Question

Assume that a sample is used to estimate a population mean μ . Find the margin of error M.E. that corresponds to a sample of size 21 with a mean of 64.4 and a standard deviation of 11.8 at a confidence level of 99.5%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. = Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

To find the margin of error (M.E.) for a confidence interval, use the formula M.E. = Z * (σ / √n), where Z is the critical value, σ is the standard deviation, and n is the sample size. For this question, the M.E. is approximately 9.4 when rounded to one decimal place.

Step-by-step explanation:

To find the margin of error (M.E.) for a confidence interval, we can use the formula:

M.E. = Z * (σ / √n)

Where Z is the critical value for the desired confidence level, σ is the population standard deviation, and n is the sample size. In this case, the sample mean is 64.4, the standard deviation is 11.8, and the sample size is 21.

First, we need to find the critical value. Since the confidence level is 99.5%, we subtract this from 100% to get 0.5%. We divide this value by 2 to account for splitting the area in the tails of the normal distribution, giving us a tail area of 0.25%. Looking up this value in a standard normal distribution table, we find that the critical value is approximately ±2.807.

Substituting the values into the formula, we get:

M.E. = 2.807 * (11.8 / √21) ≈ 9.371.

Therefore, the margin of error (M.E.) is approximately 9.4 when rounded to one decimal place.