Question

The distribution of scores on a standardized test is approximately normal with

a standard deviation of 40. Suppose a sample of 100 students had a mean
score of 180.23.
Find the margin of error for a 90% confidence interval for the mean score of all
students on this test. Round your answer to two decimal places.

Answer 1

Answer: The margin of error (ME) for a 90% confidence interval using a normal distribution with a known population standard deviation is given by:

ME = z* (σ/√n)

where z* is the z-score corresponding to the desired confidence level (in this case, 90% confidence), σ is the population standard deviation, and n is the sample size.

Using a z-score table or calculator, we find that the z-score corresponding to a 90% confidence level is 1.645.

Substituting the given values, we have:

ME = 1.645 * (40 / √100) = 6.58

Therefore, the margin of error for a 90% confidence interval for the mean score of all students on this test is 6.58. We can report the 90% confidence interval as:

180.23 ± 6.58 or (173.65, 186.81)

Explanation: