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You would like to construct a 90% confidence interval to estimate the population mean score on a nationwide examination in psychology, and for this purpose we choose a random sample of exam scores. The sample we choose has a mean of 500 and a standard deviation of 76.

(a) What is the best point estimate, based on the sample, to use for the population mean? (b) For each of the following sampling scenarios, determine which distribution should be used to calculate the critical value for the 90% confidence interval for the population mean. (In the table, Z refers to a standard normal distribution, and t refers to a *t* distribution.)

Let's analyze the problem you've presented.

### (a) Best Point Estimate

The best point estimate for the population mean, based on the sample, is simply the sample mean. Therefore:

- The best point estimate for the population mean is **500**.

### (b) Determine the Appropriate Distribution

Now, let's consider the sampling scenarios mentioned in the table, though the exact text from the table isn't entirely visible. The decision to use the **Z-distribution** or **t-distribution** depends on two factors:

1. **Sample Size (n):** 
   - If \(n\) is large (usually \(n \geq 30\)), the **Z-distribution** is typically used, even if the population standard deviation is unknown.
   - If \(n\) is small (\(n < 30\)), the **t-distribution** should be used when the population standard deviation is unknown.

2. **Population Standard Deviation (\(\sigma\)):**
   - If the population standard deviation is known, regardless of the sample size, the **Z-distribution** is used.
   - If the population standard deviation is unknown, and \(n < 30\), the **t-distribution** is used.

Based on this, I'll attempt to match these rules with the parts of the table that are visible.

#### Sampling Scenarios:
1. **Scenario 1**: The sample size \(n \geq 30\), and the population standard deviation is known.
   - **Use the Z-distribution.**

2. **Scenario 2**: The sample size \(n < 30\), and the population is normally distributed, and the population standard deviation is unknown.
   - **Use the t-distribution.**

3. **Scenario 3**: The sample size \(n \geq 30\), and the population standard deviation is unknown.
   - **Use the Z-distribution.**

If these scenarios align with the situations in the table, you would choose the corresponding distribution based on the conditions given.

Would you like to go over any specific scenario in more detail?

### Related Questions:
1. What would be the effect on the confidence interval if the confidence level was 95% instead of 90%?
2. How does the sample size influence the width of the confidence interval?
3. Why is the t-distribution used instead of the Z-distribution for small sample sizes?
4. How does knowing the population standard deviation affect the confidence interval calculation?
5. What assumptions must be met to use the t-distribution for constructing confidence intervals?
6. How would the confidence interval change if the sample standard deviation was larger?
7. Can you use the Z-distribution if the population is not normally distributed?
8. What is the interpretation of the confidence interval in the context of this problem?

### Tip:
Remember, the t-distribution is more "spread out" than the Z-distribution, especially for smaller sample sizes. This accounts for the added uncertainty in estimating the population standard deviation.

Learn how to construct a 90% confidence interval to estimate the population mean score on a nationwide psychology examination using sample data. Understand the use of the normal distribution and t-distribution based on sample size and population standard deviation. Suitable for advanced high school and college students.

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