statistics
posted by Barbara .
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 onesample ztest.
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 ztable at .05 level of significance for a onetailed test.
If the ztest statistic exceeds the critical value from the ztable, reject the null. If the ztest statistic does not exceed the critical value from the ztable, do not reject the null.
I hope this will help get you started.
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