### 统计代写|生物统计学作业代写Biostatistics代考| PROBABILITY MODELS FOR CONTINUOUS DATA

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• 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|PROBABILITY MODELS FOR CONTINUOUS DATA

In Section $3.2$ we treated the family of normal curves very informally because it was intended to reach more students and readers for whom mathematical formulas may not be very relevant. In this section we provide some supplementary information that may be desirable for those who may be more interested in the fundamentals of biostatistical inference.

A class of measurements or a characteristic on which individual observations or measurements are made is called a variable. If values of a variable may theoretically lie anywhere on a numerical scale, we have a continuous variable; examples include weight, height, and blood pressure, among others. We saw in Section $3.2$ that each continuous variable is characterized by a smooth density curve. Mathematically, a curve can be characterized by an equation of the form
$$y=f(x)$$
called a probability density function, which includes one or several parameters; the total area under a density curve is $1.0$. The probability that the variable assumes any value in an interval between two specific points $a$ and $b$ is given by
$$\int_{a}^{b} f(x) d x$$
The probability density function for the family of normal curves, sometimes referred to as the Gaussian distribution, is given by
$$f(x)=\frac{1}{\sigma \sqrt{2 \pi}} \exp \left[-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}\right] \quad \text { for }-\infty<x<\infty$$
The meaning and significance of the parameters $\mu$ and $\sigma / \sigma^{2}$ have been discussed in Section $3.2 ; \mu$ is the mean, $\sigma^{2}$ is the variance, and $\sigma$ is the standard deviation. When $\mu=1$ and $\sigma^{2}=1$, we have the standard normal distribution. The numerical values listed in Appendix B are those given by
$$\int_{0}^{z} \frac{1}{\sqrt{2 \pi}} \exp \left[-\frac{1}{2}(x)^{2}\right] d x$$
The normal distribution plays an important role in statistical inference because:

1. Many real-life distributions are approximately normal.
2. Many other distributions can be almost normalized by appropriate data transformations (e.g., taking the $\log$ ). When $\log X$ has a normal distribution, $X$ is said to have a lognormal distribution.

## 统计代写|生物统计学作业代写Biostatistics代考|Binomial Distribution

In Chapter 1 we discussed cases with dichotomous outcomes such as malefemale, survived-not survived, infected-not infected, white-nonwhite, or simply positive-negative. We have seen that such data can be summarized into proportions, rates, and ratios. In this section we are concerned with the probability of a compound event: the occurrence of $x$ (positive) outcomes $(0 \leq x \leq n)$ in $n$ trials, called a binomial probability. For example, if a certain drug is known to cause a side effect $10 \%$ of the time and if five patients are given this drug, what is the probability that four or more experience the side effect?

Let $S$ denote a side-effect outcome and $N$ an outcome without side effects. The process of determining the chance of $x S$ ‘s in $n$ trials consists of listing all the possible mutually exclusive outcomes, calculating the probability of each outcome using the multiplication rule (where the trials are assumed to be independent), and then combining the probability of all those outcomes that are compatible with the desired results using the addition rule. With five patients there are 32 mutually exclusive outcomes, as shown in Table 3.11.

Since the results for the five patients are independent, the multiplication rule produces the probabilities shown for each combined outcome. For example:

• The probability of obtaining an outcome with four $S$ ‘s and one $N$ is
$$(0.1)(0.1)(0.1)(0.1)(1-0.1)=(0.1)^{4}(0.9)$$
• The probability of obtaining all five $S$ ‘s is
$$(0.1)(0.1)(0.1)(0.1)(0.1)=(0.1)^{5}$$
Since the event “all five with side effects” corresponds to only one of the 32 outcomes above and the event “four with side effects and one without” pertains to five of the 32 outcomes, each with probability $(0.1)^{4}(0.9)$, the addition rule yields a probability
$$(0.1)^{5}+(5)(0.1)^{4}(0.9)=0.00046$$
for the compound event that “four or more have side effects.” In general, the binomial model applies when each trial of an experiment has two possible outcomes (often referred to as “failure” and “success” or “negative” and “positive”; one has a success when the primary outcome is observed). Let the probabilities of failure and success be, respectively, $1-\pi$ and $\pi$, and we “code” these two outcomes as 0 (zero successes) and 1 (one success). The experiment consists of $n$ repeated trials satisfying these assumptions:
1. The $n$ trials are all independent.
2. The parameter $\pi$ is the same for each trial.

