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 64 feet and a standard deviation of 5.4 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 69 feet and a standard deviation of 9.6 feet. Suppose that a sample of 76 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.

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Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.    Please give in format such as: Z> X or Z<x

Step 4 of 4: Make the decision for the hypothesis test.

Answer 1

1)

H0: mu1 = mu2
HA: mu1 < mu2

2)

x1 = 64 , s1 = 5.4 , n1 = 76
x2 = 69 , s2= 9,6 , n2 =76

t = (x1- x2)/sqrt(s1^2/n1+s2^2/n2)
= ( 64 - 69) /sqrt(5.4^2/76 + 9,6^2/76)
= -3.96

3)
Based on the information provided, the significance level is 5α=0.05, and the degrees of freedom are df = 118.142. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal:

Hence, it is found that the critical value for this left-tailed test is tc = -1.658

α=0.05 and df = 118.142

The rejection region for this left-tailed test is t<−1.658

4)

t is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean mu1 is less than mu2 , at the 0.05 significance level.