Denise is a professional swimmer who trains, in part, by running. she would like to estimate the average number of miles she run
s 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)
