Super Sneaker Company is evaluating two different materials, A and B, to be used to construct the soles of their new active shoe targeted to city high school students in Canada. While material B costs less than material A, the company suspects that mean wear for material B is greater than mean wear for material A. Two study designs were initially developed to test this suspicion. In both designs, Halifax was chosen as a representative city of the targeted market. In Study Design 1, 15 high school students were drawn at random from the Halifax School District database. After obtaining their shoe sizes, the company manufactured 15 pairs of shoes, each pair with one shoe having a sole constructed from material A and the other shoe, a sole constructed from material B.
After 3 months, the amount of wear in each shoe was recorded in standardized units as follows:
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | |
A | 12.43 | 9.34 | 7.88 | 10.78 | 8.54 | 8.29 | 9.73 | 9.61 | 11.61 | 8.08 | 8.77 | 12.08 | 9.98 | 9.37 | 11.51 |
B | 10.95 | 11.28 | 10.21 | 9.73 | 11.54 | 10.48 | 9.45 | 10.84 | 10.81 | 10.04 | 9.18 | 8.52 | 12.73 | 9.94 | 10.16 |
What is the test statistic value to a minimum of 5 decimal places? (not pooled, I don't believe)
A | B | Difference |
12.43 | 10.95 | 1.48 |
9.34 | 11.28 | -1.94 |
7.88 | 10.21 | -2.33 |
10.78 | 9.73 | 1.05 |
8.54 | 11.54 | -3 |
8.29 | 10.48 | -2.19 |
9.73 | 9.45 | 0.28 |
9.61 | 10.84 | -1.23 |
11.61 | 10.81 | 0.8 |
8.08 | 10.04 | -1.96 |
8.77 | 9.18 | -0.41 |
12.08 | 8.52 | 3.56 |
9.98 | 12.73 | -2.75 |
9.37 | 9.94 | -0.57 |
11.51 | 10.16 | 1.35 |
Sample mean of the difference using excel function AVERAGE(), x̅d = -0.524
Sample standard deviation of the difference using excel function STDEV.S(), sd = 1.90464
Sample size, n = 15
Test statistic:
t = (x̅d)/(sd/√n) = (-0.524)/(1.90464/√15) = -1.06552
---------------------------
If you have any doubt ask in comments.
Get Answers For Free
Most questions answered within 1 hours.