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 5050 feet and a standard deviation of 12.112.1 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 5555 feet and a standard deviation of 9.39.3 feet. Suppose that a sample of 8383 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.

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 H0H0. Round the numerical portion of your answer to three decimal places.

Step 4 of 4:

Make the decision for the hypothesis test.

Answer 1

T-test for two Means – Unknown Population Standard Deviations - Unequal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=50 and Xˉ2​=55 b. Sample Standard deviation: s1=12.1 and s2=9.3 c. Sample size: n1=83 and n2=83 (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. The degrees of freedom Assuming that the population variances are unequal, the degrees of freedom are given by (2)Test Statistics The t-statistic is computed as follows: (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 153.8185. Therefore the critical value for this Left-tailed test is tc​=-1.6549. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Left-tailed test is t<-1.6549 i.e. reject H0 if t<-1.6549 The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0017 (4) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=-2.9849 < tc​=-1.6549, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0017, and since p=0.0017≤0.05, it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is less than μ2, at the 0.05 significance level.

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