Aresearcher 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 44 feet and a standard deviation of 8.1 feet. suvs equipped with tires using compound 2 have a mean braking distance of 47 feet and a standard deviation of 11.7 feet. suppose that a sample of 51 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.

We cannot say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used

Step-by-step explanation:

If we call  

=true mean braking distance corresponding to compound 1

=true mean braking distance corresponding to compound 2

=mean of the sample for compound 1

=mean of the sample for compound 2

In this case we have the following hypothesis

The level of confidence is 0.05, so the region of rejection is represented in picture 1

(See picture attached).

Let's find the z-score corresponding to this data and if this value is < -1.64, then we can say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.

where

s1=standard deviation braking distance corresponding to the sample for compound 1

s2=standard deviation braking distance corresponding to the sample for compound 2

n=sample size

So,

Since -1.5051 is not < -1.64, we cannot refute H0 and say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.

step-by-step explanation: