### 统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

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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|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

The claim for significance tests are for those cases where alternative hypotheses are not sensible. Note that Goodness-of-Fit tests fall into this category, that is, Do the data fit a normal distribution? Here $H_{0}$ is merely a family of distributions rather than a specification of parameter values. Note also that a test of significance considers not only the event that occurred but essentially puts equal weight on more discrepent events that did not occur as opposed to a test which only considers what did occur.
A poignant criticism of Fisherian significance testing is made by Jeffreys (1961). He said

What the use of the P implies, therefore, is that a hypothesis that may be true may be rejected because it has not predicted observable results that have not occurred.

Fisher (1956b) gave as an example of a pure test of significance the following by commenting on the work of astronomer J. Michell. Michell supposed that there were in all 1500 stars of the required magnitude and sought to calculate the probability, on the hypothesis that they are individually distributed at random, that any one of them should have five neighbors within a distance of $a$ minutes of arc from it. Fisher found the details of Michell’s calculation obscure, and suggested the following argument.
“The fraction of the celestial sphere within a circle of radius $a$ minutes is, to a satisfactory approximation,
$$p=\left(\frac{a}{6875.5}\right)^{2}$$
in which the denominator of the fraction within brackets is the number of minutes in two radians. So, if $a$ is 49 , the number of minutes from Maia to its fifth nearest neighbor, Atlas, we have
$$p=\frac{1}{(140.316)^{2}}=\frac{1}{19689}$$
Out of 1499 stars other than Maia of the requisite magnitude the expected number within this distance is therefore
$$m=\frac{1499}{19689}=\frac{1}{13.1345}=.07613$$
The frequency with which five stars should fall within the prescribed area is then given approximately by the term of the Poisson series
$$e^{-m} \frac{m^{5}}{5 !}$$
or, about 1 in $50,000,000$, the probabilities of having 6 or more close neighbors adding very little to this frequency. Since 1500 stars each have this probability of being the center of such a close cluster of 6 , although these probabilities are not strictly independent,

## 统计代写|统计推断作业代写statistical inference代考|FORMS OF THE LIKELIHOOD PRINCIPLE

The model for experiment $\mathcal{E}$ consists of a sample space $S$ and a parameter space $\Theta$, a measure $\mu$, and a family of probability functions $f: S \times \Theta \rightarrow R^{+}$such that for all $\theta \in \Theta$
$$\int_{S} f d \mu=1$$

1. Unrestricted LP (ULP)
For two such experiments modeled as $\mathcal{E}=(S, \mu, \Theta, f)$ and $\mathcal{E}^{\prime}=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$, and for realization $D \in S$ and $D^{\prime} \in S^{\prime}$, if
$$f(D \mid \theta)=g\left(D, D^{\prime}\right) f^{\prime}\left(D^{\prime} \mid \theta\right) \quad \text { for } g>0$$
for all $\theta$ and the choice of $\mathcal{E}$ or $\mathcal{E}^{\prime}$ is uniformative with regard to $\theta$, then the evidence or inferential content or information concerning $\theta$, all denoted by Inf is such that
$$\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$$
Note this implies that all of the statistical evidence provided by the data is conveyed by the likelihood function. There is an often useful extension, namely, when $\theta=\left(\theta_{1}, \theta_{2}\right), \delta_{1}=h\left(\theta_{1}, \theta_{2}\right), \delta_{2}=k\left(\theta_{1}, \theta_{2}\right)$, and
$$f(D \mid \theta)=g\left(D, D^{\prime}, \delta_{1}\right) f^{\prime}\left(D^{\prime} \mid \delta_{2}\right)$$
then $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$ concerning $\delta_{2}$.
2. Weakly restricted LP (RLP)
LP is applicable whenever a) $(S, \mu, \Theta, f)=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$ and b) $(S, \mu, \Theta, f) \neq$ $\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$ when there are no structural features of $(S, \mu, \Theta, f)$ which have inferential relevance and which are not present in $\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$.
3. Strongly restricted LP (SLP)
Applicable only when $(S, \mu, \Theta, f)=\left(S^{\prime}, \mu^{\prime}, \Theta, f^{\prime}\right)$.
4. Extended $L P$ (ELP)
When $\theta=(p, \gamma)$ and $f\left(D, D^{\prime} \mid p, \gamma\right)=g\left(D, D^{\prime} \mid p\right) f^{\prime}\left(D^{\prime} \mid \gamma\right)$ it is plausible to extend LP to
$$\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$$
concerning $p$ assuming $p$ and $\gamma$ are unrelated.

