Question

A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 12 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in Table 1. Difference Car Brand 1 Brand 2 (Brand 1 - Brand 2) 1 0.018 0.001 2 4 0.000 5 0.315 0.394 0.305 0.201 0.214 0.284 0.293 0.362 0210 0.222 0.262 0.354 0.297 00300 0.207 0.163 0.330 0.240 0.190 0.24 0.105 0.280 0.161 0.015 2 9 0.004 0.078 0.024 -0.064 0.101 10 11 12

Answer 1

∑d = 0.384

∑d² = 0.06151

n = 12

Mean , x̅d = Ʃd/n = 0.384/12 = 0.032

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(0.06151-(0.384)²/12)/(12-1)] = 0.0669

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Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

Type : Matched pair t test

Test statistic:

t = (x̅d)/(sd/√n) = (0.032)/(0.0669/√12) = 1.6571

df = n-1 = 11

Critical value, t-crit = T.INV.2T(0.05, 11) = 2.201

Reject Ho if t < -2.201 or if t > 2.201

Decision:

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

No