### 统计代写|统计推断作业代写statistical inference代考|A Forerunner

statistics-lab™ 为您的留学生涯保驾护航 在代写统计推断statistical inference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。我们的专家在代写统计推断statistical inference代写方面经验极为丰富，各种代写统计推断statistical inference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|统计推断作业代写statistical inference代考|PROBABILISTIC INFERENCE

An early use of inferred probabilistic reasoning is described by Rabinovitch $(1970)$.

In the Book of Numbers, Chapter 18 , verse 5 , there is a biblical injunction which enjoins the father to redeem his wife’s first-born male child by payment of five pieces of silver to a priest (Laws of First Fruits). In the 12 th Century the theologian, physician and philosopher, Maimonides addressed himself to the following problem with a solution. Suppose one or more women have given birth to a number of children and the order of birth is unknown, nor is it known how many children each mother bore, nor which child belongs to which mother. What is the probability that a particular woman bore boys and girls in a specified sequence? (All permutations are assumed equally likely and the chances of male or female births is equal.)

Maimonides ruled as follows: Two wives of different husbands, one primiparous $(P)$ (a woman who has given birth to her first child) and one not $(\bar{P})$. Let $H$ be the event that the husband of $P$ pays the priest. If they gave birth to two males (and they were mixed up), $P(H)=1-$ if they bore a male $(M)$ and a female $(F)-P(H)=0$ (since the probability is only $1 / 2$ that the primipara gave birth to the boy). Now if they bore 2 males and a female, $P(H)=1$.
$$\begin{array}{cccccc} \frac{\text { Case 1 }}{\mathrm{M}, \mathrm{M}} & \frac{(P)}{\mathrm{M}} & \frac{(\bar{P})}{\mathrm{M}} & \frac{\text { Case } 2}{\mathrm{M}, \mathrm{F}} & \frac{(P)}{\mathrm{M}} & \frac{(\bar{P})}{\mathrm{F}} \ & & & \mathrm{F} & \mathrm{M} \ P(H)=1 & & P(H)=\frac{1}{2} \end{array}$$

$P A Y M E N T$
$\begin{array}{ccccc}\text { Case 3 } & \frac{(P)}{(\bar{P})} & Y e s & N o & \ \mathrm{M}, \mathrm{M}, \mathrm{F} & \mathrm{M}, \mathrm{M} & \mathrm{F} & \sqrt{ } & \ \mathrm{F}, \mathrm{M} & \mathrm{M} & & \sqrt{ } \ \mathrm{M}, \mathrm{F} & \mathrm{M} & \sqrt{ } & \ \mathrm{F} & \mathrm{M}, \mathrm{M} & & \sqrt{ } \ \mathrm{M} & \mathrm{F}, \mathrm{M} & \sqrt{\mathrm{M}} & \ \mathrm{M} & \mathrm{M}, \mathrm{F} & \sqrt{ } & & P(H)=\frac{2}{3}\end{array}$
Maimonides ruled that the husband of $P$ pays in Case 3 . This indicates that a probability of $2 / 3$ is sufficient for the priest to receive his 5 pieces of silver but $1 / 2$ is not. This leaves a gap in which the minimum probability is determined for payment.

What has been illustrated here is that the conception of equally likely events, independence of events, and the use of probability in making decisions were not unknown during the 12 th century, although it took many additional centuries to understand the use of sampling in determining probabilities.

## 统计代写|统计推断作业代写statistical inference代考|TESTING USING RELATIVE FREQUENCY

One of the earliest uses of relative frequency to test a Hypothesis was made by Arbuthnot $(1710)$, who questioned whether the births were equally likely to be male or female. He had available the births from London for 82 years. In every year male births exceeded females. Then he tested the hypothesis that there is an even chance whether a birth is male or female or the probability $p=\frac{1}{2}$. Given this hypothesis he calculated the chance of getting all 82 years of male exceedances $\left(\frac{1}{2}\right)^{82}$. Being that this is basically infinitesimal, the hypothesis was rejected. It is not clear how he would have argued if some other result had occurred since any particular result is small-the largest for equal numbers of male and female exceedances is less than $\frac{1}{10}$.

## 统计代写|统计推断作业代写statistical inference代考|PRINCIPLES GUIDING FREQUENTISM

Classical statistical inference is based on relative frequency considerations. A particular formal expression is given by Cox and Hinkley (1974) as follows:

Repeated Sampling Principle. Statistical procedures are to be assessed by their behavior in hypothetical repetition under the same conditions.
Two facets:

1. Measures of uncertainty are to be interpreted as hypothetical frequencies in long run repetitions.
1. Criteria of optimality are to be formulated in terms of sensitive behavior in hypothetical repetitions.
(Question: What is the appropriate space which generates these hypothetical repetitions? Is it the sample space $S$ or some other reference set?)

