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Let's go through each part of the problem based on the provided information.

### Problem Breakdown

The prompt suggests conducting a hypothesis test about the mean hours worked per week by white-collar employees in the U.S., given:
- **Sample mean** (\(\bar{x}\)) = 43.73 hours
- **Population standard deviation** (\(\sigma\)) = 3.2 hours
- **Sample size** (\(n\)) = 28
- **Null hypothesis mean** (\(\mu_0\)) = 42 hours
- **Significance level** = 1% (or 0.01)

We'll address each question in turn:

### Part (a)
**Which assumptions are required to perform a hypothesis test and construct a confidence interval for the population mean hours worked per week?**

The correct answer is:
- **Each observation is sampled randomly and independently from the same distribution**.
- **The data must be a sample from a distribution that does not deviate far from a normal distribution** (since the sample size is moderately large, the Central Limit Theorem allows us to proceed assuming normality).
- **Population standard deviation must be known**.

Thus, the answer is:
> "Exactly three of these are required."

### Part (b)
**Should you use a \( z \)-distribution or a \( t \)-distribution?**

Since the population standard deviation (\(\sigma\)) is known, we use a **\( z \)-distribution**.

### Part (c)
**Hypothesis Testing**

1. **Parameter of Interest**:
   The parameter of interest is:
   > "The true mean number of hours worked per week by all U.S. white-collar workers."

2. **Hypotheses**:
   - Null Hypothesis (\(H_0\)): \(\mu = 42\)
   - Alternative Hypothesis (\(H_a\)): \(\mu > 42\) (since we're testing if the mean hours worked is **greater** than 42)

   Therefore, the correct choice for the hypothesis is:
   > \( H_0: \mu = 42, \ H_a: \mu > 42 \)

3. **Test Statistic Calculation**:
   Since we're using a \( z \)-test, the test statistic \( z \) is calculated as:
   \[
   z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
   \]
   Plugging in the values:
   \[
   z = \frac{43.73 - 42}{3.2 / \sqrt{28}}
   \]
   Calculating the above:

   \[
   z \approx \frac{1.73}{0.604} \approx 2.864
   \]

4. **P-value Calculation**:
   For a \( z \)-score of 2.864 in a one-tailed test, we look up the corresponding p-value. For \( z = 2.864 \), the p-value is approximately 0.0021.

Since the p-value (0.0021) is less than the significance level of 0.01, we **reject the null hypothesis**. There is sufficient evidence to suggest that the true mean number of hours worked per week by U.S. white-collar workers is greater than 42.

### Summary of Answers:
1. **(a)** - "Exactly three of these are required."
2. **(b)** - Use a **\( z \)-distribution**.
3. **(c)**:
   - **Parameter**: "The true mean number of hours worked per week by all U.S. white-collar workers."
   - **Hypotheses**: \( H_0: \mu = 42, \ H_a: \mu > 42 \)
   - **Test Statistic**: \( z \approx 2.864 \)
   - **P-value**: 0.0021, which is significant at the 1% level.

Would you like further explanation on any of these steps or help with similar questions?

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### Follow-up Questions

1. How would the test change if we didn't know the population standard deviation?
2. Why do we choose a \( z \)-distribution over a \( t \)-distribution in this case?
3. What would happen if the significance level was set at 5% instead of 1%?
4. How would the sample size affect the test outcome if it were smaller?
5. How would we interpret the result if the p-value was greater than the significance level?

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### Tip

When dealing with hypothesis tests, always verify whether you know the population standard deviation, as it determines whether to use a \( z \)- or \( t \)-distribution.

Is there any evidence that the true mean number of hours worked per week by white-collar employees is greater than 42? A random sample of 28 employees worked on average 43.73 hours per week, with a population standard deviation of 3.2 hours. Perform a hypothesis test at a 1% significance level.

Conduct a hypothesis test to determine if the mean number of hours worked per week by U.S. white-collar employees is greater than 42 hours, using a sample mean of 43.73 hours and a population standard deviation of 3.2 hours. This example explores the use of the z-test in hypothesis testing at a 1% significance level.

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