Question

Use the following sample to estimate a population mean μ.

44.1

-14.6

22.2

8

29.4

-9.7

Assuming the population is normally distributed, find the 95% confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) TO TWO DECIMAL PLACES

95% C.I. = ????????

Answer 1

sample mean, xbar = 13.23
sample standard deviation, s = 22.9021
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

ME = tc * s/sqrt(n)
ME = 2.571 * 22.9021/sqrt(6)
ME = 24

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (13.23 - 2.571 * 22.9021/sqrt(6) , 13.23 + 2.571 * 22.9021/sqrt(6))
CI = (-10.81 , 37.27)

-10.81 < mu < 37.27)