Restricted (Weak) Repeated Sampling Principle. Do not use a procedure which for some possible parameter values gives, in hypothetical repetitions, misleading conclusions most of the time (too vague and imprecise to be constructive). The argument for repeated sampling ensures a physical meaning for the quantities we calculate and that it ensures a close relation between the analysis we make and the underlying model which is regarded as representing the “true” state of affairs.

An early form of frequentist inferences were Tests of Significance. They were long in use before their logical grounds were given by Fisher (1956b) and further elaborated by Barnard (unpublished lectures).
Prior assumption: There is a null hypothesis with no discernible alternatives. Features of a significance test (Fisher-Barnard)

1. A significance test procedure requires a reference set $R$ (not necessarily the entire sample space) of possible results comparable with the observed result $X=x_{0}$ which also belongs to $R$.
2. A ranking of all possible results in $R$ in order of their significance or meaning or departure from a null hypothesis $H_{0}$. More specifically we adopt a criterion $T(X)$ such that if $x_{1}>x_{2}$ (where $x_{1}$ departs further in rank than $x_{2}$ both elements of the reference set $R)$ then $T\left(x_{1}\right)>T\left(x_{2}\right)$ [if there is doubt about the ranking then there will be corresponding doubt about how the results of the significance test should be interpreted].
3. $H_{0}$ specifies a probability distribution for $T(X)$. We then evaluate the observed result $x_{0}$ and the null hypothesis.

## 统计代写|统计推断作业代写statistical inference代考|PROBABILISTIC INFERENCE

Rabinovitch 描述了推断概率推理的早期使用(1970).

情况1 米,米(磷)米(磷¯)米 案子 2米,F(磷)米(磷¯)F F米 磷(H)=1磷(H)=12

$付款\ begin {array {ccccc} \ text {案例3} & \ frac {(P) {{(\ bar {P})} & Y es & N o & \ \ mathrm {M}, \ mathrm {M , \ mathrm {F} & \ mathrm {M}, \ mathrm {M} & \ mathrm {F} & \ sqrt {} & \ \ mathrm F}, \ mathrm {M} & \ mathrm {M} & & \ sqrt} \ \ mathrm {M}, \ mathrm {F} & \ mathrm {M} & \ sqrt {} & \ \ mathrm {F} & \ mathrm {M}, \ mathrm {M} & & \ sqrt } \ \ mathrm {M} & \ mathrm {F}, \ mathrm {M} & \ sqrt {\ mathrm {M}} & \ \ mathrm {M} & \ mathrm {M}, \ mathrm {F} & \ sqrt} & & P (H) = \ frac {2} {3} \ end {数组米一种一世米这n一世d和sr在l和d吨H一种吨吨H和H在sb一种nd这F磷p一种是s一世nC一种s和3.吨H一世s一世nd一世C一种吨和s吨H一种吨一种pr这b一种b一世l一世吨是这F2 / 3一世ss在FF一世C一世和n吨F这r吨H和pr一世和s吨吨这r和C和一世在和H一世s5p一世和C和s这Fs一世l在和rb在吨1 / 2$ 不是。这留下了确定付款的最小概率的空白。

## 统计代写|统计推断作业代写statistical inference代考|TESTING USING RELATIVE FREQUENCY

Arbuthnot 最早使用相对频率来检验假设之一(1710)，谁质疑出生的男性或女性的可能性是否相同。他已经在伦敦接生了 82 年。每年男性的出生人数都超过女性。然后他检验了一个假设，即出生是男性还是女性的概率或概率是均等的。p=12. 鉴于这个假设，他计算了获得所有 82 年男性超额成绩的机会(12)82. 由于这基本上是无穷小的，因此该假设被拒绝。尚不清楚如果出现了其他结果，他会如何辩解，因为任何特定结果都很小——男性和女性数量相同的最大超标小于110.

## 统计代写|统计推断作业代写statistical inference代考|PRINCIPLES GUIDING FREQUENTISM

1. 不确定性的度量将被解释为长期重复中的假设频率。
2. 最优标准将根据假设重复中的敏感行为来制定。
（问题：产生这些假设重复的适当空间是什么？是样本空间吗？小号或其他一些参考集？）

1. 显着性检验程序需要参考集R（不一定是整个样本空间）与观察结果相当的可能结果X=X0这也属于R.
2. 所有可能结果的排名R按照它们的重要性或意义或偏离原假设的顺序H0. 更具体地说，我们采用一个标准吨(X)这样如果X1>X2（在哪里X1排名比离得更远X2参考集的两个元素R)然后吨(X1)>吨(X2)[如果对排名有疑问，那么对显着性检验的结果应该如何解释也会有相应的疑问]。
3. H0指定概率分布吨(X). 然后我们评估观察到的结果X0和原假设。

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## MATLAB代写

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