## 统计代写|广义线性模型代写generalized linear model代考|STAT3022

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Correspondence Analysis

The analysis of the hair-eye color data in the previous section revealed how hair and eye color are dependent. But this does not tell us how they are dependent. To study this, we can use a kind of residual analysis for contingency tables called correspondence analysis.

Compute the Pearson residuals $r_{P}$ and write them in the matrix form $R_{i j}$, where $i=1, \ldots, r$ and $j=1, \ldots, c$, according to the structure of the data. Perform the singular value decomposition:
$$R_{r \times c}=U_{r \times w} D_{w \times w} V_{w \times c}^{T}$$
where $r$ is the number of rows, $c$ is the number of columns and $w=\min (r, c) . U$ and $V$ are called the right and left singular vectors, respectively. $D$ is a diagonal matrix with sorted elements $d_{i}$, called singular values. Another way of writing this is:
$$R_{i j}=\sum_{k=1}^{w} U_{i k} d_{k} V_{j k}$$
As with eigendecompositions, it is not uncommon for the first few singular values to be much larger than the rest. Suppose that the first two dominate so that:
$$R_{i j} \approx U_{i 1} d_{1} V_{j 1}+U_{i 2} d_{2} V_{j 2}$$

We usually absorb the $d$ s into $U$ and $V$ for plotting purposes so that we can assess the relative contribution of the components. Thus:
\begin{aligned} R_{i j} & \approx\left(U_{i 1} \sqrt{d_{1}}\right) \times\left(V_{j 1} \sqrt{d_{1}}\right)+\left(U_{i 2} \sqrt{d_{2}}\right) \times\left(V_{j 2} \sqrt{d_{2}}\right) \ & \equiv U_{i 1} V_{j 1}+U_{i 2} V_{j 2} \end{aligned}
where in the latter expression we have redefined the $U \mathrm{~s}$ and $V \mathrm{~s}$ to include the $\sqrt{d}$.

## 统计代写|广义线性模型代写generalized linear model代考|Matched Pairs

\begin{aligned}
&\text { In the typical two-way contingency tables, we display accumulated information } \
&\text { about two categorical measures on the same object. In matched pairs, we observe } \
&\text { one measure on two matched objects. } \
&\text { In Stuart (1955), data on the vision of a sample of women is presented. The left } \
&\text { and right eye performance is graded into four categories: } \
&\text { data (eyegrade) } \
&\text { (ct c- xtabs }(y \sim \text { right+left, eyegrade)) } \
&\text { right best second third worst } \
&\text { second } 234 \text { left } 1512 \quad 432 \quad 78 \
\end{aligned}

If we check for independence:
summary (et)
Call: xtabs (formula – y right + left, data – eyegrade)
Number of cases in table: 7477
Number of factors: 2
Test for independence of all factors:
Chisq – 8097, df – 9, p-value $=0$
We are not surprised to find strong evidence of dependence. Most people’s eyes are similar. A more interesting hypothesis for such matched pair data is symmetry. Is $p_{i j}=p_{j i}$ ? We can fit such a model by defining a factor where the levels represent the symmetric pairs for the off-diagonal elements. There is only one observation for each level down the diagonal:
(symfac <- factor (apply (eyegrade $[, 2: 3], 1$, function (x) paste (sort $(x)$,
$\rightarrow$ collapse=” ” “))))
[1] best-best best-second best-third best-worst
[5] best-second second-second second-third second-worst
[9] best-third second-third third-third third-worst
10 Levels: best-best best-second best-third … worst-worst
We now fit this model:
mods <- glm(y symfac, eyegrade, familympoisson)
c (deviance (mods), df . residual (mods))
[1] $19.2496 .000$
pchisq (deviance (mods), df . residual (mods), lower=F)
[1] $0.0037629$
Here, we see evidence of a lack of symmetry. It is worth checking the residuals:
round (xtabs (residuals (mods) right+left, eyegrade), 3)
round (xtabs (residuals (mods) righ left
We see that the residuals above the diagonal are mostly positive, while they are mostly negative below the diagonal. So there are generally more poor left, good right eye combinations than the reverse. Furthermore, we can compute the marginals:
margin table $(c t, 1)$
right
$\begin{array}{lrr}\text { best second third } & \text { worst } \ 1976 & 2256 & 2456\end{array}$

## 统计代写|广义线性模型代写generalized linear model代考|Ordinal Variables

Some variables have a natural order. We can use the methods for nominal variables described earlier in this chapter, but more information can be extracted by taking advantage of the structure of the data. Sometimes we might identify a particular ordinal variable as the response. In such cases, the methods of Section $7.4$ can be used. However, sometimes we are interested in modeling the association between ordinal variables. Here the use of scores can be helpful.

Consider a two-way table where both variables are ordinal. We may assign scores $u_{i}$ and $v_{j}$ to the rows and columns such that $u_{1} \leq u_{2} \leq \cdots \leq u_{I}$ and $v_{1} \leq v_{2} \leq \cdots \leq v_{J}$. The assignment of scores requires some judgment. If you have no particular prefer-

ence, even spacing allows for the simplest interpretation. If you have an interval scale, for example, $0-10$ years old, 10-20 years old, $20-40$ years old and so on, midpoints are often used. It is a good idea to check that the inference is robust to the assignment of scores by trying some reasonable alternative choices. If your qualitative conclusions are changed, this is an indication that you cannot make any strong finding.
Now fit the linear-by-linear association model:
$$\log E Y_{i j}=\log \mu_{i j}=\log n p_{i j}=\log n+\alpha_{i}+\beta_{j}+\gamma u_{i} v_{j}$$
So $\gamma=0$ means independence while $\gamma$ represents the amount of association and can be positive or negative. $\gamma$ is rather like an (unscaled) correlation coefficient. Consider underlying (latent) continuous variables which are discretized by the cutpoints $u_{i}$ and $v_{j}$. We can then identify $\gamma$ with the correlation coefficient of the latent variables.
Consider an example drawn from a subset of the 1996 American National Election Study (Rosenstone et al. (1997)). Using just the data on party affiliation and level of education, we can construct a two-way table:
data (nes96)
xtabs ( PID + educ, nes96)
\begin{tabular}{lrrrrrrrr}
\multicolumn{8}{c}{ educ } \
PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \
strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \
weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \
indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \
indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \
indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \
weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \
strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25
\end{tabular}
Both variables are ordinal in this example. We need to convert this to a dataframe with one count per line to enable model fitting.

## 统计代写|广义线性模型代写generalized linear model代考|Correspondence Analysis

Rr×C=在r×在D在×在在在×C吨

R一世j=∑ķ=1在在一世ķdķ在jķ

R一世j≈在一世1d1在j1+在一世2d2在j2

R一世j≈(在一世1d1)×(在j1d1)+(在一世2d2)×(在j2d2) ≡在一世1在j1+在一世2在j2

## 统计代写|广义线性模型代写generalized linear model代考|Matched Pairs

在典型的双向列联表中，我们显示累积信息   关于同一对象的两个分类度量。在配对中，我们观察到   对两个匹配的对象进行一次测量。   在 Stuart (1955) 中，提供了关于女性样本视力的数据。左边   右眼表现分为四类：   数据（眼级）   (ct c-xtabs (是∼ 右+左，眼级））   对 最好 第二 第三 最差   最好的 152026612466  第二 234 剩下 151243278  第三 1173621772205  最坏的 3682179492

summary (et)
Call: xtabs (formula – y right + left, data – eyegrade)

Chisq – 8097, df – 9、p值=0

(symfac <- factor (apply (eyegrade[,2:3],1, 函数 (x) 粘贴（排序(X),
→collapse=” ” “)))))
[1] 最佳-最佳-最佳-第二-最佳-第三–最差
[5] 最佳-第二-第二-第二-第三-第二-最差
[9] 最佳-第三-第二-第三-第三-第三第三最差
10 个级别：最好最好最好第二最好第三…最差

mods <- glm(y symfac, eyegrade, familympoisson)
c (deviance (mods), df .residual (mods ))
[1]19.2496.000
pchisq (deviance (mods), df .residual (mods), lower=F)
[1]0.0037629

round (xtabs(residuals (mods) right+left, eyegrade), 3)
round (xtabs(residuals (mods) right left

margin table(C吨,1)

最好的第二第三  最坏的  197622562456

## 统计代写|广义线性模型代写generalized linear model代考|Ordinal Variables

data (nes96)
xtabs ( PID + educ, nes96)

\begin{tabular}{lrrrrrrrr} \multicolumn{8}{c}{ educ} \ PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \ strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \ indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \ indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \ indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \ strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25 \end{tabular}\begin{tabular}{lrrrrrrrr} \multicolumn{8}{c}{ educ} \ PID & MS & HSdrop HS Coll cCdeg & BAdeg MAdeg \ strDem & 5 & 19 & 59 & 38 & 17 & 40 & 22 \weakDem & 4 & 10 & 49 & 36 & 17 & 41 & 23 \ indDem & 1 & 4 & 28 & 15 & 13 & 27 & 20 \ indind & 0 & 3 & 12 & 9 & 3 & 6 & 4 \ indRep & 2 & 7 & 23 & 16 & 8 & 22 & 16 \weakRep & 0 & 5 & 35 & 40 & 15 & 38 & 17 \ strRep & 1 & 4 & 42 & 33 & 17 & 53 & 25 \end{tabular}

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|STAT3015

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Negative Binomial

Given a series of independent trials, each with probability of success $p$, let $Z$ be the number of trials until the $k^{h t h}$ success. Then:
$$P(Z=z)=\left(\begin{array}{l} z-1 \ k-1 \end{array}\right) p^{k}(1-p)^{z-k} \quad z=k, k+1, \ldots$$
The negative binomial can arise naturally in several ways. Imagine a system that can withstand $k$ hits before failing. The probability of a hit in a given time period is $p$ and we count the number of time periods until failure. The negative binomial also arises from a generalization of the Poisson where the parameter $\lambda$ is gamma distributed. The negative binomial also comes up as a limiting distribution for urn schemes that can be used to model contagion.

