Tag Archives: Sampling Theory of Surveys 代写

调查的抽样理论|STA 144/STAT 451/STAT 506/STA 317Sampling Theory of Surveys 代写

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这是一份贝叶斯统计推断作业代写的成功案

调查的抽样理论|STA 144/STAT 451/STAT 506/STA 317Sampling Theory of Surveys 代写



$$
-\sum_{i} \log \left(1-y_{1} a\right)=s_{1} a+s_{2} a_{2}^{2}+s_{3} a_{3}^{3}+\ldots
$$
where $a$ is any constant such that
$$
|a|<\frac{1}{\max y_{6}}
$$
giving us
$$
g\left(p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots\right)=\sum g_{1}(P, Q) s\left(q_{1}^{x_{1}} q_{2}{ }^{\times *} \ldots\right)
$$
and
$$
s\left(p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots\right)=\sum s_{0}(P, Q) g\left(q_{1}^{x_{1}} q_{2}^{x_{1}} \ldots\right)
$$



英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

STA 144/STAT 451/STAT 506/STA 317 COURSE NOTES ：

$N^{t} e_{j}=0$ for $i<j$
and
$$
N^{j} e_{i}=n(n-1) \ldots(n-j+1)
$$
Hence
$$
V\left(s^{2}\right)=\frac{\mu_{4}-\mu_{2}^{2}}{n}+\frac{2}{n(n-1)}{ }^{\mu_{2}^{2}} .
$$
Using the Pearsonian notation for departure from normality, this can be written as




调查的抽样理论 | Sampling Theory of Surveys 代写 STATS 3003 代考

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这是一份adelaide阿德莱德大学 STATS 3003 作业代写的成功案

调查的抽样理论 | Sampling Theory of Surveys 代写 STATS 3003 代考
问题 1.

a Suppose we ignore the fpc of a modcl-based estimator. Find
$$
V_{M}\left(\sum_{i \in S} \sum_{j \in S_{i}} b_{i j} Y_{i j}\right) \text {. }
$$
b Prove (5.39). HINT: Let
$$
c_{i j}= \begin{cases}b_{i j}-1 & \text { if } i \in \mathcal{S} \text { and } j \in \mathcal{S}{i} . \ -1 & \text { otherwise. }\end{cases} $$ Then, $\hat{T}-T=\sum{i=1}^{N} \sum_{j=1}^{M_{i}} c_{i j} Y_{i j}$.


证明 .

(Requires linear aIgebra and calculus.) Although $\hat{T}{r}$ is unbiased for model Ml, constructing an estimator with smaller variance is possible. Let $$ c{k}=\frac{m_{k}}{1+\rho\left(m_{k}-1\right)}
$$
and
$$
\hat{T}{\text {opt }}=\sum{i \in S} \sum_{j \in S_{i}} \frac{c_{i}}{m_{i}}\left[\rho M_{i}+\frac{K-\rho \sum_{k \in S} c_{k} \cdot M_{k}}{\sum_{k \in S} c_{k}}\right] Y_{i j}
$$

英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

STATS 3003 COURSE NOTES ：

Then, for one-stage pps sampling, $t_{i} / \psi_{i}=K \bar{y}{i}$, so $$ \begin{aligned} &\hat{t}{\psi}=\frac{K}{n} \sum_{i=1}^{N} Q_{i} \bar{y}{i} \ &\hat{\hat{y}}{\psi}=\frac{1}{n} \sum_{i=1}^{N} Q_{i} \bar{y}_{i}
\end{aligned}
$$