3. The Human Resources department of a large corporation wants to estimate the mean number
of unused vacation days that its employees have. They conduct a random sample of 30
employees and find a sample mean of 13.6 days. Assuming a standard deviation of 1.4 days,
find the 90% confidence level margin of error for this sample mean. Use z = 1.645. Round
your answer to the one-hundredths place.

*04.08
*1.11
*00.42
*0.08

The 90% confidence level margin of error for this sample mean is 0.42

Step-by-step explanation:

The Human Resources department conduct a random sample of 30  employees

n = 30

Standard deviation =

z = 1.645

Formula of 90% confidence level margin of error for this sample mean=

So, margin of error for this sample mean=

So, margin of error for this sample mean=0.42

Hence the 90% confidence level margin of error for this sample mean is 0.42

70,000 + 6,000 + 800 + 30

step-by-step explanation: