### 统计代写|统计推断作业代写statistical inference代考|Testing Hypotheses

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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|HYPOTHESIS TESTING VIA THE REPEATED

Neyman-Pearson (N-P) (1933) Theory of Hypotheses Testing is based on the repeated sampling principle and is basically a two decision procedure.

We will collect data $D$ and assume we have two rival hypotheses about the populations from which the data were generated: $H_{0}$ (the null hypothesis) and $H_{1}$ (the alternative hypothesis). We assume a sample space $S$ of all possible outcomes of the data $D$. A rule is then formulated for the rejection of $H_{0}$ or $H_{1}$ in the following manner. Choose a subset $s$ (critical region) of $S$ and
if $D \in s$ reject $H_{0}$ and accept $H_{1}$
if $D \in S-s$ reject $H_{1}$ and accept $H_{0}$
according to
$P\left(D \in s \mid H_{0}\right)=\epsilon$ (size) $\leq \alpha$ (level) associated with the test (Type 1 error)
$P\left(D \in s \mid H_{1}\right)=1-\beta$ (power of the test)
$P\left(D \in S-s \mid H_{1}\right)=\beta$ (Type 2 error)

The two basic concepts are size and power and N-P theory dictates that we choose a test (critical region) which results in small size and large power. At this juncture we assume size equals level and later show how size and level can be equated. Now
$$\alpha=P\left(\text { accepting } H_{1} \mid H_{0}\right) \text { or } 1-\alpha=P\left(\text { accepting } H_{0} \mid H_{0}\right)$$
and
$$1-\beta=P\left(\text { accepting } H_{1} \mid H_{1}\right) \text { or } \quad \beta=P\left(\text { accepting } H_{0} \mid H_{1}\right) .$$
Since there is no way of jointly minimizing size and maximizing power, N-P suggest choosing small size $\alpha$ and then maximize power $1-\beta$ (or minimize $\beta$ ).

## 统计代写|统计推断作业代写statistical inference代考|REMARKS ON SIZE

Before you have the data $D$ it would be reasonable to require that you should have small size (frequency of rejecting $H_{0}$ when $H_{0}$ is true). But this may mislead you once you have the data, for example, suppose you want size $=.05$ where the probabilities for the data $D=\left(D_{1}, D_{2}, D_{3}\right)$ are given in Table $4.1$.

Presumably if you want size $=.05$ you reject $H_{0}$ if event $D_{1}$ occurs and accept if $D_{2}$ or $D_{3}$ occur. However if $D_{1}$ occurs you are surely wrong to reject $H_{0}$ since $P\left(D_{1} \mid H_{1}\right)=0$. So you need more than size. Note that before making the test, all tests of the same size provide us with the same chance of rejecting $H_{0}$, but after the data are in hand not all tests of the same size are equally good. In the N-P set up we are forced to choose a test before we know what the sample value actually is, even when our interest is in evaluating hypotheses with regard to the sample data we have. Therefore if two tests $T_{1}$ and $T_{2}$ have the same size one might be led to choose the test with greater power. That this is not necessarily the best course is demonstrated in the following Example $4.1$ and its variations, Hacking (1965).

## 统计代写|统计推断作业代写statistical inference代考|NEYMAN-PEARSON FUNDAMENTAL LEMMA

Lemma 4.1 Let $F_{0}(D)$ and $F_{1}(D)$ be distribution functions possessing generalized densities $f_{0}(D)$ and $f_{1}(D)$ with respect to $\mu$. Let $H_{0}: f_{0}(D)$ vs. $H_{1}: f_{1}(D)$ and $T(D)=P\left[\right.$ rejecting $H_{0}$ ]. Then

1. Existence: For testing $H_{0}$ vs. $H_{1}$ there exists a test $T(D)$ and a constant $k$ such that for any given $\alpha, 0 \leq \alpha \leq 1$
(a) $E_{H_{0}}(T(D))=\alpha=\int_{S} T(D) f_{0}(D) d \mu$
(b) $T(D)=\left{\begin{array}{lll}1 & \text { if } & f_{0}(D)k f_{1}(D)^{*}\end{array}\right.$.
2. Sufficiency: If a test satisfies (a) and (b) then it is the most powerful (MP) for testing $H_{0}$ vs. $H_{1}$ at level $\alpha$.
1. Necessity: If $T(D)$ is most powerful $(M P)$ at level $\alpha$ for $H_{0}$ vs. $H_{1}$ then for some $k$ it satisfies (a) and (b) unless there exists a test of size less than $\alpha$ and power $1 .$

Proof: For $\alpha=0$ or $\alpha=1$ let $k=0$ or $\infty$ respectively, hence we restrict $\alpha$ such that $0<\alpha<1$.

1. Existence: Let
$$\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1} \end{array}\right.$$
If $\alpha(k)=\alpha(k-0)$ there is no need for the middle term since
$$P\left(f_{0}=k f_{1}\right)=0 .$$
Hence we can produce a test with properties (a) and (b).
2. Sufficiency: If a test satisfies (a) and (b) then it is most powerful for testing $H_{0}$ against $H_{1}$ at level $\alpha$.
Proof: Let $T^{}$ (D) be any other test such that $$E_{H_{0}}\left(T^{}(D)\right) \leq \alpha .$$
Now consider
$$E_{H_{1}}\left(T^{}(D)\right)=\int_{S} T^{} f_{1} d \mu=1-\beta^{*}$$

## 统计代写|统计推断作业代写statistical inference代考|HYPOTHESIS TESTING VIA THE REPEATED

Neyman-Pearson (NP) (1933) 假设检验理论基于重复抽样原则，基本上是一个两决定程序。

1−b=磷( 接受 H1∣H1) 或者 b=磷( 接受 H0∣H1).

## 统计代写|统计推断作业代写statistical inference代考|NEYMAN-PEARSON FUNDAMENTAL LEMMA

1. 存在：用于测试H0对比H1存在一个测试吨(D)和一个常数ķ这样对于任何给定的一种,0≤一种≤1
（一种）和H0(吨(D))=一种=∫小号吨(D)F0(D)dμ
(b) $T(D)=\左{1 如果 F0(D)ķF1(D)∗\对。$。
2. 充分性：如果一个测试满足（a）和（b），那么它是最强大的（MP）测试H0对比H1在水平一种.
3. 必要性：如果吨(D)是最强大的(米磷)在水平一种为了H0对比H1然后对于一些ķ它满足 (a) 和 (b) 除非存在规模小于一种和权力1.

1. 存在：让
\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1} \end{array}\对。\alpha(k)=P\left(f_{0} \leq k f_{1} \mid H_{0}\right)=P\left(f_{0}k f_{1} \end{array}\对。
如果一种(ķ)=一种(ķ−0)不需要中期，因为
磷(F0=ķF1)=0.
因此，我们可以生成具有属性 (a) 和 (b) 的测试。
2. 充分性：如果一个测试满足（a）和（b），那么它对测试最有效H0反对H1在水平一种.
证明：让吨(D) 是任何其他测试，使得和H0(吨(D))≤一种.
现在考虑
和H1(吨(D))=∫小号吨F1dμ=1−b∗

## 广义线性模型代考

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## MATLAB代写

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