Suppose that for data taken from a Normal distribution with unknown mean M and known standard deviation 1, we hypothesize that H0: M=0 versus H1: M != 0. Suppose that we conduct a hypothesis test at the 0.05 significance level and with sample size N=100.

Calculate the probability of a Type I error if the true value of M is 0. 
Calculate P(reject H_0 | M=1), the probability that we reject the null hypothesis if the true value of M is 1.  (This is the power of the test at M=1.)

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.

---

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.

Suppose that for data taken from a Normal distribution with unknown mean M and known standard deviation 1, we hypothesize that H0: M=0 versus H1: M != 0. Suppose that we conduct a hypothesis test at the 0.05 significance level and with sample size N=100. Calculate the probability of a Type I error if the true value of M is 0. Calculate P(reject H_0 | M=1), the probability that we reject the null hypothesis if the true value of M is 1. (This is the power of the test at M=1.)

Explore hypothesis testing for a Normal distribution with unknown mean M and known standard deviation. This problem covers calculating the probability of a Type I error at a 0.05 significance level and determining the power of the test when M=1. Detailed explanation provided with step-by-step breakdowns and relevant formulas.

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