A company is interested in estimating $$\mu$$ , 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. From historical records, the population standard deviation is about 10 days.
-How many personnel files would the director have to select in order to estimate $\mu$
To within 2 days with 99% confidence?
A)2
B)13
C)136
D)165
E)166

Question

Quizplus · Accepted Answer

To estimate the mean within a certain margin of error (E) with a given level of confidence, we use the formula for the sample size $n$ in estimating a mean with known population standard deviation $\sigma$, which is:$$n = \left(\frac{Z \cdot \sigma}{E}\right)^2$$Where:- $Z$ is the Z-value corresponding to the desired confidence level (for 99% confidence, $Z \approx 2.576$),- $\sigma$ is the population standard deviation (10 days in this case),- $E$ is the desired margin of error (2 days).Plugging in the values:$$n = \left(\frac{2.576 \cdot 10}{2}\right)^2 = \left(\frac{25.76}{2}\right)^2 = \left(12.88\right)^2 = 165.89$$Since we can't have a fraction of a file, we round up to the nearest whole number, which gives us 166.

A Company Is Interested in Estimating μ\muμ , the Mean Number of Days of Sick Leave Taken

A Company Is Interested in Estimating $\mu$ , the Mean Number of Days of Sick Leave Taken