Pembahasan Ujian PAI: A70 – No. 10 – November 2014

• \({A^2} = – N – \frac{S}{N}\)
• \(S = \sum\limits_{i = 1}^N {(2i – 1)\left[ {\ln F({X_i}) + \ln (1 – {F_{N + 1}}_{ – i})} \right]} \)

Dengan rata-rata = \(\bar X = 26,5\), standar deviasi
\(s = 12,362\) \({Z_i} = \frac{{{X_i} – \bar X}}{s}\) \(\Phi ({Z_i}) = norm.\_dist({Z_i})\) \({\rm{ }}S = \sum\limits_{i = 1}^N {(2i – 1)\left[ {\ln F({X_i}) + \ln (1 – {F_{N + 1}}_{ – i})} \right]} = \sum\limits_{i = 1}^{12} {{k_i} = – 147,461} \) \({A^2} = – N – \frac{S}{N}{\rm{ = }} – 12 + \frac{{147,461}}{{12}}\) \({A^2} = 0,288415\) \(A = 0,537043\)