We get a more convenient parameterization if we let $Y=Z-k$ and $p=(1+\alpha)^{-1}$ so that:
$$P(Y=y)=\left(\begin{array}{c} y+k-1 \ k-1 \end{array}\right) \frac{\alpha^{y}}{(1+\alpha)^{y+k}}, \quad y=0,1,2, \ldots$$
then $E Y=\mu=k \alpha$ and var $Y=k \alpha+k \alpha^{2}=\mu+\mu^{2} / k$
The log-likelihood is then:
$$\sum_{i=1}^{n}\left(y_{i} \log \frac{\alpha}{1+\alpha}-k \log (1+\alpha)+\sum_{j=0}^{y_{i}-1} \log (j+k)-\log \left(y_{i} !\right)\right)$$
The most convenient way to link the mean response $\mu$ to a linear combination of the predictors $X$ is:
$$\eta=x^{T} \beta=\log \frac{\alpha}{1+\alpha}=\log \frac{\mu}{\mu+k}$$
We can regard $k$ as fixed and determined by the application or as anditional parameter to be estimated. More on regression models for negative binomial responses may be found in Cameron and Trivedi (1998) and Lawless (1987).

## 统计代写|广义线性模型代写generalized linear model代考|Zero Inflated Count Models

Sometimes we see count response data where the number of zeroes appearing is significantly greater than the Poisson or negative binomial models would predict. Consider the number of arrests for criminal offenses incurred by individuals. A large number of people have never been arrested by the police while a smaller number have been detained on multiple occasions. Modifying the Poisson by adding a dispersion parameter does not adequately model this divergence from the standard count distributions.

We consider a sample of 915 biochemistry graduate students as analyzed by Long $(1990)$. The response is the number of articles produced during the last three years of the PhD. We are interested in how this is related to the gender, marital status, number of children, prestige of the department and productivity of the advisor of the student. The dataset may be found in the pscl package of Zeileis et al. (2008) which also provides the new model fitting functions needed in this section. We start by fitting a Poisson regression model:
$n=915 p-6$
Deviance $=1634.371$ Null Deviance $=1817.405$ (Difference $=183.034$ )
We can see that deviance is significantly larger than the degrees of freedom. Some experimentation reveals that this cannot be solved by using a richer linear predictor or by eliminating some outliers. We might consider a dispersed Poisson model or negative binomial but some thought suggests that there are good reasons why a student might produce no articles at all. We count and predict how many students produce between zero and seven articles. Very few students produce more than seven articles so we ignore these. The predprob function produces the predicted probabilities for each case. By summing these, we get the expected number for each article count.

## 统计代写|广义线性模型代写generalized linear model代考|Two-by-Two Tables

The data shown in Table $6.1$ were collected as part of a quality improvement study at a semiconductor factory. A sample of wafers was drawn and cross-classified according to whether a particle was found on the die that produced the wafer and whether the wafer was good or bad. More details on the study may be found in Hall (1994). The data might have arisen under several possible sampling schemes:

1. We observed the manufacturing process for a certain period of time and observed 450 wafers. The data were then cross-classified. We could use a Poisson model.
2. We decided to sample 450 wafers. The data were then cross-classified. We could use a multinomial model.
3. We selected 400 wafers without particles and 50 wafers with particles and then recorded the good or bad outcome. We could use a binomial model.
4. We selected 400 wafers without particles and 50 wafers with particles that also included, by design, 334 good wafers and 116 bad ones. We could use a hypergeometric model.

The first three sampling schemes are all plausible. The fourth scheme seems less likely in this example, but we include it for completeness. Such a scheme is more attractive when one level of each variable is relatively rare and we choose to oversample both levels to ensure some representation.

The main question of interest concerning these data is whether the presence of particles on the wafer affects the quality outcome. We shall see that all four sampling schemes lead to exactly the same conclusion. First, let’s set up the data in a convenient form for analysis:
$y<-\mathrm{c}(320,14,80,36)$ particle <- gl $(2,1,4$, labelsmc (“no”, “yes”) quality $<-\mathrm{g}(2,2$, labelsmc (“good”, “bad”)) (wafer <- data. frame (y, particle, quality)) y particle quality $\begin{array}{llll}1 & 320 & \text { no } & \text { good } \ 2 & 14 & \text { yes } & \text { good } \ 3 & 80 & \text { no } & \text { bad } \ 4 & 36 & \text { yes } & \text { bad }\end{array}$.

## 统计代写|广义线性模型代写generalized linear model代考|Negative Binomial

∑一世=1n(是一世日志⁡一个1+一个−ķ日志⁡(1+一个)+∑j=0是一世−1日志⁡(j+ķ)−日志⁡(是一世!))

n=915p−6

## 统计代写|广义线性模型代写generalized linear model代考|Two-by-Two Tables

1. 我们观察了一段时间的制造过程，观察了450个晶圆。然后对数据进行交叉分类。我们可以使用泊松模型。
2. 我们决定对 450 个晶圆进行采样。然后对数据进行交叉分类。我们可以使用多项式模型。
3. 我们选择了 400 个没有颗粒的晶圆和 50 个有颗粒的晶圆，然后记录了结果的好坏。我们可以使用二项式模型。
4. 我们选择了 400 个没有颗粒的晶圆和 50 个有颗粒的晶圆，按照设计，还包括 334 个好晶圆和 116 个坏晶圆。我们可以使用超几何模型。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|MAST30025

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Poisson Regression

If $Y$ is Poisson with mean $\mu>0$, then:
$$P(Y=y)=\frac{e^{-\mu} \mu^{y}}{y !}, \quad y=0,1,2, \ldots$$
Three examples of the Poisson density are depicted in Figure 5.1. In the left panel, we see a distribution that gives highest probability to $y=0$ and falls rapidly as $y$ increases. In the center panel, we see a skew distribution with longer tail on the right. Even for a not so large $\mu=5$, we see the distribution become more normally shaped. This becomes more pronounced as $\mu$ increases.
barplot (dpois $(0: 5,0.5)$, xlab=” $y$ “, ylab=”Probability”, names=0:5, main=”
$\hookrightarrow$ mean $\left.=0.5^{\prime \prime}\right)$
barplot (dpois $(0: 10,2), x 1 a b=” y ” y$ labm”Probability”, names= $0: 10$, main $=$ “
$\rightarrow$ mean $\left.=2^{\prime \prime}\right)$
barplot (dpois $(0: 15,5)$, xlab=” ” $”, y l a b=”$ Probability”, names $0: 15$, main ” “
$\rightarrow$ mean $\left.=5^{\prime \prime}\right)$
The expectation and variance of a Poisson are the same: $E Y=\operatorname{var} Y=\mu$. The Poisson distribution arises naturally in several ways:

1. If the count is some number out of some possible total, then the response would be more appropriately modeled as a binomial. However, for small success probabilities and large totals, the Poisson is a good approximation and can be used. For example, in modeling the incidence of rare forms of cancer, the number of people affected is a small proportion of the population in a given geographical area. Specifically, if $\mu=n p$ while $n \rightarrow \infty$, then $B(n, p)$ is well approximated by Pois $(\mu)$. Also, for small $p$, note that $\operatorname{logit}(p) \approx \log p$, so that the use of the Poisson with a log link is comparable to the binomial with a logit link. Where $n$ varies between cases, a rate model can be used as described in Section 5.3.

## 统计代写|广义线性模型代写generalized linear model代考|Dispersed Poisson Model

We can modify the standard Poisson model to allow for more variation in the response. But before we do that, we must check whether the large size of deviance might be related to some other cause.

In the Galápagos example, we check the residuals to see if the large deviance can be explained by an outlier:
halfnorm (residuals (modp))
The half-normal plot of the (absolute value of the) residuals shown in Figure $5.3$ shows no outliers. It could be that the structural form of the model needs some improvement, but some experimentation with different forms for the predictors will reveal that there is little scope for improvement. Furthermore, the proportion of deviance explained by this model, $1-717 / 3510=0.796$, is about the same as in the linear model above.

