# 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|MAST90085

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Boston Housing

Returning to the Boston Housing data set, we are now in a position to test if the means of the variables vary according to their location, for example, when they are located in a district with high valued houses. In Chap. 1, we built two groups of observations according to the value of $X_{14}$ being less than or equal to the median of $X_{14}$ (a group of 256 districts) and greater than the median (a group of 250 districts).
In what follows, we use the transformed variables motivated in Sect. $1.9$.
Testing the equality of the means from the two groups was proposed in a multivariate setup, so we restrict the analysis to the variables $X_1, X_5, X_8, X_{11}$, and $X_{13}$ to see if the differences between the two groups that were identified in Chap. 1 can be confirmed by a formal test. As in Test Problem 8, the hypothesis to be tested is
$$H_0: \mu_1=\mu_2 \text {, where } \mu_1 \in \mathbb{R}^5, n_1=256 \text {, and } n_2=250 .$$

$\Sigma$ is not known. The $F$-statistic given in (7.13) is equal to $126.30$, which is much higher than the critical value $F_{0.95 ; 5,500}=2.23$. Therefore, we reject the hypothesis of equal means.

To see which component, $X_1, X_5, X_8, X_{11}$, or $X_{13}$, is responsible for this rejection, take a look at the simultaneous confidence intervals defined in (7.14):
\begin{aligned} &\delta_1 \in(1.4020, \quad 2.5499) \ &\delta_5 \in(0.1315, \quad 0.2383) \ &\delta_8 \in(-0.5344,-0.2222) \ &\delta_{11} \in(1.0375,1.7384) \ &\delta_{13} \in(1.1577, \quad 1.5818) \end{aligned}
These confidence intervals confirm that all of the $\delta_j$ are significantly different from zero (note there is a negative effect for $X_8$ : weighted distances to employment centers) Q MVAsimeibh.

We could also check if the factor “being bounded by the river” (variable $X_4$ ) has some effect on the other variables. To do this compare the means of $\left(X_5, X_8, X_9, X_{12}, X_{13}, X_{14}\right)^{\top}$. There are two groups: $n_1=35$ districts bounded by the river and $n_2=471$ districts not bounded by the river. Test Problem 8 $\left(H_0: \mu_1=\mu_2\right.$ ) is applied again with $p=6$. The resulting test statistic, $F=5.81$, is highly significant $\left(F_{0.95 ; 6,499}=2.12\right)$. The simultaneous confidence intervals indicate that only $X_{14}$ (the value of the houses) is responsible for the hypothesis being rejected.

## 统计代写|多元统计分析代写Multivariate Statistical Analysis代考|Regression Models

The aim of regression models is to model the variation of a quantitative response variable $y$ in terms of the variation of one or several explanatory variables $\left(x_1, \ldots, x_p\right)^{\top}$. We have already introduced such models in Chaps. 3 and 7 where linear models were written in (3.50) as
$$y=\mathcal{X} \beta+\varepsilon,$$
where $y(n \times 1)$ is the vector of observation for the response variable, $\mathcal{X}(n \times p)$ is the data matrix of the $p$ explanatory variables and $\varepsilon$ are the errors. Linear models are not restricted to handle only linear relationships between $y$ and $x$. Curvature is allowed by including appropriate higher order terms in the design matrix $\mathcal{X}$.

Example $8.1$ If $y$ represents response and $x_1, x_2$ are two factors that explain the variation of $y$ via the quadratic response model:
$$y_i=\beta_0+\beta_1 x_{i 1}+\beta_2 x_{i 2}+\beta_3 x_{i 1}^2+\beta_4 x_{i 2}^2+\beta_5 x_{i 1} x_{i 2}+\varepsilon_i, i=1, \ldots, n .$$
This model (8.1) belongs to the class of linear models because it is linear in $\beta$. The data matrix $\mathcal{X}$ is:
$$\mathcal{X}=\left(\begin{array}{cccccc} 1 & x_{11} & x_{12} & x_{11}^2 & x_{12}^2 & x_{11} x_{12} \ 1 & x_{21} & x_{22} & x_{21}^2 & x_{22}^2 & x_{21} x_{22} \ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \ 1 & x_{n 1} & x_{n 2} & x_{n 1}^2 & x_{n 2}^2 & x_{n 1} x_{n 2} \end{array}\right)$$
For a given value of $\beta$, the response surface can be represented in a threedimensional plot as in Fig. $8.1$ where we display $y=20+1 x_1+2 x_2-8 x_1^2-$ $6 x_2^2+6 x_1 x_2$, i.e. $\beta=(20,1,2,-8,-6,+6)^{\top}$.

## 统计代写|多元统计分析代写多元统计分析代考|波士顿住房

$$H_0: \mu_1=\mu_2 \text {, where } \mu_1 \in \mathbb{R}^5, n_1=256 \text {, and } n_2=250 .$$

$\Sigma$不知道。(7.13)中给出的$F$ -statistic等于$126.30$，它比临界值$F_{0.95 ; 5,500}=2.23$高得多。因此，我们拒绝等均值假设。

\begin{aligned} &\delta_1 \in(1.4020, \quad 2.5499) \ &\delta_5 \in(0.1315, \quad 0.2383) \ &\delta_8 \in(-0.5344,-0.2222) \ &\delta_{11} \in(1.0375,1.7384) \ &\delta_{13} \in(1.1577, \quad 1.5818) \end{aligned}

## 统计代写|多元统计分析代写多元统计分析代考|回归模型

$$y=\mathcal{X} \beta+\varepsilon,$$
，其中$y(n \times 1)$是响应变量的观察向量，$\mathcal{X}(n \times p)$是$p$解释变量的数据矩阵，$\varepsilon$是误差。线性模型并不局限于处理$y$和$x$之间的线性关系。通过在设计矩阵$\mathcal{X}$中包含适当的高阶项，允许曲率。

$$y_i=\beta_0+\beta_1 x_{i 1}+\beta_2 x_{i 2}+\beta_3 x_{i 1}^2+\beta_4 x_{i 2}^2+\beta_5 x_{i 1} x_{i 2}+\varepsilon_i, i=1, \ldots, n .$$这个模型(8.1)属于线性模型的类别，因为它是线性的 $\beta$。数据矩阵 $\mathcal{X}$
$$\mathcal{X}=\left(\begin{array}{cccccc} 1 & x_{11} & x_{12} & x_{11}^2 & x_{12}^2 & x_{11} x_{12} \ 1 & x_{21} & x_{22} & x_{21}^2 & x_{22}^2 & x_{21} x_{22} \ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \ 1 & x_{n 1} & x_{n 2} & x_{n 1}^2 & x_{n 2}^2 & x_{n 1} x_{n 2} \end{array}\right)$$的给定值 $\beta$时，响应面可以用如图所示的三维图表示。 $8.1$ 我们展示的地方 $y=20+1 x_1+2 x_2-8 x_1^2-$ $6 x_2^2+6 x_1 x_2$，即。 $\beta=(20,1,2,-8,-6,+6)^{\top}$.

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: