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The confidence interval can be computed using formula

\(E(\widehat \theta ) \pm {Z_{\alpha /2}}{\sigma ^2}\)or equally, the \(100(1 - \alpha )\) confidence interval is,

\(\frac{1}{3}(\overline {{x_1}} + \overline {{x_2}} + \overline {{x_3}} ) - \overline {{x_4}} \pm {Z_{\alpha /2}} \times \sqrt {\frac{1}{9}\left( {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}} + \frac{{{s_3}^2}}{{{n_3}}}} \right) - \frac{{{s_4}^2}}{{{n_4}}}} \)

As suggested in the exercise, by replacing \({\mu _i}\) by \(\overline {{X_i}} \), the estimator becomes,

\(\widehat \theta = \frac{1}{3}({\overline x _1} + {\overline X _2} + {\overline X _3}) - {\overline X _4}\)

To compute the \(95\% \) interval the variance of estimate \(\widehat \theta \), the mean of the estimator and the \(z\)value.

The variance is,

\(\begin{aligned}{\sigma ^2}& = V(\widehat \theta )\\ &= V\left( {\frac{1}{3}({{\overline x }_1} + {{\overline x }_2} + {{\overline x }_3}) - {{\overline x }_4}} \right)\\ &= \frac{1}{9}\left( {V({{\overline x }_1}) + V({{\overline x }_2}) + V({{\overline x }_3}) - V({{\overline x }_4})} \right)\\& = \frac{1}{9}\left( {\frac{{{\sigma _1}^2}}{{{n_1}}} + \frac{{{\sigma _2}^2}}{{{n_2}}} + \frac{{{\sigma _3}^2}}{{{n_3}}}} \right) - \frac{{{\sigma _4}^2}}{{{n_4}}}\end{aligned}\)

The estimator of the standard deviation \(\sigma \)is obtained by taking a square root of the variance. By replacing \({\sigma _i}\) with \({s_i}\)the estimate is obtained.

The confidence interval can be computed using formula

\(E(\widehat \theta ) \pm {Z_{\alpha /2}}{\sigma ^2}\)or equally, the \(100(1 - \alpha )\) confidence interval is,

\(\frac{1}{3}({\overline x _1} + {\overline x _2} + {\overline x _3}) - {\overline x _4} \pm {Z_{\alpha /2}} \times \sqrt {\frac{1}{9}\left( {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}} + \frac{{{s_3}^2}}{{{n_3}}}} \right) - \frac{{{s_4}^2}}{{{n_4}}}} \)

Where,

And also the following is true,

\(\begin{aligned}{}{\sigma ^2} &= \sqrt {\frac{1}{9}\left( {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}} + \frac{{{s_3}^2}}{{{n_3}}}} \right) - \frac{{{s_4}^2}}{{{n_4}}}} \\ &= \sqrt {\frac{1}{9}\left( {\frac{{{{1.5}^2}}}{{100}} + \frac{{{{1.3}^2}}}{{90}} + \frac{{{{1.8}^2}}}{{100}}} \right) - \frac{{{{1.6}^2}}}{{120}}} \\ &= 0.1718\end{aligned}\)

And,

The z value is,

\(\begin{aligned}{}{Z_{\alpha /2}} &= {Z_{0.05/2}}\\& = {Z_{0.025}}\\& = 1.96\end{aligned}\)

The obtained form is, \(P(Z > {Z_{0.025}}) = 0.025\) and from the normal probability table in the appendix. The probability can also be computed by a software

Substituting in the above formula,

\(\begin{aligned}{}( - 0.5 - 1.96(0.1718),( - 0.5 + 1.96(0.1718)))\\ = ( - 0.84, - 0.16)\end{aligned}\)

Hence, The \(100(1 - \alpha )\% \) confidence level for \(\theta \) is \(( - 8.84, - 0.16)\).