Let's solve the questions based on the problem statement provided.

1. Describe in words what a Type II error would be in this test.

In this context, a Type II error occurs when the test fails to reject the null hypothesis $H_0: \mu = 0$ even though the alternative hypothesis $H_1: \mu > 0$ is true. In other words, a Type II error happens when we conclude that there is no evidence to suggest that the mean $\mu$ is greater than 0, when in fact, it is.

2. Suppose we want to use the estimator $\overline{X}_n$ for our test. For a sample size of $n = 20$, what is the critical value $\overline{X}_c$ with which we would reject the null hypothesis at a test level of $\alpha = 0.05$?

To find the critical value $\overline{X}_c$, we need to determine the value such that the probability of observing a sample mean greater than or equal to $\overline{X}_c$, given that $\mu = 0$, is 0.05. Since the sample size is $n = 20$, the standard error of the mean is given by: $\text{Standard error} = \frac{1}{\sqrt{20}}.$

Assuming $X_i \sim N(\mu, 1)$, under the null hypothesis $\mu = 0$, the distribution of the sample mean $\overline{X}_n$ is $N(0, \frac{1}{\sqrt{20}})$.

To find $\overline{X}_c$, we look for the z-score that corresponds to the upper 5% tail of the standard normal distribution, which is approximately 1.645. Thus, $\overline{X}_c = 1.645 \times \frac{1}{\sqrt{20}} \approx 0.367.$

3. Now let's calculate the power of the test. Recall that power refers to the probability of correctly rejecting the null hypothesis when it is false. In other words, if the alternative hypothesis is true, that is, if the true $\mu = 1$, what is the probability of getting a value equal to or greater than $\overline{X}_c$, given that $\mu = 1$? Continue to test at the level of $\alpha = 0.05$, assuming a sample size of $n = 20$.

When $\mu = 1$, the distribution of $\overline{X}_n$ is $N(1, \frac{1}{\sqrt{20}})$.

The power of the test is the probability that $\overline{X}_n \geq \overline{X}_c$, given $\mu = 1$. We need to find the z-score for $\overline{X}_c = 0.367$ under the distribution $N(1, \frac{1}{\sqrt{20}})$: $z = \frac{0.367 - 1}{1/\sqrt{20}} = \frac{0.367 - 1}{0.2236} \approx -2.83.$

The corresponding probability from the standard normal table for $z = -2.83$ is approximately 0.0023. Therefore, the power of the test is: $1 - 0.0023 = 0.9977.$

This indicates a high probability of correctly rejecting the null hypothesis when $\mu = 1$, showing that the test is very powerful under these conditions.

Would you like more details on any of these steps, or have any questions?

Here are five related questions to expand on these concepts:

1. How do you interpret the power of a test in hypothesis testing?
2. What factors can influence the power of a statistical test?
3. What is the relationship between Type I and Type II errors?
4. How would increasing the sample size affect the critical value $\overline{X}_c$ and the power of the test?
5. Can the power of a test be 1? If so, under what circumstances?

Tip: Increasing the sample size can reduce the standard error, making it easier to detect true effects and thus increasing the power of the test.