91影视

The probability density function , it is easy to obtain the following probability as

\(P\left[ {{{\left( {\frac{\alpha }{2}} \right)}^{\frac{1}{n}}} < \frac{Y}{\theta } \leqslant {{\left( {1 - \frac{\alpha }{2}} \right)}^{\frac{1}{n}}}} \right] = \int\limits_{{{\left( {\frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}^{{{\left( {1 - \frac{\alpha }{2}} \right)}^{\frac{1}{n}}}} {n \cdot {u^{n - 1}}du} \)

\(= n \cdot \left. {\frac{{{u^n}}}{n}} \right|_{{{\left( {\frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}^{{{\left( {1 - \frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}\)

\(\begin{aligned}&= 1 - \frac{\alpha }{2} - \frac{\alpha }{2} \hfill \\&= 1 - \alpha \hfill \\\end{aligned} \)

From the given probability, \(100(1 - \alpha )\% \) confidence interval for \(\theta \) can be obtained as,

\(\begin{aligned}{\left( {\frac{\alpha }{2}} \right)^{\frac{1}{n}}} < \frac{Y}{\theta } \leqslant {\left( {1 - \frac{\alpha }{2}} \right)^{\frac{1}{n}}} \hfill \\&= \frac{{{{\left( {\frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}}{Y} < \frac{1}{\theta } \leqslant \frac{{{{\left( {1 - \frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}}{Y} \hfill \\&= \frac{{\max ({X_i})}}{{{{\left( {1 - \frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}} \leqslant \theta < \frac{{\max ({X_i})}}{{{{\left( {\frac{\alpha }{2}} \right)}^{\frac{1}{n}}}}} \hfill \\\end{aligned} \)

Similarly as in part a), in order to verify that the probability is \(1 - \alpha \) look at

\(\begin{aligned}{l}P\left( {{{\left( \alpha \right)}^{\frac{1}{n}}} < \frac{Y}{\theta } \le 1} \right) & = \int\limits_{{{\left( \alpha \right)}^{\frac{1}{n}}}}^1 {n \cdot {u^{n - 1}}du} \\ & = n \cdot \left. {\frac{{{u^n}}}{n}} \right|_{{{\left( \alpha \right)}^{\frac{1}{n}}}}^1\\ & = 1 - \alpha \end{aligned}\)

The same way in part a), \(100(1 - \alpha )\% \) confidence interval for \(\theta \) can be obtained as,

\(\begin{aligned}{l}{\left( \alpha \right)^{\frac{1}{n}}} < \frac{Y}{\theta } \le 1\\ & = \max ({X_i}) \le \theta < \frac{{\max ({X_i})}}{{{{\left( \alpha \right)}^{\frac{1}{n}}}}}\end{aligned}\)

It is obvious that the interval b is shorter. You can draw \({f_U}(u)\) and notice it easier.

For given values, where, for \(95\% \) confidence level \(\alpha = 0.05,n = 5\) and the maximum value \(\max ({x_i}) = 4.2\), the \(95\% \) confidence interval is

\(\left( {4.2,\frac{{4.2}}{{{{0.05}^{1/5}}}}} \right) = (4.2,7.65)\)

Hence, The \(95\% \) confidence interval \((4.2,7.65)\).