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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.

Assume that the alternative hypothesis is \({H_a}:\mu > {\mu _0}\)

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

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

The function is continuous. This means that you can look at the parentheses and see how to express behave. The value \(n\) belong only to the expression,

\(\frac{{{\mu _0} - \mu }}{{\sigma /\sqrt n }} = \frac{{\sqrt n ({\mu _0} - \mu )}}{\sigma }\)

Obviously, because \(n\)is in the denominator, the following hold

\(\frac{{\sqrt n ({\mu _0} - \mu )}}{\sigma } \to - \infty \)

When \(n \to \infty \) because \({\mu _0} - \mu ' < 0\)(assumption that \({\mu _0} < \mu \)). So the type II error now becomes,

\(\begin{array}{l}\beta (\mu ) = \Phi \left( {{z_\alpha } + \frac{{{\mu _0} - \mu }}{{\sigma /\sqrt n }}} \right)\\\Phi ( - \infty ) = 0\end{array}\)

When \(n \to \infty \) and \(\Phi ( - \infty ) = 0\) because \(\Phi \) is confidence it standard normal random variable.

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

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

Which can be thought of as combination of the two mentioned hypotheses.

Hence, proved.