## 统计代写|统计推断作业代写statistical inference代考|LIKELIHOOD AND SIGNIFICANCE TESTING

We now compare the use of likelihood analysis with a significance test. Suppose we are only told that in a series of independent and identically distributed binary trials there were $r$ successes and $n-r$ failures, and the sampling was conducted in one of three ways:

1. The number of trials was fixed at $n$.
2. Sampling was stopped at the $r$ th success.
3. Sampling was stopped when $n-r$ failures were obtained.
Now while the three sampling probabilities differ they all have the same likelihood:
$$L=p^{r}(1-p)^{n-r}$$
The probabilities of $r$ successes and $n-r$ failures under these sampling methods are:
\begin{aligned} f_{n} &=\left(\begin{array}{l} n \ r \end{array}\right) L \quad r=0,1,2, \ldots, n \ f_{r} &=\left(\begin{array}{c} n-1 \ r-1 \end{array}\right) L \quad n=r, r+1, \ldots \ f_{n-r} &=\left(\begin{array}{c} n-1 \ n-r-1 \end{array}\right) L \quad n=r+1, r+2, \ldots \end{aligned}
where $f_{a}$ denotes the probability where subscript $a$ is fixed for sampling.

## 统计代写|统计推断作业代写statistical inference代考|FURTHER REMARKS ON TESTS OF SIGNIFICANCE

Jeffreys (1961) 对费希尔显着性检验提出了尖锐的批评。他说

“天球在半径圆内的分数一种分钟是，一个令人满意的近似值，
p=(一种6875.5)2

p=1(140.316)2=119689

## 统计代写|统计推断作业代写statistical inference代考|FORMS OF THE LIKELIHOOD PRINCIPLE

∫小号Fdμ=1

1. Unrestricted LP (ULP)
对于两个这样的实验，建模为和=(小号,μ,θ,F)和和′=(小号′,μ′,θ,F′)，并为实现D∈小号和D′∈小号′， 如果
F(D∣θ)=G(D,D′)F′(D′∣θ) 为了 G>0
对全部θ和选择和或者和′是统一的θ，然后是有关的证据或推论内容或信息θ, 所有由 Inf 表示的都是这样的
信息⁡(和,D)=信息⁡(和′,D′)
请注意，这意味着数据提供的所有统计证据都由似然函数传达。有一个经常有用的扩展，即当θ=(θ1,θ2),d1=H(θ1,θ2),d2=ķ(θ1,θ2)， 和
F(D∣θ)=G(D,D′,d1)F′(D′∣d2)
然后信息⁡(和,D)=信息⁡(和′,D′)关于d2.
2. 弱限制 LP (RLP)
LP 适用于 a)(小号,μ,θ,F)=(小号′,μ′,θ,F′)b)(小号,μ,θ,F)≠ (小号′,μ′,θ,F′)当没有结构特征时(小号,μ,θ,F)哪些具有推理相关性，哪些不存在于(小号′,μ′,θ,F′).
3. 严格限制 LP (SLP)
仅适用于(小号,μ,θ,F)=(小号′,μ′,θ,F′).
4. 扩展大号磷(ELP)
什么时候θ=(p,C)和F(D,D′∣p,C)=G(D,D′∣p)F′(D′∣C)将 LP 扩展到
信息⁡(和,D)=信息⁡(和′,D′)
关于p假设p和C是无关的。

## 统计代写|统计推断作业代写statistical inference代考|LIKELIHOOD AND SIGNIFICANCE TESTING

1. 试验次数固定为n.
2. 采样停止在r成功。
3. 采样停止时n−r失败了。
现在，虽然三个采样概率不同，但它们都具有相同的可能性：
大号=pr(1−p)n−r
的概率r成功和n−r这些抽样方法的失败是：
Fn=(n r)大号r=0,1,2,…,n Fr=(n−1 r−1)大号n=r,r+1,… Fn−r=(n−1 n−r−1)大号n=r+1,r+2,…
在哪里F一种表示下标的概率一种固定用于采样。

## 广义线性模型代考

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## MATLAB代写

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