Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population.

12.1	11.6	11.9	12.9	12.5	11.4	12.0
11.7	11.8	12.1

Answer 1

a)
sample mean, xbar = 12
sample standard deviation, s = 0.4397
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.833

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12 - 1.833 * 0.4397/sqrt(10) , 12 + 1.833 * 0.4397/sqrt(10))
CI = (11.75 , 12.25)

b)
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12 - 2.262 * 0.4397/sqrt(10) , 12 + 2.262 * 0.4397/sqrt(10))
CI = (11.69 , 12.31)

c)
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.25

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12 - 3.25 * 0.4397/sqrt(10) , 12 + 3.25 * 0.4397/sqrt(10))
CI = (11.55 , 12.45)