For a Poisson distribution, the mean is equal to the variance. Let’s investigate this relationship for this model. It is difficult to estimate the variance for a given value of the mean, but $(y-\hat{\mu})^{2}$ does serve as a crude approximation. We plot this estimated variance against the mean, as seen in the second panel of Figure 5.3:
plot ( $\log ($ fitted (modp) ), $\log (($ gala\$Species-fitted (modp) ) 2$), \quad x l a b=\hookrightarrow$expression (hat (mu)), ylab=expression$\left.\left((y-h a t(m u))^{\wedge} 2\right)\right)$abline$(0,1)$We see that the variance is proportional to, but larger than, the mean. When the variance assumption of the Poisson regression model is broken but the link function and choice of predictors are correct, the estimates of$\beta$are consistent, but the standard er- rors will be wrong. We cannot determine which predictors are statistically significant in the above model using the output we have. The Poisson distribution has only one parameter and so is not very flexible for empirical fitting purposes. We can generalize by allowing ourselves a dispersion parameter. Over- or underdispersion can occur in various ways in Poisson models. For example, suppose the Poisson response$Y$has rate$\lambda$which is itself a random variable. The tendency to fail for a machine may vary from unit to unit even though they are the same model. We can model this by letting$\lambda$be gamma distributed with$E \lambda=\mu$and var$\lambda=\mu / \phi$. Now$Y$is negative binomial with mean$E Y=\mu$. The mean is the same as the Poisson, but the variance var$Y=\mu(1+\phi) / \phi$which is not equal to$\mu$. In this case, overdispersion would occur and could be modeled using a negative binomial model as demonstrated in Section 5.4. If we know the specific mechanism, as in the above example, we could model the response as a negative binomial or other more flexible distribution. If the mechanism is not known, we can introduce a dispersion parameter$\phi$such that var$Y=\phi E Y=\phi \mu$.$\phi=1$is the regular Poisson regression case, while$\phi>1$is overdispersion and$\phi<1$is underdispersion. The dispersion parameter may be estimated using: $$\hat{\phi}=\frac{X^{2}}{n-p}=\frac{\sum_{i}\left(y_{i}-\hat{\mu}{i}\right)^{2} / \hat{\mu}{i}}{n-p}$$ ## 统计代写|广义线性模型代写generalized linear model代考|Rate Models The number of events observed may depend on a size variable that determines the number of opportunities for the events to occur. For example, if we record the number of burglaries reported in different cities, the observed number will depend on the number of households in these cities. In other cases, the size variable may be time. For example, if we record the number of customers served by a sales worker, we must take account of the differing amounts of time worked. Sometimes, it is possible to analyze such data using a binomial response model. For the burglary example above, we might model the number of burglaries out of the number of households. However, if the proportion is small, the Poisson approxima- tion to the binomial is effective. Furthermore, in some examples, the total number of potential cases may not be known exactly. The modeling of rare diseases illustrates this issue as we may know the number of cases but not have precise population data. Sometimes, the binomial model simply cannot be used. In the burglary example, some households may be robbed more than once. In the customer service example, the size variable is not a count. An alternative approach is to model the ratio. However, there are often difficulties with normality and unequal variance when taking this approach, particularly if the counts are small. In Purott and Reeder (1976), some data is presented from an experiment conducted to determine the effect of gamma radiation on the numbers of chromosomal abnormalities (ca) observed. The number (cells), in hundreds of cells exposed in each run, differs. The dose amount (doseamt) and the rate (doserate) at which the dose is applied are the predictors of interest. We may format the data for observation like this: data (dicentric, package=”faraway”) round (xtabs (ca/cells doseamt+doserate, dicentric),2) ## 广义线性模型代考 ## 统计代写|广义线性模型代写generalized linear model代考|Poisson Regression 如果是是泊松的均值μ>0， 然后： 磷(是=是)=和−μμ是是!,是=0,1,2,… 图 5.1 描述了泊松密度的三个示例。在左侧面板中，我们看到一个分布，它给出的概率最高是=0并迅速下降是增加。在中心面板中，我们看到右侧有较长尾部的偏斜分布。即使对于一个不是那么大的μ=5，我们看到分布变得更正常。这变得更加明显μ增加。 条形图（dpois(0:5,0.5), xlab =”是“, ylab=”概率”, 名称=0:5, main=” 意思是=0.5′′) 条形图（dpois(0:10,2),X1一个b=”是”是实验室“概率”，名称=0:10， 主要的= “ →意思是=2′′) 条形图（dpois(0:15,5), xlab =” ””,是l一个b=”概率”，名称0:15， 主要的 ” ” →意思是=5′′) 泊松的期望和方差是相同的：和是=曾是⁡是=μ. 泊松分布以多种方式自然产生： 1. 如果计数是某个可能总数中的某个数字，则响应将更适合建模为二项式。但是，对于较小的成功概率和较大的总数，泊松是一个很好的近似值，可以使用。例如，在模拟罕见癌症的发病率时，受影响的人数只是特定地理区域内人口的一小部分。具体来说，如果μ=np尽管n→∞， 然后乙(n,p)由 Pois 很好地逼近(μ). 另外，对于小p， 注意罗吉特⁡(p)≈日志⁡p，因此使用对数链接的泊松与使用对数链接的二项式相当。在哪里n不同情况下的不同，可以使用第 5.3 节中描述的费率模型。 ## 统计代写|广义线性模型代写generalized linear model代考|Dispersed Poisson Model 我们可以修改标准泊松模型以允许响应的更多变化。但在我们这样做之前，我们必须检查较大的偏差是否与其他原因有关。 在加拉帕戈斯的例子中，我们检查残差，看看是否可以用异常值来解释大偏差： halfnorm (residuals (modp)) 残差（ 的绝对值）的半正态图如图所示5.3显示没有异常值。可能是模型的结构形式需要一些改进，但是对预测变量的不同形式进行一些实验会发现改进的余地很小。此外，该模型解释的偏差比例，1−717/3510=0.796, 与上述线性模型中的大致相同。 对于泊松分布，均值等于方差。让我们研究这个模型的这种关系。对于给定的均值，很难估计方差，但是(是−μ^)2确实可以作为粗略的近似值。 我们将这个估计的方差与平均值作图，如图 5.3 的第二个面板所示：日志⁡(安装（modp）），日志⁡((晚会$物种拟合 (modp) ) 2),Xl一个b=

rors 将是错误的。我们无法使用我们拥有的输出确定哪些预测变量在上述模型中具有统计显着性。

φ^=X2n−p=∑一世(是一世−μ^一世)2/μ^一世n−p

## 统计代写|广义线性模型代写generalized linear model代考|Rate Models

data (dicentric, package=”faraway”)
round (xtabs (ca/cells doseamt​​+doserate, dicentric),2)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|MAST90084

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Prospective and Retrospective Sampling

Consider the data shown in Table $4.1$ from a study on infant respiratory disease which shows the proportions of children developing bronchitis or pneumonia in their first year of life by type of feeding and sex, which may be found in Payne (1987):

\begin{tabular}{llll}
& Bottle Only & Some Breast with Supplement & Breast Only \
\hline Boys & $77 / 458$ & $19 / 147$ & $47 / 494$ \
Girls & $48 / 384$ & $16 / 127$ & $31 / 464$
\end{tabular}
Table $4.1$ Incidence of respiratory disease in infants to the age of 1 year.
We can recover the layout above with the proportions as follows:
data (babyfood, package=” faraway”)
xtabs (disease/ (disease + nondisease) $\sim$ sex + food, babyfood)
food
sex $\quad$ Bottle Breast Suppl
G1rl $0.125000 .066810 \quad 0.12598$
In prospective sampling, the predictors are fixed and then the outcome is observed. This is also called a cohort study. In the infant respiratory disease example shown in Table 4.1, we would select a sample of newborn girls and boys whose parents had chosen a particular method of feeding and then monitor them for their first year.

In retrospective sampling, the outcome is fixed and then the predictors are observed. This is also called a case-control study. Typically, we would find infants coming to a doctor with a respiratory disease in the first year and then record their sex and method of feeding. We would also obtain a sample of respiratory diseasefree infants and record their information. The method for obtaining the samples is important – we require that the probability of inclusion in the study is independent of the predictor values.

## 统计代写|广义线性模型代写generalized linear model代考|Prediction and Effective Doses

Sometimes we wish to predict the outcome for given values of the covariates. For binomial data this will mean estimating the probability of success. Given covariates $x_{0}$, the predicted response on the link scale is $\hat{\eta}=x_{0} \hat{\beta}$ with variance given by $x_{0}^{T}\left(X^{T} W X\right)^{-1} x_{0}$. Approximate confidence intervals may be obtained using a normal approximation. To get an answer in the probability scale, it will be necessary to transform back using the inverse of the link function. We predict the response for the insect data:
data (bliss, packagem” faraway”)
$1 \mathrm{mod}<-$ glm(cbind (dead, alive) conc, familymbinomial, data=bliss)
lmodsum <- summary (lmod)
We show how to predict the response at a dose of $2.5$ :
$x 0<-c(1,2.5)$
eta0 $<-\operatorname{sum}(x 0 * \operatorname{coe}(1 \mathrm{mod}))$
$1 \log i t(\mathrm{eta0})$
[1) $0.64129$
A $64 \%$ predicted chance of death at this dose – now compute a $95 \%$ confidence interval (CI) for this probability. First, extract the variance matrix of the coefficients:
(cm \&- lmodsum\$cov. unscaled) (Intercept)$\begin{array}{rr}\text { (Intercept) } & \text { conc } \ \text { conc } & -0.065823\end{array}$se <- sqrt$(t(x 0)$왛은$\mathrm{cm}$화화$x 0)$so the CI on the probability scale is: ilogit (c (eta0$-1.96 *$se, eta0$1.96 * \mathrm{se})$) [1)$0.534300 .73585$A more direct way of obtaining the same result is: predict (lmod, newdata data. frame (conc=2.5), se=$T$) [1]$0.58095$\$se.fit
[1] $0.2263$
1logit (c (0.58095-1.960.2263,0.58095+1.960.2263))
[1] $0.534300 .73585$
Note that in contrast to the linear regression situation, there is no distinction possible between confidence intervals for a future observation and those for the mean response. Now we try predicting the response probability at the low dose of $-5$ :
$x 0<-c(1,-5)$
se $<-\operatorname{sqrt}(t(x 0)$ 왛의 $\mathrm{cm} \mathrm{~ ㅇ}$
eta0 <- sum $(x 0 * 1 \mathrm{mod}$ scoef $)$
ilogit (c (eta0 -1.96*se, eta0 $0+1.96 * s e)$ )
[1) $2.3577 \mathrm{e}-053.6429 \mathrm{e}-03$

## 统计代写|广义线性模型代写generalized linear model代考|Matched Case-Control Studies

In a case-control study, we try to determine the effect of certain risk factors on the outcome. We understand that there are other confounding variables that may affect the outcome. One approach to dealing with these is to measure or record them, include them in the logistic regression model as appropriate and thereby control for

their effect. But this method requires that we model these confounding variables with the correct functional form. This may be difficult. Also, making an appropriate adjustment is problematic when the distribution of the confounding variables is quite different in the cases and controls. So we might consider an alternative where the confounding variables are explicitly adjusted for in the design.

