Question

3. A sample of 230 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 18. (20 pts)
4. 1. Determine the standard error of the mean.
5. 2. Determine the 98% confidence interval for the population mean

Answer 1

a)

std.error = s/sqrt(n)
= 26/sqrt(230)
= 1.7144

b)

sample mean, xbar = 18
sample standard deviation, σ = 26
sample size, n = 230

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)
ME = 2.33 * 26/sqrt(230)
ME = 3.99

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (18 - 2.33 * 26/sqrt(230) , 18 + 2.33 * 26/sqrt(230))
CI = (14.01 , 21.99)