应用回归分析|STAT 423/BUS 41100/MATH 3113/STAT 4230/6230/STAT 415/615/NHM 726 Applied Regression Analysis代写

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这是一份应用回归分析作业代写的成功案

应用回归分析|STAT 423/BUS 41100/MATH 3113/STAT 4230/6230/STAT 415/615/NHM 726 Applied Regression Analysis代写


$$
\hat{\boldsymbol{\beta}}{O L S}=\frac{n}{n-1} \hat{\boldsymbol{\Sigma}}{\boldsymbol{x}}^{-1} \hat{\boldsymbol{\Sigma}}{\boldsymbol{x} Y} \stackrel{D}{\rightarrow} \boldsymbol{\beta}{O L S} \text { as } \mathrm{n} \rightarrow \infty
$$
and
$$
\hat{\boldsymbol{\Sigma}} \boldsymbol{x} Y=\frac{1}{n} \sum_{i=1}^{n} \boldsymbol{x}{i} Y{i}-\overline{\boldsymbol{x}} \bar{Y}
$$
Thus
$$
\hat{\Sigma}{\boldsymbol{x} Y}=\frac{1}{n}\left[\sum{j: Y_{j}=1} \boldsymbol{x}{j}(1)+\sum{j: Y_{j}=0} \boldsymbol{x}{j}(0)\right]-\overline{\boldsymbol{x}} \hat{\pi}{1}=
$$



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MTH 412/NURS 629/STAT 4530/STA 111L/QTM 100/STA 290 COURSE NOTES :

The discriminant function estimator
$$
\hat{\boldsymbol{\beta}}{D}=\frac{n(n-1)}{N{0} N_{1}} \hat{\Sigma}^{-1} \hat{\Sigma}{\boldsymbol{x}} \hat{\boldsymbol{\beta}}{O L S} .
$$
Now when the conditions of Definition $10.3$ are met and if $\mu_{1}-\mu_{0}$ is small enough so that there is not perfect classification, then
$$
\boldsymbol{\beta}{L R}=\Sigma^{-1}\left(\mu{1}-\mu_{0}\right) .
$$
Empirically, the OLS ESP and LR ESP are highly correlated for many LR data sets where the conditions are not met, eg when some of the predictors are factors. This suggests that $\boldsymbol{\beta}{L R} \approx d \boldsymbol{\Sigma}{\boldsymbol{x}}^{-1}\left(\boldsymbol{\mu}{1}-\boldsymbol{\mu}{0}\right)$ for many LR data sets where $d$ is some constant depending on the data. Results from Haggstrom (1983) suggest that if a binary regression model is fit using OLS software for MLR, then a rough approximation is $\hat{\boldsymbol{\beta}}{L R} \approx \hat{\boldsymbol{\beta}}{O L S} / M S E$. So a rough approximation is LR ESP $\approx(\mathrm{OLS}$ ESP $) / M S E$.




应用回归分析|Applied Regression Analysis 代写 STAT*3240代考

0

这是一份uoguelph圭尔夫大学学院STAT*3240作业代写的成功案

应用回归分析|Applied Regression Analysis 代写 STAT*3240代考
问题 1.

Solution: a) $E S P=\hat{\alpha}+\hat{\beta}{1} x{1}+\hat{\beta}{2} x{2}=-6.26111-0.0536078(65)+$ $0.0028215(3500)=0.1296$. So
$$
\hat{\rho}(\boldsymbol{x})=\frac{e^{E S P}}{1+e^{E S P}}=\frac{1.1384}{1+1.1384}=0.5324 .
$$


证明 .

b) i) Ho the reduced model is good $H_{A}$ use the full model
ii) $G^{2}(R \mid F)=313.457-234.792=78.665$
iii) Now $d f=264-257=7$, and comparing $78.665$ with $\chi_{7,0.999}^{2}=24.32$ shows that the pval $=0<1-0.999=0.001$.
iv) Reject Ho, use the full model.

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STAT*3240 COURSE NOTES :


a) Predict $\hat{\rho}(x)$ if $x=40.0$.
b) Find a $95 \%$ CI for $\beta$.
c) Perform the 4 step Wald test for $H_{o}: \beta=0$.
Solution: a) $\exp [E S P]=\exp [\hat{\alpha}+\hat{\beta}(40)]=\exp [-5.84211+0.103606(40)]=$ $\exp [-1.69787]=0.1830731$. So
$$
\hat{\rho}(\boldsymbol{x})=\frac{e^{E S P}}{1+e^{E S P}}=\frac{0.1830731}{1+0.1830731}=0.1547
$$