A sample of 25 different payroll depts. found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days. What is the 90% confidence interval for the average days worked by employees in all payroll departments?
To find the 90% confidence interval for the average days worked by employees in all payroll departments, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
1. Finding the Critical Value:
Since we want a 90% confidence interval, we need to find the critical value corresponding to a confidence level of 90%. The critical value will depend on the sample size and the desired confidence level.
Since we have a sample size of 25, the degrees of freedom for a t-distribution will be 25 - 1 = 24. Checking a t-distribution table or using a calculator, we find that the critical value for a 90% confidence level with 24 degrees of freedom is approximately 1.711.
2. Calculating the Standard Error:
The standard error represents the standard deviation of the sample mean and is calculated by dividing the population standard deviation by the square root of the sample size.
Standard Error (SE) = Standard Deviation / √(Sample Size)
In this case, the standard deviation is 23.8 days, and the sample size is 25. Hence,
SE = 23.8 / √(25) = 23.8 / 5 = 4.76
3. Calculating the Confidence Interval:
Now we can substitute the values into the confidence interval formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
The sample mean is given as 310.3 days. Hence,
Confidence Interval = 310.3 ± (1.711 * 4.76) = 310.3 ± 8.149
Therefore, the 90% confidence interval for the average days worked by employees in all payroll departments is (302.151, 318.449).
90% = mean ± 1.645 SEm
SEm = SD/√n