In a matched case-control study, we match each case (diseased person, defective object, success, etc.) with one or more controls that have the same or similar values of some set of potential confounding variables. For example, if we have a 56-year-old, Hispanic male case, we try to match him with some number of controls who are also 56-year-old Hispanic males. This group would be called a matched set. Obviously, the more confounding variables one specifies, the more difficult it will be to make the matches. Loosening the matching requirements, for example, accepting controls who are 50-60 years old, might be necessary. Matching also gives us the possibility of adjusting for confounders that are difficult to measure. For example, suppose we suspect an environmental effect on the outcome. However, it is difficult to measure exposure, particularly when we may not know which substances are relevant. We could match subjects based on their place of residence or work. This would go some way to adjusting for the environmental effects.

Matched case-control studies also have some disadvantages apart from the difficulties of forming the matched sets. One loses the possibility of discovering the effects of the variables used to determine the matches. For example, if we match on sex, we will not be able to investigate a sex effect. Furthermore, the data will likely be far from a random sample of the population of interest. So although relative effects may be found, it may be difficult to generalize to the population.

Sometimes, cases are rare but controls are readily available. A $1: M$ design has $M$ controls for each case. $M$ is typically small and can even vary in size from matched set to matched set due to difficulties in finding matching controls and missing values. Each additional control yields a diminished return in terms of increased efficiency in estimating risk factors – it is usually not worth exceeding $M=5$.

## 统计代写|广义线性模型代写generalized linear model代考|Prospective and Retrospective Sampling

\begin{tabular}{llll} & 瓶装 & 一些含补充剂的乳房 & 仅乳房 \ \hline 男孩 & $77 / 458$ & $19 / 147$ & $47 / 494$ \ 女孩 & $48 / 384$ & $16 / 127$ & $31 / 464$ \end{表格}\begin{tabular}{llll} & 瓶装 & 一些含补充剂的乳房 & 仅乳房 \ \hline 男孩 & $77 / 458$ & $19 / 147$ & $47 / 494$ \ 女孩 & $48 / 384$ & $16 / 127$ & $31 / 464$ \end{表格}

data (babyfood, package=”faraway”)
xtabs (disease/ (disease + nondisease)∼性+食物，婴儿食品）

## 统计代写|广义线性模型代写generalized linear model代考|Prediction and Effective Doses

1米○d<−glm(cbind (dead, alive) conc, familymbinomial, data=bliss)
lmodsum <- summary (lmod)

X0<−C(1,2.5)

1日志⁡一世吨(和吨一个0)
[1) 0.64129

(cm \&- lmodsum $cov. unscaled) (Intercept) （截距） 浓 浓 −0.065823 se <- sqrt(吨(X0)哇C米华华X0) 所以概率尺度上的CI为： ilogit (c (eta0−1.96∗硒, eta01.96∗s和) ) [1) 0.534300.73585 获得相同结果的更直接的方法是： predict (lmod, newdata data.frame (conc=2.5), se=吨 ) [1] 0.58095$ se.fit
[1]0.2263
1logit (c (0.58095-1.960.2263,0.58095+1.960.2263))
[1]0.534300.73585

X0<−C(1,−5)

eta0 <- 总和(X0∗1米○d斯科夫)
ilogit (c (eta0 -1.96*se, eta00+1.96∗s和) )
[1) 2.3577和−053.6429和−03

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|STAT8111

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Beta Regression

Beta regression is useful for responses that are bounded in $(0,1)$ such as proportions. It could also be used for variables that are bounded in some other finite interval simply by rescaling to $(0,1)$. A Beta-distributed random variable $Y$ has density:
$$f(y \mid a, b)=\frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} y^{a-1}(1-y)^{b-1}$$
for parameters $a, b$ and Gamma function $\Gamma()$. It is more convenient to transform the parameters so $\mu=a /(a+b)$ and $\phi=a+b$ so that $E Y=\mu$ and $\operatorname{var} Y=\mu(1-\mu) /(1+$ $\emptyset)$. We can then link the linear predictor $\eta$ using $\eta=g(\mu)$ using a link function $g$ where any of the choices used for the binomial model would be suitable.

An implementation of the Beta regression model can be found in the macv package of Wood (2006). We can apply this to the mammalsleep also used in the previous section.

The default choice of link is the logit function. The estimated value of $\phi$ is $8.927$. A comparison of the fitted values of this model and the quasi-binomial model fitted earlier reveals no substantial difference. The advantage of the Beta-based model is the full distributional model which would allow the construction of full predictive distributions rather than just a point estimate and standard error.

## 统计代写|广义线性模型代写generalized linear model代考|Latent Variables

Suppose that students answer questions on a test and that a specific student has an aptitude $T$. A particular question might have difficulty $d$ and the student will get the answer correct only if $T>d$. Now if we consider $d$ fixed and $T$ as a random variable with density $f$ and distribution function $F$, then the probability that the student will get the answer wrong is:
$$p=P(T \leq d)=F(d)$$
$T$ is called a latent variable. Suppose that the distribution of $T$ is logistic:
$$F(y)=\frac{\exp (y-\mu) / \sigma}{1+\exp (y-\mu) / \sigma}$$
$\mathrm{SO}$
$$\operatorname{logit}(p)=-\mu / \sigma+d / \sigma$$
If we set $\beta_{0}=-\mu / \sigma$ and $\beta_{1}=1 / \sigma$, we now have a logistic regression model. We can illustrate this in the following example where we set $d=1$ and let $T$ have mean $-1$ and $\sigma=1$ :
$x<-\operatorname{seq}(-6,4,0.1)$
$y<-d \log _{\text {is }}(x$, location $=-1)$
plot (x,y, type=”1″, ylab=”density”, $x l a b=” t “)$
i1 $<-(x<1)$
polygon (c $(x[11], 1,-6), c(y[11], 0,0)$, col=’ gray’)
The plot in Figure $4.1$ shows a logistically distributed latent variable. We can see that this distribution is apparently very similar to the normal distribution. The shaded area represents the probability of getting an answer wrong. As the mean aptitude of this student is somewhat less than the difficulty of the question, this probability is substantially greater than one half.

This idea also arises in a bioassay where we might treat an animal, plant or person with some concentration of a treatment and observe the outcome. For example, suppose we are interested in the concentration of insecticide to be used in exterminating a pest. Insects will have varying tolerances for the toxin and will survive if their tolerance is greater than the dose. In this context, the term tolerance distribution for $T$ is used. Applications in several other areas exist where we observe only a binary outcome but believe this to be generated by some continuous but unobserved variable.

Until now we have used logit link function to connect the probability and the linear predictor. But other choices of link function are reasonable. We need a function that bounds the probability between zero and one. We also expect the link function to be monotone. It is conceivable that the success probability may go up and down as the linear predictor increases but this circumstance is best modeled by adding nonlinear components to the linear predictors such as quadratic terms rather than modifying the link function. The latent variable formulation suggests some other possibilities for the link function. Here are some choices which are implemented in the glm () function:

1. Probit: $\eta=\Phi^{-1}(p)$ where $\Phi$ is the normal cumulative distribution function. This arises from a normally distributed latent variable.
2. Complementary $\log -\log : \eta=\log (-\log (1-p))$. A Gumbel-distributed latent variable will lead to this.
3. Cauchit: $\eta=\tan ^{-1}(\pi(p-1 / 2))$ which is motivated by a Cauchy-distributed latent variable.

We can illustrate the choices using some data from Bliss (1935) on the numbers of insects dying at different levels of insecticide concentration. We fit all four link functions:

These are not very different, but now look at a wider range from $[-4,8]$. We apply the predict function to each of the four models forming a matrix of predicted values. We label the columns and add the information about the dose. The tidyr package is useful for reformatting data from a wide format of multiple measured values per row to a long format where there is only one response value per row. This is accomplished using the gather () function. This format, where each row is an observation and each column is a variable, is the most convenient form for many $R$ analyses. Finally, the ggplot 2 package is useful for completing and well-labeled plot.

