Question

Engineers at a large automobile manufacturing company are trying to decide whether to purchase brand A or brand B tires for the company’s new models. To help them arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are as follows: Brand A: x ̅_1= 37,900 kilometers,s_1 = 5100 kilometers. Brand B: x ̅_2 = 39,800 kilometers,s_2 = 5900 kilometers. Test the hypothesis that there is no difference in the average wear of the two brands of tires. Assume the populations to be approximately normally distributed with equal variances.

Answer 1

Let µ₁ be the average kms run by Brand A.

Let µ₂ be the average kms run by Brand B.

Given:

For Brand A: x̅₁ = 37900 kms, s₁ = 5100 kms, n₁ = 12

For Brand B: x̅₂ = 39800 kms, s₂ = 5900 kms, n₂ = 12

The degrees of freedom used is n1 + n2 - 2 = 12 + 12 -2 = 22 (since pooled variance is used)

Sp = 5514.5263

The Hypothesis:

H0: µ₁ = µ₂: There is no difference in the average wear of the 2 brands of tyres.

Ha: µ₁ ≠ µ₂: There is a difference in the average wear of the 2 brands of tyres.

This is a Two tailed test.

The Test Statistic: We use the students t test as population standard deviations are unknown.

The p Value:    The p value (2 Tail) for t = -0.844, df = 22, is; p value = 0.4077

The Critical Value:   The critical value (2 tail) at α = 0.05(default), df = 22, tcritical = +2.0739 and -2.0739

The Decision Rule:    If tobserved is > tcritical or if tobserved is < -tcritical, Then Reject H0.

Also If the P value is < α, Then Reject H0

The Decision: Since t(-0.844) lies in between +2.0739 and -2.0739, We Fail To Reject H0

Also since P value (0.4077) is > α (0.05), We Fail to Reject H0.

The Conclusion: There is insufficient evidence at the 95% significance level to conclude that there is a difference in the average wear of the 2 brands of tyres.