A tire manufacturer guarantess that its tires will last for at least 60,000 miles. If the tire fails earlier, the manufacturer offers a prorated replacement. Too many such replacements are bad for business, so the manufacturer tests tires at random to see whether corrective action is needed on the production line. A random sample of (n = 49 tires). is examined. The average durability is (xbar = 58,000 miles), with a standard deviation of (s = 7000 miles). At the .05 level should the manufacturer be satisfied that its claim is valid? Be sure to state your hypothesis and your basis for rejecting or not rejecting the null hypothesis (sample Z test statistic and rejection threshold)

Question

A tire manufacturer guarantess that its tires will last for at least 60,000 miles.  If the tire fails earlier, the manufacturer offers a prorated replacement.  Too many such replacements are bad for business, so the manufacturer tests tires at random to see whether corrective action is needed on the production line.  A random sample of (n = 49 tires). is examined.  The average durability is (xbar = 58,000 miles), with a standard deviation of (s = 7000 miles).  At the .05 level should the manufacturer be satisfied that its claim is valid?  Be sure to state your hypothesis and your basis for rejecting or not rejecting the null hypothesis (sample Z test statistic and rejection threshold)

MathGuru · Answer

Use a one-sample z-test.

Formula:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With the data in your problem:
z = (58000 - 60000)/(7000/√49)

Finish the calculation. If the test statistic exceeds the critical value you find in a z-table, reject the null. If the test statistic does not exceed the critical value, do not reject the null.

I hope this will help get you started.