Question

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. SUVs equipped with tires using compound 1 have a mean braking distance of 7272 feet and a standard deviation of 11.811.8 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 7575 feet and a standard deviation of 11.311.3 feet. Suppose that a sample of 5757 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.050.05 level of significance.

State null hypothesys

Reject if

Reject or Fail to reject

Answer 1

(1)
H0:Null Hypothesis: (Braking distance for SUVs equipped with tires using compound 1 is greater than or equal to the braking distance when compound 2 is used)

HA:Alternative Hypothesis: (Braking distance for SUVs equipped with tires using compound 1 is less than the braking distance when compound 2 is used) (Claim)

(2)

= 0.05

ndf = n1 + n2 - 2 = 57 + 57 - 2 = 112

One Tail - Left Side

Test

From Table,critical value of t = - 1.6586

Rejection Rule:
Reject null hypothesis:

if t < - 1.6586

(3)

Test statistic is:

t = (72-75)/2.1640 = - 1.3863

Since the calculated value of t = - 1.3863 is greater than critical value of t = - 6586, the difference is not significant. Fail to reject null hypothesis.

Answers to questions asked:

(1)

H0:Null Hypothesis: (Braking distance for SUVs equipped with tires using compound 1 is greater than or equal to the braking distance when compound 2 is used)

HA:Alternative Hypothesis: (Braking distance for SUVs equipped with tires using compound 1 is less than the braking distance when compound 2 is used) (Claim)

(2)

Reject null hypothesis: if t < - 1.6586

(3)

Fail to reject null hypothesis.