Math Problem Statement

Two popular brands of tires for tractor-trailers are the Puma and the Eternal. Carmen is a buyer for a major shipping company and wants to determine if there is any difference between the two brands of tire in the mean distance (in thousands of km) driven on them before they need to be replaced. In the company's testing lab, Carmen tests a random sample of 14 Puma tires and 15 Eternal tires. Perform a two-tailed test at the 0.01 significance level to determine if there is a difference in mean distances.

Solution

Let's solve this problem step by step:

Part (a): State the null hypothesis H0H_0 and the alternate hypothesis H1H_1.

We are testing whether there is a difference between the mean distances (in thousands of km) driven before the tires need replacement for the two brands, Puma and Eternal.

  • Null Hypothesis H0H_0: There is no difference in the population mean distances driven before replacement for Puma and Eternal tires. H0:μ1μ2=0H_0: \mu_1 - \mu_2 = 0 where μ1\mu_1 is the mean distance for Puma tires and μ2\mu_2 is the mean distance for Eternal tires.

  • Alternative Hypothesis H1H_1: There is a difference in the population mean distances driven before replacement for Puma and Eternal tires. H1:μ1μ20H_1: \mu_1 - \mu_2 \neq 0

Part (b): Determine the type of test statistic to use.

Since we are comparing the means of two independent samples and we know the sample variances, we will use a two-sample t-test. The appropriate test statistic for comparing two means is given by: t=(xˉ1xˉ2)s12n1+s22n2t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} where:

  • xˉ1=61.23\bar{x}_1 = 61.23 (sample mean for Puma)
  • xˉ2=52.28\bar{x}_2 = 52.28 (sample mean for Eternal)
  • s12=5.44s_1^2 = 5.44 (sample variance for Puma)
  • s22=78.88s_2^2 = 78.88 (sample variance for Eternal)
  • n1=14n_1 = 14 (sample size for Puma)
  • n2=15n_2 = 15 (sample size for Eternal)

The degrees of freedom for this test can be calculated using the formula: df=(s12n1+s22n2)2(s12n1)2n11+(s22n2)2n21df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}

Would you like me to continue with the calculations to complete the hypothesis test?

Additional questions:

  1. How do we calculate the p-value for this two-tailed test?
  2. Why do we use a two-tailed test in this scenario?
  3. What would be the conclusion if the test statistic falls outside the critical values at the 0.01 significance level?
  4. How does the sample size affect the power of the hypothesis test?
  5. What assumptions must be met for the two-sample t-test to be valid?

Tip: When conducting hypothesis testing, always check if the assumptions (normality, independence, etc.) are met to ensure valid results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-sample t-test
Sample Variance
Degrees of Freedom

Formulas

t = (x̄1 - x̄2) / √[(s1^2 / n1) + (s2^2 / n2)]
Degrees of Freedom: df = [(s1^2 / n1) + (s2^2 / n2)]^2 / [(s1^2 / n1)^2 / (n1-1) + (s2^2 / n2)^2 / (n2-1)]

Theorems

Central Limit Theorem
t-distribution

Suitable Grade Level

College level, Introductory Statistics