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 representative city of the targeted market. In Study Design 1, 16 high school students were drawn at random from the Halifax School District database. After obtaining their shoe sizes, the company manufactured 16 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.8 10 11 12 13 14 15 16 A 14.08 11.46 10.21 12.68 10.77 10.5611.79 11.69 13.38 10.38 10.97 13.79 12.00 11.48 13.30 13.02 в 12.48|12.75|11.841143|1297|12 07|11.191238|1235|11.69|10.97|10.40 113.9911.61|11.80|12.92what is the 90% confidence interval for the difference in wear between material B and material A (use B-A)? Use software to get a more precise critical value, but confirm its roughtly the same value you get from the table. Use at least 5 digits to the right of the decimal. Lower bound: -0.56418 Upper bound 0.72418 3 pt(s)]

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Answer #1

Solution:-

90% confidence interval for the difference in wear between material B and material A is C.I = ( 0.40464, 2.15536).

14.08 11.46 10.21 12.48 1.6 8.2944 12.75 11.84 1.29 0.0000999 1.63 0.1224999 12.97 12.07 11.19 12.38 0.690 12.35 11.69 2.2 0.846 1.51 0.0528999 10.77 10.56 11.79 11.69 13.38 10.38 3.5344 0.3480999 1.03 5.3361 1.31 0.0008999 13.79 12.00 10.40 13.99 3.39 1.99 21.8089 0.5041 11.48 11.61 0.1299999 1.3225 13.02 191.56 12.92 -0.099 Sum Mean 192.84 11.9725 2.0525 5 1.9044 59.8434 0.08 3.7402125 1.28

s = sqrt{rac{sum (d-ar{d})^{2}}{n-1}}

s = 1.9974

SE = s / sqrt(n)

S.E = 0.49935

C.I = ar{d}pm t_{alpha /2}ast rac{sigma }{sqrt{n}}

C.I = 1.28 + 1.753*0.49935

C.I = 1.28 + 0.8754

C.I = ( 0.40464, 2.15536)

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