Which type of confidence interval should you use in this situation? A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days. The distribution is approximately normally distributed. Find the 90% confidence interval for the average days worked of all payroll departments.
Since the given information contains the sample mean, sample standard deviation, and the sample size, we can use a t-confidence interval to estimate the population mean.
The formula for the t-confidence interval is:
CI = x̄ ± (t_(α/2) * (s/√n))
where:
CI = Confidence interval
x̄ = Sample mean
t_(α/2) = t-value for a given level of confidence (in this case, 90% confidence interval) and degrees of freedom (n-1)
s = Sample standard deviation
n = Sample size
Here, x̄ = 310.3, s = 23.8, and n = 25.
To find the t-value for a 90% confidence level with 24 degrees of freedom (25-1), we can consult a t-distribution table or use a statistical calculator. For this problem, let's assume the t-value to be 1.711.
CI = 310.3 ± (1.711 * (23.8/√25))
Simplifying further:
CI = 310.3 ± (1.711 * 4.76)
Calculating:
CI = 310.3 ± 8.16
Therefore, the 90% confidence interval for the average days worked of all payroll departments is (302.14, 318.46).