Question

Order to analyze the salaries of former students, 15 former students were randomly surveyed.

The average and standard deviation of monthly salaries of these students were $3,598 and $550, respectively. What is the margin of error of the 99% confidence interval for the true mean monthly salary of all former students? Assume the population is normally distributed. Round your answer to two decimal places (the hundredths place).

Answer 1

384.67

Explanation:

The margin of error for the 99% confidence interval for the true mean monthly salary of all former students can be calculated using the formula:

Margin of error = z * (standard deviation / sqrt(sample size))

where z is the z-score associated with the desired confidence level (in this case, 99%), standard deviation is the population standard deviation (given as $550), and sqrt(sample size) is the square root of the sample size (15).

The z-score for a 99% confidence level can be found using a standard normal distribution table or calculator, and it is approximately 2.576.

Plugging in the values, we get:

Margin of error = 2.576 * ($550 / sqrt(15))

≈ $384.67

Therefore, the margin of error of the 99% confidence interval for the true mean monthly salary of all former students is approximately $384.67.