## 统计代写|广义线性模型代写generalized linear model代考|Beta Regression

Beta 回归对于有界的响应很有用(0,1)比如比例。它也可以用于限制在其他有限区间内的变量，只需重新缩放到(0,1). 一个 Beta 分布的随机变量是有密度：

F(是∣一个,b)=Γ(一个+b)Γ(一个)Γ(b)是一个−1(1−是)b−1

Beta 回归模型的实现可以在 Wood (2006) 的 macv 包中找到。我们可以将其应用于上一节中使用的哺乳动物睡眠。

## 统计代写|广义线性模型代写generalized linear model代考|Latent Variables

p=磷(吨≤d)=F(d)

F(是)=经验⁡(是−μ)/σ1+经验⁡(是−μ)/σ

X<−序列⁡(−6,4,0.1)

i1<−(X<1)

1. 概率：这=披−1(p)在哪里披是正态累积分布函数。这是由一个正态分布的潜变量引起的。
2. 补充日志−日志:这=日志⁡(−日志⁡(1−p)). 一个 Gumbel 分布的潜在变量将导致这一点。
3. 考希特：这=棕褐色−1⁡(圆周率(p−1/2))这是由柯西分布的潜在变量驱动的。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|OLET5608

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Pearson’s χ 2 Statistic

The deviance is one measure of how well the model fits the data, but there are alternatives. The Pearson’s $X^{2}$ statistic takes the general form:
$$X^{2}=\sum_{i=1}^{n} \frac{\left(O_{i}-E_{i}\right)^{2}}{E_{i}}$$

where $O_{i}$ is the observed count and $E_{i}$ is the expected count for case $i$. For a binomial response, we count the number of successes for which $O_{i}=y_{i}$ while $E_{i}=n_{i} \hat{p}{i}$ and failures for which $O{i}=n_{i}-y_{i}$ and $E_{i}=n_{i}\left(1-\hat{p}{i}\right)$, which results in: $$X^{2}=\sum{i=1}^{n} \frac{\left(y_{i}-n_{i} \hat{p}{i}\right)^{2}}{n{i} \hat{p}{i}\left(1-\hat{p}{i}\right)}$$
If we define Pearson residuals as:
$$r_{i}^{P}=\left(y_{i}-n_{i} \hat{p}{i}\right) / \sqrt{\operatorname{var} \hat{y}{i}}$$
which can be viewed as a type of standardized residual, then $X^{2}=\sum_{i=1}^{n}\left(r_{i}^{P}\right)^{2}$. So the Pearson’s $X^{2}$ is analogous to the residual sum of squares used in normal linear models.

The Pearson $X^{2}$ will typically be close in size to the deviance and can be used in the same manner. Alternative versions of the hypothesis tests described above might use the $X^{2}$ in place of the deviance with the same approximate null distributions. However, some care is necessary because the model is fit to minimize the deviance and not the Pearson’s $X^{2}$. This means that it is possible, although unlikely, that the $X^{2}$ could increase as a predictor is added to the model. $X^{2}$ can be computed like this:
[1] $28.067$
Compare this to:
deviance (lmod)
[1] $16.912$
In this case there is more than the typical small difference between $X^{2}$ and the deviance. However, a test for model fit:
1 -pchisq $(28.067,21)$
[1] $0.13826$
results in a moderate sized $p$-value which would not reject this model which agrees with decision based on the deviance statistic.

## 统计代写|广义线性模型代写generalized linear model代考|Overdispersion

If the binomial model specification is correct, we expect that the residual deviance will be approximately distributed $\chi^{2}$ with the appropriate degrees of freedom. Sometimes, we observe a deviance that is much larger than would be expected if the model were correct. We must then determine which aspect of the model specification is incorrect.

The most common explanation is that we have the wrong structural form for the model. We have not included the right predictors or we have not transformed or combined them in the correct way. We have a number of ways of determining the importance of potential additional predictors and diagnostics for determining better transformations – see Section 8.4. Suppose, however, that we are able to exclude this explanation. This is difficult to achieve, but when we have only one or two predictors, it is feasible to explore the model space quite thoroughly and be sure that there is not a plausible superior model formula.

Another common explanation for a large deviance is the presence of a small

number of outliers. Fortunately, these are easily checked using diagnostic methods. When larger numbers of points are identified as outliers, they become unexceptional, and we might more reasonably conclude that there is something amiss with the error distribution.

Sparse data can also lead to large deviances. In the extreme case of a binary response, the deviance is not even approximately $\chi^{2}$. In situations where the group sizes are simply small, the approximation is poor. Because we cannot judge the fit using the deviance, we shall exclude this case from further consideration in this section.
Having excluded these other possibilities, we might explain a large deviance by deficiencies in the random part of the model. A binomial distribution for $Y$ arises when the probability of success $p$ is independent and identical for each trial within the group. If the group size is $m$, then var $Y=m p(1-p)$ if the binomial assumptions are correct. However, if the assumptions are broken, the variance may be greater. This is overdispersion. In rarer cases, the variance is less and underdispersion results.

## 统计代写|广义线性模型代写generalized linear model代考|Quasi-Binomial

In the previous section, we have demonstrated ways to model data where the supposedly binomial response is more variable than should be expected. A quasi-binomial model is another way to allow for extra-binomial variation. We will explain the method in greater generality than immediately necessary because the idea can be used across a wider range of response types.

The idea is to specify only how the mean and variance of the response are connected to the linear predictor. The method of weighted least squares, as used for standard linear models, would be a simple example of this. An examination of the fitting of the binomial model reveals that this only requires the mean and variance information and does not use any additional information about the binomial distribution. Hence, we can obtain the parameter estimates $\hat{\beta}$ and standard errors without making the full binomial assumption.

The problem arises when we attempt to do inference. To construct a confidence interval or perform an hypothesis test, we need some distributional assumptions. Previously we have used the deviance, but for this we need a likelihood and to compute a likelihood we need a distribution. Now we need a suitable substitute for a likelihood that can be computed without assuming a distribution.

Let $Y_{i}$ have mean $\mu_{i}$ and variance $\phi V\left(\mu_{i}\right)$. We assume that $Y_{i}$ are independent. We

define a score, $U_{i}$ :
$$U_{i}=\frac{Y_{i}-\mu_{i}}{\phi V\left(\mu_{i}\right)}$$
Now:
$$\begin{gathered} E U_{i}=0 \ \operatorname{var} U_{i}=\frac{1}{\phi V\left(\mu_{i}\right)} \ -E \frac{\partial U_{i}}{\partial \mu_{i}}=-E \frac{-\phi V\left(\mu_{i}\right)-\left(Y_{i}-\mu_{i}\right) \phi V^{\prime}\left(\mu_{i}\right)}{\left[\phi V\left(\mu_{i}\right)\right]^{2}}=\frac{1}{\phi V\left(\mu_{i}\right)} \end{gathered}$$
These properties are shared by the derivative of the log-likelihood, $l^{\prime}$. This suggests that we can use $U$ in place of $l^{\prime}$. So we define:
$$Q_{i}=\int_{y_{i}}^{\mu_{i}} \frac{y_{i}-t}{\phi V(t)} d t$$
The intent is that $Q$ should behave like the log-likelihood. We then define the log quasi-likelihood for all $n$ observations as:
$$Q=\sum_{i=1}^{n} Q_{i}$$

## 统计代写|广义线性模型代写generalized linear model代考|Pearson’s χ 2 Statistic

X2=∑一世=1n(○一世−和一世)2和一世

X2=∑一世=1n(是一世−n一世p^一世)2n一世p^一世(1−p^一世)

r一世磷=(是一世−n一世p^一世)/曾是⁡是^一世

[1]28.067

[1]进行比较16.912

1 -pchisq(28.067,21)
[1] 0.13826

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|STA441H5

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Estimation Problems

Estimation of the logistic regression model using the Fisher scoring algorithm, described in Section 8.2, is usually fast. However, difficulties can sometimes arise. When convergence fails, it is sometimes due to a problem exhibited by the following dataset. We take a subset of the famous Fisher Iris data to consider only two of the three species of Iris and use only two of the potential predictors:
ibrary (dplyr)
irisr <- filter(iris, Species ! “virginica”) to ? select (Sepal. Width,
$\rightarrow$ Sepal. Length, Species)
We plot the data using a different shape of plotting symbol for the two species:
(p <- ggplot(irisr, aes (x=Sepa1. Width, y=Sepal. Length, shape=Species))
$\hookrightarrow$ tgeom point ())
We now fit a logistic regression model to see if the species can be predicted from the two sepal dimensions.

$n-100 p-3$ Deviance $-0.000$ Null Deviance $-138.629$ (Difference $-138.629$ ) Notice that the residual deviance is zero indicating a perfect fit and yet none of the predictors are significant due to the high standard errors. A look at the data reveals the reason for this. We see that the two groups are linearly separable so that a perfect fit is possible. We suffer from an embarrassment of riches in this example – we can fit the data perfectly. Unfortunately, this results in unstable estimates of the parameters and their standard errors and would (probably falsely) suggest that perfect predictions can be made. An alternative fitting approach might be considered in such cases called exact logistic regression. See Cox (1970) or Mehta and Patel (1995). Implementations can be found in the elrm and logistix packages in R.

## 统计代写|广义线性模型代写generalized linear model代考|Binomial Regression Model

Suppose the response variable $Y_{i}$ for $i=1, \ldots, n$ is binomially distributed $B\left(m_{i}, p_{i}\right)$ so that:
$$P\left(Y_{i}=y_{i}\right)=\left(\begin{array}{c} m_{i} \ y_{i} \end{array}\right) p_{i}^{y_{i}}\left(1-p_{i}\right)^{m_{i}-y_{i}}$$
We further assume that the $Y_{i}$ are independent. The individual outcomes or trials that compose the response $Y_{i}$ are all subject to the same $q$ predictors $\left(x_{i 1}, \ldots, x_{i q}\right)$. The group of trials is known as a covariate class. For example, we might record whether customers of a particular type make a purchase or not. Conventionally, one outcome is labeled a success (say, making purchase in this example) and the other outcome is labeled as a failure. No emotional meaning should be attached to success and failure in this context. For example, success might be the label given to a patient death with survival being called a failure. Because we need to have multiple trials for each covariate class, data for binomial regression models is more likely to result from designed experiments with a few predictors at chosen values rather than observational data which is likely to be more sparse.
As in the binary case, we construct a linear predictor:
$$\eta_{i}=\beta_{0}+\beta_{1} x_{i 1}+\cdots+\beta_{q} x_{i q}$$
We can use a logistic link function $\eta_{i}=\log \left(p_{i} /\left(1-p_{i}\right)\right)$. The log-likelihood is then given by:
$$l(\beta)=\sum_{i=1}^{n}\left[y_{i} \eta_{i}-m_{i} \log \left(1+e_{i}^{\eta}\right)+\log \left(\begin{array}{c} m_{i} \ y_{i} \end{array}\right)\right]$$
Let’s work through an example to see how the analysis differs from the binary response case.

