### 统计代写|生物统计学作业代写Biostatistics代考| BRIEF NOTES ON THE FUNDAMENTALS

statistics-lab™ 为您的留学生涯保驾护航 在代写生物统计学Biostatistics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写生物统计学Biostatistics方面经验极为丰富，各种代写生物统计学Biostatistics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|生物统计学作业代写Biostatistics代考|Mean and Variance

As seen in Sections $3.3$ and $3.4$, a probability density function $f$ is defined so that:
(a) $f(k)=\operatorname{Pr}(X=k)$ in the discrete case
(b) $f(x) d x=\operatorname{Pr}(x \leq X \leq x+d x)$ in the continuous case
For a continuous distribution, such as the normal distribution, the mean $\mu$ and variance $\sigma^{2}$ are calculated from:
(a) $\mu=\int x f(x) d x$
(b) $\sigma^{2}=\int(x-\mu)^{2} f(x) d x$
For a discrete distribution, such as the binomial distribution or Poisson distribution, the mean $\mu$ and variance $\sigma^{2}$ are calculated from:

(a) $\mu=\sum x f(x)$
(b) $\sigma^{2}=\sum(x-\mu)^{2} f(x)$
For example, we have for the binomial distribution,
\begin{aligned} \mu &=n p \ \sigma^{2} &=n p(1-p) \end{aligned}
and for the Poisson distribution,
\begin{aligned} \mu &=\theta \ \sigma^{2} &=\theta \end{aligned}

## 统计代写|生物统计学作业代写Biostatistics代考|Pair-Matched Case–Control Study

Data from epidemiologic studies may come from various sources, the two fundamental designs being retrospective and prospective (or cohort). Retrospective studies gather past data from selected cases (diseased individuals) and controls (nondiseased individuals) to determine differences, if any, in the exposure to a suspected risk factor. They are commonly referred to as case-control studies. Cases of a specific disease, such as lung cancer, are ascertained as they arise from population-based disease registers or lists of hospital admissions, and controls are sampled either as disease-free persons from the population at risk or as hospitalized patients having a diagnosis other than the one under investigation. The advantages of a case-control study are that it is economical and that it is possible to answer research questions relatively quickly because the cases are already available. Suppose that each person in a large population has been classified as exposed or not exposed to a certain factor, and as having or not having some disease. The population may then be enumerated in a $2 \times 2$ table (Table 3.12), with entries being the proportions of the total population.
Using these proportions, the association (if any) between the factor and the disease could be measured by the ratio of risks (or relative risk) of being disease positive for those with or without the factor:

\begin{aligned} \text { relative risk } &=\frac{P_{1}}{P_{1}+P_{3}} \div \frac{P_{2}}{P_{2}+P_{4}} \ &=\frac{P_{1}\left(P_{2}+P_{4}\right)}{P_{2}\left(P_{1}+P_{3}\right)} \end{aligned}
since in many (although not all) situations, the proportions of subjects classified as disease positive will be small. That is, $P_{1}$ is small in comparison with $P_{3}$, and $P_{2}$ will be small in comparison with $P_{4}$. In such a case, the relative risk is almost equal to
\begin{aligned} \theta &=\frac{P_{1} P_{4}}{P_{2} P_{3}} \ &=\frac{P_{1} / P_{3}}{P_{2} / P_{4}} \end{aligned}
the odds ratio of being disease positive, or
$$=\frac{P_{1} / P_{2}}{P_{3} / P_{4}}$$
the odds ratio of being exposed. This justifies the use of an odds ratio to determine differences, if any, in the exposure to a suspected risk factor.

As a technique to control confounding factors in a designed study, individual cases are matched, often one to one, to a set of controls chosen to have similar values for the important confounding variables. The simplest example of pair-matched data occurs with a single binary exposure (e.g., smoking versus nonsmoking). The data for outcomes can be represented by a $2 \times 2$ table (Table 3.13) where $(+,-)$ denotes (exposed, unexposed).

For example, $n_{10}$ denotes the number of pairs where the case is exposed, but the matched control is unexposed. The most suitable statistical model for making inferences about the odds ratio $\theta$ is to use the conditional probability of the number of exposed cases among the discordant pairs. Given $n=n_{10}+n_{01}$ being fixed, it can be seen that $n_{10}$ has $B(n, p)$, where
$$p=\frac{\theta}{1+\theta}$$

## 统计代写|生物统计学作业代写Biostatistics代考|NOTES ON COMPUTATIONS

In Sections $1.4$ and $2.5$ we covered basic techniques for Microsoft’s Excel: how to open/form a spreadsheet, save it, retrieve it, and perform certain descriptive statistical tasks. Topics included data-entry steps, such as select and drag, use of formula bar, bar and pie charts, histograms, calculations of descritive statistics such as mean and standard deviation, and calculation of a coefficient of correlation. In this short section we focus on probability models related to the calculation of areas under density curves, especially normal curves and $t$ curves.
Normal Curves The first two steps are the same as in obtaining descriptive statistics (but no data are needed now): (1) click the paste function icon, $\mathrm{f}^{*}$, and (2) click Statistical. Among the functions available, two are related to normal curves: NORMDIST and NORMINV. Excel provides needed information for any normal distribution, not just the standard normal distribution as in Appendix B. Upon selecting either one of the two functions above, a box appears asking you to provide (1) the mean $\mu,(2)$ the standard deviation $\sigma$, and (3) in the last row, marked cumulative, to enter TRUE (there is a choice $F A L S E$, but you do not need that). The answer will appear in a preselected cell.

