Question

In a random sample of five ​people, the mean driving distance to work was 20.2 miles and the standard deviation was 5.8 miles. Assuming the population is normally distributed and using the​ t-distribution, a 95​% confidence interval for the population mean μ is (13.0, 27.4) (and the margin of error is 7.2​). Through​ research, it has been found that the population standard deviation of driving distances to work is 6.6.Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, find the margin of error and construct a 95​% confidence interval for the population mean μ. Interpret and compare the results.

Answer 1

sample mean, xbar = 20.2
sample standard deviation, σ = 6.6
sample size, n = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 6.6/sqrt(5)
ME = 5.79

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (20.2 - 1.96 * 6.6/sqrt(5) , 20.2 + 1.96 * 6.6/sqrt(5))
CI = (14.4 , 26.0)

Here, populatio standard deviation is given so we use z test