### 统计代写|实验设计作业代写experimental design代考|GOODNESS OF FIT OF THE MODEL

statistics-lab™ 为您的留学生涯保驾护航 在代写实验设计experimental designatistical Modelling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。我们的专家在代写实验设计experimental design代写方面经验极为丰富，各种代写实验设计experimental design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|实验设计作业代写experimental design代考|AN EXAMPLE – VALUE OF A POSTAGE STAMP OVER TIME

The value of an Australian stamp ( $1963 \& 2$ sepia colored) given, in $\boldsymbol{L}$ sterling, by the Stanley Gibbons Catalogues is shown in Table $1.7 .1$ for the years 1972-1980. The aim is to $f$ it a model to the value of the stamp over time.

Let the time be $x=0$ for $1972, x=1$ for 1973 , etc. In this example, interest probably centres on the relative value of the stamp from one year to the next. Alternatively, if there is interest in the investment value of the stamp over the period 1972-1980, it may be useful to express the value as $y$, the value relative to that of 1972. The values of $y$ and $x$ are also shown above and are graphed in Figure 1.7.1. The relationship between $y$ and $x$ is not a 1 inear one. When $y$ is transformed by taking natural logarithms, a strong linear trend is apparent as shown by Figure 1.7.2. The predicted values of in $y$ and the residuals are shown in Table 1.7.2. The prediction equation is

Of course, the constant term could be omitted and the graph forced to pass through the orlgin.
The residuals fall into a reasonable horizontal band so that there is 1 ttle evidence to contradict the assumptions that the deviations are distributed with mean zero and constant variance.
However, there is a marked pattern in the residuals for their signs are:
Clearly, the residuals are positively correlated (as a positive residual is often followed by a positive residual, and a negative residual followed by a negative). With the small number of observations in the sample, we should hesitate to make dogmatio statements about the model but the pattern in the residuals may suggest that
(i) The population mean is not correct.
The residuals could indicate that a cubed term would remove.

## 统计代写|实验设计作业代写experimental design代考|INTRODUCTION

In Chapter 1, we considered fitting models to data. To answer the question of how well a model fits the data, we need to develop statistical tests and, to do so, we lean heavily on the assumption that the deviations have independent normal distributions. For our purposes, the main benefit of the normality assumption is that we can find the distribution of estimates and perform test of significance. The theorems listed in Appendix B will be of assistance in this. For the general linear model
$$y=1 \alpha+X_{\beta}+\varepsilon$$
where $\alpha$ is a constant term, $X$ is an $n \times k$ matri $x$ of known constants, deviations from their mean, and $\varepsilon$ are the deviations, assumed to be independent and normally distributed with mean zero and constant variance of $\sigma^{2}$, that is $\varepsilon-N\left(0, \sigma^{2} I\right)$. I is an $n \times n$ identity matrix with all diagonal elements equal to 1 and off diagonal elements equal to zero. Thus, the covariances are zero and the variances are all equal.

With these assumptions, we have, in effect, made assumptions about $y$. The mean, or expected value, of $y$ is $X$ a as $E(\varepsilon)$ is 0 . The variance of $y$ is the variance of $\varepsilon_{,} \delta^{2} I$.
2.2 COEEFICIENT ESTIMATES FOR UNIVARIATE REGRESSION
Consider the simple univariate regression with a constant term included, that is
$y=a 1+B x+\varepsilon$ With $\varepsilon \sim N\left(0, \sigma^{2} I\right)$
or $\boldsymbol{y} \sim \mathrm{N}\left(\alpha 1+\beta x, \theta^{2} I\right)$
The least squares estimate of $B$ is
$$b=\left(x^{T} x\right)^{-1} x^{T} y$$
Note that since the $x^{\prime} s$ are deviations from thefr means
\begin{aligned} b &=S_{x y} / S_{x x} \quad(\text { from } 1+6.2) \ \text { and } S_{x y} &=\sum x_{1}\left(y_{1}-y\right)=\sum x_{i} y_{1}=x^{T} y \end{aligned}
As $b 1 s$ a linear combination of the observed $y$ values, then Appendix B 1 can be invoked to give
$d^{2}$ can be estimated by the sample variance of the residuals, namely
$$s^{2}=\sum e_{1}^{2} /(n-2)$$
This has $n-2$ degrees of freedom as there are $n$ observations, but two parameters, $\alpha$ and $\beta$, to be estimated.

## 统计代写|实验设计作业代写experimental design代考|ANOVA TABLES

The analysis of variance table, ANOVA, represents the components of the variation of the dependent variable, $y$. In section $1.5$, we

showed that, for any number of predictor variables, we can divide the vector of observations, $y$, into two orthogonal parts consisting of the predicted values and the residuals, which we can represent by Eigure 1.5.1.
As the vectors e and $\boldsymbol{y}$ are orthogonal, e $\boldsymbol{9}$, then from Pythagoras’ Theorem we know that
$(\operatorname{length} y)^{2}=(\text { ength } \mathbf{y})^{2}+(1 \text { ength e })^{2}\left(r^{2}\right.$
$$y^{T} y=\boldsymbol{y}^{T} \boldsymbol{y}=e^{T} e$$
If there are k predictor terms in the model, but no constant (intercept) term, then the sums of squares and degrees of freedom, d.f., can be displayed as in Table $2.4 .1 .$
Almost invariably it is deviations from the mean which the model is required to explain, so a constant term is included in the model. We have seen from section $1.6 .2$, this has the effect of adjusting each var iable for its mean, and the sum of squares for the mean is given by
$$S S(M e a n)=y^{T} 1\left(1_{1}^{T}\right)^{-1} 1^{T} y=\left(\Sigma y_{j}\right)^{2} / n=n y^{2}$$

（i）总体均值不正确。

## 统计代写|实验设计作业代写experimental design代考|INTRODUCTION

2.2 单变量回归的系数估计

b=(X吨X)−1X吨是的

b=小号X是的/小号XX( 从 1+6.2)  和 小号X是的=∑X1(是的1−是的)=∑X一世是的1=X吨是的

d2可以通过残差的样本方差来估计，即
s2=∑和12/(n−2)

## 统计代写|实验设计作业代写experimental design代考|ANOVA TABLES

(长度⁡是的)2=( 长度 是的)2+(1 英语 )2(r2

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。