91Ó°ÊÓ

There is a random sample \({X_1}, \ldots ,{X_n}\) \(normal\left( {\mu ,{\sigma ^2}} \right)\). Here \(\mu \)unknown and \({\sigma ^2}\) is known. \(\Phi \) is the cumulative distribution function of the random sample.

Consider the confidence interval,

\(\left( {{{\bar X}_n} - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }},{{\bar X}_n} + {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }}} \right)\)

Where \(\gamma \) is the coefficient for\(\mu \) , and \({\bar X_n}\) is the sample mean

Let us consider

\(\Pr \left( {{{\bar X}_n} - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }} < \mu < {{\bar X}_n} + {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\frac{\sigma }{{\sqrt n }}} \right)\)

Now by subtracting\({\bar X_n}\)from all the sides and the divide the sides by\(\frac{\sigma }{{\sqrt n }}\), we get,

\(\Pr \left[ { - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right) < \frac{{\mu - {{\bar X}_n}}}{{{\raise0.7ex\hbox{$\sigma $} \!\mathord{\left/ {\vphantom {\sigma {\sqrt n }}}\right. \ } \!\lower0.7ex\hbox{${\sqrt n }$}}}} < {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)} \right]\)

Therefore, the probability that the variable lies between\( - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\;and\;{\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)\)is,

\(\begin{align}\Pr \left( { - {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right) < \frac{{\mu - {{\bar X}_n}}}{{{\raise0.7ex\hbox{$\sigma $} \!\mathord{\left/ {\vphantom {\sigma {\sqrt n }}}\right. \ } \!\lower0.7ex\hbox{${\sqrt n }$}}}} < {\Phi ^{ - 1}}\left( {\frac{{1 + \gamma }}{2}} \right)} \right) &= \frac{{1 + \gamma }}{2} - \left( {1 - \frac{{1 + \gamma }}{2}} \right)\\ &= \frac{{1 + \gamma }}{2} + \frac{{1 + \gamma }}{2} - 1\\ &= \frac{{2\left( {1 + \gamma } \right)}}{2} - 1\\ &= 1 + \gamma - 1\\ &= \gamma \end{align}\)

Thus, it is proved.