Let's break down and solve this problem in two parts:

Part 1: Probability of a Type I error (i.e., rejecting the null hypothesis $H_0$ when it's true)

• Hypotheses:
  • $H_0: M = 0$
  • $H_1: M \neq 0$

The Type I error occurs when we reject $H_0$ (the true hypothesis) incorrectly. The significance level (α) is the probability of making a Type I error.

Given:

• The significance level $\alpha = 0.05$,
• The sample size $N = 100$,
• The standard deviation $\sigma = 1$.

When testing a hypothesis about the mean, we use the test statistic:

$Z = \frac{\bar{X} - M_0}{\frac{\sigma}{\sqrt{N}}}$

Where:

• $\bar{X}$ is the sample mean,
• $M_0$ is the hypothesized value of the mean (0 in this case),
• $\sigma = 1$,
• $N = 100$.

Substituting the known values, the test statistic becomes:

$Z = \frac{\bar{X}}{\frac{1}{\sqrt{100}}} = \bar{X} \times 10$

The significance level of 0.05 means the critical values of the standard normal distribution correspond to the 2.5th percentile and the 97.5th percentile for a two-tailed test. Thus, the rejection region occurs when $Z > 1.96$ or $Z < -1.96$.

Since we are given $\alpha = 0.05$, the probability of a Type I error is simply $\alpha$, which is:

$P(\text{Type I error}) = 0.05$

Part 2: Power of the test at $M = 1$ (i.e., $P(\text{reject } H_0 \mid M = 1)$)

Now, let's compute the power of the test when $M = 1$. The power of the test is the probability that we reject $H_0$ when the true mean is $M = 1$.

Using the same test statistic formula:

$Z = \frac{\bar{X} - M_0}{\frac{\sigma}{\sqrt{N}}}$

For $M = 1$ and $N = 100$:

$Z = \frac{\bar{X} - 0}{\frac{1}{\sqrt{100}}} = \bar{X} \times 10$

But when the true mean is $M = 1$, the distribution of $\bar{X}$ is centered at 1, not 0. So we shift the test statistic by subtracting 1:

$Z = \frac{\bar{X} - 1}{\frac{1}{\sqrt{100}}} = (\bar{X} - 1) \times 10$

The rejection region for the null hypothesis is still $Z > 1.96$ or $Z < -1.96$.

We now compute the power by finding the probability that the shifted $Z$-score falls in the rejection region. Since the true mean is 1, we have:

$P(\text{reject } H_0 \mid M = 1) = P(Z > 1.96 \text{ or } Z < -1.96 \mid M = 1)$

To calculate this, we compute the shift in the $Z$-distribution. Under $M = 1$, the new test statistic becomes:

$Z = \frac{\bar{X} - 1}{\frac{1}{\sqrt{100}}} = (\bar{X} - 1) \times 10$

This is a normal distribution with mean 1 and standard deviation 0.1. The rejection region is still $Z > 1.96$ or $Z < -1.96$, but now the true distribution is centered at 1.

Thus, we compute:

$P(\text{reject } H_0 \mid M = 1) = P\left( Z > 1.96 \text{ under } M=1 \right) + P\left( Z < -1.96 \text{ under } M=1 \right)$

To find this, we calculate the corresponding probabilities using the standard normal distribution.

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If you'd like further clarifications or details on how to continue calculating these probabilities, let me know!

Related questions:

1. What is the Type II error, and how can it be minimized?
2. How does increasing the sample size affect the power of a hypothesis test?
3. What is the role of standard deviation in hypothesis testing?
4. How does the choice of significance level influence Type I and Type II errors?
5. Can we determine the exact distribution of sample means?

Tip:

In hypothesis testing, a larger sample size reduces the margin of error and increases the test's power without changing the significance level.