A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year. Assume a population standard deviation of 23.8 days. What is the 90% confidence interval for the
Average days worked, $\mu$ of all payroll departments?
A) $301.0

Question

A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year. Assume a population standard deviation of 23.8 days. What is the 90% confidence interval for the
Average days worked, $\mu$ of all payroll departments? 
A) $301.0 < \mu < 319.6$
B) $298.0 < \mu < 322.6$
C) $302.4 < \mu < 318.2$
D) $314.1 < \mu < 316.8$

Quizplus · Accepted Answer

We can use the formula:
Confidence interval = sample mean +/- z* (standard deviation/ square root of sample size)
Here, sample size (n) = 25, sample mean ($\bar{x}$) = 310.3, population standard deviation ($\sigma$) = 23.8 and the level of confidence (C) = 90%.
The corresponding z-value for 90% confidence interval can be found from the standard normal distribution table or calculator, and it is approximately 1.645.
Plugging in these values, we get:
Confidence interval = 310.3 +/- 1.645 * (23.8/ sqrt(25))
Confidence interval = 310.3 +/- 9.32
Confidence interval = (310.3 - 9.32, 310.3 + 9.32)
Confidence interval = (301.0, 319.6)
Therefore, the 90% confidence interval for the average days worked by all payroll departments is (302.4 < $\mu$ < 318.2), which is option C.

A Sample of 25 Different Payroll Departments Found That the Employees