In January 1986, the space shuttle Challenger exploded shortly after launch. An investigation was launched into the cause of the crash and attention focused on the rubber O-ring seals in the rocket boosters. At lower temperatures, rubber becomes more brittle and is a less effective sealant. At the time of the launch, the temperature was $31^{\circ} \mathrm{F}$. Could the failure of the O-rings have been predicted? In the 23 previous shuttle missions for which data exists, some evidence of damage due to blow by and erosion was recorded on some O-rings. Each shuttle had two boosters, each with three O-rings. For each mission, we know the number of $\mathrm{O}$-rings out of six showing some damage and the launch temperature. This is a simplification of the problem see Dalal et al. (1989) for more details.

## 统计代写|广义线性模型代写generalized linear model代考|Inference

We use the same likelihood-based methods as in Section $2.3$ to derive the binomial deviance:
$$D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\right) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right}$$ where $\hat{y}{i}$ are the fitted values from the model.
Provided that $Y$ is truly binomial and that the $m_{i}$ are relatively large, the deviance is approximately $\chi^{2}$ distributed with $n-q-1$ degrees of freedom if the model is correct. Thus we can use the deviance to test whether the model is an adequate fit. For the logit model of the Challenger data, we may compute:
pchisq (deviance (1mod), df . residual (1mod), lower FALSE)
[1] $0.71641$
Since this $p$-value is well in excess of $0.05$, we conclude that this model fits sufficiently well. Of course, this does not mean that this model is correct or that a simpler model might not also fit adequately. Even so, for the null model:
pchisq $(38.9,22$, lower FALSE)
[1] $0.014489$
We see that the fit is inadequate, so we cannot ascribe the response to simple variation not dependent on any predictor. Note that a $\chi_{d}^{2}$ variable has mean $d$ and standard deviation $\sqrt{2 d}$ so that it is often possible to quickly judge whether a deviance is large or small without explicitly computing the $p$-value. If the deviance is far in excess of the degrees of freedom, the null hypothesis can be rejected.

The $\chi^{2}$ distribution is only an approximation that becomes more accurate as the $m_{i}$ increase. The approximation is very poor for small $m_{i}$ and fails entirely in binary cases where $m_{i}=1$. Although it is not possible to say exactly how large $m_{i}$ should be for an adequate approximation, $m_{i} \geq 5 \forall i$ has often been suggested. Permutation or bootstrap methods might be considered as an alternative.

We can also use the deviance to compare two models, with smaller model $S$ representing a subspace (usually a subset) of a larger model $L$. The likelihood ratio test statistic becomes $D_{S}-D_{L}$. This test statistic is asymptotically distributed $\chi_{l-s}^{2}$, assuming that the smaller model is correct and the distributional assumptions hold. We can use this to test the significance of temperature by computing the difference in the deviances between the model with and without temperature. The model without temperature is just the null model and the difference in degrees of freedom or parameters is one:
pchisq (38.9-16.9,1, lower=FALSE)
[1] $2.7265 \mathrm{e}-06$

## 统计代写|广义线性模型代写generalized linear model代考|Estimation Problems

ibrary (dplyr)
irisr <- filter(iris, Species ! “virginica”) to ? 选择（萼片。宽度，
→萼片。长度，物种）

（p <- ggplot(irisr, aes (x=Sepa1. Width, y=Sepal.Length, shape=Species)）
tgeom point ())

n−100p−3偏差−0.000零偏差−138.629（区别−138.629) 请注意，残差为零表示完美拟合，但由于高标准误差，所有预测变量均不显着。看一下数据就可以发现其中的原因。我们看到这两组是线性可分的，因此可以完美拟合。在这个例子中，我们遭受了财富的尴尬——我们可以完美地拟合数据。不幸的是，这会导致参数及其标准误差的估计不稳定，并且会（可能错误地）表明可以做出完美的预测。在这种情况下，可以考虑另一种拟合方法，称为精确逻辑回归。参见 Cox (1970) 或 Mehta 和 Patel (1995)。可以在 R 的 elrm 和 logistix 包中找到实现。

## 统计代写|广义线性模型代写generalized linear model代考|Binomial Regression Model

l(b)=∑一世=1n[是一世这一世−米一世日志⁡(1+和一世这)+日志⁡(米一世 是一世)]

1986 年 1 月，挑战者号航天飞机在发射后不久爆炸。对坠机原因进行了调查，并将注意力集中在火箭助推器中的橡胶 O 形密封圈上。在较低温度下，橡胶变得更脆，并且是一种不太有效的密封剂。发射时气温为31∘F. O 形圈的故障是否可以预测？在有数据的之前的 23 次航天飞机任务中，一些 O 形环上记录了一些因吹漏和腐蚀而损坏的证据。每个航天飞机有两个助推器，每个助推器有三个 O 形环。对于每个任务，我们知道○- 六环显示一些损坏和发射温度。这是对问题的简化，请参见 Dalal 等人。(1989) 了解更多详情。

## 统计代写|广义线性模型代写generalized linear model代考|Inference

D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\右) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right}D=2 \sum_{i=1}^{n}\left{y_{i} \log y_{i} / \hat{y}{i}+\left(m{i}-y_{i}\右) \log \left(m_{i}-y_{i}\right) /\left(m_{i}-\hat{y}{i}\right)\right}在哪里是^一世是模型的拟合值。

pchisq (deviance (1mod), df .residual (1mod), lower FALSE)
[1]0.71641

pchisq(38.9,22, 下 FALSE)
[1]0.014489

pchisq (38.9-16.9,1, lower=FALSE)
[1]2.7265和−06

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|ST411

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Diagnostics

Regression diagnostics are useful in checking the assumptions of the model and in identifying any unusual points. As with linear models, residuals are the most important means of determining how well the data fits the model and where any changes or

improvements might be advisable. We can compute residuals as a difference between observed and fitted values. There are two kinds of fitted (or predicted) values:
linpred <- prediet (lmod)
predprob <- predict (lmod, type=” response”)
The former is the predicted value in the linear predictor scale, $\eta$, while the latter is the
predicted probability $p=\operatorname{logit}^{-1}(\eta)$. Here are the first few values and a confirmation
of the relationship between them:
These can also be obtained as residuals (lmod, type $=$ “response”). Following
the standard practice for diagnostics in linear models, we plot the residuals against
the fitted values:

plot (rawres linpred, xlab” “linear predictor”, ylab=” residuals”)
The plot, as seen in the first panel of Figure 2.6, is not very helpful. Because $y=0$ or 1 , the residual can take only two values given a fixed linear predictor. The upper line in the plot corresponds to $y=1$ and the lower line to $y=0$. We gain no insight into the fit of the model. We have chosen to plot the linear predictor rather than the predicted probability on the horizontal axis because the former provides a better spacing of the points in this direction.

## 统计代写|广义线性模型代写generalized linear model代考|Model Selection

The analysis thus far has used only two of the predictors available but we might construct a better model for the response if we used some of the other predictors. We might find that not all these predictors are helpful in explaining the response. We would like to identify a subset of the predictors that model the response well without including any superfluous predictors.

We could use the inferential methods to construct hypothesis tests to compare various candidate models and use this as a mechanism for choosing a model. Back-

ward elimination is one such method which is relatively easy to implement. The method proceeds sequentially:

1. Start with the full model including all the available predictors. We can add derived predictors formed from transformations or interactions between two or more predictors.
2. Compare this model with all the models consisting of one less predictor. Compute the $p$-value corresponding to each dropped predictor. The dropl function in $\mathrm{R}$ can be used for this purpose.
3. Eliminate the term with largest $p$-value that is greater than some preset critical value, say $0.05$. Return to the previous step. If no such term meets this criterion, stop and use the current model.

Thus predictors are sequentially eliminated until a final model is settled upon. Unfortunately, this is an inferior procedure. Although the algorithm is simple to use, it is hard to identify the problem to which it provides a solution. It does not identify the best set of predictors for predicting future responses. It is not a reliable indication of which predictors are the best explanation for the response. Even if one believes the fiction that there is a true model, this procedure would not be best for identifying such a model.

The Akaike information criterion (AIC) is a popular way of choosing a model see Section A.3 for more. The criterion for a model with likelihood $L$ and number of parameters $q$ is defined by
$$A I C=-2 \log L+2 q$$
We select the model with the smallest value of AIC among those under consideration. Any constant terms in the definition of log-likelihood can be ignored when comparing different models that will have the same constants. For this reason we can use $\mathrm{AIC}=$ deviance $+2 q$

## 统计代写|广义线性模型代写generalized linear model代考|Goodness of Fit

As mentioned earlier, we cannot use the deviance for a binary response GLM as a measure of fit. We can use diagnostic plots of the binned residuals to help us identify inadequacies in the model but these cannot tell us whether the model fits or not. Even so the process of binning can help us develop a test for this purpose. We divide the observations up into $J$ bins based on the linear predictor. Let the mean response in the $j^{t h}$ bin be $y_{j}$ and the mean predicted probability be $\hat{p}{j}$ with $m{j}$ observations within the bin. We compute these values:
wogsm $<-$ na.omit (wogs)
wegsm \&- mutate (wcgsm, predprob”predict (lmod, type=” response”))
gdf <- group by (wcgsm, cut (linpred, breaksmunique (quantile (linpred,
$\hookrightarrow(1: 100) / 101))))$
hldf <- summarise (gde, y=sum (y), ppredmean (predprob), count=n()) There are a few missing values in the data. The default method is to ignore these cases. The na. omit command drops these cases from the data frame for the purposes of this calculation. We use the same method of binning the data as for the residuals but now we need to compute the number of observed cases of heart disease and total observations within each bin. We also need the mean predicted probability within
each bin. When we make a prediction with probability $p$, we would hope that the event oc-
When we make a prediction with probability $p$, we would hope that the event occurs in practice with that proportion. We can check that by plotting the observed proportions against the predicted probabilities as seen in Figure 2.9. For a wellcalibrated prediction model, the observed proportions and predicted probabilities should be close.

