Chapter 9.1, Problem 26E

(a)

To determine

To explain in simple language what “power $=0.23$ ” means in this setting.

Explanation of Solution

It is given in the question, that a manufacturer sells a liquid product whose electrical conductivity is supposed to be five and for this claim the test was conducted, which gives that:

  $\begin{array}{l}{H}_0:\mu =5\\ \text{Alternative: }\mu =5.1\\ \text{Power}=0.23=23\%\end{array}$

Thus, the power is the probability of rejecting the null hypothesis when the null hypothesis is false. So, if the population mean is $\mu =5.1$ , then the probability of rejecting the null hypothesis $\mu =5$ is $0.23$ or $23\%$ .

(b)

To determine

To explain should you make more measurements or fewer to increase power.

Answer to Problem 26E

We should make more measurements to increase power.

Explanation of Solution

It is given in the question, that a manufacturer sells a liquid product whose electrical conductivity is supposed to be five and for this claim the test was conducted, which gives that:

  $\begin{array}{l}{H}_0:\mu =5\\ \text{Alternative: }\mu =5.1\\ \text{Power}=0.23=23\%\end{array}$

Thus, if you want to increase the power, then you require more information about the data, which can be obtained by increasing the number of measurement. Thus, we should make more measurements to increase power.

(c)

To determine

To find out will the power increase or decrease if you decide to use $\alpha =0.10$ in place of $\alpha =0.05$ with no other changes in the test.

Answer to Problem 26E

The power will increase.

Explanation of Solution

It is given in the question, that a manufacturer sells a liquid product whose electrical conductivity is supposed to be five and for this claim the test was conducted, which gives that:

  $\begin{array}{l}{H}_0:\mu =5\\ \text{Alternative: }\mu =5.1\\ \text{Power}=0.23=23\%\end{array}$

Now, if $\alpha$ increases from $0.05\text{ to 0}\text{.10}$ and thus the probability of making a Type I error increases. And if the probability of making Type I error increases, then the probability of making Type II error decreases. Since the power is one decreased by the probability of making a Type II error is:

  $\text{Power}=1-\beta$

Then we know that if $\beta$ decreases then the power has to increase.

(d)

To determine

To explain will the power increase or decrease if you shift your interest to the alternative $\mu =5.2$ with no other changes.

Answer to Problem 26E

The power has increased.

Explanation of Solution

It is given in the question, that a manufacturer sells a liquid product whose electrical conductivity is supposed to be five and for this claim the test was conducted, which gives that:

  $\begin{array}{l}{H}_0:\mu =5\\ \text{Alternative: }\mu =5.1\\ \text{Power}=0.23=23\%\end{array}$

Now, the alternative changes from $\mu =5.1\text{ to }\mu =5.2$ . This means that the alternative is further from the null hypothesis $\mu =5$ . Then it is easier to detect difference between the null hypothesis $\mu =5$ and the alternative $\mu =5.1$ , which means that the power has increased.