Final answer:
The margin of error at a 95% confidence level for a sample size of n = 20 and standard deviation s = 8 is approximately 3.507 when rounded to three decimal places.
Step-by-step explanation:
To calculate the margin of error (ME) at a 95% confidence level, we will use the formula ME = z * (s/√n), where 'z' is the z-score corresponding to the desired confidence level, 's' is the sample standard deviation, and 'n' is the sample size.
Given that n=20, x-bar=38 (which is not used in the calculation of the margin of error but rather for the confidence interval), and s=8, we first need to find the corresponding z-score for a 95% confidence level. The z-score that corresponds to a 95% confidence level is approximately 1.96 (please note that this value is typically found using a z-table or standard normal distribution table).
Using the formula, ME = 1.96 * (8/√20). Now, calculating the square root of 20 gives us approximately 4.472. So, we have ME = 1.96 * (8/4.472) which simplifies to ME = 1.96 * 1.789, and this gives us ME ≈ 3.507.
Therefore, the margin of error at a 95% confidence level is approximately 3.507 (rounded to three decimal places).