A local tire store suspects that the mean life of a new discount tire is less that 39,000 miles. To check the claim, the store selects randomly 18 of these new discount tires. When they are tested, it is found that the mean life is 38,250 miles with a sample standard deviation s=1200 miles. Assume the distribution is normally distributed.
Use the critical value to method from the normal distribution to test for the population mean u. Test the company's claim at the level of significance a=0.05
Use a one-sample z-test.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (38250 - 39000)/(1200/√18) = ?
Finish the calculation.
Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.
I hope this will help get you started.