Question

The manager of a grocery store has taken a random sample of 100 customers. the average length of time it took these 100 customers to check out was 3.0 minutes. it is known that the standard deviation of the population of checkout times is one minute. the 95% confidence interval for the true average checkout time (in minutes) is

Answer 1

The confidence interval is given by:
m +/- z · σ/√n

For a 95% confidence level the z-score is z = 1.96

Therefore:
z · σ/√n = 1.96 · 1/√100 = 0.196

Hence the confidence interval is:
(3-0.196 , 3+0.196)
which gives:
(2.804 , 3.196)