In this problem, you're dealing with hypothesis testing, and you're tasked with calculating the probability of committing a Type II error (β) for different sample sizes. A Type II error occurs when you fail to reject the null hypothesis ($H_0$) even though the alternative hypothesis ($H_A$) is true.

Hypotheses:

• Null Hypothesis: $H_0: \mu \leq 208$
• Alternative Hypothesis: $H_A: \mu > 208$

Given:

• Population mean: $\mu_0 = 209$
• Null hypothesis population mean: $\mu = 208$
• Population standard deviation: $\sigma = 22$
• Significance level: $\alpha = 0.05$

Now, to calculate the probability of a Type II error, we follow these steps:

Step-by-Step Process

1. Calculate the critical value for $\alpha = 0.05$:

   Since the hypothesis is one-tailed, we find the critical value of $z$ corresponding to $\alpha = 0.05$ for a one-tailed test. Using a standard normal distribution table:

   $z_{\alpha} = 1.645$ This means that if the test statistic exceeds this value, we reject $H_0$.

2. Compute the standard error of the mean:

   The standard error depends on the sample size $n$:

   $\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{n}}$

3. Find the critical value in terms of the sample mean:

   Using the critical value $z_{\alpha} = 1.645$, we convert it into the corresponding value for the sample mean:

   $\text{Critical Value for } \bar{x} = 208 + z_{\alpha} \cdot SE$ $\text{Critical Value for } \bar{x} = 208 + 1.645 \cdot \frac{22}{\sqrt{n}}$

4. Find the probability of committing a Type II error:

   Now, under the assumption that the true population mean $\mu_0 = 209$, we calculate the test statistic based on this mean:

   $z_{\beta} = \frac{\text{Critical Value for } \bar{x} - 209}{SE}$ Then, look up the corresponding probability (or area) in the standard normal distribution table to get $\beta$, the probability of Type II error.

a. For $n = 10$:

• Standard Error: $SE = \frac{22}{\sqrt{10}} \approx 6.96$
• Critical Value for $\bar{x}$: $208 + 1.645 \cdot 6.96 \approx 219.45$

Now calculate the $z$-score for the true mean $\mu = 209$:

$z_{\beta} = \frac{219.45 - 209}{6.96} \approx 1.50$

Looking up $z = 1.50$ in the standard normal distribution table gives us a probability of approximately 0.9332. The area to the right of this is $1 - 0.9332 = 0.0668$.

Thus, the probability of a Type II error for $n = 10$ is approximately $\beta = 0.0668$.

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Would you like to proceed with parts b and c? Let me know if you want further clarification or a detailed explanation of any step!

Here are 5 follow-up questions to consider:

1. How does increasing the sample size affect the standard error and Type II error?
2. Why do we use the z-distribution in this hypothesis test?
3. How would the calculation differ if the population standard deviation were unknown?
4. What would happen if we decreased the significance level $\alpha$ from 0.05 to 0.01?
5. Can you explain the relationship between Type I and Type II errors?

Tip: Increasing the sample size reduces the standard error, which makes it easier to detect smaller differences from the hypothesized mean, thus reducing the probability of a Type II error.