这是一份uoguelph圭尔夫大学学院STAT*3240作业代写的成功案
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
$$
\hat{\rho}(\boldsymbol{x})=\frac{e^{E S P}}{1+e^{E S P}}=\frac{1.1384}{1+1.1384}=0.5324 .
$$
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.
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
$$
\hat{\rho}(\boldsymbol{x})=\frac{e^{E S P}}{1+e^{E S P}}=\frac{0.1830731}{1+0.1830731}=0.1547
$$