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 |