Answer:
CI (95%)lb = 576,51
Explanation:
Sample Information:
sample size n = 12
sample mean x = 614 h
sample standard deviation s = 59 h
Population with normal distribution, sample size n < 30 in order to find out CI 95% significance level is α = 5 % α = 0,05 α/2 = 0,025
t(c) for α/2 = 0,025 and degree of freedom df = n - 1 df = 12 - 1
df = 11 is:
t(c) = 2,201
CI = x ± t(c) * s / √n
CI (95%) = 614 ± 2,201 * 59 / √12
CI (95%)lb = 614 - 37,49
CI (95%)lb = 576,51