A company is interested in estimating μ,the mean number of days of sick leave taken by its employees.The firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee.The sample mean is 12.2 days,and the sample standard deviation is 10 days.How many personnel files would the statistician have to select in order to estimate μ to within 2 days with a 99 percent confidence interval?
A)2
B)13
C)136
D)165
E)173

Question

Quizplus · Accepted Answer

We need to use the formula for the margin of error in a confidence interval:

Margin of Error = Z* (standard deviation / sqrt(sample size))

We want the margin of error to be 2 days and we want a 99% confidence interval. From a standard normal distribution table, we know that the Z-score for a 99% confidence interval is 2.58.

2 = 2.58 * (10 / sqrt(n))

Solving for n, we get n = 173. Therefore, the statistician would have to randomly select 173 personnel files in order to estimate μ to within 2 days with a 99% confidence interval.

The answer is E: 173.

A Company Is Interested in Estimating μ,The Mean Number of Days