## 统计代写|生物统计学作业代写Biostatistics代考|Poisson Distribution

The next discrete distribution that we consider is the Poisson distribution, named after a French mathematician. This distribution has been used extensively in health science to model the distribution of the number of occurrences $x$ of some random event in an interval of time or space, or some volume of matter. For example, a hospital administrator has been studying daily emergency admissions over a period of several months and has found that admissions have averaged three per day. He or she is then interested in finding the probability that no emergency admissions will oocur on a particular day. The Poisson distribution is characterized by its probability density function:
$$\operatorname{Pr}(X=x)=\frac{\theta^{x} e^{-\theta}}{x !} \quad \text { for } x=0,1,2, \ldots$$
It turns out, interestingly enough, that for a Poisson distribution the variance is equal to the mean, the parameter $\theta$ above. Therefore, we can answer probability questions by using the formula for the Poisson density above or by converting the number of occurrences $x$ to the standard normal score, provided that $\theta \geq 10$ :
$$z=\frac{x-\theta}{\sqrt{\theta}}$$
In other words, we can approximate a Poisson distribution by a normal distribution with mean $\theta$ if $\theta$ is at least 10 .

Here is another example involving the Poisson distribution. The infant mortality rate (IMR) is defined as
$$\operatorname{IMR}=\frac{d}{N}$$
for a certain target population during a given year, where $d$ is the number of deaths during the first year of life and $N$ is the total number of live births. In the studies of IMRs, $N$ is conventionally assumed to be fixed and $d$ to follow a Poisson distribution.

Example 3.9 For the year 1981 we have the following data for the New England states (Connecticut, Maine, Massachusetts, New Hampshire, Rhode Island, and Vermont):
\begin{aligned} d &=1585 \ N &=164,200 \end{aligned}
For the same year, the national infant mortality rate was $11.9$ (per 1000 live

births). If we apply the national IMR to the New England states, we would have
\begin{aligned} \theta &=(11.9)(164.2) \ & \simeq 1954 \text { infant deaths } \end{aligned}
Then the event of having as few as 1585 infant deaths would occur with a probability
\begin{aligned} \operatorname{Pr}(d \leq 1585) &=\operatorname{Pr}\left(z \leq \frac{1585-1954}{\sqrt{1954}}\right) \ &=\operatorname{Pr}(z \leq-8.35) \ & \simeq 0 \end{aligned}
The conclusion is clear: Either we observed an extremely improbable event, or infant mortality in the New England states is lower than the national average. The rate observed for the New England states was $9.7$ deaths per 1000 live births.

## 统计代写|生物统计学作业代写Biostatistics代考|PROBABILITY MODELS FOR CONTINUOUS DATA

∫一种bF(X)dX

F(X)=1σ2圆周率经验⁡[−12(X−μσ)2] 为了 −∞<X<∞

∫0和12圆周率经验⁡[−12(X)2]dX

1. 许多现实生活中的分布是近似正态的。
2. 许多其他分布几乎可以通过适当的数据转换来归一化（例如，取日志）。什么时候日志⁡X具有正态分布，X据说服从对数正态分布。

## 统计代写|生物统计学作业代写Biostatistics代考|Binomial Distribution

• 获得四个结果的概率小号的和一个ñ是
(0.1)(0.1)(0.1)(0.1)(1−0.1)=(0.1)4(0.9)
• 获得所有五个的概率小号是
(0.1)(0.1)(0.1)(0.1)(0.1)=(0.1)5
由于事件“所有五个都有副作用”仅对应于上述 32 个结果中的一个，而事件“四个有副作用，一个没有”事件与 32 个结果中的五个相关，每个结果都有概率(0.1)4(0.9), 加法规则产生一个概率
(0.1)5+(5)(0.1)4(0.9)=0.00046
对于“四个或更多有副作用”的复合事件。一般而言，当实验的每个试验都有两种可能的结果（通常称为“失败”和“成功”或“消极”和“积极”；当观察到主要结果时，一个成功）时，适用二项式模型。让失败和成功的概率分别为1−圆周率和圆周率，我们将这两个结果“编码”为 0（零成功）和 1（成功）。该实验包括n满足这些假设的重复试验：
1. 这n试验都是独立的。
2. 参数圆周率每次试验都是一样的。

## 统计代写|生物统计学作业代写Biostatistics代考|Poisson Distribution

d=1585 ñ=164,200

θ=(11.9)(164.2) ≃1954 婴儿死亡

## 广义线性模型代考

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## MATLAB代写

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