alpha信度系数的95%置信水平的区间估计公式如下：

[img]http://latex.codecogs.com/gif.latex?\inline {\color{red} \blacksquare } \hat{\alpha }\pm \left ( 1.96\right

)\begin{bmatrix} \sqrt{\frac{Q}{m}} \end{bmatrix}" title="\inline {\color{red} \blacksquare } \hat{\alpha }\pm

\left ( 1.96\right )\begin{bmatrix} \sqrt{\frac{Q}{m}} \end{bmatrix}" [/img]
Q为
[img]http://latex.codecogs.com/gif.latex?\begin{bmatrix} \frac{2n^{2}}{\left ( n-1 \right )^{2}\left ( \mathbf

{{j}'\mathbf{S}}\mathbf{j} \right )^{3}} \end{bmatrix}\left [ \left ( \mathbf{{j}'\mathbf{S}}\mathbf{j} \right )

\left ( tr\mathbf{S}^{2}+tr^{2} \mathbf{S}\right )-2\left ( tr\mathbf{S} \right )\left ( \mathbf{{j}'\mathbf{S^

{2}}}\mathbf{j} \right ) \right ]" title="\begin{bmatrix} \frac{2n^{2}}{\left ( n-1 \right )^{2}\left ( \mathbf

{{j}'\mathbf{S}}\mathbf{j} \right )^{3}} \end{bmatrix}\left [ \left ( \mathbf{{j}'\mathbf{S}}\mathbf{j} \right )

\left ( tr\mathbf{S}^{2}+tr^{2} \mathbf{S}\right )-2\left ( tr\mathbf{S} \right )\left ( \mathbf{{j}'\mathbf{S^

{2}}}\mathbf{j} \right ) \right ][/img]<br>其中：n为题目数，m为考生数，j为n*1的向量，S为n*n题目与题目之间的协方差矩阵。