Pembahasan Ujian PAI: A70 – No. 25 – November 2014

• \(F(x) = 1 – {e^{ – {\rm{ }}{{\left( {\frac{x}{\theta }} \right)}^{0,2}}}},x > 0\)
• Sampel dari empat buah klaim diketahui sebesar 130, 240, 300, 540 dan dua buah klaim lainnya diketahui lebih dari 1000

• \(L(\theta ) = f({x_1}) \cdot f({x_2}) \cdot f({x_3}) \cdot f({x_4}) \cdot {\left[ {1 – F({x_5})} \right]^2}\)
• \(l(\theta ) = \ln L(\theta )\)

Klaim sebesar 130, 240, 300, 540, 1.000
\(L(\theta ) = f(130) \cdot f(240) \cdot f(300) \cdot f(540) \cdot {\left[ {1 – F(1.000)} \right]^2}\) \(L(\theta ) = 0,{2^4} \cdot {\theta ^{ – 0,8}} \cdot \left( {\frac{1}{{{{130}^{0,8}}}} \cdot … \cdot \frac{1}{{{{540}^{0,8}}}}} \right) \cdot {e^{ – \left[ {{{\left( {\frac{{130}}{\theta }} \right)}^2} + {{\left( {\frac{{240}}{\theta }} \right)}^2} + {{\left( {\frac{{300}}{\theta }} \right)}^2} + {{\left( {\frac{{540}}{\theta }} \right)}^2} + 2{{\left( {\frac{{1000}}{\theta }} \right)}^2}} \right]}}\)

Dengan menghilangkan konstanta yang tidak digunakan:
\(l(\theta ) = \ln L(\theta ) \propto – 0,8\ln (\theta ) – \left[ {{\theta ^{ – 0,2}}({{130}^{0,2}} + {{240}^{0,2}} + … + 2{{(1.000)}^{0,2}})} \right]\) \(l'(\theta ) = – \frac{{0,8}}{\theta } + 0,2{\theta ^{ – 1,2}}({130^{0,2}} + {240^{0,2}} + … + 2{(1.000)^{0,2}}) = 0\) \(\hat \theta = 3.325,69\)