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 481 and a standard deviation of 73.

(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 at distribution. It could use either Z or t. Or it could be unclear.)

Sampling scenario

Z

Could use either Z or t

Unclear

The sample has size 11, and it is from a population with a distribution about which we know very little.

The sample has size 19, and it is from a normally distributed population with an unknown standard deviation.

The sample has size 95, and it is from a non-normally distributed population with a known standard deviation of 75.

Let's go through the questions step by step.

### Part (a) 
**Best Point Estimate:**

The best point estimate for the population mean \(\mu\) is the sample mean. 

Given:
- Sample mean \(\bar{x} = 481\)

So, the best point estimate for the population mean is **481**.

### Part (b) 
**Determine the Distribution to Use:**

To determine which distribution (Z or t) to use when calculating the critical value for the 90% confidence interval, we consider the sample size, whether the population standard deviation is known, and whether the population is normally distributed.

#### 1. **Sampling scenario:**
   - **Sample size:** 11
   - **Population distribution:** Unknown distribution

   **Explanation:** 
   - The sample size is small (\(n < 30\)).
   - We know very little about the population distribution.
   - When the sample size is small and the population distribution is unknown, we should use the **t-distribution**.
   
   **Answer:** **t**

#### 2. **Sampling scenario:**
   - **Sample size:** 19
   - **Population distribution:** Normally distributed population with an unknown standard deviation

   **Explanation:** 
   - The sample size is small (\(n < 30\)).
   - The population is normally distributed.
   - The population standard deviation is unknown, so the **t-distribution** should be used.
   
   **Answer:** **t**

#### 3. **Sampling scenario:**
   - **Sample size:** 95
   - **Population distribution:** Non-normally distributed population with a known standard deviation of 75

   **Explanation:** 
   - The sample size is large (\(n \geq 30\)), which typically allows the Central Limit Theorem to apply, meaning the sampling distribution of the sample mean will be approximately normal even if the population is not normal.
   - The population standard deviation is known, so the **Z-distribution** can be used.
   
   **Answer:** **Z**

### Summary of the Answers:
- **Part (a):** Best point estimate = 481
- **Part (b):**
  1. **t**
  2. **t**
  3. **Z**

Would you like further details on any part? Here are some related questions you might be interested in:

1. How do you calculate the margin of error for a confidence interval?
2. What are the assumptions of the Central Limit Theorem?
3. How does sample size affect the choice of distribution for confidence intervals?
4. How do you interpret a confidence interval in context?
5. How does the confidence level affect the width of the confidence interval?

**Tip:** When dealing with small sample sizes, the t-distribution is often used because it accounts for the additional uncertainty due to the smaller sample.

Learn how to construct a 90% confidence interval to estimate the population mean score on a nationwide examination in psychology using a sample mean and standard deviation. Includes step-by-step instructions and sampling scenarios with appropriate distribution selections (t-distribution or Z-distribution).

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