91影视

Object depend on random event is a random variable.

From the exercise 5, mean square error is

\(MSE = 0.066\)

Using the fact that random variable

\(\theta = 0.5.\,\,{\overline x _1}\, + 0.5.\,\,{\overline x _2} - \,\,{\overline x _{3.}}\,\,\)

has approximately student distribution with \(I\left( {J - 1} \right) = 27\) degrees of freedom, from the

\(P\left( {{t_\alpha }_{/2,27} < 0.5{{\overline x }_1} + 0.5.{{\overline x }_2} - {{\overline x }_3} < {t_\alpha }_{/2,27}} \right) = 1 - \alpha \)

where, in this case, and from the table in the appendix the

\({t_\alpha }_{/2,27} = {t_{0.025}}_{,27} = 2.052\)

confidence interval

\(95\% \)

\(\sum\limits_{i = 1}^3 {{c_i}{{\overline x }_i}. \pm {t_{\alpha /2,I\left( {J - 1} \right).}}} \sqrt {\frac{{MSE.\sum {{{^3}_{i = 1}}{c^2}_i} }}{J}} \)

MSE is estimate of variance\({\sigma ^2}\) .

\({{\rm{C}}_1} = {{\rm{C}}_2} = 0.5{\rm{,}}{{\rm{C}}_3} = - 1\)

\(\sum\limits_{i = 1}^3 {{c_i}^2. = 1.5} \)

Estimate\(\theta \) IS

\(\widehat \theta = \sum\limits_{i = 1}^3 {{c_i}{{\overline x }_i}. = 0.175} \)

where all values are given in the table in the exercise 5. Therefore, the\(95\% \) confidence interval for is\(\theta \)

\(0.175 \pm 2.052.\sqrt {\frac{{0.066.15}}{{10}}} = 0.175 \pm 0.204\)

Or equally

\(\left( { - 0.029,0.379} \right)\)

Hence,

\(\left( { - 0.029,0.379} \right)\)