Question

Restaurant power Refer to Exercise $86$ Determine if each of the following changes would increase or decrease the power of the test. Explain your answers.

a. Use a random sample of $30$ people instead of $50$ people.

b. Try to detect that $\mu =\backslash (85,500$ instead of $\mu =\backslash )86,000$

c. Change the significance level to $\alpha =0.10$

Short answer

Part (a) Power decrease.

Part (b) Power decrease.

Part (c) Power increases.

Step-by-step solution

Part (a) Step 1: Given information

$$\begin{gathered} H_0:\mu =\$85000 \\ {H}_1:\mu >\$85000 \\ {\mu }_A=alternativemean=\$86000 \\ \alpha =significancelevel=0.05 \end{gathered}$$

Power $=0.64=64\%$

Part (a) Step 2: Explanation

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis. The sample size was reduced from $50$to $30$

Because the sample size has been reduced, there is less knowledge on the population, therefore estimates will be less accurate. Because estimations are less precise, it is less likely that the null hypothesis will be appropriately rejected (once the alternative hypothesis is true), and so the power will be reduced.

Part (b) Step 1: Explanation

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis. Changing the genuine mean to $\mu =\$85,500$instead of $\mu =\$86,000$the mean will be closer to the hypothesized mean of $\mu =\$85,000$Because the gap between the genuine mean and the hypothesized mean is less, detecting that the hypothesized mean is not the true mean is more difficult, and so the power is reduced.

Part (c) Step 1: Explanation

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis. Increasing the significance threshold from $\alpha =0.05to\alpha =0.10$ is a good way to start. Because the significance level measures the likelihood of making a type I error, as the significance level rises, the likelihood of making a type I error rises, and the likelihood of making a type II error decreases. The probability of a type II error, on the other hand, reduces the power by one, hence the power grows.

$Power=1-P(TypeIIerror)$