高等线性模型|MATH2931 Higher Linear Models代写 unsw代写

Statistics is about using probability models to make decisions from data in
the face of uncertainty. This course gives an introduction to the process of
building statistical models using an important class of models (linear models).
In a linear model we try to predict or explain variation in a response variable
in terms of related quantities (predictors). The relationship between the
expected response and predictors is assumed to be linear. Topics covered
in the course include how to estimate parameters in linear models, how to
compare models, how to select a good model or models when prediction of
the response is the goal. Concepts are illustrated with applications from the
natural and social sciences.

这是一份unsw新南威尔斯大学MATH2931的成功案例

高等线性模型|MATH2931 Higher Linear Models代写 unsw代写


问题 1.

The regression coefficients $\hat{\beta}{1}, \hat{\beta}{2}, \ldots, \hat{\boldsymbol{\beta}}{k}$ in $\hat{\boldsymbol{\beta}}{1}$ can be standardized so as to show the effect of standardized $x$ values (sometimes called $z$ scores). We illustrate this for $k=2$. The model in centered form
$$
\hat{y}{i}=\bar{y}+\hat{\beta}{1}\left(x_{i 1}-\bar{x}{1}\right)+\hat{\beta}{2}\left(x_{i 2}-\bar{x}_{2}\right)
$$


证明 .

This can be expressed in terms of standardized variables as
$$
\frac{\hat{y}{i}-\bar{y}}{s{y}}=\frac{s_{1}}{s_{y}} \hat{\beta}{1}\left(\frac{x{i 1}-\bar{x}{1}}{s{1}}\right)+\frac{s_{2}}{s_{y}} \hat{\beta}{2}\left(\frac{x{i 2}-\bar{x}{2}}{s{2}}\right),
$$
where $s_{j}=\sqrt{s_{j j}}$ is the standard deviation of $x_{j}$. We thus define the standardized coefficients as
$$
\hat{\beta}{j}^{*}=\frac{s{j}}{s_{y}} \hat{\beta}_{j}
$$







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MATH2931 COURSE NOTES :

The hypothesis $H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}$ can be expressed as $H_{0}: \alpha_{1}-\alpha_{2}=0$ and $\alpha_{1}-\alpha_{3}=0$. Thus $H_{0}$ is testable if $\alpha_{1}-\alpha_{2}$ and $\alpha_{1}-\alpha_{3}$ are estimable. To check $\alpha_{1}-\alpha_{2}$ for estimability, we write it as
$$
\alpha_{1}-\alpha_{2}=(0,1,-1,0,0,0) \boldsymbol{\beta}=\boldsymbol{\lambda}{1}^{\prime} \boldsymbol{\beta} $$ and then note that $\boldsymbol{\lambda}{1}^{\prime}$ can be obtained from $\mathbf{X}$ as
$$
(1,0,-1,0,0,0) \mathbf{X}=(0,1,-1,0,0,0)
$$
and from $\mathbf{X}^{\prime} \mathbf{X}$ as
$$
\left(0, \frac{1}{2},-\frac{1}{2}, 0,0,0\right) \mathbf{X}^{\prime} \mathbf{X}=(0,1,-1,0,0,0)
$$




















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