Question

Noise levels at 3 concerts were measured in decibels yielding the following data: 124,134,134 Construct the 98% confidence interval for the mean noise level at such locations. Assume the population is approximately normal.

Answer 1

sample mean, xbar = 130.67
sample standard deviation, s = 5.77
sample size, n = 3
degrees of freedom, df = n - 1 = 2

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 6.965

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (130.67 - 6.965 * 5.77/sqrt(3) , 130.67 + 6.965 * 5.77/sqrt(3))
CI = (107.47 , 153.87)