Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Dari tabel nomor 22. Tentukan interval kepercayaan 95% berdistribusi log transformed untuk H(3) berdasarkan estimasi Nelson-Aalen
- \(\left( {0,3:0,9} \right)\)
- \(\left( {0,31:1,54} \right)\)
- \(\left( {0,39:0,99} \right)\)
- \(\left( {0,56:0,79} \right)\)
- \(\left( {0,44:1,07} \right)\)
| Diketahui |
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| Rumus yang digunakan | \(\hat H\left( {{t_k}} \right) = \sum\limits_{j = 1}^k {\frac{{{d_j}}}{{{r_j}}}} ,{\rm{ }}\) \({\rm{ }}{t_k} \le t < {t_{k + 1}}\)
\(\widehat {Var\left( {\hat H\left( {{t_k}} \right)} \right)} = \sum\limits_{j = 1}^k {\frac{{{d_j}}}{{r_j^2}}} ,{\rm{ }}{t_k} \le t < {t_{k + 1}}\) Log-Transformed confidence untuk estimasi Nelson-Aaen \(\left( {\frac{{\hat H\left( {{t_k}} \right)}}{U}:U \cdot \hat H\left( {{t_k}} \right)} \right)\) dengan \(U = \exp \left( {\frac{{{z_{\frac{{p + 1}}{2}}}\sqrt {\widehat {Var}\left[ {\hat H\left( {{t_k}} \right)} \right]} }}{{\hat H\left( {{t_k}} \right)}}} \right)\) | ||||||||||||||||||
| Proses pengerjaan | \(\hat H\left( 3 \right) = \sum\limits_{j = 1}^3 {\frac{{{d_j}}}{{{r_j}}}} = \frac{5}{{30}} + \frac{9}{{27}} + \frac{6}{{32}} = 0,6875\) | ||||||||||||||||||
| \(\widehat {Var}\left[ {\hat H\left( 3 \right)} \right] = \sum\limits_{j = 1}^3 {\frac{{{d_j}}}{{r_j^2}}} = \frac{5}{{{{30}^2}}} + \frac{9}{{{{27}^2}}} + \frac{6}{{{{32}^2}}} = 0,0238\) | |||||||||||||||||||
| \(U = \exp \left( {\frac{{{z_{\frac{{0,95 + 1}}{2}}}\sqrt {\widehat {Var}\left[ {\hat H\left( 3 \right)} \right]} }}{{\hat H\left( 3 \right)}}} \right) = \exp \left( {\frac{{1,96\sqrt {0,0238} }}{{0,6875}}} \right) = 1,5519\) | |||||||||||||||||||
| \(\left( {\frac{{\hat H\left( 3 \right)}}{U}:U \cdot \hat H\left( 3 \right)} \right) = \left( {\frac{{0,6875}}{{1,5519}}:\left( {0,6875} \right)\left( {1,5519} \right)} \right) = \left( {0,443:1,0669} \right)\) | |||||||||||||||||||
| Jawaban | e. \(\left( {0,44:1,07} \right)\) |