Fifty randomly selected individuals were timed completing a tax form. the sample mean was 23.6 minutes; the sample standard deviation was 2.4 minutes. a 99% confidence interval for the mean time required by all individuals to complete the form is about:

Given:
n = 50, sample size
, sample mean
s = 2.4 min, sample standard deviation.

The confidence interval is


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)

(22.7, 24.5)

25% increaseplus i think decrease i only know increase: ((

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