## 统计代写|广义线性模型代写generalized linear model代考|Diagnostics

linpred <- prediet (lmod)
predprob <- predict (lmod, type=”response”)

plot (rawres linpred, xlab” “linear predictor”, ylab=”residuals”)

## 统计代写|广义线性模型代写generalized linear model代考|Model Selection

1. 从包含所有可用预测变量的完整模型开始。我们可以添加由两个或多个预测变量之间的转换或交互形成的派生预测变量。
2. 将此模型与包含一个较少预测变量的所有模型进行比较。计算p- 对应于每个丢弃的预测器的值。中的 dropl 函数R可用于此目的。
3. 消除最大的项p- 大于某个预设临界值的值，例如0.05. 返回上一步。如果没有此类术语符合此标准，请停止并使用当前模型。

Akaike 信息标准 (AIC) 是一种流行的选择模型的方法，请参阅第 A.3 节了解更多信息。具有似然性的模型的标准大号和参数数量q定义为

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|广义线性模型代写generalized linear model代考|STAT 7430

statistics-lab™ 为您的留学生涯保驾护航 在代写广义线性模型generalized linear model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。我们的专家在代写广义线性模型generalized linear model代写方面经验极为丰富，各种代写广义线性模型generalized linear model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|广义线性模型代写generalized linear model代考|Heart Disease Example

What might affect the chance of getting heart disease? One of the earliest studies addressing this issue started in 1960 and used 3154 healthy men, aged from 39 to 59 , from the San Francisco area. At the start of the study, all were free of heart disease. Eight and a half years later, the study recorded whether these men now suffered from heart disease along with many other variables that might be related to the chance of developing this disease. We load a subset of this data from the Western Collaborative Group Study described in Rosenman et al. (1975):
data (wcgs, package=”faraway”)
We start by focusing on just three of the variables in the dataset:
We see that only 257 men developed heart disease as given by the factor variable chd. The men vary in height (in inches) and the number of cigarettes (cigs) smoked per day. We can plot these data using $R$ base graphics:
plot (height chd, wcgs)
wcgs\$y <- ifelse (wogs\$chd = “no”, 0, 1)
plot (jitter $(y, 0.1) \sim$ jitter (height), wegs, $x l a b=$ “Height”, $y$ lab=”Heart
$\hookrightarrow$ Disease”, peh=”.”)
The first panel in Figure $2.1$ shows a boxplot. This shows the similarity in the distribution of heights of the two groups of men with and without heart disease. But the heart disease is the response variable so we might prefer a plot which treats it as such. We convert the absence/presence of disease into a numerical $0 / 1$ variable and plot this in the second panel of Figure 2.1. Because heights are reported as round numbers of inches and the response can only take two values, it is sensible to add a small amount of noise to each point, called jittering, so that we can distinguish them. Again we can see the similarity in the distributions. We might think about fitting a line to this plot.

More informative plots may be obtained using the ggplot2 package of Wickham (2009). In the first panel of Figure 2.2, we see two histograms showing the distribution of heights for both those with and without heart disease. The dodge option ensures that the two histograms are interleaved. We see that the two distributions are similar. We also had to set the bin width of the histogram. It was natural to use one inch as all the height measurements are rounded to the nearest inch. In the second panel of Figure $2.2$, we see the corresponding histograms for smoking. In this case, we have shown the frequency rather than the count version of the histogram. We see that smokers are more likely to get heart disease.

## 统计代写|广义线性模型代写generalized linear model代考|Logistic Regression

Suppose we have a response variable $Y_{i}$ for $i=1, \ldots, n$ which takes the values zero or one with $P\left(Y_{i}=1\right)=p_{i}$. This response may be related to a set of $q$ predictors $\left(x_{i 1}, \ldots, x_{i q}\right)$. We need a model that describes the relationship of $x_{1}, \ldots, x_{q}$ to the probability $p$. Following the linear model approach, we construct a linear predictor:
$$\eta_{i}=\beta_{0}+\beta_{1} x_{i 1}+\cdots+\beta_{q} x_{i q}$$
Since the linear predictor can accommodate quantitative and qualitative predictors with the use of dummy variables and also allows for transformations and combinations of the original predictors, it is very flexible and yet retains interpretability. The idea that we can express the effect of the predictors on the response solely through the linear predictor is important. The idea can be extended to models for other types of response and is one of the defining features of the wider class of generalized linear models (GLMs) discussed later in Chapter 8 .

We have seen previously that the linear relation $\eta_{i}=p_{i}$ is not workable because we require $0 \leq p_{i} \leq 1$. Instead we shall use a link function $g$ such that $\eta_{i}=g\left(p_{i}\right)$. We need $g$ to be monotone and be such that $0 \leq g^{-1}(\eta) \leq 1$ for any $\eta$. The most popular choice of link function in this situation is the logit. It is defined so that:
$$\eta=\log (p /(1-p))$$
or equivalently:
$$p=\frac{e^{\eta}}{1+e^{\eta}}$$
Combining the use of the logit link with a linear predictor gives us the term logistic regression. Other choices of link function are possible but we will defer discussion of these until later. The logit and its inverse are defined as logit and ilogit in the faraway package. The relationship between $p$ and the linear predictor $\eta$ is shown in Figure 2.4.

## 统计代写|广义线性模型代写generalized linear model代考|Consider two models

Consider two models, a larger model with $l$ parameters and likelihood $L_{L}$ and a smaller model with $s$ parameters and likelihood $L_{S}$ where the smaller model represents a subset (or more generally a linear subspace) of the larger model. Likelihood

methods suggest the likelihood ratio statistic:
$$2 \log \frac{L_{L}}{L_{S}}$$
as an appropriate test statistic for comparing the two models. Now suppose we choose a saturated larger model – such a model typically has as many parameters as cases and has fitted values $\hat{p}{i}=y{i}$. The test statistic becomes:
$$D=-2 \sum_{i=1}^{n} \hat{p}{i} \operatorname{logit}\left(\hat{p}{i}\right)+\log \left(1-\hat{p}{i}\right)$$ where $\hat{p}{i}$ are the fitted values from the smaller model. $D$ is called the deviance and is useful in making hypothesis tests to compare models.

In other examples of GLMs, the deviance is a measure of how well the model fit the data but in this case, $D$ is just a function of the fitted values $\hat{p}$ so it cannot be used for that purpose. Other methods must be used to judge goodness of fit for binary data – for example, the Hosmer-Lemeshow test described in Section 2.6.
In the summary output previously, we had:
Deviance – 1749.049 Nu11 Deviance $=1781.244$ (Difference – 32.195)
The Deviance is the deviance for the current model while the Nu1l Deviance is the deviance for a model with no predictors and just an intercept term.

We can use the deviance to compare two nested models. The test statistic in (2.1) becomes $D_{S}-D_{L}$. This test statistic is asymptotically distributed $\chi_{l-s}^{2}$, assuming that the smaller model is correct and the distributional assumptions hold. For example, we can compare the fitted model to the null model (which has no predictors) by considering the difference between the residual and null deviances. For the heart disease example, this difference is $32.2$ on two degrees of freedom (one for each predictor). Hence, the $p$-value for the test of the hypothesis that at least one of the predictors is related to the response is:
1-pchisq $(32,2,2)$
(1) $1.0183 \mathrm{e}-07$
Since this value is so small, we are confident that there is some relationship between the predictors and the response. Note that the expected value of a $\chi^{2}$-variate with $d$ degrees of freedom is simply $d$ so we knew the $p$-value would be small before even calculating it.

## 统计代写|广义线性模型代写generalized linear model代考|Heart Disease Example

data (wcgs, package=”faraway”)

plot (height chd, wcgs)
wcgs $y <- ifelse (wogs$ chd = “no”, 0, 1)
plot (jitter(是,0.1)∼抖动（高度），wegs，Xl一个b=“高度”，是实验室=”心脏

p=和这1+和这

## 统计代写|广义线性模型代写generalized linear model代考|Consider two models

2日志⁡大号大号大号小号

Deviance – 1749.049 Nu11 Deviance=1781.244（差值 – 32.195）

1-pchisq(32,2,2)
(1) 1.0183和−07

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|Generalized linear model代考广义线性模型代写|Null Hypothesis Statistical Significance Testing

statistics-lab™ 为您的留学生涯保驾护航 在代写Generalized linear model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Generalized linear model代写方面经验极为丰富，各种代写Generalized linear model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|Generalized linear model代考广义线性模型代写|Null Hypothesis Statistical Significance Testing

The main purpose of this chapter is to transition from the theory of inferential statistics to the application of inferential statistics. The fundamental process of inferential statistics is called null hypothesis statistical significance testing (NHST). All procedures in the rest of this textbook are a form of NHST, so it is best to think of NHSTs as statistical procedures used to draw conclusions about a population based on sample data.
There are eight steps to NHST procedures:

1. Form groups in the data.
2. Define the null hypothesis $\left(\mathrm{H}_{0}\right)$. The null hypothesis is always that there is no difference between groups or that there is no relationship between independent and dependent variables.
3. Set alpha $(\alpha)$. The default alpha $=.05$.
4. Choose a one-tailed or a two-tailed test. This determines the alternative hypothesis $\left(\mathrm{H}_{\mathrm{l}}\right)$.
5. Find the critical value, which is used to define the rejection region. Calculate the observed value.
6. Compare the observed value and the critical value. If the observed value is more extreme than the critical value, then the null hypothesis should be rejected. Otherwise, it should be retained.
7. Calculate an effect size.
8. Readers who have had no previous exposure to statistics will find these steps confusing and abstract right now. But the rest of this chapter will define the terminology and show how to put these steps into practice. To reduce confusion, this book starts with the simplest possible NHST: the $z$-test.

