91影视

A \(100(1 - \alpha )\% \) confidence interval for mean \(1/\lambda \) can be obtained from the fact (because the given random variable has chi square distribution)

\(P({X^2}_{1 - \alpha /2,2r} < 2\lambda {T_r} < {X^2}_{\alpha /2,2r}) = 1 - \alpha \)

With adequate transformations, this becomes

\(P(\frac{{{\chi ^2}_{1 - \alpha /2,2r}}}{{2{T_r}}} < \lambda < \frac{{{\chi ^2}_{\alpha /2,2r}}}{{2{T_r}}}) = 1 - \alpha \)

And the CI for \(1/\lambda \) can be obtained from,

\(P(\frac{{2{T_r}}}{{{\chi ^2}_{\alpha /2,2r}}} < \lambda < \frac{{2{T_r}}}{{{\chi ^2}_{1 - \alpha /2,2r}}}) = 1 - \alpha \)

Where, for \({t_r} = {y_1} + {y_2} + ... + {y_r} + (n - r){y_r}\), the \(100(1 - \alpha )\) confidence interval is,

\((\frac{{2{t_r}}}{{{\chi ^2}_{\alpha /2,2r}}},\frac{{2{t_r}}}{{{\chi ^2}_{1 - \alpha /2,2r}}})\)

It is given that random variable \(2\lambda {T_r}\) has a chi square distribution with \(v = 2r\) degrees of freedom.

From the example 6.8, the given data is \(n = 20,r = 10\)and the first \(10\)failure times are\(11,15,29,33,35,40,47,55,58,72\) from which \({t_r}\) becomes

\(\begin{array}{c}{t_{10}} = 11 + 15 + ... + 58 + 72 + (20 - 10)(72)\\ = 1115\end{array}\)

The given random variable has \(2.\)

\(r = 2(10)\)

\(r = 20\)degrees of freedom.

From ,

\(\begin{aligned}100(1 - \alpha ) & = 95\\\alpha & = 0.05\\\alpha /2 & = 0.025\end{aligned}\)

Therefore, the critical values are,

\(\begin{aligned}{l}{\chi ^2}_{\alpha /2,2r} & = {\chi ^2}_{0.025,20}\\{\chi ^2}_{\alpha /2,2r} & = 34.17\\{\chi ^2}_{1 - \alpha /2,2r} & = {\chi ^2}_{0.975,20}\\{\chi ^2}_{1 - \alpha /2,2r} & = 9.591\end{aligned}\)

Finally, the confidence interval for \(1\lambda \)is,

\(\begin{aligned}{l}(\frac{{2{t_r}}}{{{\chi ^2}_{\alpha /2,2r}}},\frac{{2{t_r}}}{{{\chi ^2}_{1 - \alpha /2,2r}}}) & = \left( {\frac{{2(1115)}}{{34.17}},\frac{{2(1115)}}{{9.591}}} \right)\\ & = (65.3,232.5)\end{aligned}\)

Hence, the required confidence interval is \((65.3,232.5)\).