One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles. A Sample 49 tires revealed that the mean number of miles is 45,000 miles with a standard deviation of 15,000 miles. Test the hypothesis with a 0.05 significance level. A) The decision rule is to reject if Z 1.645. The calculated value of z is +2.33. B) The decision rule is to reject if Z 1.96. The calculated value of z is +2.33. D) The decision rule is to reject if Z

Question

One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles. A Sample 49 tires revealed that the mean number of miles is 45,000 miles with a standard deviation of 15,000 miles. Test the hypothesis with a 0.05 significance level. A) The decision rule is to reject if Z > 1.645. The calculated value of z is +2.33. B) The decision rule is to reject if Z < 1.645. The calculated value of z is -2.33. C) The decision rule is to reject if Z > 1.96. The calculated value of z is +2.33. D) The decision rule is to reject if Z < 1.96. The calculated value of z is +2.33. E) None of these statements is correct.

Quizplus · Accepted Answer

The hypothesis test is one-tailed (right-tailed) since the company believes the tire lasts more than 40,000 miles. The critical value for a one-tailed test at a 0.05 significance level is 1.645. The calculated z-value is +2.33, which is greater than 1.645, so we reject the null hypothesis.

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