引进假设${\displaystyle H:}$该女士没有鉴别力. 如果${\displaystyle H}$正确, 则女士只是在进行随机挑选, 全部挑对的概率为$$1/(\genfrac{}{}{0ex}{}{8}{4})=\frac{1}{70},$$这是小概率的事件, 因此拒绝${\displaystyle H}$.
如果女士只说对了${\displaystyle 3}$杯, 发生的概率是$$(\genfrac{}{}{0ex}{}{4}{3})\cdot 4/(\genfrac{}{}{0ex}{}{8}{4})=\frac{17}{70},$$这并不稀奇, 因此无法拒绝${\displaystyle H}$.

如果${\displaystyle H}$成立, 则$$P({\nu }_1=n_1,\cdots ,{\nu }_r=n_r)=\frac{n!}{n_1!\cdots n_r!}{p}_1^{n_1}\cdots p_r^{n_r},$$从而${\displaystyle ({\nu }_1,\cdots ,{\nu }_r)}$的 特征函数 为$$g(t_1,\cdots ,t_r)=(p_1{\mathrm{e}}^{\mathrm{i}{t}_1}+\cdots +p_r{\mathrm{e}}^{\mathrm{i}{t}_r}{)}^n.$$令${\displaystyle Y_j=\frac{({\nu }_j-n{p}_j)}{\sqrt{n{p}_j}},1\le j\le r}$, 则${\displaystyle k=\sum _{j=1}^r{Y}_j^2}$. 而${\displaystyle (Y_1,\cdots ,Y_r)}$有特征函数$$\phi (t_1,\cdots ,t_r)=\exp (-\mathrm{i}\sqrt{n}\sum _{j=1}^r\sqrt{p_j}{t}_j)\times g(\frac{t_1}{\sqrt{n{p}_1}},\cdots ,\frac{t_r}{\sqrt{n{p}_r}}),$$两边取对数, 作 Taylor 展开, 得$$\begin{array}{rl}\ln \phi (t_1,\cdots ,t_r)=& n\ln [1+\mathrm{i}\frac{1}{\sqrt{n}}\sum _{j=1}^r\sqrt{p_j}{t}_j-\frac{1}{2n}\sum _{j=1}^r{t}_j^2+O(n^{-\frac{3}{2}})]-\mathrm{i}\sqrt{n}\sum _{j=1}^r\sqrt{p_j}{t}_j\\ =& -\frac{1}{2}\sum _{j=1}^r{t}_j^2+\frac{1}{2}{\left(\sum _{j=1}^r\sqrt{p_j}{t}_j\right)}^2+O(n^{-\frac{1}{2}}),\end{array}$$故$$\begin{array}{rl}\text{(1.3)}& \underset{n\to \mathrm{\infty }}{lim}\phi (t_1,\cdots ,t_r)=& \exp (-\frac{1}{2}Q(t_1,\cdots ,t_r)),Q(t_1,\cdots ,t_r)=& \sum _{j=1}^r{t}_j^2-{\left(\sum _{j=1}^r\sqrt{p_j}{t}_j\right)}^2.\end{array}$$
将二次型${\displaystyle Q(t_1,\cdots ,t_r)}$的方阵记为${\displaystyle A=I_r-p{p}^{\mathrm{T}},p=(\sqrt{p_1},\cdots ,\sqrt{p_r}{)}^{\mathrm{T}}}$.
记${\displaystyle Y=(Y_1,\cdots ,Y_r{)}^{\mathrm{T}}}$, 作正交变换$$Z=(Z_1,\cdots ,Z_r{)}^{\mathrm{T}}=BY,$$使${\displaystyle B}$的第一行为${\displaystyle p^{\mathrm{T}}}$, 则$$Z_1=p^{\mathrm{T}}=\frac{1}{\sqrt{n}}\sum _{j=1}^r({\nu }_j-n{p}_i)=0.$$由变换的正交性,$$k=Y_1^2+\cdots +Y_r^2=Z_1^2+\cdots +Z_r^2=\sum _{j=2}^r{Z}_j^2.$$ ${\displaystyle Z}$的特征函数为${\displaystyle \psi (u)=\phi (B^{\prime }u)}$, 其中${\displaystyle u=(u_1,\cdots ,u_r{)}^{\mathrm{T}}}$. 从而由 (1.3),$$\underset{n\to \mathrm{\infty }}{lim}\psi (u)=\exp (-\frac{1}{2}Q(B^{\mathrm{T}}u)).$$但$$Q(B^{\mathrm{T}}u)=(B^{\mathrm{T}}u{)}^{\mathrm{T}}I(B^{\mathrm{T}}u)-(B^{\mathrm{T}}u{)}^{\mathrm{T}}p{p}^{\mathrm{T}}(B^{\mathrm{T}}u)=u_2^2+\cdots +u_r^2,$$因此${\displaystyle (B^{\mathrm{T}}u{)}^{\mathrm{T}}I(B^{\mathrm{T}}u)=u^{\mathrm{T}}B{B}^{\mathrm{T}}u=u^{\mathrm{T}}u}$. 另一方面, ${\displaystyle B}$正交且第一行为${\displaystyle p^{\mathrm{T}}}$, 因此${\displaystyle p^T{B}^{\mathrm{T}}}$除第一个元为${\displaystyle 1}$外, 其余都是 ${\displaystyle 0}$, 从而${\displaystyle p^{\mathrm{T}}(B^{\mathrm{T}}u)=u_1\Rightarrow (B^{\mathrm{T}}u{)}^{\mathrm{T}}p{p}^{\mathrm{T}}(B^{\mathrm{T}}u)=u_1^2}$, 从而回代入${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\psi (u)}$得$$\underset{n\to \mathrm{\infty }}{lim}\psi (u)=\exp (-\frac{1}{2}(u_2^2+\cdots +u_r^2)).$$因此${\displaystyle n\to \mathrm{\infty }}$时${\displaystyle (Z_2,\cdots ,Z_r)}$趋于一个${\displaystyle r-1}$维的各分量独立同分布于${\displaystyle N(0,1)}$的分布, 故${\displaystyle k=\sum _{j=2}^r{Z}_j^2\to {\chi }_{r-1}^2}$.

