A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of $$\begin{gathered} H_0:\mu =300 \\ {H}_a:\mu <300 \end{gathered}$$where μ is the true mean breaking strength of this company’s classroom chairs.

a. The power of the test to detect that μ=294 based on a random sample of 30

chairs and a significance level of α=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

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

$$\begin{gathered} {\mu }_A=294 \\ n=30 \\ \alpha =0.05 \\ Power=0.71=71\% \end{gathered}$$

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$H_1:\mu <300$

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

$P(TypeIerror)=\alpha =0.05=5\%$

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)