Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Untuk sebuha regresi 2 vaiabel berdasakan 8 pengamatan, diperoleh informasi:
\(\sum {{{\left( {{X_i} – \overline X } \right)}^2}} = 2000\) \(\sum {\widehat \varepsilon _i^2 = 957} \)Hitunglah \({s_\beta }\), yaitu standar error untuk \({s_\beta }\)
- 0.22
- 0.25
- 0.28
- 0.31
- 0.34
| Diketahui | \(n = 8\) \(\sum {{{\left( {{X_i} – \overline X } \right)}^2}} = 2000\) \(\sum {\widehat \varepsilon _i^2 = 957} \) |
| Rumus yang digunakan | \(\sum\limits_{}^{} {x_i^2} = \sum {{{({X_i} – \overline X )}^2}} \) \(\widehat \sigma = \sqrt {\frac{{\sum {\widehat \varepsilon _i^2} }}{{n – 2}}} \) \({s_{_{\widehat \beta }}} = \frac{\sigma }{{\sqrt {\sum\limits_{}^{} {x_i^2} } }}\) |
| Proses pengerjaan | \(\sum\limits_{}^{} {x_i^2} = \sum {{{({X_i} – \overline X )}^2}} \) \(= 2000\) \(\widehat \sigma = \sqrt {\frac{{\sum {\widehat \varepsilon _i^2} }}{{n – 2}}} \) \(= \sqrt {\frac{{957}}{{8 – 2}}} \) \(= 12.63\) \({s_{_{\widehat \beta }}} = \frac{\sigma }{{\sqrt {\sum\limits_{}^{} {x_i^2} } }}\) \(= \frac{{12.63}}{{\sqrt {2000} }}\) \(= 0.28\) |
| Jawaban | b. 0.28 |