Question

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 repres ntativ city of the targeted market. In Study Design 1, 11 high school students were drawn at random from the Halifax School District database. After obtaining their shoe sizes, the company manufactured 11 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 The researcher in charge of this design asked the company to randomly assign material A to the right shoe or to the left shoe, for each of the 11 pairs. Why? Right shoe wear is generally different than left shoe wear O Right shoe wear is always less than left shoe wear Right shoe wear is always greater than left shoe wean [1 pt(s)1 Submit Answer Tries 0/1 After 3 months, the amount of wear in each shoe was record d in standardized units as follows: 10 11 A 14.63 11.54 10.08 12.98 10.74 10.49 11.93 11.81 13.81 10.28 10.97 В 12.63 1296|1 1.89| 11.411322| 12.16 11.13 12.52|12.49|11.721086 A null hypothesis for the test is: O X<14.63 0 μ 11.92 [1 pt(s)] Submit Answer Tries 0/1 An alternative hypothesis is: [1 pt s)] Submit Answer Tries 0/1 To test whether mean wear for material B is greater than mean wear for material A, calculate a test statistic. Use at least 5 digits to the right of the decimal 3 pt(s1 Submit Answer Tries 0/3

Answer 1

data

A	B
14.63	12.63
11.54	12.96
10.08	11.89
12.98	11.41
10.74	13.22
10.49	12.16
11.93	11.13
11.81	12.52
13.81	12.49
10.28	11.72
10.97	10.86

null hypothesis
option A)

alternative hypothesis
option A)

using Excel

t-Test: Paired Two Sample for Means
	A	B
Mean	11.75090909	12.09
Variance	2.229689091	0.5749
Observations	11	11
Pearson Correlation	0.160830269
Hypothesized Mean Difference	0
df	10
t Stat	-0.719914917
P(T<=t) one-tail	0.244031079
t Critical one-tail	2.763769458
P(T<=t) two-tail	0.488062157
t Critical two-tail	3.169272673

TS =

-0.719914917

statistical   table

C) t-distribution

df = n-1 = 10

this is 1-sided

p-value = 0.244031079

option F) p-value > 0.1