91Ó°ÊÓ

$$\begin{gathered} H_0:\mathrm{μ}=300 \\ {H}_a:\mathrm{μ}<300 \end{gathered}$$

$$\begin{gathered} {\mathrm{μ}}_A=294 \\ n=30 \\ \mathrm{Î\pm }=0.05 \\ Power=0.71=71\% \end{gathered}$$

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis. If the genuine mean breaking strength of this company's classroom chairs is $294$ pounds, then the alternative hypothesis has a $71$ percent chance of being supported$H_1:\mathrm{μ}<300$

The significance level determines the likelihood of a type I error:

$P(TypeIerror)=\mathrm{Î\pm }=0.05=5\%$

The power reduces the probability of a type II error by one.

$P(TypeIIerror)=1−Power=1-0.71=0.29=29\%$

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis.

Increase the power by:

Boost the level of significance (because this increases the probability of making a Type I error and decreases the probability of making a Type II error; since the power is $1$decreased by the probability of making a Type II error, the power increases).

Making the alternative more harsh entails reducing the size of the alternative (since more extreme alternatives are easier to prove)