Question

Students in a semester-long health-fitness program have their percentage of body fat measured at the beginning of the semester and at the end of the semester. Assume that the percentages of body fat follow normal distributions. Find a 95% confidence interval for, μ_D, the mean of the difference in the percentages for the students based on the following measurements of 9 students.

Pre-program % 11   12 19 12 15 14 11 13 14

Post-program % 9 10 16 9 15 12 10 11 11

Answer 1

Pre	Post	Post - Pre
11	9	-2
12	10	-2
19	16	-3
12	9	-3
15	15	0
14	12	-2
11	10	-1
13	11	-2
14	11	-3

sample mean, xbar = -2
sample standard deviation, s = 1
sample size, n = 9
degrees of freedom, df = n - 1 = 8

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.306

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (-2 - 2.306 * 1/sqrt(9) , -2 + 2.306 * 1/sqrt(9))
CI = (-2.77 , -1.23)