91影视

The null hypothesis, denoted by H₀, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by H_a, is the assertion that is contradictory to H₀.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H₀ is false. If the sample does not strongly contradict H₀, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H₀.

Using the two-tailed level \(0.01\) test, the

\(\begin{array}{l}{z_{\alpha /2}} = {z_{0.005}}\\{z_{\alpha /2}} = 2.58\end{array}\)

Type II error probability \(\beta (\mu ')\) for a level \(\alpha \) test, when alternative hypothesis is \({H_a}:\mu \ne {\mu _0}\) is,

\(\beta (\mu ') = \Phi \left( {{z_{\alpha /2}} + \frac{{{\mu _0} - \mu '}}{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} + \frac{{{\mu _0} - \mu '}}{{\sigma /\sqrt n }}} \right)\)

Therefore the following is true,

\(\beta ({\mu _0} - \Delta ) = \Phi \left( {{z_{\alpha /2}} + \frac{{{\mu _0} - {\mu _0} + \Delta }}{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} + \frac{{{\mu _0} - {\mu _0} + \Delta }}{{\sigma /\sqrt n }}} \right)\)

\( = \Phi \left( {{z_{\alpha /2}} + \frac{\Delta }{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} + \frac{\Delta }{{\sigma /\sqrt n }}} \right)\)

\( = \left( {1 - \Phi \left( { - {z_{\alpha /2}} - \frac{\Delta }{{\sigma /\sqrt n }}} \right)} \right) - \left( {1 - \Phi \left( {{z_{\alpha /2}} - \frac{\Delta }{{\sigma /\sqrt n }}} \right)} \right)\)

\(\begin{array}{l} = \Phi \left( {{z_{\alpha /2}} - \frac{\Delta }{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} - \frac{\Delta }{{\sigma /\sqrt n }}} \right)\\ = \Phi \left( {{z_{\alpha /2}} + \frac{{{\mu _0} - ({\mu _0} + \Delta )}}{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} + \frac{{{\mu _0} - ({\mu _0} + \Delta )}}{{\sigma /\sqrt n }}} \right)\\ = \beta ({\mu _0} + \Delta )\end{array}\)

Hence, proved the claim \(\beta ({\mu _0} - \Delta ) = \beta ({\mu _0} + \Delta )\)