Answer to Question #290168 in Statistics and Probability for Gaskaw

Question

The manufacturer of the Horizon truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Comet Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Comet’s experience different from that claimed by the manufacturer at the .05 significance level?

Required: Use both the critical value method and p-value method to test the hypothesis

Answer 1

$a)$ Critical value method.

$\sigma=5000\\n=48\\\bar x=59500$

The hypotheses to be tested are,

$H_0:\mu=60000\\vs\\H_1:\mu\not=60000$

The test statistic is,

$Z={(\bar x-\mu)\over{\sigma\over\sqrt{n}}}={59500-60000\over{5000\over\sqrt{48}}}={-500\over721.687837}=-0.69282032\approx-0.693$

$Z$ is compared with the standard normal table value at $\alpha=0.05$ given as, $Z_{0.05\over2}=Z_{0.025}=1.96$

The null hypothesis is rejected if $|Z|\gt Z_{0.025}$.

Since $|Z|=|-0.693|=0.693\lt Z_{0.025}=1.96$, we fail to reject the null hypothesis which means that at the 5% significance level the data do not provide sufficient evidence to conclude that the mean mileage the tire can be driven before the thread wears out differs from 60,000 miles.

$b)$ The p-value approach

Using this approach, we determine the p-value related to the test statistic, $Z=-0.693$ obtained in part a above.

The p-value is given as,

$p-value={\phi(-0.693)\times2}=2\times 0.2451=0.4902$. The reason we multiply $\phi(-0.693)$ by 2 is because we are performing a two-sided test.

The null hypothesis is rejected if, $p-value\lt\alpha$ .

Since $p-value=0.4902\gt\alpha=0.05,$ we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the mean mileage differs from 60,000 miles at 5 % significance level.