-1)$

# 抽样调查代考

## 统计代写|抽样调查作业代写sampling theory of survey代考|Estimation of Population Proportion

$A ; \bar{\gamma}(s)=n_A / n=\widehat{\pi}A=$ 样本中属于该组的单位比例 $A$ 和 $n_A$ 是样本中属于该组的单元总数 $A$. 现在注意到 (一世) $S_Y^2=\frac{1}{(N-1)} \sum{i=1}^N\left(y_i-\bar{Y}\right)^2=\frac{1}{(N-1)}\left(\sum_{i=1}^N y_i-N \bar{Y}^2\right)$

$(-) \hat{\pi}_A$ 是人口比例的无偏估计量 $\pi_A$.
$$V\left(\widehat{\pi}_A\right)=\frac{N-n}{n(N-1)} \pi_A\left(1-\pi_A\right)$$
iii) 无偏估计 $V\left(\widehat{\pi}_A\right)$ 是
$$\widehat{V}\left(\widehat{\pi}_A\right)=\frac{N-n}{N(n-1)} \widehat{\pi}_A\left(1-\widehat{\pi}_A\right) .$$

## 统计代写|抽样调查作业代写sampling theory of survey代考|Estimation of Domain Mean and Total

$=\left[\sum_{i \in D} \gamma_i^2-N\left(\pi_D \bar{Y} D\right)^2\right] /(N-1)$
$=\left(N_D-1\right) S \gamma D^2 /(N-1)+N_D \bar{Y} D^2\left(1-\pi_D\right) /(N-1)$
$=\left[\left(N \pi_D-1\right) S \gamma D^2+N \pi_D\left(1-\pi_D\right) \bar{Y} D^2\right] /(N-1)$

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