Question

A random sample of 40 people who owned their own business reported a yearly mean profit of $11,290 with a standard deviation of $1,251. Find the 92% confidence interval for the population mean. (Round your answers to the nearest dollar) Preview < μ< Preview What does this mean?

Answer 1

sample mean, xbar = 11290
sample standard deviation, s = 1251
sample size, n = 40
degrees of freedom, df = n - 1 = 39

Given CI level is 92%, hence α = 1 - 0.92 = 0.08
α/2 = 0.08/2 = 0.04, tc = t(α/2, df) = 1.8

CI = (x-te x 18 +te x )

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (11290 - 1.8 * 1251/sqrt(40) , 11290 + 1.8 * 1251/sqrt(40))
CI = (10934 , 11646)

One can be 92% confident that the mean population profit lies within the CI.
Therefore, based on the data provided, the 92% confidence interval for the population mean is 10934 < μ < 11646 which indicates that we are 92% confident that the true population mean μ is contained by the interval (10934 , 11646)