Question

A tire manufacturer has been producing tires with an average life expectancy of 26,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data.

Answer 1

Given that the life expectancy of the 6 tested sample are: 28,000; 27,000; 25,000; 28,000; 29,000; 25,000

Part A:

The mean is given by:

$\bar{x}= \frac{28,000+27,000+25,000+28,000+29,000+25,000}{6} \\ \\ = \frac{162,000}{6} =27,000$

The standard deviation is given by:

$s \ in \ thousand=\sqrt{\frac{(28-27)^2+(27-27)^2+(25-27)^2+(28-27)^2+(29-27)^2+(25-27)^2}{6-1}} \\ \\ = \sqrt{\frac{1^2+0^2+(-2)^2+1^2+2^2+(-2)^2}{5}} =\sqrt{\frac{1+4+1+4+4}{5}} \\ \\ =\sqrt{\frac{14}{5}}=\sqrt{2.8}=1.67$

Thus, the standard deviation is 1,670



Part B:

For 99% confidence interval $z_{\alpha/2}=2.58$.

Test statistics is given by:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{27,000-26,000}{1,670/\sqrt{6}} \\ \\ = \frac{1,000}{681.77} =1.47$

Rejection region is given by

$z\ \textgreater \ z_{\alpha/2}$

Since the test statistics is not greater than $z_{\alpha/2}$, we fail to reject the null hypothesis and we conclude that there is no sufficient evidence to reject the claim that the life expectance of its new tires has increased.



Part C:

Taking α = 0.05.

The p-value of the test statistics is given by:

$P(z \ \textgreater \ 1.47)=1-P(z\ \textless \ 1.47) \\ \\ =1-0.9292=0.0708$

Rejection region is given by:

$p-value\leq\alpha$

Since the p-value of 0.0708 is not less that the alpha of 0.05, thus we cannot reject the null hypothesis and we conclude that there is no sufficient evidence to reject the claim that the life expectance of its new tires has increased.