统计代写|概率与统计作业代写Probability and Statistics代考|EAS305

统计代写|概率与统计作业代写Probability and Statistics代考|Mean Squared Error

A small bias of an estimator under a sampling plan does not guarantee that individual sample results $\hat{\theta}_k$ are actually close to the population parameter $\theta$; it just states that they are close on average, if we were to sample over and over again. However, we often only collect one sample, and thus the performance of an estimator on average is not our only concern. We are also concerned with the variability of the estimator across different samples. To capture the variability in the sample results $\hat{\theta}1, \hat{\theta}_2, \ldots, \hat{\theta}_K$ with respect to the true value $\theta$, we use the so-called mean squared error (MSE). This is defined as $$\mathrm{MSE}=\sum{k=1}^K\left(\hat{\theta}_k-\theta\right)^2 \pi_k$$
The MSE measures the weighted average squared distance of the sample results $\hat{\theta}_1, \hat{\theta}_2, \ldots, \hat{\theta}_K$ from the population parameter $\theta$. The weights are again determined by the sampling probabilities. Often, the smaller the MSE the better the sampling plan. Sometimes the root mean square error (RMSE) is reported, which is simply $\mathrm{RMSE}=\sqrt{\mathrm{MSE}}$.

Another measure that is relevant to sampling plans, and closely related to the bias and MSE, is the standard error (SE). The standard error is defined as
$$\mathrm{SE}=\sqrt{\sum_{k=1}^K\left(\hat{\theta}_k-\mathbb{E}(T)\right)^2 \pi_k}$$
It represents the variability of the sampling plan with respect to the expected population parameter $\mathbb{E}(T)$ instead of using the true population parameter $\theta$. Note that the standard error of an estimator is used as a measure to represent our uncertainty regarding an estimate. In many examples the standard error contains population parameters (see Sect. 2.6) that we do not know. To use standard errors in practice we have to estimate the standard error as well, and this is what researchers and professionals typically do. We will explore this in more detail below when we derive analytical expressions for the bias and standard error of the sample mean, sample variance, and sample proportion for simple random sampling, systematic sampling, stratified sampling, and one-stage and two-stage cluster random sampling, respectively.

Figure $2.2$ shows how bias, MSE, and SE relate: if the bias is small, $\mathbb{E}(T)$ is close to the parameter value $\theta$. On the other hand, if the bias is large, $\mathbb{E}(T)$ is not close to $\theta$. If the MSE is small, the variability of the $\hat{\theta}_k$ ‘s around $\theta$ is small, while if the MSE is large, the variability around $\theta$ is large. If the SE is small, the variability of the $\hat{\theta}_k$ ‘s around $\mathbb{E}(T)$ is small. Finally, note that if the sampling plan is unbiased and thus $\mathbb{E}(T)=\theta$, the RMSE and the SE are identical. More generally, it can be demonstrated that
$$\mathrm{RMSE}=\sqrt{\mathrm{SE}^2+(\mathbb{E}(T)-\theta)^2}$$
Thus the RMSE is never smaller than the SE.

统计代写|概率与统计作业代写Probability and Statistics代考|Illustration of a Comparison of Sampling Plans

Here we will assume that we have full knowledge about the population, so that we can evaluate the bias, standard error, and mean squared error for different sampling plans. Clearly, in practice we never have this information, otherwise the sampling becomes obsolete. However, we often do have some knowledge of the population in practice, using information from registries or historical data, and this information can be used to evaluate different strategies, often in combination with simulations (see Sect. 2.5.4).

Our population of interest is provided in Table 2.1, which is taken from Table $2.3$ of Levy and Lemeshow (2013). The population consists of six schools in a community with in each school the total number of students and the number of students that were not immunized for measles. In total there are 314 students, of which 30 students are not immunized for measles. The population parameter of interest is $\theta=30 / 314=$ $0.09554$, the proportion of students not being immunized for measles. We assume that schools 1,3, and 4 are located in the north of the community and the schools 2 , 5 , and 6 are located in the south. Two sampling approaches are being considered: a single-stage cluster sample with a simple random sample of two clusters (schools) and a single-stage cluster sample with stratified sampling of two clusters. For the stratified sampling, the strata are north and south.

For the single-stage cluster sampling with simple random sampling, there are $K=6 ! /[2 ! \times 4 !]=15$ possible samples of two schools: $S_1=(1,2) ; S_2=$ $(1,3) ; \ldots ; S_6=(1,6) ; S_7=(2,3) ; \ldots ; S_{10}=(2,6) ; \ldots ; S_{15}=(5,6)$. Each pair of schools has the same probability of being collected, i.e., $\pi_k=1 / K=1 / 15$ for $k=1,2, \ldots, 15$. The expected population parameter for this sampling approach is given by
\begin{aligned} \mathbb{E}(T)=& \sum_{k=1}^K \hat{\theta}_k \pi_k \ =& {\left[\frac{9}{87}+\frac{7}{149}+\frac{7}{103}+\frac{11}{95}+\frac{12}{116}+\frac{8}{118}+\frac{8}{72}+\frac{12}{64}\right.} \ \left.\quad+\frac{13}{85}+\frac{6}{134}+\frac{10}{126}+\frac{11}{147}+\frac{10}{80}+\frac{11}{101}+\frac{15}{93}\right] \times \frac{1}{15} \ =& 0.10341 . \end{aligned}
The bias is therefore determined by bias $=0.10341-0.09554=0.00787$. Thus the simple random sample of two schools (single stage cluster sample) is not fully unbiased, but the bias is relatively small.

概率与统计代考

统计代写|概率与统计作业代写概率与统计代考|均方误差

。MSE度量样本结果$\hat{\theta}_1, \hat{\theta}_2, \ldots, \hat{\theta}_K$到总体参数$\theta$的加权平均平方距离。权重由抽样概率决定。通常，均方误差越小，抽样方案越好。有时报告的均方根误差(RMSE)是$\mathrm{RMSE}=\sqrt{\mathrm{MSE}}$ .

$$\mathrm{SE}=\sqrt{\sum_{k=1}^K\left(\hat{\theta}_k-\mathbb{E}(T)\right)^2 \pi_k}$$

$$\mathrm{RMSE}=\sqrt{\mathrm{SE}^2+(\mathbb{E}(T)-\theta)^2}$$

统计代写|概率与统计作业代写Probability and Statistics代考|Illustration of a Comparison of Sampling Plans

\begin{aligned} \mathbb{E}(T)=& \sum_{k=1}^K \hat{\theta}_k \pi_k \ =& {\left[\frac{9}{87}+\frac{7}{149}+\frac{7}{103}+\frac{11}{95}+\frac{12}{116}+\frac{8}{118}+\frac{8}{72}+\frac{12}{64}\right.} \ \left.\quad+\frac{13}{85}+\frac{6}{134}+\frac{10}{126}+\frac{11}{147}+\frac{10}{80}+\frac{11}{101}+\frac{15}{93}\right] \times \frac{1}{15} \ =& 0.10341 . \end{aligned}

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