Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | April 2019 |
| Nomor Soal | : | 1 |
SOAL
Diketahui persamaan regresi linear berikut:
\(\hat Y = 1,245 + 0,17X\)
Jika diberikan data pengamatan sebagai berikut,
| \({x_i}\) | 7,5 | 4 | 3 | 1,25 |
| \({y_i}\) | 0,5 | 1 | 0 | -1,5 |
Hitunglah nilai standard error dari persamaan regeresi linear tersebut.
- 1,245
- 1,425
- 1,396
- 1,963
- 1,546
| Diketahui | \(\hat Y = 1,245 + 0,17X\) dan data| \({x_i}\) | 7,5 | 4 | 3 | 1,25 | | \({y_i}\) | 0,5 | 1 | 0 | -1,5 |
|
| Rumus yang digunakan | \({\hat y_i} = \hat \beta {x_i} + {\hat \varepsilon _i} \Leftrightarrow {\hat \varepsilon _i} = {\hat y_i} – \hat \beta {x_i} = {\hat y_i} – 0,17{x_i}\)
\({\hat y_i} = 0,17{x_i}\)
\({s^2} = \frac{{\sum {\hat \varepsilon _i^2} }}{{N – 2}}\) |
| Proses pengerjaan | | \({x_i}\) | \({y_i}\) | \({\hat y_i}\) | \({\hat \varepsilon _i}\) | \(\hat \varepsilon _i^2\) | \({\left( {{{\hat \varepsilon }_t} – {{\hat \varepsilon }_{t – 1}}} \right)^2}\) | | 7.5000 | 0.5000 | 1.2750 | -0.7750 | 0.6006 | 0.0000 | | 4.0000 | 1.0000 | 0.6800 | 0.3200 | 0.1024 | 1.1990 | | 3.0000 | 0.0000 | 0.5100 | -0.5100 | 0.2601 | 0.6889 | | 1.2500 | -1.5000 | 0.2125 | -1.7125 | 2.9327 | 1.4460 | | Total | 3.8958 | 3.3339 |
\({s^2} = \frac{{\sum {\hat \varepsilon _i^2} }}{{N – 2}}\)
\(= \frac{{3,8958}}{{4 – 2}}\)
\(= 1,9479\)
\(s = \sqrt {1,9479} = 1,3957\) |
| Jawaban | c. 1,396 |
