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.
-A 93% confidence interval for $\mu$
,the mean number of days of sick leave would be _____.
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

Quizplus · Accepted Answer

The confidence interval is calculated using the formula: $\bar{x} \pm Z \times \frac{\sigma}{\sqrt{n}}$. For a 93% confidence level, the Z-score is approximately 1.81. Plugging in the values: $12.2 \pm 1.81 \times \frac{10}{\sqrt{100}}$ gives the interval [10.390, 14.010].

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