Question

In a random sample of five

​people, the mean driving distance to work was 22.9

miles and the standard deviation was 4.9

miles. Assuming the population is normally distributed and using the​ t-distribution, a 95​%

confidence interval for the population mean is (16.8, 29.0)

​(and the margin of error is

6.1​). Through​ research, it has been found that the population standard deviation of driving distances to work is 6.2.

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.

Interpret and compare the results.

Answer 1

sample mean, xbar = 22.9
sample standard deviation, σ = 6.2
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 * 6.2/sqrt(5)
ME = 5.43

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (22.9 - 1.96 * 6.2/sqrt(5) , 22.9 + 1.96 * 6.2/sqrt(5))
CI = (17.47 , 28.33)

Therefore, based on the data provided, the 95% confidence interval for the population mean is 17.47 < μ < 28.33 which indicates that we are 95% confident that the true population mean μ is contained by the interval (17.47 , 28.33)