Given:
n = 50, sample size
[tex]\bar{x} = 23.6 \, min[/tex], sample mean
s = 2.4 min, sample standard deviation.
The confidence interval is
[tex]\bar{x} \pm t^{*} \frac{s}{ \sqrt{n} } [/tex]
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