提出假设${\displaystyle H}$: 事故可能性与班次无关, 也即${\displaystyle F(\{i\})=1/3,i=1,2,3}$. 对本例, ${\displaystyle n=15,p_i=1/3,{\nu }_1=6,{\nu }_2=3,{\nu }_3=6}$; 则由 (1.1) 算出${\displaystyle k_0=1.2}$; 查自由度为${\displaystyle 3-1=2}$的${\displaystyle {\chi }^2}$分布表, 知${\displaystyle P(k\ge k_0)>\frac{1}{3}}$, 这并不稀奇, 因此并不能认为和班次有关.

定义${\displaystyle p_j({\theta }_1,\cdots ,{\theta }_t)=F(a_j,{\theta }_1,\cdots ,{\theta }_t)-F(a_{j-1},{\theta }_1,{\theta }_t)}$(类似 (1.4)). 定义$$\begin{array}{rl}k({\theta }_1,\cdots ,{\theta }_t)=& \sum _{j=1}^r\frac{[{\nu }_j-n{p}_j({\theta }_1,\cdots ,{\theta }_t){]}^2}{n{p}_j({\theta }_1,\cdots ,{\theta }_t)},\\ k^{\ast }=& k({\hat{\theta }}_1,\cdots ,{\hat{\theta }}_t).\end{array}$$若满足以下条件

1. ${\displaystyle \mathrm{\Theta }}$为${\displaystyle {\mathbb{R}}^t}$的开集, ${\displaystyle \mathrm{\exists }({\theta }_{10},\cdots ,{\theta }_{t0})\in \mathrm{\Theta }}$, 使得${\displaystyle X}$的分布为${\displaystyle F(x,{\theta }_{10},\cdots ,{\theta }_{t0})}$;
2. 对任意不同的${\displaystyle ({\theta }_{11},\cdots ,{\theta }_{t1}),({\theta }_{12},\cdots ,{\theta }_{t2})\in \mathrm{\Theta }}$, 有$$\sum _{j=1}^r|p_j({\theta }_{11},\cdots ,{\theta }_{t1})-p_j({\theta }_{12},\cdots ,{\theta }_{t2})|>0;$$
3. ${\displaystyle \frac{\partial p_j({\theta }_1,\cdots ,{\theta }_t)}{\partial {\theta }_i}(i=1,\cdots ,t;j=1,\cdots ,r)}$在${\displaystyle \mathrm{\Theta }}$内连续;
4. 令$$I_{rs}({\theta }_1,\cdots ,{\theta }_t)=\sum _{i=1}^r\frac{1}{p_i({\theta }_1,\cdots ,{\theta }_t)}\frac{\partial p_i({\theta }_1,\cdots ,{\theta }_t)}{\partial {\theta }_r}\frac{\partial p_i({\theta }_1,\cdots ,{\theta }_t)}{\partial {\theta }_s}(r,s=1,\cdots ,t),$$ ${\displaystyle I({\theta }_1,\cdots ,{\theta }_t)=(I_{rs})}$为${\displaystyle t}$阶方阵; ${\displaystyle |I|\ne 0,\mathrm{\forall }({\theta }_1,\cdots ,{\theta }_t)\in \mathrm{\Theta }}$;
5. 方程组$$\sum _{j=0}^r\frac{{\nu }_j}{p_j}\frac{\partial p_j}{\partial {\theta }_i}=0(i=1,\cdots ,t)$$的解${\displaystyle ({\hat{\theta }}_1,\cdots ,{\hat{\theta }}_t)}$是${\displaystyle ({\theta }_1,\cdots ,{\theta }_t)}$的 弱相合估计.

则${\displaystyle n\to \mathrm{\infty }}$时, ${\displaystyle k^{\ast }}$依分布收敛于${\displaystyle {\chi }_{r-1-t}^2}$.

取对数似然函数$$\log L=\sum _{i=1}^r\sum _{j=1}^s{n}_{ij}(\ln p_i^{(1)}+\ln p_j^{(2)}).$$ 在约束条件${\displaystyle \sum _{i=1}^r{p}_i^{(1)}=\sum _{i=1}^s{p}_i^{(2)}=1}$下构造 Lagrange 函数$$f(p_1^{(1)},\cdots ,p_r^{(1)},p_1^{(2)},\cdots ,p_s^{(2)};{\lambda }_1,{\lambda }_2)=\log L-{\lambda }_1(\sum _{i=1}^r{p}_i^{(1)}-1)-{\lambda }_2(\sum _{i=1}^s{p}_i^{(2)}-1).$$
求偏导得$$\begin{array}{r}\{\begin{array}{rl}& \frac{\partial f}{\partial p_i^{(1)}}=\frac{1}{p_i^{(1)}}\sum _{j=1}^s{n}_{ij}-{\lambda }_1=0,\\ & \frac{\partial f}{\partial p_i^{(2)}}=\frac{1}{p_j^{(2)}}\sum _{i=1}^r{n}_{ij}-{\lambda }_2=0\end{array}\Rightarrow \{\begin{array}{rl}& {\hat{p}}_i^{(1)}=\frac{1}{n}\sum _{j=1}^s{n}_{ij},\\ & {\hat{p}}_j^{(2)}=\frac{1}{n}\sum _{i=1}^r{n}_{ij}.\end{array}\end{array}$$