What is a 95% confidence interval for mue, the population mean time the postal service employees have spent with the postal service?

n=100
mean=7
S=2 years

Standard error of the mean (SE) = S/square root of n

95% confidence interval for μ = mean ± 2SE

I hope this helps. Thanks for asking.

To calculate a 95% confidence interval for the population mean (mue) of the time postal service employees have spent with the postal service, you will need the sample size (n), sample mean (mean), and sample standard deviation (S).

The formula for calculating the confidence interval is:

Confidence Interval = mean ± (z * (S / √n))

where 'z' is the z-score corresponding to the desired confidence level.

For a 95% confidence level, the z-score is approximately 1.96.

Plugging in the values you provided:

n = 100
mean = 7
S = 2

Confidence Interval = 7 ± (1.96 * (2 / √100))

Now, let's calculate the lower and upper bounds of the confidence interval:

Lower bound = 7 - (1.96 * (2 / √100))
Upper bound = 7 + (1.96 * (2 / √100))

Calculating these values, we get:

Lower bound ≈ 6.604
Upper bound ≈ 7.396

Therefore, the 95% confidence interval for the population mean time the postal service employees have spent with the postal service is approximately (6.604, 7.396) years.