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 45 feet and a standard deviation of 11.6 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 49 feet and a standard deviation of 6.1 feet. Suppose that a sample of 76 braking tests is 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 0.1 level of significance.

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 two decimal places.

Step 4 of 4: Make the decision for the hypothesis test. Fail or reject to fail.

Step 2 of 4 :  

Compute the value of the test statistic. Round your answer to two decimal places.

Answer 1

The provided sample means are shown below:

Also, the provided sample standard deviations are:

and the sample sizes are

n1​=76 and n2​=76.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

An F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.682 and FU​=1.466, and since F=3.616, then the null hypothesis of equal variances is rejected.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the degrees of freedom are df=113.533.

Hence, it is found that the critical value for this left-tailed test is tc​=−1.289, for α=0.1 and df=113.533.

The rejection region for this left-tailed test is R={t:t<−1.289}.

(3) Test Statistics

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=−2.661<tc​=−1.289, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0045, and since p=0.0045<0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is less than μ2​, at the 0.1 significance level.

Graphically

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