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### Given Information:
- Sample mean \( \bar{x} = 21.2 \)
- Sample standard deviation \( s = 3.8 \)
- The population is normally distributed.
- For different sample sizes, we will construct confidence intervals.

### (a) Construct an 80% confidence interval when \( n = 44 \).

1. **Formula for confidence interval**:
   \[
   \text{CI} = \bar{x} \pm t_{\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}}
   \]
   Where:
   - \( t_{\frac{\alpha}{2}} \) is the critical value from the t-distribution.
   - \( \alpha = 1 - 0.80 = 0.20 \), so \( \frac{\alpha}{2} = 0.10 \).

2. **Find \( t_{\frac{\alpha}{2}} \)**:
   For \( n = 44 \), degrees of freedom \( df = 43 \). Use a t-table or calculator to find the critical value for 80% confidence with 43 degrees of freedom.

3. **Calculate margin of error (E)**:
   Once you have the \( t \)-value, plug it into the formula along with \( s = 3.8 \) and \( n = 44 \).

### (b) Construct an 80% confidence interval when \( n = 76 \).

The process is the same as part (a), except with \( n = 76 \) (which changes the degrees of freedom to 75).

**How does increasing the sample size affect the margin of error (E)?**

- **Effect of sample size**: As \( n \) increases, the margin of error \( E = t_{\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}} \) decreases because \( \sqrt{n} \) is in the denominator. Larger sample sizes reduce the variability of the estimate.

### (c) Construct a 90% confidence interval when \( n = 44 \).

Same process as (a), but now for a 90% confidence level, the critical value \( t_{\frac{\alpha}{2}} \) will be different since \( \alpha = 1 - 0.90 = 0.10 \).

**How does increasing the level of confidence affect the size of the margin of error (E)?**

- **Effect of confidence level**: As the confidence level increases, the critical value \( t_{\frac{\alpha}{2}} \) increases, which makes the margin of error larger. Higher confidence levels result in wider confidence intervals.

### (d) If the sample size is 27, what conditions must be satisfied to compute the confidence interval?

- The population must be normally distributed, or the sample size should be large enough (typically \( n > 30 \)) for the Central Limit Theorem to apply.
- The sample must be random and independent.

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Would you like me to calculate the actual confidence intervals for parts (a) through (c), or explain any concept in more detail?

### Related Questions:
1. How do you interpret a confidence interval in a real-world context?
2. What happens to the width of a confidence interval if you increase the sample size further?
3. How would you determine if a population is normally distributed?
4. What is the relationship between standard deviation and margin of error?
5. How does the choice of confidence level affect the probability of including the population mean?

### Tip:
Always ensure the sample size is adequate and the population assumptions (normality or large sample size) hold when constructing confidence intervals.

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 21.2, and the sample standard deviation, s, is found to be 3.8. (a) Construct a 80% confidence interval about μ if the sample size, n, is 44. (b) Construct a 80% confidence interval about μ if the sample size, n, is 76. How does increasing the sample size affect the margin of error, E? (c) Construct a 90% confidence interval about μ if the sample size, n, is 44. How does increasing the level of confidence affect the size of the margin of error, E? (d) If the sample size is 27, what conditions must be satisfied to compute the confidence interval?

Learn how to construct 80% and 90% confidence intervals for a normally distributed population, and understand how sample size and confidence level impact the margin of error. This problem involves confidence intervals, margin of error, and t-distribution critical values.

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