A company is interested in estimating ,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.Calculate a 93% confidence interval for ,the mean number of days of sick leave.
A)[10.725 13.675]
B)[10.390 14.010]
C)[12.019 12.381]
D)[12.053 12.348]
E)[11.735 12.665]

Question

A company is interested in estimating  ,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.Calculate a 93% confidence interval for  ,the mean number of days of sick leave.
A)[10.725 13.675] 
B)[10.390 14.010] 
C)[12.019 12.381] 
D)[12.053 12.348] 
E)[11.735 12.665]

Quizplus · Accepted Answer

First, we need to calculate the margin of error:
ME = z* (sigma/sqrt(n))
where z* is the z-score for a 93% confidence interval, which can be found using a z-table or calculator. For a 93% confidence interval, z* = 1.8119

sigma is the population standard deviation, but since it's unknown, we use the sample standard deviation as an estimate. So, sigma = 10

n is the sample size, so n = 100

Now we can plug in the values:
ME = 1.8119 * (10/sqrt(100)) = 1.8119

Next, we can use the formula:
CI = x̄ ± ME
where x̄ is the sample mean, which is 12.2

CI = 12.2 ± 1.8119

CI = [10.3881, 13.0119]

Therefore, the closest answer is B: [10.390, 14.010].

A Company Is Interested in Estimating ,The Mean Number