91影视

Upper confidence bound for and lower confidence bound for are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \) . Then distribution the random sample is taken from is normal.

The interval width \(({z_\gamma } + {z_{\alpha - \gamma }}) \times \frac{s}{{\sqrt n }}\) is the shortest when the difference \(({z_\gamma } + {z_{\alpha - \gamma }})\) is the shortest. This difference can be represented as

\(\begin{gathered}({z_\gamma } + {z_{\alpha - \gamma }})&= {\phi ^{ - 1}}(1 - \gamma ) + {\phi ^{ - 1}}(1 - (\alpha - \gamma )) \\&= {\phi ^{ - 1}}(1 - \gamma ) + {\phi ^{ - 1}}(1 - \alpha + \gamma ) \\\end{gathered} \)

It can be minimized by taking derivative

\(\frac{d}{{d\gamma }}({\phi ^{ - 1}}(1 - \gamma ) + {\phi ^{ - 1}}(1 - \alpha + \gamma )) = - \frac{1}{{\phi (1 - \gamma )}} + \frac{1}{{\phi (1 - \alpha + \gamma )}}\)

Where the hint was used. Therefore, from

\(\begin{aligned}- \frac{1}{{\phi (1 - \gamma )}} + \frac{1}{{\phi (1 - \alpha + \gamma )}} &= 0 \\\frac{1}{{\phi (1 - \alpha + \gamma )}} &= \frac{1}{{\phi (1 - \gamma )}} \\\end{aligned} \)

The \(\phi \) is confidence of standard normal distribution. From the last equality, the following is true

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

This was needed to prove.

Hence, the given statement is true.