• NORMDIST gives the area under the normal curve (with mean and variance provided) all the way from the far-left side (minus infinity) to the value $x$ that you have to specify. For example, if you specify $\mu=0$ and $\sigma=1$, the return is the area under the standard normal curve up to the point specified (which is the same as the number from Appendix B plus $0.5)$.
• NORMINV performs the inverse process, where you provide the area under the normal curve (a number between 0 and 1), together with the mean $\mu$ and standard deviation $\sigma$, and requests the point $x$ on the horizontal axis so that the area under that normal curve from the far-left side (minus infinity) to the value $x$ is equal to the number provided between 0 and 1. For example, if you put in $\mu=0, \sigma=1$, and probability $=0.975$, the return is $1.96$; unlike Appendix $B$, if you want a number in the right tail of the curve, the input probability should be a number greater than $0.5$.
The $t$ Curves: Procedures TDIST and TINV We want to learn how to find the areas under the normal curves so that we can determine the $p$ values for statistical tests (a topic starting in Chapter 5). Another popular family in this category is the $t$ distributions, which begin with the same first two steps: (1) click the paste function icon, $\mathrm{f}^{*}$, and (2) click Statistical. Among the functions available, two related to the $t$ distributions are TDIST and TINV. Similar to the case of NORMDIST and NORMINV, TDIST gives the area under the $t$ curve, and TINV performs the inverse process where you provide the area under the curve and request point $x$ on the horizontal axis. In each case you have to provide the degrees of freedom. In addition, in the last row, marked with tails, enter:
• (Tails=) $I$ if you want one-sided
• (Tails $=) 2$ if you want $t$ wo-sided
(More details on the concepts of one- and two-sided areas are given in Chapter
5.) For example:
• Example 1: If you enter $(\mathrm{x}=) 2.73,($ deg freedom $=) 18$, and, (Tails $=) 1$, you’re requesting the area under a $t$ curve with 18 degrees of freedom and to the right of $2.73$ (i.e., right tail); the answer is $0.00687$.
• Example 2: If you enter $(\mathrm{x}=) 2.73$, (deg freedom $=) 18$, and (Tails $=) 2$, you’re requesting the area under a $t$ curve with 18 degrees of freedom and to the right of $2.73$ and to the left of $-2.73$ (i.e., both right and left tails); the answer is $0.01374$, which is twice the previous answer of $0.00687$.

## 统计代写|生物统计学作业代写Biostatistics代考|Mean and Variance

(a)F(ķ)=公关⁡(X=ķ)在离散情况下
(b)F(X)dX=公关⁡(X≤X≤X+dX)在连续情况下

(a)μ=∫XF(X)dX
(二)σ2=∫(X−μ)2F(X)dX

（一种）μ=∑XF(X)
(二)σ2=∑(X−μ)2F(X)

μ=np σ2=np(1−p)

μ=θ σ2=θ

## 统计代写|生物统计学作业代写Biostatistics代考|Pair-Matched Case–Control Study

θ=磷1磷4磷2磷3 =磷1/磷3磷2/磷4

=磷1/磷2磷3/磷4

p=θ1+θ

## 统计代写|生物统计学作业代写Biostatistics代考|NOTES ON COMPUTATIONS

• NORMDIST 给出从最左侧（负无穷大）一直到值的正态曲线下面积（提供均值和方差）X您必须指定。例如，如果您指定μ=0和σ=1，返回是标准正态曲线下到指定点的面积（与附录 B 中的数字相同加上0.5).
• NORMINV 执行逆过程，您提供正态曲线下的面积（0 到 1 之间的数字）以及平均值μ和标准差σ, 并请求点X在水平轴上，使得从最左侧（负无穷大）到该值的正态曲线下的面积X等于在 0 和 1 之间提供的数字。例如，如果您输入μ=0,σ=1, 和概率=0.975，回报是1.96; 不同于附录乙，如果你想在曲线的右尾有一个数字，输入概率应该是一个大于0.5.
这吨曲线：过程 TDIST 和 TINV 我们想学习如何找到正态曲线下的区域，以便我们可以确定p统计检验的值（从第 5 章开始的主题）。此类别中另一个受欢迎的家庭是吨发行版，前两个步骤相同：（1）单击粘贴功能图标，F∗，和 (2) 单击统计。在可用的功能中，有两个与吨分布是 TDIST 和 TINV。与 NORMDIST 和 NORMINV 的情况类似，TDIST 给出了吨曲线，TINV 执行逆过程，您提供曲线下面积和请求点X在水平轴上。在每种情况下，您都必须提供自由度。此外，在最后一行标有尾巴的地方，输入：
• （尾巴=）一世如果你想要一面
• （尾巴=)2如果你想吨双面
（有关单面和双面区域概念的更多详细信息，请参见第
5 章。）例如：
• 示例 1：如果您输入(X=)2.73,(你自由=)18, 并且, (尾巴=)1，您正在请求 a 下的区域吨具有 18 个自由度和右侧的曲线2.73（即右尾）；答案是0.00687.
• 示例 2：如果您输入(X=)2.73,（你的自由=)18, 和 (尾巴=)2，您正在请求 a 下的区域吨具有 18 个自由度和右侧的曲线2.73和左边−2.73（即左右尾巴）；答案是0.01374，这是先前答案的两倍0.00687.

## 广义线性模型代考

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。