Riemann surface
matlab
不必忧虑！我们的抽样调查专家团队会以相同的承诺为您解决问题。我们拥有广泛的专业知识和丰富的经验，可以帮助您应对在抽样调查学习中遇到的各种困难。无论是复杂的作业还是论文，我们都有能力为您提供协助，确保您在学习过程中取得顺利的进展！
以下是一些我们可以帮助您解决的问题：
抽样调查的基本概念：包括抽样调查的基础知识，如总体、样本、参数和统计量等。
抽样方法和设计：探究各种抽样方法，包括随机抽样、系统抽样、分层抽样和整群抽样，以及如何设计一个有效的抽样方案。
数据收集和数据处理：介绍在抽样调查中如何进行数据收集和数据处理，包括调查问卷的设计和调查数据的清洗。
抽样误差和非抽样误差：研究抽样误差和非抽样误差的概念，以及如何减少这些误差。
样本估计和假设检验：研究如何通过样本数据对总体参数进行估计和假设检验，包括点估计、区间估计和各种假设检验方法。
抽样调查的案例分析：介绍一些实际的抽样调查案例，包括成功的和失败的，以及从中可以学到的经验和教训。
复杂抽样设计：探讨更复杂的抽样设计，如多阶段抽样和抽样权重的计算。
无论您在抽样调查方面遇到什么问题，我们都会全力为您提供专业的帮助，确保您的学习之路顺利无阻！
Exercise 2.1 Cultivated surface area
We want to estimate the surface area cultivated on the farms of a rural township. Of the $N=2010$ farms that comprise the township, we select 100 using simple random sampling. We measure $y_k$, the surface area cultivated on the farm $k$ in hectares, and we find
$$
\sum_{k \in S} y_k=2907 \text { ha and } \sum_{k \in S} y_k^2=154593 \mathrm{ha}^2 .
$$
Give the value of the standard unbiased estimator of the mean
$$
\bar{Y}=\frac{1}{N} \sum_{k \in U} y_k
$$
Give a $95 \%$ confidence interval for $\bar{Y}$.
Solution
In a simple design, the unbiased estimator of $\bar{Y}$ is
$$
\widehat{\bar{Y}}=\frac{1}{n} \sum_{k \in S} y_k=\frac{2907}{100}=29.07 \text { ha. }
$$
The estimator of the dispersion $S_y^2$ is
$$
s_y^2=\frac{n}{n-1}\left(\frac{1}{n} \sum_{k \in S} y_k^2-\widehat{\bar{Y}}^2\right)=\frac{100}{99}\left(\frac{154593}{100}-29.07^2\right)=707.945 .
$$
82 Simple Random Sampling
The sample size $n$ being ‘sufficiently large’, the $95 \%$ confidence interval is estimated in hectares as follows:
$$
\begin{aligned}
{\left[\hat{\bar{Y}} \pm 1.96 \sqrt{\frac{N-n}{N} \frac{s_y^2}{n}}\right] } & =\left[29.07 \pm 1.96 \sqrt{\frac{2010-100}{2010} \times \frac{707.45}{100}}\right] \
& =[23.99 ; 34.15]
\end{aligned}
$$
Exercise 2.2 Occupational sickness
We are interested in estimating the proportion of men $P$ affected by an occupational sickness in a business of 1500 workers. In addition, we know that three out of 10 workers are usually affected by this sickness in businesses of the same type. We propose to select a sample by means of a simple random sample.
What sample size must be selected so that the total length of a confidence interval with a 0.95 confidence level is less than 0.02 for simple designs with replacement and without replacement ?
What should we do if we do not know the proportion of men usually affected by the sickness (for the case of a design without replacement) ?
To avoid confusions in notation, we will use the subscript $W R$ for estimators with replacement, and the subscript $W O R$ for estimators without replacement.
Solution
a) Design with replacement.
If the design is of size $m$, the length of the (estimated) confidence interval at a level $(1-\alpha)$ for a mean is given by
$$
\mathrm{CI}(1-\alpha)=\left[\hat{\bar{Y}}-z_{1-\alpha / 2} \sqrt{\frac{\tilde{s}y^2}{m}}, \hat{\bar{Y}}+z{1-\alpha / 2} \sqrt{\frac{\tilde{s}y^2}{m}}\right], $$ where $z{1-\alpha / 2}$ is the quantile of order $1-\alpha / 2$ of a random normal standardised variate. If we denote $\widehat{P}{W R}$ as the estimator of the proportion for the design with replacement, we can write $$ \begin{aligned} & C I(1-\alpha)=\left[\widehat{P}{W R}-z_{1-\alpha / 2} \sqrt{\frac{\widehat{P}{W R}\left(1-\hat{P}{W R}\right)}{m-1}} .\right. \
& \left.\widehat{P}{W R}+z{1-\alpha / 2} \sqrt{\frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1}}\right] \
&
\end{aligned}
$$
Exercise 2.2
9
Indeed, in this case,
$$
\widehat{\operatorname{var}}\left(\widehat{P}{W R}\right)=\frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{(m-1)} . $$ So that the total length of the confidence interval does not exceed 0.02 , it is necessary and sufficient that $$ 2 z{1-\alpha / 2} \sqrt{\frac{\hat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1}} \leq 0.02 .
$$
By dividing by two and squaring, we get
$$
z_{1-\alpha / 2}^2 \frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1} \leq 0.0001,
$$
which gives
$$
m-1 \geq z_{1-\alpha / 2}^2 \frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{0.0001}
$$
For a $95 \%$ confidence interval, and with an estimator of $P$ of 0.3 coming from a source external to the survey, we have $z_{1-\alpha / 2}=1.96$, and
$$
m=1+1.96^2 \times \frac{0.3 \times 0.7}{0.0001}=8068.36 .
$$
The sample size $(m=8069)$ is therefore larger than the population size, which is possible (but not prudent) since the sampling is with replacement.
E-mail: help-assignment@gmail.com 微信:shuxuejun
help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务