To determine the sample size needed for a specified margin of error in a confidence interval, we can use the formula:

$n = \left( \frac{Z \cdot \sigma}{E} \right)^2$

where:

• $n$ is the sample size,
• $Z$ is the Z-score corresponding to the desired confidence level,
• $\sigma$ is the population standard deviation,
• $E$ is the margin of error.

Given:

• $\sigma = 17.2$
• $E = 3.8$

Assuming you want a 95% confidence interval (which corresponds to a Z-score of approximately 1.96), the calculation would be:

$n = \left( \frac{1.96 \cdot 17.2}{3.8} \right)^2$

Now, let's calculate it step by step.

1. Calculate $\frac{1.96 \cdot 17.2}{3.8}$: $\frac{1.96 \cdot 17.2}{3.8} = \frac{33.712}{3.8} \approx 8.86$

2. Square the result: $n \approx (8.86)^2 \approx 78.5$

Since the sample size must be a whole number, we round up to ensure we meet the margin of error. Thus, the required sample size $n$ is approximately 79.

Summary

To achieve a margin of error of 3.8 with a standard deviation of 17.2 at a 95% confidence level, a sample size of 79 is needed.

Would you like more details or have any questions? Here are some related questions you might consider:

1. What is the effect of increasing the confidence level on the required sample size?
2. How does the population standard deviation impact the sample size needed?
3. What happens to the margin of error if the sample size increases?
4. Can you explain the difference between a population and a sample in statistics?
5. How would you calculate the sample size for different confidence levels (e.g., 90% or 99%)?

Tip: Always round up your sample size to ensure the margin of error is met.