Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
Mei 2017 |
| Nomor Soal |
: |
23 |
SOAL
Anda mencocokkan model berikut dalam empat pengamatan:
\({Y_i} = {\beta _1} + {\beta _2}{X_{2i}} + {\beta _3}{X_{3i}} + {\varepsilon _i}\) , \(i = 1,2,3,4\)
Diberikan data sebagai berikut:
| \(i\) |
\({X_{2i}}\) |
\({X_{3i}}\) |
| 1 |
-4 |
-2 |
| 2 |
-2 |
4 |
| 3 |
2 |
-4 |
| 4 |
4 |
2 |
Estimasi least square dari \({\beta _3}\) dinyatakan sebagai \({\hat \beta _3} = \sum\nolimits_{i = 1}^4 {{w_i}{Y_i}} \)
Tentukan nilai dari \(\left( {{w_1},{w_2},{w_3},{w_4}} \right)\)
- \(\left( { – \frac{1}{{20}},\frac{3}{{20}}, – \frac{3}{{20}},\frac{1}{{20}}} \right)\)
- \(\left( { – \frac{1}{{20}}, – \frac{3}{{20}},\frac{3}{{20}},\frac{1}{{20}}} \right)\)
- \(\left( {\frac{1}{{20}}, – \frac{2}{{20}},\frac{2}{{20}}, – \frac{1}{{20}}} \right)\)
- \(\left( { – \frac{1}{{20}},\frac{2}{{20}}, – \frac{2}{{20}},\frac{1}{{20}}} \right)\)
- \(\left( {\frac{1}{4},\frac{1}{4}, – \frac{1}{4}, – \frac{1}{4}} \right)\)
| Diketahui |
Anda mencocokkan model berikut dalam empat pengamatan:
\({Y_i} = {\beta _1} + {\beta _2}{X_{2i}} + {\beta _3}{X_{3i}} + {\varepsilon _i}\) , \(i = 1,2,3,4\)
Diberikan data sebagai berikut
| \(i\) |
\({X_{2i}}\) |
\({X_{3i}}\) |
| 1 |
-4 |
-2 |
| 2 |
-2 |
4 |
| 3 |
2 |
-4 |
| 4 |
4 |
2 |
Estimasi least square dari \({\beta _3}\) dinyatakan sebagai \({\hat \beta _3} = \sum\nolimits_{i = 1}^4 {{w_i}{Y_i}} \) |
| Rumus yang digunakan |
\({{\hat \beta }_3} = \sum\nolimits_{i = 1}^n {{w_i}{Y_i}} \)
\(= \sum\nolimits_{i = 1}^n {\left[ {\frac{{\left( {{x_i} – \bar x} \right)}}{{\sum\nolimits_{i = 1}^n {{{\left( {{x_i} – \bar x} \right)}^2}} }}} \right]} {Y_i}\) |
| Proses pengerjaan |
\({w_i} = \frac{{\left( {{x_{3i}} – {{\bar x}_3}} \right)}}{{\sum\nolimits_{i = 1}^n {{{\left( {{x_{3i}} – {{\bar x}_3}} \right)}^2}} }}\) , sehingga
\({\bar x_3} = \frac{{ – 2 + 4 – 4 + 2}}{4} = 0\) dan \(\sum\nolimits_{i = 1}^n {{{\left( {{x_{3i}} – {{\bar x}_3}} \right)}^2}} = {\left( { – 2} \right)^2} + {\left( 4 \right)^2} + {\left( { – 4} \right)^2} + {\left( 2 \right)^2} = 40\)
diperoleh
\({w_1} = \frac{{ – 2}}{{40}} = – \frac{1}{{20}}\)
\({w_2} = \frac{4}{{40}} = \frac{2}{{20}}\)
\({w_3} = \frac{{ – 4}}{{40}} = – \frac{2}{{20}}\)
\({w_4} = \frac{2}{{40}} = \frac{1}{{20}}\) |
| Jawaban |
D. \(\left( { – \frac{1}{{20}},\frac{2}{{20}}, – \frac{2}{{20}},\frac{1}{{20}}} \right)\) |
