Denise is a professional swimmer who trains, in part, by running. she would like to estimate the average number of miles she runs in each week. for a random sample of 20 weeks, the mean is = 17.5 miles with standard deviation s = 3.8 miles. find a 99% confidence interval for the population mean number of miles denise runs. use a graphing calculator for this one and not the t chart from the book.

Given:

n = 20, sample size

xbar = 17.5, sample mean

s = 3.8, sample standard deiation

99% confidence interval

The degrees of freedom is

df = n-1 = 19

We do not know the population standard deviation, so we should determine t * that corresponds to df = 19.

From a one-tailed distribution, 99% CI means using a p-value of 0.005.

Obtain

t * = 2.8609.

The 99% confidence interval is

xbar + / - t * (s/√n)

t * (s/√n) = 2.8609 * (3.8/√20) = 2.4309

The 99% confidence interval is

(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)

Answer: The 99% confidence interval is (15.07, 19.93)