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 62 feet and a standard deviation of 10.6 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 68 feet and a standard deviation of 13.9 feet. Suppose that a sample of 77 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 μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance.

Answer 1

We can Use Two sample t test assuming unequal variance.

Given: = 0.05

compound 1:

n1 = 77, = 62, S1 = 10.6

compound 2:

n2 = 77, = 68, S2 = 13.9

Hypothesis:

Degrees of Freedom:

Critical value:

   …………………..….using t table

Test statistic:

Conclusion:

Test statistic (t) < Critical value i.e -3.01 < -1.6557, That is Reject Ho at 5% level of sinificance.

Therefore, there is Sufficient evidence that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.