# 统计代写|概率与统计作业代写Probability and Statistics代考|CEE305

## 统计代写|概率与统计作业代写Probability and Statistics代考|Systematic Sampling

Recall that for systematic sampling we split our population into $n$ groups consisting of $m$ units, where from each group we select the $k$ th unit $(k \in{1,2, \ldots, m})$. Here we will assume that the ratio of the population size $N$ and sample size $n$ is an integer, i.e., $N / n=m \in{1,2,3, \ldots ., N}$ to keep calculations mathematically more simple. The population is now perfectly split into $n$ groups of size $m$. The sampling plan is $S_1$, $S_2, \ldots, S_m$, with $S_k={k, k+m, k+2 m, \ldots, k+(n-1) m}$, and each sample $S_k$ having probability $1 / m$ of being collected. The sample average can now be written as $\bar{x}k=\sum{i=1}^n x_{k+m(i-1)} / n$.

If the population can be perfectly split up into $n$ groups of $m$ units, the sample average $\bar{x}k$ is an unbiased estimator of the population mean $\mu=$ $\sum{h=1}^m \sum_{i=1}^n x_{h+m(i-1)} / N$, with $N=m n$. The mean square error is given by (see Table $2.2)$
$$\sigma^2-\frac{1}{N} \sum_{h=1}^n \sum_{i=1}^m\left(x_{h+m(i-1)}-\bar{x}h\right)^2,$$ with $\sigma^2$ the population variance given by $\sigma^2=\sum{k=1}^m \sum_{i=1}^n\left(x_{k+m(i-1)}-\mu\right)^2 / N$ and with $\bar{x}_h$ the sample average for sample set $S_h$. It is obvious that the MSE of the sample average under systematic sampling is different from the MSE of the sample average under simple random sampling (just compare Eq. (2.4) with Eq. (2.2)). Systematic sampling can be more efficient than simple random sampling, in particular when the variance in the systematic samples is larger than the population variance (which is impossible to verify in practice).

## 统计代写|概率与统计作业代写Probability and Statistics代考|Stratified Sampling

To discuss the properties of the weighted sample average under stratified sampling we will change the notation for the index of units. Instead of using index $i$ for a unit in the population, we will use the indices ( $h, i)$ to indicate the unit $i \in\left{1,2, \ldots, N_h\right}$ for units in stratum $h \in{1,2, \ldots, M}$, and with $N_1+N_2+\cdots+N_M=N$ the total number of population units. Thus the variable $x_{h i}$ represents the value of unit $i$ in stratum $h$. The population mean can then be rewritten into
$$\mu=\frac{1}{N} \sum_{h=1}^M \sum_{i=1}^n x_{h i}=\sum_{h=1}^m w_h \mu_h$$
with $w_h=N_h / N$ and $\mu_h=\sum_{i=1}^{N_h} x_{h i} / N_h$ the population average in stratum $h$ or the strata mean. Note that the weights add up to one, i.e., $w_1+w_2+\cdots+w_M=1$. Thus the population mean is a weighted mean of the strata means $\mu_h$.

The variance $\sigma_h^2$ in stratum $h$ is defined as $\sigma_h^2=\sum_{i=1}^{N_h}\left(x_{h i}-\mu_h\right)^2 / N_h$. The relationship between the population variance and the strata variances is given by
$$\sigma^2 \equiv \frac{1}{N} \sum_{h=1}^M \sum_{i=1}^{N_h}\left(x_{h i}-\mu\right)^2=\sum_{h=1}^M w_h \sigma_h^2+\sum_{h=1}^M w_h\left(\mu_h-\mu\right)^2$$
Thus the population variance is the sum of two parts. The first part represents a weighted mean of the within strata variances and the second part represents a weighted mean of the squared distances of the strata means to the population mean. They may be referred to as within and between (strata) variances.

Now let’s assume that we have determined in some way the sample size $n_h$ in stratum $h$, such that the sum is equal to the total sample size $n=n_1+n_2+$ $\cdots+n_m$. In case we draw a simple random sample in each stratum, the possible samples for stratum $h$ are now denoted by $S_{h, 1}, S_{h, 2}, \ldots, S_{h, K_h}$, with $K_h=$ $N_{h} ! /\left[n_{h} !\left(N_h-n_h\right) !\right] . S_{h, k}$ is the collected sample in stratum $h$, the sample mean is given by $\bar{x}{h, k}-\sum{i \in S_{h, k}} x_{h i} / n_h$. The samplé variance in stratum $h_h$ is thên denoted by $s_{h, k}^2=\sum_{i \in S_{h, k}}\left(x_{h i}-\bar{x}_{h, k}\right)^2 /\left(n_h-1\right)$.

# 概率与统计代考

## 统计代写|概率与统计作业代写概率统计代考|系统抽样

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