这是一份Sydney悉尼大学STAT4026的成功案例
Once we obtain the estimates of ED50s we calculate contidence intervals for them. For this purpose we apply Fieller’s theorem which enables us to obtain confidence intervals of ratios of Gaussian random variables. Let $\gamma=z_{a} / 2 V_{11} / \beta^{2}$; then the $100(1-\alpha) \%$ confidence interval for $E D 50$ is
$$
\widehat{E D 50}+\frac{\gamma}{1-\gamma}\left(\widehat{E D 50}+\frac{V_{10}}{V_{11}}\right) \pm \frac{z_{a / 2}}{(1-\gamma) \hat{\beta}_{1}} K,
$$
where
$$
K^{2}=V_{00}+2 \widehat{E D 5} 0 V_{10}+\widehat{E D 5} 0^{2} V_{11}-\gamma\left(V_{00}-V_{10}^{2} / V_{11}\right) .
$$
The matrix of $V_{(i j)^{5}}$ is obtained from the logistic regression output. This equation can also be extended to the case of general $p$ by replacing $E D 50$ by $E D 100 p$ in the equation.
STAT4026 COURSE NOTES ：
Since the $\tau_{i}$ are randomly selected, testing individual effects is meaningless and the hypotheses of interest are: $H_{0}: \sigma_{T}^{2}=0$ versus $H_{1} ; \sigma_{T}^{2}>0$. Under $H_{0}$, the test statistic
$$
F_{o}=\frac{\mathrm{SS}(\mathrm{TRT}) /(a-1)}{\mathrm{SSE} /(N-a)}=\frac{\mathrm{MS}(\mathrm{TRT})}{\mathrm{MSE}} \sim \mathcal{F}{a-1, N-a} $$ is constructed in exactly the same manner as in the fixed effects case. However, the expected mean squares associated with the random effects model are different and are needed to construct estimators of the variance components. For the model $(8.1)$, it can be shown that: $$ E[\mathrm{MS}(\mathrm{TRT})]=\sigma^{2}+n \sigma{T}^{2} \quad E[\mathrm{MSE}]=\sigma^{2}
$$