Chapter 9.3, Problem 91E

(a)

To determine

To Explain: that the power of the test to detect that mean is 294 based on a random sample of 30 chairs and a significance level of $\alpha =0.05$ is 0.71 and interpret this value.

Answer to Problem 91E

If the true mean breaking strength of this company’s classroom chairs is 294 pounds, then have a 71% probability that find the convincing proof to help the alternative hypothesis $H_1:\mu <300$ .

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =300\\ H_1:\mu <300\\ {\mu }_A=294\\ n=30\\ \alpha =0.05\\ \text{Power = 0}\text{.71=71\%}\end{array}$

The power is the probability of rejecting the null hypothesis once the alternative hypothesis is true.

If the true mean breaking strength of this company’s classroom chairs is 294 pounds, then have a 71% probability that find the convincing proof to help the alternative hypothesis $H_1:\mu <300$ .

(b)

To determine

To find: the probability of a Type-I error and the probability of a Type-II error for the test in part (a).

Answer to Problem 91E

  $P\left(\text{Type I error}\right)=\alpha =0.05=5\%$

  $P\left(\text{Type II error}\right)\text{=0}\text{.29=29\%}$

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =300\\ H_1:\mu <300\\ {\mu }_A=294\\ n=30\\ \alpha =0.05\\ \text{Power = 0}\text{.71=71\%}\end{array}$

Calculation:

The probability of a type I error is given by the significance level:

  $P\left(\text{Type I error}\right)=\alpha =0.05=5\%$

The probability of a type II error is 1 decreased by the power.

  $P\left(\text{Type II error}\right)=1-\text{Power =1-0}\text{.71=0}\text{.29=29\%}$

(c)

To determine

To Explain: the ways to increase the power of the test in part (a).

Answer to Problem 91E

Increase the significance level

Increase sample size

Making the alternative mean more extreme

Explanation of Solution

The power is the probability of rejecting the null hypothesis once the alternative hypothesis is true.

Increase the power by:

Increase the significance level (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 mean $\mu$ more extreme, therefore smaller (since more extreme alternatives are easier to prove)