Question

The authors of a paper compared two different methods for measuring body fat percentage. One method uses ultrasound, and the other method uses X-ray technology. The table gives body fat percentages for 16 athletes using each of these methods (a subset of the data given in a graph that appeared in the paper). For purposes of this exercise, you can assume that the 16 athletes who participated in this study are representative of the population of athletes Ultrasound Athlete X ray 15.25 2 6.75 3 9.00 4 12.25 17.00 29.50 5.50 6.25 8.50 9.00 10.75 13 12.25 12.00 15.50 11.75 14.25 16.75 17.75 18.50

Answer 1

X-ray	Ultrasound	Difference
5.25	4.50	0.75
6.75	3.75	3.00
9.00	9.25	-0.25
12.25	11.75	0.50
17.00	17.25	-0.25
29.50	27.50	2.00
5.50	6.75	-1.25
6.25	6.75	-0.50
7.75	8.50	-0.75
8.50	9.75	-1.25
9.00	9.75	-0.75
10.75	12.25	-1.50
11.75	12.00	-0.25
14.25	15.50	-1.25
16.75	18.25	-1.50
17.75	18.50	-0.75

Sample mean of the difference using excel function AVERAGE(), x̅d = -0.2500

Sample standard deviation of the difference using excel function STDEV.S(), sd = 1.2682

Sample size, n = 16

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

Test statistic:

t = (x̅d)/(sd/√n) = (-0.25)/(1.2682/√16) = -0.7885 = -0.8

df = n-1 = 15

Two tailed p-value = T.DIST.2T(ABS(-0.7885), 15) = 0.443

Decision:

p-value > α, Do not reject the null hypothesis

We fail to reject Ho. the data do not provide convincing evidence that the mean body fat percentage measurement differs for the two months.