## 统计代写|Generalized linear model代考广义线性模型代写|z-Test

Recall from the previous chapter that social scientists are often selecting samples from a population that they wish to study. However, it is usually impossible to know how representative a single sample is of the population. One possible solution is to follow the process shown at the end of Chapter 6 where a researcher selects many samples from the same population in order to build a probability distribution. Although this method works, it is not used in the real world because it is too expensive and time-consuming. (Moreover, nobody wants to spend their life gathering data from an infinite number of populations in order to build a sampling distribution.) The alternative is to conduct a $z$-test. A z-test is an NHST that scientists use to determine whether their sample is typical or representative of the population it was drawn from.

In Chapter 6 we learned that a sample mean often is not precisely equal to the mean of its parent population. This is due to sampling error, which is also apparent in the variation in mean values from sample to sample. If several samples are taken from the parent population, the means from each sample could be used to create a sampling distribution of means. Because of the principles of the central limit theorem (CLT), statisticians know that with an infinite number of sample means, the distribution will be normally distributed (if the $n$ of each sample is $\geq 25$ ) and the mean of means will always be equal to the population mean.

Additionally, remember that in Chapter 5 we saw that the normal distribution theoretically continues on from $-\infty$ to $+\infty$. This means that any $\bar{X}$ value is possible. However, because the sampling distribution of means is tallest at the population mean and shortest at the tails, the sample means close to the population mean are far more likely to occur than sample means that are very far from $\mu$.

Therefore, the question of inferential statistics is not whether a sample mean is possible but whether it is likely that the sample mean came from the population of interest. That requires a researcher to decide the point at which a sample mean is so different from the population mean that obtaining that sample mean would be highly unlikely (and, therefore, a more plausible explanation is that the sample really does differ from the population). If a sample mean $(\bar{X})$ is very similar to a population mean ( $\mu$ ), then the null hypothesis $(\bar{X}=\mu)$ is a good model for the data. Conversely, if the sample mean and population mean are very different, then the null hypothesis does not fit the data well, and it is reasonable to believe that the two means are different.

This can be seen in Figure 7.1, which shows a standard normal distribution of sampling means. As expected, the population mean $(\mu)$ is in the middle of the distribution, which is also the peak of the sampling distribution. The shaded regions in the tails are called the rejection region. If, when we graph the sample mean, it falls within the rejection region, the sample is so different from the mean that it is unlikely that sampling error alone could account for the differences between $\bar{X}$

and $\mu$ – and it is more likely that there is an actual difference between $\bar{X}$ and $\mu$. If the sample mean is outside the rejection region, then $\bar{X}$ and $\mu$ are similar enough that it can be concluded that $\bar{X}$ is typical of the population (and sampling error alone could possibly account for all of the differences between $\bar{X}$ and $\mu$ ).

Judging whether differences between $\bar{X}$ and $\mu$ are “close enough” or not due to just sampling error requires following the eight steps of NHST. To show how this happens in real life, we will use an example from a UK study by Vinten et al. (2009).

## 统计代写|Generalized linear model代考广义线性模型代写|Cautions for Using NHSTs

NHST procedures dominate quantitative research in the behavioral sciences (Cumming et al., 2007; Fidler et al., 2005; Warne, Lazo, Ramos, \& Ritter, 2012). But it is not a flawless procedure, and NHST is open to abuse. In this section, we will explore three of the main problems with NHST: (1) the possibility of errors, (2) the subjective decisions involved in conducting an NHST, and (3) NHST’s sensitivity to sample size.

Type I and Type II Errors. In the Vinten et al. (2009) example, we rejected the null hypothesis because the $z$-observed value was inside the rejection region, as is apparent in Figure $7.4$ (where the z-observed value was so far below zero that it could not be shown on the figure). But this does not mean that the null hypothesis is definitely wrong. Remember that theoretically – the probability distribution extends from $-\infty$ to to . Therefore, it is possible that a random sample could have an $\bar{X}$ value as low as what was observed in Vinten et al.’s (2009) study. This is clearly an unlikely event – but, theoretically, it is possible. So even though we rejected the null hypothesis and the $\bar{X}$ value in this example had a very extreme z-observed value, it is still possible that Vinten et al. just had a weird sample (which would produce a large amount of sampling error). Thus, the results of this z-test do not prove that the anti-seizure medication is harmful to children in the womb.

Scientists never know for sure whether their null hypothesis is true or not – even if that null is strongly rejected, as in this chapter’s example (Open Science Collaboration, 2015; Tukey, 1991). There is always the possibility (no matter how small) that the results are just a product of sampling error. When researchers reject the null hypothesis and it is true, they have made a Type I error. We can use the $z$-observed value and Appendix Al to calculate the probability of Type I error if the null hypothesis were perfectly true and the researcher chose to reject the null hypothesis (regardless of the $\alpha$ level). This probability is called a $p$-value (abbreviated as $p$ ). Visually, it can be represented, as in Figure $7.6$, as the region of the sampling distribution that starts at the observed value and includes everything beyond it in the tail.

To calculate $p$, you should first find the $z$-observed value in column $\mathrm{A}$. (If the $z$-observed value is not in Appendix A1, then the Price Is Right rule applies, and you should select the number in column A that is closest to the $z$-observed value without going over it.) The number in column $\mathrm{C}$ in the same row will be the $p$-value. For example, in a one-tailed test, if $z$-observed were equal to $+2.10$, then the $p$-value (i.e., the number in column $\mathrm{C}$ in the same row) would be 0179 . This $p$-value means that in this example the probability that these results could occur through purely random sampling error is .0179 (or $1.79 \%$ ). In other words, if we selected an infinite number of samples from the population, then $1.79 \%$ of $\bar{X}$ values would be as different as or more different than $\mu$. But remember that this probability only applies if the null hypothesis is perfectly true (see Sidebar 10.3).

In the Vinten et al. $(2009)$ example, the $z$-observed value was $-9.95$. However, because Appendix Al does not have values that high, we will select the last row (because of the Price Is Right rule), which has the number $\pm 5.00$ in column A. The number in column $\mathrm{C}$ in the same row is $.0000003$, which is the closest value available for the $p$-value. In reality, $p$ will be smaller than this tiny number. (Notice how the numbers in column $C$ get smaller as the numbers in column $A$ get bigger. Therefore, a $z$-observed value that is outside the $\pm 5.00$ range will have smaller $p$-values than the values in the table.) Thus, the chance that – if the null hypothesis were true – Vinten et al. (2009) would obtain a random sample of 41 children with such low VABS scores is less than .0000003-or less than 3 in 10 million. Given this tiny probability of making a Type I error if the null hypothesis were true, it seems more plausible that these results are due to an actual difference between the sample and the population – and not merely to sampling error.

## 统计代写|Generalized linear model代考广义线性模型代写|Null Hypothesis Statistical Significance Testing

NHST 程序有八个步骤：

1. 在数据中形成组。
2. 定义零假设(H0). 零假设始终是组之间没有差异，或者自变量和因变量之间没有关系。
3. 设置阿尔法(一种). 默认阿尔法=.05.
4. 选择单尾或双尾测试。这决定了备择假设(Hl).
5. 找到用于定义拒绝区域的临界值。计算观察值。
6. 比较观察值和临界值。如果观测值比临界值更极端，则应拒绝原假设。否则，应保留。
7. 计算效应量。
8. 以前没有接触过统计数据的读者现在会发现这些步骤令人困惑和抽象。但本章的其余部分将定义术语并展示如何将这些步骤付诸实践。为了减少混淆，本书从最简单的 NHST 开始：和-测试。

## 统计代写|Generalized linear model代考广义线性模型代写|Cautions for Using NHSTs

NHST 程序主导了行为科学的定量研究（Cumming 等人，2007；Fidler 等人，2005；Warne，Lazo，Ramos，\& Ritter，2012）。但这不是一个完美的程序，NHST 很容易被滥用。在本节中，我们将探讨 NHST 的三个主要问题：(1) 错误的可能性，(2) 进行 NHST 所涉及的主观决定，以及 (3) NHST 对样本量的敏感性。

I 型和 II 型错误。在文滕等人。（2009）的例子，我们拒绝了原假设，因为和- 观察值在拒绝区域内，如图所示7.4（其中 z 观测值远低于零，无法在图中显示）。但这并不意味着零假设肯定是错误的。请记住，理论上——概率分布从−∞到 到 。因此，一个随机样本可能有一个X¯值与 Vinten 等人 (2009) 研究中观察到的值一样低。这显然是一个不太可能发生的事件——但从理论上讲，这是可能的。所以即使我们拒绝了原假设和X¯这个例子中的值有一个非常极端的 z 观察值，Vinten 等人仍然有可能。只是有一个奇怪的样本（这会产生大量的抽样误差）。因此，该 z 检验的结果并不能证明抗癫痫药物对子宫内的儿童有害。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。