Exercise
May 13'23
Answer
Key: B
[[math]]
\begin{aligned}
L(\theta) &= \left( \frac{\theta}{10}\right)^9 \left( \frac{\theta}{10} - \frac{\theta}{25} \right)^6 \left( \frac{\theta}{25}\right)^5 \propto (10-\theta)^9 \theta^{11} \\
l(\theta) &= 9 \ln(10 − \theta ) + 11 \ln(\theta ) \\
l^{'}(\theta) &= - \frac{9}{10-\theta} + \frac{11}{\theta} = 0 \\
11(10-\theta) &= 9 \theta \\
110 &= 20\theta \\
\theta &= 110/20 = 5.5
\end{aligned}
[[/math]]